Basic question in Heisenberg's uncertainity principle

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1 "madhusoodhanan" Basic question in Heisenberg's uncertainity principle donderdag 21 maart 2002 12:39
2 "Harold Ensle" Re: Basic question in Heisenberg's uncertainity principle maandag 25 maart 2002 6:27
3 "Mike Hanson" Re: Basic question in Heisenberg's uncertainity principle maandag 25 maart 2002 14:44
4 "Harold Ensle" Re: Basic question in Heisenberg's uncertainity principle dinsdag 26 maart 2002 5:49
5 "Bilge" Re: Basic question in Heisenberg's uncertainity principle woensdag 27 maart 2002 11:31
6 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle woensdag 27 maart 2002 22:14
7 "Jon Bell" Re: Basic question in Heisenberg's uncertainity principle woensdag 27 maart 2002 22:50
8 "Bilge" Re: Basic question in Heisenberg's uncertainity principle donderdag 28 maart 2002 0:42
9 "brian a m stuckless" Re: Basic question in Heisenberg's uncertainity principle donderdag 28 maart 2002 14:58
10 "Bilge" Re: Basic question in Heisenberg's uncertainity principle donderdag 28 maart 2002 18:21
11 "Bilge" Re: Basic question in Heisenberg's uncertainity principle donderdag 28 maart 2002 19:37
12 "Gregory L. Hansen" Re: Basic question in Heisenberg's uncertainity principle donderdag 28 maart 2002 20:18
13 "Gordon D. Pusch" Re: Basic question in Heisenberg's uncertainity principle donderdag 28 maart 2002 23:11
14 "brian a m stuckless" Re: Basic question in Heisenberg's uncertainity principle vrijdag 29 maart 2002 3:10
15 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle vrijdag 29 maart 2002 10:54
16 "Bilge" Re: Basic question in Heisenberg's uncertainity principle vrijdag 29 maart 2002 18:53
17 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle maandag 1 april 2002 22:05
18 "Bilge" Re: Basic question in Heisenberg's uncertainity principle dinsdag 2 april 2002 12:07
19 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle dinsdag 2 april 2002 22:18
20 "Tom Clarke" Re: Basic question in Heisenberg's uncertainity principle dinsdag 2 april 2002 23:32
21 "Bilge" Re: Basic question in Heisenberg's uncertainity principle woensdag 3 april 2002 3:42
22 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle woensdag 3 april 2002 12:03
23 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle woensdag 3 april 2002 14:04
24 "Tom Clarke" Re: Basic question in Heisenberg's uncertainity principle woensdag 3 april 2002 18:20
25 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle woensdag 3 april 2002 20:09
26 "Bilge" Re: Basic question in Heisenberg's uncertainity principle woensdag 3 april 2002 23:44
27 Re: Basic question in Heisenberg's uncertainity principle donderdag 4 april 2002 5:59
28 "Bilge" Re: Basic question in Heisenberg's uncertainity principle zaterdag 6 april 2002 5:42
29 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle vrijdag 12 april 2002 16:04
30 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle vrijdag 12 april 2002 16:04
31 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle vrijdag 12 april 2002 16:05
32 "Bilge" Re: Basic question in Heisenberg's uncertainity principle zaterdag 13 april 2002 16:55
33 "Frank Wappler" Re: Basic question in Heisenberg's uncertainity principle zondag 14 april 2002 9:42
34 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle zondag 14 april 2002 14:56
35 "ca314159" Re: Basic question in Heisenberg's uncertainity principle zondag 14 april 2002 16:29
36 "Frank Wappler" Re: Basic question in Heisenberg's uncertainity principle zondag 14 april 2002 20:58
37 "Frank Wappler" Re: Basic question in Heisenberg's uncertainity principle zondag 14 april 2002 21:15
38 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle maandag 15 april 2002 10:48
39 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle maandag 15 april 2002 15:44
40 "Bilge" Re: Basic question in Heisenberg's uncertainity principle dinsdag 16 april 2002 22:34
41 "Bilge" Re: Basic question in Heisenberg's uncertainity principle dinsdag 16 april 2002 22:46
42 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle woensdag 17 april 2002 15:59
43 "Bilge" Re: Basic question in Heisenberg's uncertainity principle woensdag 17 april 2002 20:15
44 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle donderdag 18 april 2002 13:19
45 "Bilge" Re: Basic question in Heisenberg's uncertainity principle vrijdag 19 april 2002 7:24
46 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle vrijdag 19 april 2002 10:30
47 "Bilge" Re: Basic question in Heisenberg's uncertainity principle zaterdag 20 april 2002 20:13
48 "Frank Wappler" Re: Basic question in Heisenberg's uncertainity principle zondag 21 april 2002 9:21
49 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle maandag 22 april 2002 12:33
50 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle maandag 22 april 2002 15:12
51 "Frank Wappler" Re: Basic question in Heisenberg's uncertainity principle maandag 22 april 2002 23:07
52 "Frank Wappler" Re: Basic question in Heisenberg's uncertainity principle dinsdag 23 april 2002 22:27
53 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle woensdag 24 april 2002 22:36
54 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle vrijdag 26 april 2002 16:26
55 "Nicolaas Vroom" Re: Basic question in Heisenberg's uncertainity principle zaterdag 27 april 2002 17:22
56 "Bilge" Re: Basic question in Heisenberg's uncertainity principle dinsdag 30 april 2002 9:49


1 Basic question in Heisenberg's uncertainity principle

Van: "madhusoodhanan"
Onderwerp: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 21 maart 2002 12:39

hello,
I have a very elementary question in Heisenberg's Uncertainity principle. I have read a hypothetical thought experiment in a book prooving uncertainity principle .(The Photon gun Expt.) i have my own doubts in it. Please clarify.

It says that in order to measure the location of the particle P1 , we need another atomic particle say photon . This can be produced by a photon gun which bombards photons to locate P1. These when hit P1 either change the position when it collides or changes its velocity .

now my question is , why can't we have 2 photon guns bombarding photons exactly opposite to each other and allowing the particle P1 to come exactly in centre of them . Also , we could move the guns with a relative velocity 0 . This would mean by changing the velocity of the gun , we could measure P1's velocity and as P1 is in centre , we could measure its position by the reflecting photon.

This situation is similar to 2 tennis balls thrown opposte to each other and letting a soccer ball go in between them. Now bcos both balls nully their affects , there is no change in position , further as the source of tennis balls are in motion we can deduce the velocity of the soccer ball .

This means we could measure both velocity and position simultaneously ????

I dont know if my question is right , but please clarify if i am wrong.

thank you in advance Madhu


2 Basic question in Heisenberg's uncertainity principle

Van: "Harold Ensle"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 25 maart 2002 6:27

madhusoodhanan wrote in message news:8f18fda4.0203210239.2cc819e0@posting.google.com...
> hello,
I have a very elementary question in Heisenberg's Uncertainity principle. I have read a hypothetical thought experiment in a book prooving uncertainity principle .(The Photon gun Expt.) i have my own doubts in it. Please clarify.

Uncertainty is a huge lie.

It was not enough that they didn't understand what was going on, but in their arrogance they believed that it could not be known.

There is, of course, experimental uncertainty, but there is no intrinsic uncertainty. In every case of supposed intrinsic uncertainty, there is an unknown variable. It is that simple.

H.Ellis Ensle


3 Basic question in Heisenberg's uncertainity principle

Van: "Mike Hanson"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 25 maart 2002 14:44

"Harold Ensle" wrote in message news:...
> madhusoodhanan wrote in message news:8f18fda4.0203210239.2cc819e0@posting.google.com...
> > hello,
I have a very elementary question in Heisenberg's Uncertainity principle. I have read a hypothetical thought experiment in a book prooving uncertainity principle .(The Photon gun Expt.) i have my own doubts in it. Please clarify.
>

Uncertainty is a huge lie.

It was not enough that they didn't understand what was going on, but in their arrogance they believed that it could not be known.

No. Experimental fact forced these conclusions on them, which they found to be quite as absurd as you find them. Being good scientists, though, they did not insist that the universe must behave the way they thought it ought to, no matter how distasteful it must have been to them - classical physicists as they were.

> There is, of course, experimental uncertainty, but there is no intrinsic uncertainty.

Well, if that be the case, then such "experimental uncertainty" seems to be remarkably uniform: the amount of uncertainty can be measured with great precision and found to be exactly in accord with what HUP predicts. I would have expected something rather more arbitrary coming from experimental inaccuracy - bearing in mind that we are talking about different equipment set up in different ways in thousands of laboratories across the world, all of which produce results that accord with HUP to an equally high degree of precision.

> In every case of supposed intrinsic uncertainty, there is an unknown variable. It is that simple.

Not quite that simple. Hidden-variable models work, but most of them do have the embarrassing feature of a faster-than-light undetectable pilot wave - which is at least as absurd as the intrinsic uncertainty that you are trying to get rid of.

Mike.


4 Basic question in Heisenberg's uncertainity principle

Van: "Harold Ensle"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 26 maart 2002 5:49

Mike Hanson wrote in message news:11d49fef.0203250444.9c230c4@posting.google.com...
> "Harold Ensle" wrote in message news:...
> > madhusoodhanan wrote in message news:8f18fda4.0203210239.2cc819e0@posting.google.com...
> > > hello, I have a very elementary question in Heisenberg's Uncertainity principle. I have read a hypothetical thought experiment in a book prooving uncertainity principle .(The Photon gun Expt.) i have my own doubts in it. Please clarify.
> >

Uncertainty is a huge lie.

It was not enough that they didn't understand what was going on, but in their arrogance they believed that it could not be known.

>

No. Experimental fact forced these conclusions on them, which they found to be quite as absurd as you find them. Being good scientists, though, they did not insist that the universe must behave the way they thought it ought to, no matter how distasteful it must have been to them - classical physicists as they were.

It is not a question of "distasteful". It is a question of absurdity. If the experimental evidence seems to indicate that there is an intrinsic uncertainty, then one must be misinterpreting the experiments. It is that simple.

> > There is, of course, experimental uncertainty, but there is no intrinsic uncertainty.
>

Well, if that be the case, then such "experimental uncertainty" seems to be remarkably uniform: the amount of uncertainty can be measured with great precision and found to be exactly in accord with what HUP predicts. I would have expected something rather more arbitrary coming from experimental inaccuracy - bearing in mind that we are talking about different equipment set up in different ways in thousands of laboratories across the world, all of which produce results that accord with HUP to an equally high degree of precision.

> >

In every case of supposed intrinsic uncertainty, there is an unknown variable. It is that simple.

>

Not quite that simple. Hidden-variable models work, but most of them do have the embarrassing feature of a faster-than-light undetectable pilot wave - which is at least as absurd as the intrinsic uncertainty that you are trying to get rid of.

This is not true. No faster than light object is required. The mistake they made was not realizing when the information was transmitted. The information of that famous experiment was being carried by the photon itself.

H.Ellis Ensle


5 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 27 maart 2002 11:31

Harold Ensle said some stuff about

> It is not a question of "distasteful". It is a question of absurdity. If the experimental evidence seems to indicate that there is an intrinsic uncertainty, then one must be misinterpreting the experiments. It is that simple.

It's an intrinsic uncertainty is certain simultaneous measurements. In particular, variables that are canonically conjugate. You can measure either p or x to arbitrary precision, just not both simultaneously. The problem occured before quantum mechanics - it just wasn't obvious. But it did lead to quantum mechanics. In E&M the fields are given as some superposition f(x,t) = \sum_n a_n*exp(-i(k_nz - w_nt)). But, you can invert that to find a function of A(k,w). The quantities k,x have an intrinsic uncertainty \Delta k\Delta x = 2pi and likewise for w,t. That's where the interference comes from. The difference kx - wt is the phase of the wave and is a constant. Quantum mechanics merely notes that d/dx exp(-(ikx-iw)t) = ik exp((ik-wt)) and p = \hbar k, so that p = (hbar/i)d/dx and similarly for E,t. So basically, you can look at quantum mechanics as a logical outgrowth of what already existed in E&M, once planck's constant appeared. All relativistic qm does is take E^2=p^2 + m^2 and replace E and p by the differentials above.


6 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 27 maart 2002 22:14

"Bilge" schreef in bericht news:slrnaa34kc.5a3.root@radioactivex.lebesque-al.net...
> Harold Ensle said some stuff about
> >

It is not a question of "distasteful". It is a question of absurdity. If the experimental evidence seems to indicate that there is an intrinsic uncertainty, then one must be misinterpreting the experiments. It is that simple.

>

It's an intrinsic uncertainty is certain simultaneous measurements. In particular, variables that are canonically conjugate. You can measure either p or x to arbitrary precision, just not both simultaneously.

How do you measure momentum p = mv ? In ONE measurement ?

A particle has many parameters. You can measure each of those parameters to arbitrary precision but not simultaneous. What is so special about selecting p and x ?

You can also not measure x twice.

> The problem occured before quantum mechanics - it just wasn't obvious. But it did lead to quantum mechanics. In E&M the fields are given as some superposition f(x,t) = \sum_n a_n*exp(-i(k_nz - w_nt)). But, you can invert that to find a function of A(k,w). The quantities k,x have an intrinsic uncertainty \Delta k\Delta x = 2pi and likewise for w,t. That's where the interference comes from. The difference kx - wt is the phase of the wave and is a constant. Quantum mechanics merely notes that d/dx exp(-(ikx-iw)t) = ik exp((ik-wt)) and p = \hbar k, so that p = (hbar/i)d/dx and similarly for E,t. So basically, you can look at quantum mechanics as a logical outgrowth of what already existed in E&M, once planck's constant appeared. All relativistic qm does is take E^2=p^2 + m^2 and replace E and p by the differentials above.

Can some one explain the equation: E^2=p^2 + m^2 ? E means energie but that is not true for p^2 and m^2

>


7 Basic question in Heisenberg's uncertainity principle

Van: "Jon Bell"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 27 maart 2002 22:50

In article , Nicolaas Vroom wrote:
>

"Bilge" schreef in bericht news:slrnaa34kc.5a3.root@radioactivex.lebesque-al.net...

>> [...] All relativistic qm does is take E^2=p^2 + m^2 and replace E and p by the differentials above.
>

Can some one explain the equation: E^2=p^2 + m^2 ? E means energie but that is not true for p^2 and m^2

He's using a system of units in which c = 1. Theorists like to do this because then they don't have to worry about factors of c. When they actually need to do practical calculations, they can put back the missing factors of c by doing some dimensional analysis, in this case replacing p with pc and m with mc^2.

-- Jon Bell Presbyterian College Dept. of Physics and Computer Science Clinton, South Carolina USA


8 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 28 maart 2002 0:42

Nicolaas Vroom said some stuff about
>

How do you measure momentum p = mv ? In ONE measurement ?

That is really not relavent, but just for general information, I'll mention a few examples. It depends upon what you have to work with. Charged particles are easy. The product of the magnetic field and the radius of curvature for the trajectory of a charged particle in the field, B*r, is called the magnetic rigidity. It's related to the momentum p and charge, q, of the particle by B*r = p/q, so that p = B*r/q.

It's harder for uncharged particles. There, you have to rely on some type of scattering that allows you to measure p via conservation of momentum. That's why the neutrino mass is so hard to determine. The obvious way to determine it would be from \beta endpoint spectra, but it's too small so the difference between E = pc and E = sqrt((pc)^2 + (mc^2)^2) is hard to utilize.

> A particle has many parameters. You can measure each of those parameters to arbitrary precision but not simultaneous.

"Simultaneously" really means knowing both at the same time. I.e., the second measurement doesn't change the value of the first. Or better, that you get the same result regardless of the order you measure them, so that measuring AB = measuring BA. If AB - BA = 0, then you can measure both A & B "simultaneously". This is denoted: [A,B] == AB - BA = 0 and A & B are said to commute. If [A,B] != 0, then it means the value of A & B depend upon the order you measure them. In QM, measuring p,x or E,t and others give a difference of i\hbar.

> What is so special about selecting p and x ?

x,p are called canonical conjugates. (more generally, this would be the canonical variables p,q just like in classical mechanics). They are the variables from which one obtains conservation laws through the lagrangian which minimizes the value of a quantity called the action, which has units of angular momenta. In classical mechanics, this relationship is called a poisson bracket: {x,p} = 1. It's a statement about symmetry. A change in coordinates that preserves this relationship leads to a conserved quantity. In quantum mechanics, the commutator [x,p] = i\hbar, leads to quantization. In relativistic quantum mechanics, this leads to a quantized field. In maxwell's equation's it leads to the relationship called the continuity equation (conserved charge).

>You can also not measure x twice.

Sure, I can. [Possibly not in GR, but I'm not sure how that would work and that's a different issue. Ssomeone else might be able to answer it though]

[...]
>>
>

Can some one explain the equation: E^2=p^2 + m^2 ? E means energie but that is not true for p^2 and m^2

I think someone already got this one, but, since I'm here... Fill in 'c' as appropriate. 'c' just acts a scale factor, so it's not convenient to constantly write it down, since it appears so often and usually doesn't add anything to the result that is any more illuminating. Depending upon what type of physicist you are talking with, the choices of units may be c==1, so that mass, energy and momentum are all given in eV's (usually, MeV or GeV), velocities are given by v/c. Then there is the choice, c == 1, hbar == 1, so that the previous applies and length is measured in 1/GeV and finally, c==1, hbar==1 G==1, so that everything is measured in lengths and masses become the schwarzchild radius, i.e., mass sun = 1.6 km, or else is dimensionless. Usually, the charge of the electron is given as e = sqrt([4pi]\alpha\hbar c) so that it has either units of sqrt(MeV-fm) or is dimensionless. The [4pi] is optional. Since the only time these things matter is when you get to the end, it's a lot simpler to keep track of things if don't carry around constants independently of the physics they are attached to.


9 Basic question in Heisenberg's uncertainity principle

Van: "brian a m stuckless"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 28 maart 2002 14:58

"Nicolaas Vroom" wrote in message news:...
> "Bilge" schreef in bericht news:slrnaa34kc.5a3.root@radioactivex.lebesque-al.net...
> > Harold Ensle said some stuff about
> > >

If the experimental evidence seems to indicate that there is an intrinsic uncertainty, then one must be misinterpreting the experiments. It is that simple.

> >

It's an intrinsic uncertainty is certain simultaneous measurements. In particular, variables that are canonically conjugate. You can measure either p or x to arbitrary precision, just not both simultaneously.

You are talking nonsense, Bilge. Momentum & position are NOT 'canonically conjugate', by definition! MOMENTUM & POSITION 'CANNOT be' canonically conjugate: Two(2) positions A and B are fundamentally PRE-requisite of ANY assigned vector. Momentum depends DIRECTLY on your 2nd coordinate position choice. Your 2nd coordinate position choice can ONLY be another abritrary position on the path of the particle, --even if ALL of the coordinate positions located on the path are known.

Momentum UNCERTAINTY depends DIRECTLY on the path's STRAiGHTness. Clearly MOMENTUM is NOT an 'intrinsic' property of any particle.

''Instantaneous'' MOMENTUM or MOTION or ACTION is another OXYMORON. There is NO momentum at ANY point, --only BETWEEN ADJACENT points, on ANY path. This is very, very fundamental physics, Tin-bit.

Sincerely,
```brian < bastuckl@avalon.nf.ca > mar 28, 2002
p.s.
Google group search: < http://groups.google.com >
< {GUESS} Integrated Standard System{ISS} >
< electron magnetic moment FORMER 'anomaly' >
< momentum vector ABCs and quantum gravity >
< {ISS}.wpd; the fundamental CONSTANT set >
< Avagadro's Law > < creation Vs vacuum >
< Conservation of information >

> Can some one explain the equation: E^2=p^2 + m^2 ?

No.
That's GR-tivity NOT ER-tivity (i.e. NOT Einstein's).
^^^^^end of post.


10 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 28 maart 2002 18:21

Bilge said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
> Nicolaas Vroom said some stuff about
> >

How do you measure momentum p = mv ? In ONE measurement ?

>

That is really not relavent, but just for general information, I'll mention a few examples. It depends upon what you have to work with. Charged particles are easy. The product of the magnetic field and the radius of curvature for the trajectory of a charged particle in the field, B*r, is called the magnetic rigidity. It's related to the momentum p and charge, q, of the particle by B*r = p/q, so that p = B*r/q.

That should be p = Brq not Br/q.


11 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 28 maart 2002 19:37

brian a m stuckless, australopithicus at large wrote: Re: Basic question in Heisenberg's uncertainity principle to usenet:
>

You are talking nonsense, Bilge.

Would it help if I simplified a bit more than that for you? I had forgotten someone might be reading this to you.

> Momentum & position are NOT 'canonically conjugate', by definition!

I'd say, don't be a moron, but I'm sure you'd only wipe out another brain cell or two trying to figure out what that means.

> MOMENTUM & POSITION 'CANNOT be' canonically conjugate:

OF CoUrSE nOt. It's POSITION & MOMENTUM. AEe YOuDafT?

> Two(2) positions A and B are fundamentally PRE-requisite of ANY assigned vector.

You point being?

> Momentum UNCERTAINTY depends DIRECTLY on the path's STRAiGHTness. Clearly MOMENTUM is NOT an 'intrinsic' property of any particle.

Hmmm. I was under the impression that:

the path's UNCERTAINTY depends DIRECTLY on Momentum STRAiGHTness. any particle is NOT Clearly MOMENTUM of an 'intrinsic' property

> ''Instantaneous'' MOMENTUM or MOTION or ACTION is another OXYMORON. There is NO momentum at ANY point, --only BETWEEN ADJACENT points, on ANY path. This is very, very fundamental physics, Tin-bit.

You must be learning a lot there at Olduvai Gorge U, but you should take a break from volcano diving and bobbing for cape buffalo to broaden your horizons. Maybe go to new york and get a full time job as a discovery of the leaky's in some museum.


12 Basic question in Heisenberg's uncertainity principle

Van: "Gregory L. Hansen"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 28 maart 2002 20:18

In article <6150b7f6.0203280458.126d4d96@posting.google.com>, brian a m stuckless wrote:
> "Nicolaas Vroom" wrote in message news:...
>> "Bilge" schreef in bericht news:slrnaa34kc.5a3.root@radioactivex.lebesque-al.net...
>> > Harold Ensle said some stuff about
>> > >

If the experimental evidence seems to indicate that there is an intrinsic uncertainty, then one must be misinterpreting the experiments. It is that simple.

>> >

It's an intrinsic uncertainty is certain simultaneous measurements. In particular, variables that are canonically conjugate. You can measure either p or x to arbitrary precision, just not both simultaneously.

>

You are talking nonsense, Bilge. Momentum & position are NOT 'canonically conjugate', by definition! MOMENTUM & POSITION 'CANNOT be' canonically conjugate: Two(2) positions A and B are fundamentally PRE-requisite of ANY assigned vector. Momentum depends DIRECTLY on your 2nd coordinate position choice. Your 2nd coordinate position choice can ONLY be another abritrary position on the path of the particle, --even if ALL of the coordinate positions located on the path are known.

Given a Lagrangian L(q,dq/dt), the canonical conjugate is defined as

k = dL/d(dq/dt)

Do the math. -- "'No user-serviceable parts inside.' I'll be the judge of that!"


13 Basic question in Heisenberg's uncertainity principle

Van: "Gordon D. Pusch"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 28 maart 2002 23:11

glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:

> In article <6150b7f6.0203280458.126d4d96@posting.google.com>, brian a m stuckless wrote:
> > "Nicolaas Vroom" wrote in message news:...
> >> "Bilge" schreef in bericht news:slrnaa34kc.5a3.root@radioactivex.lebesque-al.net...
> >> > Harold Ensle said some stuff about
> >> > >

If the experimental evidence seems to indicate that there is an intrinsic uncertainty, then one must be misinterpreting the experiments. It is that simple.

> >> >

It's an intrinsic uncertainty is certain simultaneous measurements. In particular, variables that are canonically conjugate. You can measure either p or x to arbitrary precision, just not both simultaneously.

> >

You are talking nonsense, Bilge. Momentum & position are NOT 'canonically conjugate', by definition! MOMENTUM & POSITION 'CANNOT be' canonically conjugate: Two(2) positions A and B are fundamentally PRE-requisite of ANY assigned vector. Momentum depends DIRECTLY on your 2nd coordinate position choice. Your 2nd coordinate position choice can ONLY be another abritrary position on the path of the particle, --even if ALL of the coordinate positions located on the path are known.

>

Given a Lagrangian L(q,dq/dt), the canonical conjugate is defined as

k = dL/d(dq/dt)

Do the math.

Stuckbrain's `math' is limited to numerology and bad dimensional analysis.


14 Basic question in Heisenberg's uncertainity principle

Van: "brian a m stuckless"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 29 maart 2002 3:10

root@radioactivex.lebesque-al.net (Bilge) wrote in message news:...

Dear Bilge, MOMENTUM 'CANNOT be' a conjugate of POSiTiON: TWO(2) POSiTiONs A and B are fundamentally PRE-requisite of ANY assigned VECTOR. A VECTOR, Bilge, is a fundamental PRE-requisite of ANY VELOCiTY. And guess what, Bilge! VELOCiTY is a very, very, fundamental PRE-requisite of ANY MOMENTUM. So it follows, Dimwit, that MOMENTUM depends DIRECTLY on your 2nd coordinate POSiTiON choice. Your selected 2nd coordinate POSiTiON choice can ONLY be another abritrary POSiTiON on the path of the particle, --even if ALL of the coordinate POSiTiONs located on the path are known.

Momentum UNCERTAINTY depends DIRECTLY on the path's STRAiGHTness. Clearly then, MOMENTUM is NOT 'intrinsic' to any particle.

''Instantaneous'' MOMENTUM or MOTION or ACTION is another OXYMORON. There is NO momentum at ANY point, --indeed, only BETWEEN ADJACENT points, on ANY path. You ought to know basic physics, Bilge.

> Can some one explain the equation: E^2=p^2 + m^2 ?

No. That's dated GR-tivity --NOT ER-tivity (--NOT Einstein's).

The {GUESS} {ISS} Hamiltonian Enthalpy energy equation:
E = m*c^2 + pL*c + pA*f1
= m*c^2 + h*f + nA*hbar*f1
= rest energy + LaGrangian L + volt*amp*sec energy eV

You seem to forget that MOMENTUM is a VECTOR quantity. All VECTORs have TWO (2) ends --BOTH of them on the path.

> > Two(2) positions A and B are fundamentally PRE-requisite of ANY assigned vector.
>

You point being?

Do you know the meaning of the word 'PRE-requisite'? A 2nd VECTOR 'end-point' (on the path) is PRE-required,.. 'FIRST', 'BEFORE', 'PRIOR TO' your arbitrary choice of a pre-MOMENTUM velocity VECTOR.

You are putting the ass backwards before the cart, Dimwit.

If the following 'radial and lateral angular accelerations' is over your head, then scroll on down to 'momentum vector ABCs and quantum gravity' (you missed this in school).

Take your time with it, Bilge. Try to be ONE with it.

Sincerely,
```brian < bastuckl@avalon.nf.ca > mar 28, 2002
p.s.
Google group search: < http://groups.google.com >
< {GUESS} Integrated Standard System{ISS} >
< electron magnetic moment FORMER 'anomaly' >
< momentum vector ABCs and quantum gravity >
< {ISS}.wpd; the fundamental CONSTANT set >
< Avagadro's Law > < creation Vs vacuum >
< Conservation of information >

^^^^references follow..

1     5    10    15    20    25    30    35    40    45    50   55
Radial and lateral angular accelerations: COORDINATEs are always POSITIONs of points located on a path which are LATER selected from to arbitrarily choose a VECTOR to define a VELOCITY to subsequently define an ACCELERATION.

Acceleration is independent of any coordinate system but not independent of your vector end-point coordinate choices.

Its easy for some to get very confused about the difference between 'coordinate system choice' and ‘coordinate choice'.

Please note that NO ‘different' coordinate SYSTEMs are suggested or implied here!

ANY acceleration is arbitrarily DEPENDENT on your prior choice of pre-vector end-point COORDINATES on the path. Even if every point on the path is measured or otherwise known, ..for any CURVED path.

Acceleration = (DIFFERENCE in velocity) / duration. Velocity = (DIFFERENCE in cooridinates) / duration. Displacement due to motion = DIFFERENCE in coordinates.

Acceleration is only either RADIAL or LATERAL ANGULAR. Acceleration arbritrarily depends on a VECTOR.

DISPLACEMENT = (DIFFERENCE in path coordinates). VELOCITY = (DIFFERENCE in coordinates) / duration. ACCELERATION = (DIFFERENCE in coordinates) / (duration)^2.

Gravitational -->force --> mass*displacement / (duration)^2 Inertial -->force --> SAME mass*displacement / (SAME duration)^2 Therefore, since the test mass is the SAME, and the duration is the SAME, then only the ICON or MAGNITUDE or METHOD of the DISPLACEMENT can distinguish any DIFFERENCE between ‘gravitational' force and ‘inertial' force. Obviously the ‘magnitude' has to be the same also, in any mathematical proof. This leaves only the ICON or METHOD to fully distinguish the same MAGNITUDE of displacement. Sample experimental demonstrations follow..

1. Drop a cannonball and olive off the Leaning Tower of Pisa together.
2. Accelerate cannonball and olive the same together LATERALLY.

Clearly ‘radial gravitational' and ‘inertial otherwise lateral angular' accelerations ARE DIFFERENT concepts of 'acceleration'.

Gravitational ---> radial force is an energy SOURCE. Inertial ---> lateral angular force is an energy SINK.

CONCLUSIONs: 1. ‘Gravitational' radial acceleration entails a DIFFERENT ‘method' of accelerating the SAME test mass for the SAME duration than does ‘inertial' lateral angular acceleration.

2. Mass times velocity / duration:
--> - gravitational radial force --> + inertial lateral angular force.

3. Radial gravitational acceleration is AUTOMATIC acceleration, ..while otherwise lateral angular acceleration is NOT AUTOMATIC.

Promorphologically speaking, fundamentals CANNOT be arbitrary. Vector fundamentals NEVER change, and ACCELERATION is surely STILL a VECTOR quantity.

1     5    10    15    20    25    30    35    40    45    50    55
          momentum vector ABCs and quantum gravity

momentum vector ABC's: Momentum is a vector quantity ..not a conjugate of position. Vectors cannot be 'measured'. Vectors can only be 'imagined' between TWO 'located' positions on a path, in retrospect. Any other two points on the same path will constitute another imaginary vector even though they are also on the same path.

Therefore, the 'uncertainty of the momentum' of a particle decreases with the increased straightness of the path.

          o o                          o  o
                o                   o         o
                 o                o             o
                  A - - - - - vector - - - - - - B - -> C
                   o            o                 o
                     o        o
                        o  o
                     variable path

Note that velocity AB / t is independent of the speed A^^B / t.

Any ''as-measured'' velocity between any other two points (located on the same path) will indicate a 'different' velocity, --even though the path does not change. Again, this is exactly why the 'uncertainty of the momentum' of a particle decreases with the increased straightness of the path.

If the required pre-vector of momentum position A is known, the required pre-vector position B on the path is arbitrary and is NOT a conjugate of the subsequent momentum. A momentum vector is 'directly' dependent on said position B --NOT conjugate. Vector momentum depends directly on THAT imagined vector, --again, unless you know (for sure) the path is very straight.

Vector projection is independent of position B on a straight path. Vectors and vector projections are always straight, like CHORDs, --for any TWO points A and B located on ANY path.

quantum gravity: A GR-'geodesic' is the SHORTEST ARC(along the surface) between any TWO points located on a sphere --NOT the straight CHORD vector, ..(-- and NOT ANY ‘'other LONGER way round'‘ ..either):

a. quantum gravity = (geodesic - straight chord) / integer
b. curved path = vector + integer*(quantum GR-tivity)

NOTE that * is a multiplication sign which separates icons.

The ONLY path between any ADJACENT points on any path is the vector between them. Therefore, any non-zero GR-tivity quantum would be an AFFINE GAP between ADJACENT points, --mathematically speaking.

No GR-tivity cracked-pot would EVER call them Tin-bit points. GR-tivity BECOMES a Tin-bit.

1     5    10    15    20    25    30    35    40    45    50    55
^^^^^end of post.


15 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 29 maart 2002 10:54

"Bilge" schreef in bericht news:slrnaa4iu4.997.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff about
> >

How do you measure momentum p = mv ? In ONE measurement ?

>

That is really not relavent,

Why is that not relevant ? If a law or principle makes a claim about a parameter, in this case p, you have to state how you calculate it, (which you do below) or how you measure it.

> but just for general information, I'll mention a few examples. It depends upon what you have to work with. Charged particles are easy. The product of the magnetic field and the radius of curvature for the trajectory of a charged particle in the field, B*r, is called the magnetic rigidity. It's related to the momentum p and charge, q, of the particle by B*r = p/q, so that p = B*r/q.

That means if you need p (=m*v) you have to measure B, r and q. Again r !!!! If that is correct than the whole principle becomes like magic.

>
> >

You can also not measure x twice.

>

Sure, I can. [Possibly not in GR, but I'm not sure how that would work and that's a different issue. Ssomeone else might be able to answer it though]

Try a different one:
You can also not measure p twice precisely.

If you don't agree, (that means that you agree with the statement that you can measure p twice precisely) why can not you than in stead of the second measurement of p, measure x precisely ?

> > Can some one explain the equation: E^2=p^2 + m^2 ?
E means energie but that is not true for p^2 and m^2
>

I think someone already got this one, but, since I'm here... Fill in 'c' as appropriate. 'c' just acts a scale factor, so it's not convenient to constantly write it down, since it appears so often and usually doesn't add anything to the result that is any more illuminating. Depending upon what type of physicist you are talking with, the choices of units may be c==1, so that mass, energy and momentum are all given in eV's (usually, MeV or GeV), velocities are given by v/c. Then there is the choice, c == 1, hbar == 1, so that the previous applies and length is measured in 1/GeV and finally, c==1, hbar==1 G==1, so that everything is measured in lengths and masses become the schwarzchild radius, i.e., mass sun = 1.6 km, or else is dimensionless. Usually, the charge of the electron is given as e = sqrt([4pi]\alpha\hbar c) so that it has either units of sqrt(MeV-fm) or is dimensionless. The [4pi] is optional. Since the only time these things matter is when you get to the end, it's a lot simpler to keep track of things if don't carry around constants independently of the physics they are attached to.

It seems to me that in the above equation "many" "constants" are set to one. IMO that is tricky way to do science.


16 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 29 maart 2002 18:53

Nicolaas Vroom said some stuff about

>

Why is that not relevant ?

Because quantum mechanics didn't say that the uncertainty principle only applied to specific measuring techniques. Besides, I provided an answer for you anyway, did I not?

> If a law or principle makes a claim about a parameter, in this case p, you have to state how you calculate it, (which you do below) or how you measure it.

The uncertainty principle states that you can't measure something to better precision than it provides. If you wish to build a device that does, I'm not going to stop you.

>> but just for general information, I'll mention a few examples. It depends upon what you have to work with. Charged particles are easy. The product of the magnetic field and the radius of curvature for the trajectory of a charged particle in the field, B*r, is called the magnetic rigidity. It's related to the momentum p and charge, q, of the particle by B*r = p/q, so that p = B*r/q.
>

That means if you need p (=m*v) you have to measure B, r and q. Again r !!!!

Can you divide? Br = p/q = mv/q = m\omega r/q => \omega = qB/m. \omega is the cyclotron resonance. Now, build a velocity filter from F/q = E + v x B = 0. I'm afraid you are stuck having to start somewhere and define something for a standard set of references and then derive additional units from your definitions and rely on self-consistency. I can always choose to pick one type of charged particle, one specific piece of apparatus and define a single constant of proportionality which corresponds to something I can measure while fixing the others. I can define Bq/m, by using a specific magnet with a specific current meter and a specific partcile, say a deuteron, and define a unit like 1 zog (which would be proportional to I/2 since q/A for a deuteron is 1/2). Then I can find particles that are also 1 zog, 2 zogs, etc.

It should be obvious that how you measure something is pretty much irrelavent, so long as you are consistent and choose something that can act as a reference.

> If that is correct than the whole principle becomes like magic.

Self-consistency is not magic, or else it wouldn't be possible to measure anything at all. I'm sorry if self-consistency is not good enough, but that is all you get. Feel free to explain (1) how to do better than that, (2) why it isn't sufficient for any degree of precision obtainable in this universe, given that the universe has to allow you measure up to the resolution it allows you to measure or be an inconsistent universe.

>> > You can also not measure x twice.
>>

Sure, I can. [Possibly not in GR, but I'm not sure how that would work and that's a different issue. Ssomeone else might be able to answer it though]

> Try a different one:
You can also not measure p twice precisely.

You said measure x twice. You did not say measure x twice, precisely. For that matter, I don't even know exactly what you mean by measuring something "twice precisely".

>
> If you don't agree, (that means that you agree with the statement that you can measure p twice precisely) why can not you than in stead of the second measurement of p, measure x precisely ?

Since I don't know exactly what "twice precisely" means, I can't answer this. If you mean two successive measurements, then you can certainly make successive measurements with the same precision as the first, but whether of not the value in the second measurement is the same, is a different issue.

> It seems to me that in the above equation "many" "constants" are set to one. IMO that is tricky way to do science.

It's no different than having f(x) = sin(2x) and deciding to set y = 2x. You can always recover the constants afterword if they are important. The point is that for the most part, there is not any physics in those constants. Just like setting y = 2x, all those constants do is act as a scale factor. Unless you've forgotton, the units in those constants were defined by people to begin with. We could have just as easily started out with units in which all those things really were 1.


17 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 1 april 2002 22:05

"Bilge" schreef in bericht news:slrnaa978d.si2.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff abou
> >

Try a different one: You can also not measure p twice precisely.

>

You said measure x twice. You did not say measure x twice, precisely. For that matter, I don't even know exactly what you mean by measuring something "twice precisely".

> >

If you don't agree, (that means that you agree with the statement that you can measure p twice precisely) why can not you than in stead of the second measurement of p, measure x precisely ?

>

Since I don't know exactly what "twice precisely" means, I can't answer this. If you mean two successive measurements, then you can certainly make successive measurements with the same precision as the first, but whether of not the value in the second measurement is the same, is a different issue.

Accordingly to HUP you can:

Measure p very precise and than measure x roughly Measure x very precise and than measure p roughly

Accordingly to HUP you can not:
Measure p very precise and than measure x very precise Measure x very precise and than measure p very precise.

Accordingly to Hup you can
Measure p precise and than measure x precise Measure x precise and than measure p precise.

Assuming the reader agrees with this definition my question is: Why can not you measure p of a particle twice very precise ? Why can not you measure x of a particle twice very precise ? I expect if you measure p twice (very precise) you could get a different answer but I am not sure if that is always the case.

The same for for x.

Nick


18 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 2 april 2002 12:07

Nicolaas Vroom said some stuff about

> Accordingly to HUP you can:

Measure p very precise and than measure x roughly
Measure x very precise and than measure p roughly

That is _NOT_ what the uncertainty principle states at all. It states that you cannot measure them both in the same measurement. You wanted an example of how to measure both in a single measurement, presumably because you didn't think ot was possible, even classically. I gave you an example.

> Assuming the reader agrees with this definition my question is:

The reader does not agree with the definition.

> Why can not you measure p of a particle twice very precise ?
Why can not you measure x of a particle twice very precise ?
I expect if you measure p twice (very precise) you could get a different answer but I am not sure if that is always the case.

Again, what "measure twice" mean? In the same measurement? Measuring x or p twice in the same measurement means x^2 or p^2. Measuring them in two measurements is just two measurements and it makes no difference if you measure x then x; x then p; p then x; p then p. The measurements are independent so long as you don't try to violate the uncertainty principle by reducing the time between measurements to less that it allows.


19 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 2 april 2002 22:18

"Bilge" schreef in bericht news:slrnaaj10k.700.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff abou
> >

Accordingly to HUP you can:

Measure p very precise and than measure x roughly
Measure x very precise and than measure p roughly

>

That is _NOT_ what the uncertainty principle states at all. It states that you cannot measure them both in the same measurement.

I agree that you cannot measure the position and the momentum of a particle in one measurement ie simultaneous but that is not (IMO ?) what HUP is all about.

HUP makes the following "prediction" If you want to know momemtum at infinite precision than you can not say anything about its position If you want to know position at infinite precision than you can not say anything about its momentum This is expressed by the following equation dp*dx<= h/2pi

For me the question is how do you demonstrate this equation in reality. and show that that equation is correct. How do you measure the position of a particle. How do you measure the momentum of a particle. How do you measure the position first and than the momentum How do you measure the momentum first and than the position.

All those measurements I want to do as precise as possible but accordingly to HUP this is not possible.

So apparently something of my understanding is wrong.

I also want to measure the position first and than again I also want to measure the momentum first and than again Also as precise as possible.

In order to measure the position you need a target with a hole. (Mike Hanson in sci.physics) The smaller the target the more accurate you can determine the position. Accordingly to HUP you can than not determine its momentum accurately. IMO you can also than also not determine its position again accurately.

A similar reasoning should also consists if you measure p first. Two different cases are possible: The particle has a charge The particle has no charge. In order to measure the momentum of a charged particle (Mike Hanson in sci.physics) you need a magnetic field Ofcourse when you do that you change the direction of the particle. I do not know if HUP is all about that ?

> > Why can not you measure p of a particle twice very precise ? Why can not you measure x of a particle twice very precise ? I expect if you measure p twice (very precise) you could get a different answer but I am not sure if that is always the case.
>

Again, what "measure twice" mean? In the same measurement?

No - See above

> Measuring x or p twice in the same measurement means x^2 or p^2. Measuring them in two measurements is just two measurements and it makes no difference if you measure x then x; x then p; p then x; p then p. The measurements are independent so long as you don't try to violate the uncertainty principle by reducing the time between measurements to less that it allows.

What means: so long as you don't try to violate the uncertainty principle ? How can I violate the uncertainty principle ?

Nick


20 Basic question in Heisenberg's uncertainity principle

Van: "Tom Clarke"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 2 april 2002 23:32

Nicolaas Vroom wrote:

> For me the question is how do you demonstrate this equation in reality. and show that that equation is correct. How do you measure the position of a particle. How do you measure the momentum of a particle. How do you measure the position first and than the momentum How do you measure the momentum first and than the position.

All those measurements I want to do as precise as possible but accordingly to HUP this is not possible.

So apparently something of my understanding is wrong.

I also want to measure the position first and than again I also want to measure the momentum first and than again Also as precise as possible.

You have to be clever about the experimental arrangement. Consider light going through a pinhole. At the point when tthe light passes the pinhole you know the position accurately, so HUP says the momentum is unknown.

But when you put a photosensor (film) beyond the pinhole you get a large diffuse cloud of light - this the the result of the large spread in (unknown) momentum casued by the pinhole.

But if you open up the aperture of the pinhole, the cloud of light gets smaller (down to size of hole as the hole gets big) becuase the momentum is not as unknown.

Momentum of light is related to the color or wavelength of the photons. If you pass the light through a narrow bandpass filter, it is the same as measuring the momentum accurately. If you put a narrow pulse of light (known position) into such a bandpass filter, the result that comes out is a long pulse (unknown position) of the known frequency.

Similar sorts of experiments can be set up with charged particles.

Tom Clarke


21 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 3 april 2002 3:42

Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>

"Bilge" schreef in bericht

>> Nicolaas Vroom said some stuff abou
>> >

Accordingly to HUP you can:

Measure p very precise and than measure x roughly
Measure x very precise and than measure p roughly

>>

That is _NOT_ what the uncertainty principle states at all. It states that you cannot measure them both in the same measurement.

>

I agree that you cannot measure the position and the momentum of a particle in one measurement ie simultaneous but that is not (IMO ?) what HUP is all about.

That is what it's about. Or at least the most obvious ramifications.

> HUP makes the following "prediction" If you want to know momemtum at
[*snip*]

What exactly do you think it means to say you can't measure p,x simultaneously? All you did was repeat what I said without knowing it.

> For me the question is how do you demonstrate this equation in reality. and show that that equation is correct.

You go try to measure p,x to better than the uncertainty principle allows. Just like I explained.

[...]
> All those measurements I want to do as precise as possible but accordingly to HUP this is not possible.

Then you have a choice. Either accept the uncertainty principle or set out to do an experiment to show it's wrong.

> So apparently something of my understanding is wrong.

Apparently. The uncertainty principle says that any pair of canonically conjugate variables, p,q have an uncertainty given by their commutator, [p,q] = -ihbar

> I also want to measure the position first and than again
I also want to measure the momentum first and than again
Also as precise as possible.

Those are diffrernt measurements.

>

In order to measure the position you need a target with a hole. (Mike Hanson in sci.physics) The smaller the target the more accurate you can determine the position. Accordingly to HUP you can than not determine its momentum accurately. IMO you can also than also not determine its position again accurately.

Use a 2 targets.

[...]
>> Measuring x or p twice in the same measurement means x^2 or p^2. Measuring them in two measurements is just two measurements and it makes no difference if you measure x then x; x then p; p then x; p then p. The measurements are independent so long as you don't try to violate the uncertainty principle by reducing the time between measurements to less that it allows.
>

What means: so long as you don't try to violate the uncertainty principle ?

Never mind, until you know what the principle means.

> How can I violate the uncertainty principle ?

As far as I know, you can't.


22 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 3 april 2002 12:03

"Tom Clarke" schreef in bericht news:3CAA238B.17C3ECC1@ist.ucf.edu...
> Nicolaas Vroom wrote
> >

For me the question is how do you demonstrate this equation in reality. and show that that equation is correct. How do you measure the position of a particle. How do you measure the momentum of a particle. How do you measure the position first and than the momentum How do you measure the momentum first and than the position.

All those measurements I want to do as precise as possible but accordingly to HUP this is not possible.

So apparently something of my understanding is wrong.

I also want to measure the position first and than again I also want to measure the momentum first and than again Also as precise as possible.

>

You have to be clever about the experimental arrangement. Consider light going through a pinhole. At the point when tthe light passes the pinhole you know the position accurately, so HUP says the momentum is unknown.

This is a rather "empty" sentence (Sorry to say) Can you clarify this by using numbers ? If position is indicate as x and momentum is p and to claim that in this case dx = 0 and that as such dx*dp = 0 *dp = 0 < h/2pi does not clarify it.

> But when you put a photosensor (film) beyond the pinhole you get a large diffuse cloud of light - this the the result of the large spread in (unknown) momentum casued by the pinhole.

In fact what you are doing is a single slit experiment. What this experiment shows is that if you use 1000 single photons (of all the same frequency) through a single hole and you have a matrix of CCD's the photons will not hit the CCD directly after the hole (centre CCD) but they will be spread out (normal distribution) The average distance towards the centre CCD is a function of the size of the hole. Large size small average distance. Small size large average distance.

Is this HUP in action ?

(Assuming that all photons are rather similar and that the results of each experiment with one photon is so much different tells me more about the effect/disturbance that the hole has on the photon than on the photon itself)

You can also do a rather similar experiment but now using to sets of photons with different frequency's f1 and f2 but with the same size of hole. I expect you will get something like f1 * average distance 1 = f2 * average distance 2

Is this HUP in action ?

> But if you open up the aperture of the pinhole, the cloud of light gets smaller (down to size of hole as the hole gets big) becuase the momentum is not as unknown.

Momentum of light is related to the color or wavelength of the photons. If you pass the light through a narrow bandpass filter, it is the same as measuring the momentum accurately. If you put a narrow pulse of light (known position) into such a bandpass filter, the result that comes out is a long pulse (unknown position) of the known frequency.

Similar sorts of experiments can be set up with charged particles.

I think there is a big difference. Is there any one who has tried this and can tell me his or hers experience ?

> Tom Clarke

Nick


23 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 3 april 2002 14:04

"Bilge" schreef in bericht news:slrnaaknpd.8v9.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff abou Re: Basic question in Heisenberg's uncertainity principle to usenet:
> >

"Bilge" schreef in bericht

> >> Nicolaas Vroom said some stuff abou
> >> >

Accordingly to HUP you can:

Measure p very precise and than measure x roughly Measure x very precise and than measure p roughly

> >>

That is _NOT_ what the uncertainty principle states at all. It states that you cannot measure them both in the same measurement.

> >

I agree that you cannot measure the position and the momentum of a particle in one measurement ie simultaneous but that is not (IMO ?) what HUP is all about.

>

That is what it's about. Or at least the most obvious ramifications.

I expect, what you mean is, that in general of a particle you want to know (and to measure) as much as possible when that particle is in the same state (position) at the same instance i.e. "you" want to know all those parameters instantaneous. and or in one experiment (measurement)

IMO you can not. If you want to know momentum than at least you need v which you can only measure in two experiments i.e. over a long period.

>
> >

HUP makes the following "prediction" If you want to know momemtum at

> [*snip*]

What exactly do you think it means to say you can't measure p,x simultaneously? All you did was repeat what I said without knowing it.

> >

For me the question is how do you demonstrate this equation in reality. and show that that equation is correct.

>

You go try to measure p,x to better than the uncertainty principle allows.

I do not want to do it better than HUP allows. I want to see it demonstrated. I want to see what I should expect based on actual experiments performed.

> Just like I explained. [...]
> > All those measurements I want to do as precise as possible but accordingly to HUP this is not possible.
>

Then you have a choice. Either accept the uncertainty principle

Before I accept something I first must fully understand it. You do not understand it when someone asks: How much is dp*dx (d = delta) And you answer: less than or equal to h/2pi

The same "if" you say the following formula is true: dp1*dx1 = dp2*dx2

There is much more involved.

> or set out to do an experiment to show it's wrong.
Before I can do that, I have to understand it. (I hope some one can explain it clearly)

> > So apparently something of my understanding is wrong.
>

Apparently. The uncertainty principle says that any pair of canonically conjugate variables, p,q have an uncertainty given by their commutator, [p,q] = -ihbar

Just some questions in order to understand: What is p. How do you measure p. What is q. How do you measure q What is = -ihbar (h/2pi ?) What means the operator commutator ? In order to demonstrate this equation: Do you have to measure p first and than q ? Please give an example. With actual numbers.

> > What means: so long as you don't try to violate the uncertainty principle ?
>

Never mind, until you know what the principle means.

> >

How can I violate the uncertainty principle ?

>

As far as I know, you can't.

But then why do you mention it ?

Nick


24 Basic question in Heisenberg's uncertainity principle

Van: "Tom Clarke" Onderwerp: Re: Basic question in Heisenberg's uncertainity principle Datum: woensdag 3 april 2002 18:20

Nicolaas Vroom wrote:

> "Tom Clarke" schreef in bericht
> > Nicolaas Vroom wrote
> > >

For me the question is how do you demonstrate this equation in reality. and show that that equation is correct. How do you measure the position of a particle. How do you measure the momentum of a particle. How do you measure the position first and than the momentum How do you measure the momentum first and than the position.

All those measurements I want to do as precise as possible but accordingly to HUP this is not possible.

So apparently something of my understanding is wrong.

I also want to measure the position first and than again I also want to measure the momentum first and than again Also as precise as possible.

> >

You have to be clever about the experimental arrangement. Consider light going through a pinhole. At the point when tthe light passes the pinhole you know the position accurately, so HUP says the momentum is unknown.

>

This is a rather "empty" sentence (Sorry to say) Can you clarify this by using numbers ? If position is indicate as x and momentum is p and to claim that in this case dx = 0 and that as such dx*dp = 0 *dp = 0 < h/2pi does not clarify it.

An extreme pinhole might have a diameter of one micron. If green light of wavelength .5 microns goes through, it will be diffracted over an angle of +/- 30 degrees. (arcsine 0.5)

In particle language the (trasnverse) momentum of the photon would be +/- h/(2pi*(1 micron)). (6.6x10-34 J.s/2pi/1micron = del-p =6.6x10-28 Js/m) The forward momentum of a photon is p=h/wavelength, since wavelength is 0.5 microns, p=13.2x10-27 Js/m) (or kgm/s)

So transversely del-p is half p along the direction of propagation so the same +/- 30 degrees emerges.

> > But when you put a photosensor (film) beyond the pinhole you get a large diffuse cloud of light - this the the result of the large spread in (unknown) momentum casued by the pinhole.
>

In fact what you are doing is a single slit experiment.

Sure.

> What this experiment shows is that if you use 1000 single photons (of all the same frequency) through a single hole and you have a matrix of CCD's the photons will not hit the CCD directly after the hole (centre CCD) but they will be spread out (normal distribution)

Not precisely normal distribution. But an interference pattern with peaks and nodes.

> The average distance towards the centre CCD is a function of the size of the hole. Large size small average distance. Small size large average distance.

Is this HUP in action ?

Yes.

> (Assuming that all photons are rather similar and that the results of each experiment with one photon is so much different tells me more about the effect/disturbance that the hole has on the photon than on the photon itself)

You can also do a rather similar experiment but now using to sets of photons with different frequency's f1 and f2 but with the same size of hole. I expect you will get something like f1 * average distance 1 = f2 * average distance 2

Is this HUP in action ?

Yes. Since p along direction of motion is h/wavelength your formula emerges.

> > But if you open up the aperture of the pinhole, the cloud of light gets smaller (down to size of hole as the hole gets big) becuase the momentum is not as unknown.
>
> >

Momentum of light is related to the color or wavelength of the photons. If you pass the light through a narrow bandpass filter, it is the same as measuring the momentum accurately. If you put a narrow pulse of light (known position) into such a bandpass filter, the result that comes out is a long pulse (unknown position) of the known frequency.

>
> >

Similar sorts of experiments can be set up with charged particles.

> I think there is a big difference. Is there any one who has tried this and can tell me his or hers experience ?

I saw a video with elecron diffraction but I can't recall the name. Here is a URL that shows an apparatus and an image: http://www.ph.utexas.edu/~phy453/lab/electron_diff/elec_diff.html

Tom Clarke


25 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 3 april 2002 20:09

"Tom Clarke" schreef in bericht news:3CAB2BE2.E0A17FD1@ist.ucf.edu...
>

An extreme pinhole might have a diameter of one micron. If green light of wavelength .5 microns goes through, it will be diffracted over an angle of +/- 30 degrees. (arcsine 0.5)

In particle language the (trasnverse) momentum of the photon would be +/- h/(2pi*(1 micron)). (6.6x10-34 J.s/2pi/1micron = del-p =6.6x10-28 Js/m) The forward momentum of a photon is p=h/wavelength, since wavelength is 0.5 microns, p=13.2x10-27 Js/m) (or kgm/s)

So transversely del-p is half p along the direction of propagation so the same +/- 30 degrees emerges.

Congratulations. Thanks for this detailed answer.

> > Similar sorts of experiments can be set up with charged particles. I think there is a big difference. Is there any one who has tried this and can tell me his or hers experience ?
>

I saw a video with elecron diffraction but I can't recall the name. Here is a URL that shows an apparatus and an image: http://www.ph.utexas.edu/~phy453/lab/electron_diff/elec_diff.html

Tom Clarke

Nick http://users.pandora.be/nicvroom/


26 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 3 april 2002 23:44

Nicolaas Vroom said some stuff about [...]
> I expect, what you mean is, that in general of a particle you want to know (and to measure) as much as possible when that particle is in the same state (position) at the same instance i.e. "you" want to know all those parameters instantaneous. and or in one experiment (measurement)

Look, I don't why you are making this more complex than it is. I _deliberately_ did NOT use the words "instantaneous" or "simultaneous" in that last post for a reason. A particle is described by a wavefunction \Psi. A measurement of postion is given by:

x|\psi>

A measurment of the momentum is given by:

p|\psi>

A measurement of both is given by either:

px|\psi> or xp|\psi>

if (px - xp)|\psi> were zero, then you could measure both as precisely as possible. But they aren't. Measuring xp differs from measuring px.

>

IMO you can not.

Then why did you argue with me when I told you that in response to your first post?

[...]
> I do not want to do it better than HUP allows.
I want to see it demonstrated.
I want to see what I should expect based on actual experiments performed.

Any interference pattern demonstrates it.

[...]

>> Then you have a choice. Either accept the uncertainty principle
> Before I accept something I first must fully understand it. You do not understand it when someone asks: How much is dp*dx (d = delta) And you answer: less than or equal to h/2pi

It's _greater_ than, not less than.

> The same "if" you say the following formula is true:
dp1*dx1 = dp2*dx2

You can't say that. All you can say is that those represent the minimum.

[...]
>

Just some questions in order to understand: What is p. How do you measure p.

For what? I mean, p is a momentum. How do you measure a momentum? Well, are we talking about a truck, planet, a photon, an electron, a proton in a nucleus, what? I gave you some examples, but I'm not going to go collect millions of pages describing every experiment ever performed. How you do it depends upon circumstances. Quantum mechanics says it doesn't matter how you do it, so that is really not relavent.

> What is q. How do you measure q What is = -ihbar (h/2pi ?)

sqrt(-1) h/2pi

> What means the operator commutator ? In order to demonstrate this equation: Do you have to measure p first and than q ? Please give an example. With actual numbers.

OK. It appears the problem is that you need to first read some quantum mechanics. What you are asking is to cover the material in the first couple of chapters of a quantum mechanics text in a usenet post without the benefit of being able to write equations and assign homework. If you don't know what an operator or a commutator is, you really need to start by obtaining a book on quantum mechanics. Right off hand, I don't know what is considered to be a good undergraduate level book these days and I have no idea whether the one I used is even in print.

>> > What means: so long as you don't try to violate the uncertainty principle ?
>>

Never mind, until you know what the principle means.

>> >

How can I violate the uncertainty principle ?

>>

As far as I know, you can't.

>

But then why do you mention it ?

Nick

--


27 Basic question in Heisenberg's uncertainity principle

Van:
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 4 april 2002 5:59

Bilge wrote:
>

Nicolaas Vroom said some stuff abou [...]

> > I expect, what you mean is, that in general of a particle you want to know (and to measure) as much as possible when that particle is in the same state (position) at the same instance i.e. "you" want to know all those parameters instantaneous. and or in one experiment (measurement)
>

Look, I don't why you are making this more complex than it is.

Has he ever done anything else in any of the threads he has taken part in?

I am beginning to think that the "misunderstanding" is an act designed to waste peoples' time.

John Anderson


28 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zaterdag 6 april 2002 5:42

ande452@attglobal.net said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
> Bilge wrote
>>

Nicolaas Vroom said some stuff abou [...]

>> > I expect, what you mean is, that in general of a particle you want to know (and to measure) as much as possible when that particle is in the same state (position) at the same instance i.e. "you" want to know all those parameters instantaneous. and or in one experiment (measurement)
>>

Look, I don't why you are making this more complex than it is.

>

Has he ever done anything else in any of the threads he has taken part in?

I am beginning to think that the "misunderstanding" is an act designed to waste peoples' time.

I hadn't realized he was a repeat, but it's not too much of a surprise.


29 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 12 april 2002 16:04

"Tom Clarke" schreef in bericht news:3CAB2BE2.E0A17FD1@ist.ucf.edu...
> Nicolaas Vroom wrote
> >

Can you clarify this by using numbers ? If position is indicate as x and momentum is p and to claim that in this case dx = 0 and that as such dx*dp = 0 *dp = 0 < h/2pi does not clarify it.

>

An extreme pinhole might have a diameter of one micron. If green light of wavelength .5 microns goes through, it will be diffracted over an angle of +/- 30 degrees. (arcsine 0.5)

In particle language the (transverse) momentum of the photon would be +/- h/(2pi*(1 micron)). (6.6x10-34 J.s/2pi/1micron = del-p =6.6x10-28 Js/m) The forward momentum of a photon is p=h/wavelength, since wavelength is 0.5 microns, p=13.2x10-27 Js/m) (or kgm/s)

So transversely del-p is half p along the direction of propagation so the same +/- 30 degrees emerges.

This is a good start of a discussion. In general what "you" are saying is that the diffraction angle = transverse momentum/forward momentum = h/(2pi*diameter) / (h/wave length) = wave length / (2pi * diameter) = arcsin (wave length/ diameter) = acrsin (0.5 / 1) = 30 degrees What this formula also shows is that when you make the diameter smaller the diffraction angle gets larger. I agree that the single slit experiment shows that if you perform this experiment 1000 times (See below) However IMO that experiment does not measure the momentum of a single photon. Nor it does not explain how "uncertainty is involved" except that you have to perform the experiment many times. ("comparing" matter with a wave introduces uncertainty)

> > > But when you put a photosensor (film) beyond the pinhole you get a large diffuse cloud of light - this the the result of the large spread in (unknown) momentum casued by the pinhole.
> >

In fact what you are doing is a single slit experiment.

>

Sure.

> >

What this experiment shows is that if you use 1000 single photons (of all the same frequency) through a single hole and you have a matrix of CCD's the photons will not hit the CCD directly after the hole (centre CCD) but they will be spread out (normal distribution)

>

Not precisely normal distribution. But an interference pattern with peaks and nodes.

IMO that is not true. You get only an interference pattern when you use many photons "simultaneous" or when you use single photons with a double slit

> > The average distance towards the centre CCD is a function of the size of the hole. Large size small average distance. Small size large average distance.

Is this HUP in action ?

>

Yes.

> >

(Assuming that all photons are rather similar and that the results of each experiment with one photon is so much different tells me more about the effect/disturbance that the hole has on the photon than on the photon itself)

You can also do a rather similar experiment but now using to sets of photons with different frequency's f1 and f2 but with the same size of hole. I expect you will get something like f1 * average distance 1 = f2 * average distance 2

Is this HUP in action ?

>

Yes. Since p along direction of motion is h/wavelength your formula emerges.

> > >

But if you open up the aperture of the pinhole, the cloud of light gets smaller (down to size of hole as the hole gets big) because the momentum is not as unknown.

> >
> > >

Momentum of light is related to the color or wavelength of the photons. If you pass the light through a narrow bandpass filter, it is the same as measuring the momentum accurately. If you put a narrow pulse of light (known position) into such a bandpass filter, the result that comes out is a long pulse (unknown position) of the known frequency.

> >
> > >

Similar sorts of experiments can be set up with charged particles.

> > I think there is a big difference. Is there any one who has tried this and can tell me his or hers experience ?
>

I saw a video with elecron diffraction but I can't recall the name. Here is a URL that shows an apparatus and an image: http://www.ph.utexas.edu/~phy453/lab/electron_diff/elec_diff.html

My physics by Kronig (in Dutch) describes the same. However this book does not use that information to explain the HUP. (It does not make the link) This book "describes" page 459: It is impossible to do measurements of systems of atomaire dimensions without disturbing those systems because the tools we use (light quanta, and particles) are of the same size as the object involved. As a consequence we can not determine the initial conditions accurately nor can we predict its future behaviour accurately. We can only use probabilities.

>

Tom Clarke


30 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 12 april 2002 16:04

schreef in bericht news:3CABCFB3.214E@attglobal.net...
> Bilge wrote
> >

Nicolaas Vroom said some stuff abou [...]

> > > I expect, what you mean is, that in general of a particle you want to know (and to measure) as much as possible when that particle is in the same state (position) at the same instance i.e. "you" want to know all those parameters instantaneous. and or in one experiment (measurement)
> >

Look, I don't why you are making this more complex than it is.

>

Has he ever done anything else in any of the threads he has taken part in?

Hum. I have learned a lot by raising questions using Usenet http://users.pandora.be/nicvroom/ Select Usenet (IMO the people who stay silent are the losers)

> I am beginning to think that the "misunderstanding" is an act designed to waste peoples' time.

Every one has the right for his or her own opinion.

Related to HUP there are three issues: Mathematics, interpretations and experiments

IMO the most important part are the experiments specific because the subject of HUP is measurements.

The problem is that most people (books, articles) do not give much information about the experiments involved. How difficult they are, what possible errors are, what the actual results are etc.

Most people want to explain the mathematics involved and the interpretations (hidden variables, many worlds)

My questions are mainly related towards experiments. How do you actual prove/demonstrate HUP As such I ask questions like: How do you actual measure p of a particle (photon, charged particle) A formula like p = m*v is not "enough"

I would like to thank any one who helps me to improve my understanding related to this subject.

Nick

> John Anderson


31 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 12 april 2002 16:05

"Bilge" schreef in bericht news:slrnaamu89.8v9.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff abou

> Look, I don't why you are making this more complex than it is. I _deliberately_ did NOT use the words "instantaneous" or "simultaneous" in that last post for a reason.
Okay

> A particle is described by a wavefunction \Psi. A measurement of postion is given by:

x|\psi>

A measurment of the momentum is given by:

p|\psi>

That is the mathematical part of HUP But how do you do that in practice.

> A measurement of both is given by either:

px|\psi> or xp|\psi>

if (px - xp)|\psi> were zero, then you could measure both as precisely as possible.
But they aren't. Measuring xp differs from measuring px.

With measuring xp I assume you mean: measuring x first and measuring p second. With measuring px I assume you mean:
measuring p first and measuring x second. According to "you" (HUP) the outcome of those experiments is different.
(I agree) However and that is the main reason of my questions:
How do you do that in practice. How do you describe such an experiment in detail ?
Is that the one hole (slit) experiment with single photons ?
How do you do that with charged particles.

> > I do not want to do it better than HUP allows. I want to see it demonstrated. I want to see what I should expect based on actual experiments performed.
>

Any interference pattern demonstrates it.

Interference patterns only show up with photons when you send many photons through a single slit or a single photon through a double slit.

They do not allow you to measure "something" (p, x, px) of a (one) particle as precisily as possible.

If you use a single hole with single photons you must repeat the experiment many times in order to get an average result.

> > Just some questions in order to understand: What is p. How do you measure p.
>

For what? I mean, p is a momentum. How do you measure a momentum? Quantum mechanics says it doesn't matter how you do it, so that is really not relavent.

That is a strong sentence. Quantum mechanics (HUP) says something about measurements. It claims in a quantified way that you can not measure two parameters precisely. dp*dq <= h/2pi (dE*dt <= h/2pi) If you make such a claim than IMO it is important to demonstrate with an actual experiment that this claim is correct. IMO this is very difficult.

> > What is q. How do you measure q What is = -ihbar (h/2pi ?)
>

sqrt(-1) h/2pi

Thanks

Nick


32 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zaterdag 13 april 2002 16:55

Nicolaas Vroom said some stuff about

> That is the mathematical part of HUP But how do you do that in practice.

Any diffraction experiment.

> Bilge wrote
>> A measurement of both is given by either:

px|\psi> or xp|\psi>

if (px - xp)|\psi> were zero, then you could measure both as precisely as possible. But they aren't. Measuring xp differs from measuring px.

> With measuring xp I assume you mean: measuring x first and measuring p second. With measuring px I assume you mean: measuring p first and measuring x second.

Not exactly, but that's close enough.

> According to "you" (HUP) the outcome of those experiments is different. (I agree)

Then why is this a problem?

> However and that is the main reason of my questions: How do you do that in practice. How do you describe such an experiment in detail?

When you shine light on an aperature and see an interference pattern, the uncertainty principle is exactly what you are seeing at work.

> Is that the one hole (slit) experiment with single photons?

That's a good example. In principle, you could move the source of those single photons as far back as you wish so that, the angle formed by the location of the source and the edges of the aperature -> 0. Since p is the total momentum, it's a vector sum, p = p_x + p_y, so that the vector, p = |p|[cos(A)i + sin(A)j], i,j are unit vectors in x,y direction. The vector r from the source to the screen is given similarly by: r = |r|[cos(A)i + sin(A)j].

         |
         |      p_y
           p . |
       . 'A    +--p_x
S : '__________
    ' .   A
         '  .
         |
         |
As the angle A->0, you would expect classically to preferentially select only those photons that have p_y very small, so that the photons follow a straight line through the aperature and strike the screen, just like a bullet would do. But that is not what happens. The aperature constitutes _both_ a position measurement and a momentum measurement. If you choose to view it as a position measurement, then the momentum is completely undetermined, so that it points just like you would expect from a huygens construction. You add up the phases of all the rays to a particlular point on the screen. If you view it as a momentum measurement, i.e., since p = hv/c, you are trying to force p to point along a line from the source to the screen, p == p_x, then the position across the aperature is left completely undetermined, and again you take the rays from every point. However, I'm not going to do all of your work for you.

> How do you do that with charged particles.

The same way. With an interference pattern. The most blatantly non-classical example being the aharanov-bohm effect, which not only produces an interference pattern for electrons which pass on either side of a solenoid (and therefore acts as a double slit), but the interference pattern shifts if you change the field in the solenoid, despite the fact that E=0 everwhere and B=0 everywhere outside the solenoid (i.e., where the electrons are).

> Interference patterns only show up with photons when you send many photons through a single slit or a single photon through a double slit.

They do not allow you to measure "something" (p, x, px) of a (one) particle as precisily as possible.

Then, contrary to what you get from the outline above, you should be able to predict precisely where each photon strikes the screen if you move the source far enough back from the slit and you don't get an interference pattern. Do you think you'll get an interference pattern if you shoot bullets through a 1 cm x 1cm aperature 100 meters away?

> If you use a single hole with single photons you must repeat the experiment many times in order to get an average result.

The "average result" doesn't give you interference. You only get interference if _each_ photon interferes with itself. Try averaging bullets shot through a small hole.

[...]
>> Quantum mechanics says it doesn't matter how you do it, so that is really not relavent.
>

That is a strong sentence. Quantum mechanics (HUP) says something about measurements. It claims in a quantified way that you can not measure two parameters precisely. dp*dq <= h/2pi (dE*dt <= h/2pi) If you make such a claim than IMO it is important to demonstrate with an actual experiment that this claim is correct.

Don't be silly. Since quantum mechanics tells you this is true for any experiment, how do you expect to find an experiment for which this is not true?

> IMO this is very difficult.

It's not difficult at all. Quantum mechanics claims this is true for ANY experiment. I claim every experiment you can possibly imagine will demonstrate the uncertainty principle, regardless of how precise you can make the instruments from materials in this universe. All you have to do is find a _single_ example that proves me wrong. So, I've just stated that every experiment ever done will support what I claim. Find a counter example.

Now, before you come back and say "no it isn't" or ask about the above, make a stab at finding some information (in a qm book or on some website) and try working through some examples and some problems. No one has ever learned anything by having someone tell them everything so that they don't have to think. I've done this already, and don't need the practice. So, if you really have an interest in understanding the material, ask a question about a real excercise you've tried to work and have gotten stuck on. So far, I'm not sure you've done much more than casually peruse any of the information several people have provided for you.

I can come up with lots of experiments that show quantum effects explicitly, but at the moment you aren't in a position to see why the difference between a classical result and quantum result exists, in which case, if you wanted to argue against quantum mechanics, you'd end up like a couple of others who use arguments which are themselves quantum mechanical effects without classical explanations, like parametric down conversion which will only make your argument look as funny those.


33 Basic question in Heisenberg's uncertainity principle

Van: "Frank Wappler"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zondag 14 april 2002 9:42

Bilge wrote:

> Nicolaas Vroom [wrote:]

> > Quantum mechanics (HUP) says something about measurements. It claims in a quantified way that you can not measure two parameters precisely. dp*dq <= h/2pi (dE*dt <= h/2pi)

Actually, the formalism of QM allows to prove the (general) uncertainty principle (as you can find worked out for instance in any decent QM text book):

Delta( A ) * Delta( B ) >= 1/2 sqrt( ((A B - B A)~(A B - B A)) )

for any two measurement operators A and B.

Then for instance with momentum and distance operators, A == p == hbar/i d/dq(), and B == q:

A B - B A == hbar/i (d/dq( q _ ) - q d/dq( _ )) = hbar/i (d/dq( q ) + q d/dq( _ ) - q d/dq( _ )) = hbar/i d/dq( q ) = hbar/i,

and thereby

Delta( p ) * Delta( q ) >= 1/2 hbar = 1/2 h/2Pi.

> > If you make such a claim than IMO it is important to demonstrate with an actual experiment that this claim is correct

No: the proof of the claim is completely indicated above. In order to conduct and analyze any actual experiment you'd have to define in the first place the exact measurement procedures by which you'd determine the result values "Delta( A )" and "Delta( B )". Those exact measurement procedures can be unambiguously represented as mathematical operators which satisfy the shown (general) inequality with mathematical certainty. The only role of quantum mechanics is thereby the realization that the inequality can indeed be proven and therefore needs to be taken seriously whenever the actual definition of measurement procedures/operators is taken seriously.

Not surprisingly, the definition of measurement procedures predetermines the range of result values, and aspects of their statistical distributions (the "Delta" indicates "spread" determined from a sufficiently large set of results) In particular with the above uncertainty principle example: the interdependence of statistical distributions of measurement procedures/operators which involve the same geometric concept, "distance q", in differently defined physical quantities (namely as distance itself, vs. momentum, which is essentially defined as "invariant wrt. change in distance").

In short: any experiment can be only as compelling and demonstrative as your corresponding understanding and description in terms of a thought experiment; and all thought experimentally defined measurement procedures and corresponding operators satisfy the general uncertainty principle with mathematical certainty.

If any actual experiments correspond to their description, then so must the results.

Regards, Frank W ~@) R

p.s.

Or put noticibly pedagogical and efficient and reasoned:

> Don't be silly. Since quantum mechanics tells you this is true for any experiment, how do you expect to find an experiment for which this is not true?

> Quantum mechanics claims this is true for ANY experiment. I claim every experiment you can possibly imagine will demonstrate the uncertainty principle, regardless of how precise you can make the instruments from materials in this universe. All you have to do is find a _single_ example that proves me wrong. So, I've just stated that every experiment ever done will support what I claim. Find a counter example.


34 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zondag 14 april 2002 14:56

"Frank Wappler" schreef in bericht news:s4or8lik9ph.fsf@acunix2.albany.edu...
>

Bilge wrote:

> >

Nicolaas Vroom [wrote:]

>
> > >

Quantum mechanics (HUP) says something about measurements. It claims in a quantified way that you can not measure two parameters precisely. dp*dq <= h/2pi (dE*dt <= h/2pi)

>

Actually, the formalism of QM allows to prove the (general) uncertainty principle (as you can find worked out for instance in any decent QM text book):

Delta( A ) * Delta( B ) >= 1/2 sqrt( ((A B - B A)~(A B - B A)) )

for any two measurement operators A and B.

Then for instance with momentum and distance operators, A == p == hbar/i d/dq(), and B == q:

A B - B A == hbar/i (d/dq( q _ ) - q d/dq( _ )) = hbar/i (d/dq( q ) + q d/dq( _ ) - q d/dq( _ )) = hbar/i d/dq( q ) = hbar/i,

and thereby

Delta( p ) * Delta( q ) >= 1/2 hbar = 1/2 h/2Pi.

You are 100% right. My questions are not related to that.

> > > If you make such a claim than IMO it is important to demonstrate with an actual experiment that this claim is correct
>

No: the proof of the claim is completely indicated above.

The mathematical derivation is supplied.

> In order to conduct and analyze any actual experiment you'd have to define in the first place the exact measurement procedures by which you'd determine the result values "Delta( A )" and "Delta( B )".

Who is this you ? If someone comes up with a claim (a theory) than IMO that same person should also try, indicate how to demonstrate that claim in reality.

If someone claims that "you" can not measure p accurately than that same person should also indicate what means measure, what is p, of what (object) and what means accurately Even better what is accurately ie how much.

If I claim that cold fusion is possible (theory) than at least I should also TRY to demonstrate it.

> Not surprisingly, the definition of measurement procedures predetermines the range of result values, and aspects of their statistical distributions (the "Delta" indicates "spread" determined from a sufficiently large set of results) In particular with the above uncertainty principle example: the interdependence of statistical distributions of measurement procedures/operators which involve the same geometric concept, "distance q", in differently defined physical quantities (namely as distance itself, vs. momentum, which is essentially defined as "invariant wrt. change in distance").

In short: any experiment can be only as compelling and demonstrative as your corresponding understanding and description in terms of a thought experiment; and all thought experimentally defined measurement procedures and corresponding operators satisfy the general uncertainty principle with mathematical certainty.

If any actual experiments correspond to their description, then so must the results.

Regards, Frank W ~@) R

p.s.

Or put noticibly pedagogical and efficient and reasoned:

> >

Don't be silly. Since quantum mechanics tells you this is true for any experiment, how do you expect to find an experiment for which this is not true?

In dutch we say: Mijn naam is haas en ik weet van niets. Translated: My name is hare and I know nothing. Meaning: I keep silent.

> > Quantum mechanics claims this is true for ANY experiment. I claim every experiment you can possibly imagine will demonstrate the uncertainty principle, regardless of how precise you can make the instruments from materials in this universe. All you have to do is find a _single_ example that proves me wrong. So, I've just stated that every experiment ever done will support what I claim. Find a counter example.

One of the best examples (brillant demonstrations) is at: http://www.aip.org/history/heisenberg/p08b.htm

At this url a thought experiment is demonstrated in which one photon (gamma ray) collides with one electron. There is written (parts between quotes) "Heisenberg imagined using this microscope to see an electron and to measure its position. " "He found that the electron's position and momentum did indeed obey the uncertainty relation he had derived mathematically." IMO you can not write "he found" for a thought experiment

"Bohr pointed out some flaws in the experiment, but once these were corrected the demonstration was fully convincing." More or less the same comment as above.

Problem one I have with this thought experiment is that what you measure in the microscope in each run is the position dx of the reflected/ diffracted (one) gamma ray. What is not clear how you measure dp and as such that: Dpx *Dx ~ h.

At the end of the url is written: "The true quantum interaction, and the true uncertainty associated with it, cannot be demonstrated with any kind of picture that looks like everyday colliding objects. " "To get the actual result you must work through the formal mathematics that calculates probabilities for abstract quantum states. " " Clever experiments on such interactions are still being done today. " Can some one point me to an url where those are described ? Do those experiments measure dx and dp in such a way that they are in agreement with dx*dp~h ?

"So far the experiments all confirm Heisenberg's conviction that there is no "real" microscopic classical collision at the bottom." Hum....

At page 35 of the book Physics and Philosophy by Werner Heisenberg about the above experiment he writes: page 35 peguin isbn 0-14-014660-1

"The electron may have been practically at rest before the observation" The question that immediates pops up is the following: how do you know that? You can not use light to establish that.

Next he writes: "But in the act of observation at least one light quantum of the g-ray must have past the microscope and must first have been deflected bt the electron" "therefore the electron has been pushed by the light quantum it has changed its momentum and its velocity and one can SHOW that the uncertainty of this change is just big enough to guarantee the validity of the uncertainty relations." What means show ? How ? (IMO to demonstrate HUP by experiment is impossible)

I have no doubt that if a photon hits an electron that the v of the electron before and after is different, but that does not mean (show) that the amount of change is in agreement with dp*dx ~ h Accordingly to the url the frequency (L) of the photon is different before and after. (Later they declaire that this difference is zero) That seems (?) IMO to be more a measure of how much the v of the electron changes.

Nick


35 Basic question in Heisenberg's uncertainity principle

Van: "ca314159"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zondag 14 april 2002 16:29

Nicolaas Vroom wrote:

> That is the mathematical part of HUP But how do you do that in practice. It claims in a quantified way that you can not measure two parameters precisely. dp*dq <= h/2pi (dE*dt <= h/2pi) If you make such a claim than IMO it is important to demonstrate with an actual experiment that this claim is correct. IMO this is very difficult.

Simplify by dividing the energy-time relationship by h.

The simpler statement is then: How does one measure frequency at a specific time ?

If you have a camera shutter in front of a prism and you let through some sun light, the spectrum is only determined for the time interval for which the shutter is open. The faster the shutter the more precise the time resolution is, but the spectrum becomes more uncertain since you're letting through less light.

Then Einstein said to Bohr[3], if the shutter lets through only one photon then I know the exact frequency of the photon and the time it was measured. But Bohr replied: but, we are uncertain about the whole spectrum of the source of that photon and therefore we can't be certain about the motion of the source since we don't have enough samples to determine a relativistic Doppler shift and so we really don't know then the exact time the photon was released from the source.

A holographer never uses a mechanical shutter mounted on an optical isolation table because it would vibrate the table. Instead, they might block the beam of light with a piece of black cardboard sitting on the table. When it comes time to make the exposure they lift the cardboard "shutter" up off the table but still holding it in front of the beam to block it. Then after waiting for awhile for the vibrations caused by lifting the shutter off the table to reside, they remove it quickly from in front of the beam to release the light with less ill effects on the sensitive hologram.

But to some extent even the this method of eliminating the effects of measurement is not perfect since removing the shutter from in front of the beam causes a minute amount edge scattering. Happily though this is much less statistically significant that table vibration on the results.

A longer exposure time can wash out small noises but it also exposes the result to more noise, so there is a window of signal to noise ratios that needs to be achieved. That's where the uncertainty principle really shows itself in terms of a "sweet spot". For the macrocosmic holographer the uncertainty relationship is very broad and he can make a reasonable hologram. In the microcosm though, the sweet spot is on the order of h.

But it very important to keep clear exactly what and when one is referring to when using the words of position, time, energy and momentum otherwise you'll wind up like Bohr and Einstein quibbling whether "time" meant "the time the photon left the source" or "the time the photon hit the detector" or "the time the photon was Fourier transformed by the prism" etc.

[1] Introduction to Digital Signal Processing, John Karl
[2] Digital Signal Processing in C, Reid and Passin
[3] http://www.nobel.se/physics/educational/tools/quantum/interpretation-2.html


36 Basic question in Heisenberg's uncertainity principle

Van: "Frank Wappler"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zondag 14 april 2002 20:58

Nicolaas Vroom wrote:

> Frank Wappler [wrote:]
> > In order to conduct and analyze any actual experiment you'd have to define in the first place the exact measurement procedures by which you'd determine
^^^

> > the result values "Delta( A )" and "Delta( B )".

> Who is this you ?
^^^ (just trying to point out what took me a moment's effort to track down :)

> [...] If someone claims that "you" can not measure p accurately [then] that same person should also indicate what [it] means [to] measure, what is p, of what (object) and what [is meant by] accurately Even better what [it] is accurately [...]

That's exactly my point; and that's what QM emphasizes by making this requirement explicit in the formalism. Namely: "momentum p" or any other physical quantity of which we'd like to obtain and agree on measurements must be defined through a measurement procedure

which guarantees that (only) _real number_ result values are obtained

(in successful trials; or otherwise, that the trial can be identified and rejected as unsuccessful/not_properly_set_up/inconclusive/etc.).

However, the claim and (seemingly somewhat paradoxial) implication are _both_ momentum p as well as distance q can be measured with in prinicple unlimited accuracy in _separate_ sets of trials,

but that their _definitional interdependence_ implies a certain statistical interdependence of result values if those result values were obtained jointly from observations collected in one and the same set of trials.

The most straightforward explicit example I know would be the attempt of a joint measurement of distances and of (average) velocity (... you'll believe that velocity is quite closely related to momentum, I presume) from one particular small set of trials:

Consider given:

- first A met K (thus measuring distance A to K accurately as x( who met K, A )_when( A met K ) = 0
- then B met K, and
- A and B succeeded measuring their distance wrt each other as some nonzero value, accurately equal throughout K's travelling

x( who met K, A )_when( B met K ) = x( B, A )_always

- A and B succeeded measuring the interval "tK_B_from_A" between their individual observations of B having met K and A having met K.

Given these observations and basic measurements A (and B) and K can obtain _one accurate value_ of K's average velocity wrt A (and B),

namely v_averageK_A == (x( B, A )_always - 0) / tK_B_from_A.

But, necessarily in the process of deriving v_averageK_A, there were obtained _two distinct values_ of x( who met K, A ), namely 0 as well as x( B, A )_always (which was supposed to be nonzero).

Taking these two distinct (though individually perfectly accurate) values as input in the determination of Delta( x( who met K, A ) ) as defined over all result values through the entire experimental trial then, not surprisingly, it is found Delta( x( who met K, A ) ) > 0.

> If someone comes up with a claim (a theory) than IMO that same person should also try, indicate how to demonstrate that claim in reality.

Agreed; for any experimentally falsifiable theory one must define to begin with how to measure the experimental quantities in terms of which the (prediction of the) theory is expressed.

But therefore, inevitably: QM is _not_ an experimentally falsifiable theory itself, but instead equivalent to the statement agreed just above, namely _that_ experimental quantities need to be defined. (This emphasized point of) QM is a _prerequisit_ to expressing and recognizing any particular falsifiable theory.

Instead, QM is therefore to be considered a (general) measurement procedure, which are definitions that are prerequisit to any experiemental results and which therefore cannot in turn be falsified by experimental results. But of course there are still criteria which measurement procedures must satisfy: they must be unambiguously understandable and agreeable before they're being employed to derive experimental results/measurments from the various observations that will be collected (or that may have been collected already).

Anyone who asks for "experimental evidence for or against QM" or tries to "explain" aspects of QM (such as the uncertainty principle) _only through_ particular experimental results thereby disquailfies him/her/itself from having understood QM generally.

> One of the best examples (brillant demonstrations) is at: http://www.aip.org/history/heisenberg/p08b.htm

> Problem one I have with this thought experiment is that what you measure in the microscope in each run is the position dx of the reflected/ diffracted (one) gamma ray.

No, AFAIU, the gamma ray is to be observed behind the microscope (in order to define a successful trial/run), and the measurement derived from observing the gamma ray e.g. on a screen, is that the electron had been in front of the aperture anywhere in regions of the order delta_x = wave_length / (2 sin( half_opening_angle ));

where further details of the observations on/by the screen might be used to decide just in which of these possibly many approximate regions the electron had been contained.

> What is not clear how you measure dp and as such that: Dpx *Dx ~ h.

IIUC, what's been measured about the electron's momentum is (only) that its x momentum before being illuminated must have been around zero plus/minus 2 h sin( half_opening_angle ) / wave_length

(and perhaps that afterwards it may have x momentum anywhere about within twice that range).

> At the end of the url is written:
"The true quantum interaction, and the true uncertainty associated with it, cannot be demonstrated with any kind of picture that looks like everyday colliding objects. "

I'd hope that this expresses my above statement about QM in general; but then one shouldn't have to ask for ever "more true" quantum interactions "underneath" what's been described already.

> "To get the actual result you must work through the formal mathematics that calculates probabilities for abstract quantum states. "

Which you indicated being able to do, at least in principle.

> "Clever experiments on such interactions are still being done today." Can some one point me to an url where those are described ?

Well, I've always found Paul Kwiat's

http://www.physics.uiuc.edu/People/Faculty/profiles/Kwiat/Interaction-Free-Measurements.htm>

(which used to be the now defunct http://p23.lanl.gov/Quantum/kwiat/ifm-folder/ifmtext.html

pretty clever; though not directly connected to the uncertainty relation and not presented well/general enough to exhibit this connection nevertheless.

> "So far the experiments all confirm Heisenberg's conviction that there is no "real" microscopic classical collision at the bottom."

> Hum ....

... and a loud echo from here. What, if anything particular, might the authos mean by "classical collision" ??

> []

> How ? (IMO to demonstrate HUP by experiment is impossible)

As explained, I agree in principle; instead, the HUP can be motivated and subsequently be exactly derived based on having asked "How ?" in the first place, i.e. by taking the question as serious as can be matematicall expressed.

> I have no doubt that if a photon hits an electron that the v of the electron before and after is different,

_How_ would you define and measure "the v of the electron", to be so free of doubt ? (picked up below, at least for what it is worth for now.)

> Accordingly to the url the frequency (L) of the photon is different before and after.

I haven't quite found that spelled out (the drawing don't seem to come up, and L denotes of course the wave_length which equals c/frequency),

by I figure that, yes, L always denotes the wave_length only upon observation back at the "screen".

> (Later they declaire that this difference is zero)

No, the url uses the approximation (for small enough apertures) L_left_extreme =~= L_right_extreme =~= L_any_other_way.

The wave_length upon emission and before scattering doesn't seem to appear at all.

> That

... i.e. the difference between initial and final photon frequency/energy/momentum/wavelength ...

> seems (?) IMO to be more a measure of how much the v of the electron changes.

But by design of the setup only for the v component along the (general/approximate) direction "source/screen - aperture - electron", not for the direction x _perpendicular_ to the "line of sight".

The distance component of the electron perpendicular to the aperture (e.g. wrt the rim of the aperture) and the velocity of the electron along the "line of sight" are _not_ necessarily interrelated by their very definitions (of how to measure them).

Regards, Frank W ~@) R

p.s.

> In dutch we say: Mijn naam is haas en ik weet van niets.

Or, as the German's dubbed Buggs Bunny (for the purpose of rhyming, as well as to translate BB correctly):

Mein Name ist Hase -- ich wei$ bescheid!
(My name is hare -- I know what's going on!)

p.p.s.

> > Or put noticibly pedagogical and efficient and reasoned:

Following rather unintentionally the premise that "less is more".


37 Basic question in Heisenberg's uncertainity principle

Van: "Frank Wappler"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zondag 14 april 2002 21:15

Frank Wappler wrote:

> Anyone who asks for "experimental evidence for or against QM" or tries to "explain" aspects of QM (such as the uncertainty principle) _only through_ particular experimental results thereby

... timeout:

> [disqualifies]

continue ...

> him/her/itself from having understood QM generally.

(Hope this still edged into extended/overtime play. :)


38 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 15 april 2002 10:48

"Bilge" schreef in bericht news:slrnabgi3v.tlk.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff about
> >

However and that is the main reason of my questions: How do you do that in practice. How do you describe such an experiment in detail?

>

When you shine light on an aperature and see an interference pattern, the uncertainty principle is exactly what you are seeing at work.

Interference patterns you get when you perform the experiments by Fresnel and Young. (I have done that one at University) However I doubt if that experiment allows you to validate HUP ie dp*dx ~ h

> > Interference patterns only show up with photons when you send many photons through a single slit or a single photon through a double slit.

They do not allow you to measure "something" (p, x, px) of a (one) particle as precisily as possible.

>

Then, contrary to what you get from the outline above, you should be able to predict precisely where each photon strikes the screen if you move the source far enough back from the slit and you don't get an interference pattern. Do you think you'll get an interference pattern if you shoot bullets through a 1 cm x 1cm aperature 100 meters away?

> >

If you use a single hole with single photons you must repeat the experiment many times in order to get an average result.

>

The "average result" doesn't give you interference. You only get interference if _each_ photon interferes with itself.

I do not understand why you mention interference patterns twice. If you shoot single photons through a double slit you get an interference pattern by repeating the experiment many times. However in each single instance how much is x (dx) and how much is p (dp) ? I doubt if those results allow you to validate HUP.

> Try averaging bullets shot through a small hole.

I agree. If you shoot single photons through a single slit you get a normal distribution by repeating the experiment many times. However in each single instance how much is x (dx) and how much is p (dp) ? IMO it is impossible to validate HUP.

See also my comments to Frank Wappler
http://www.aip.org/history/heisenberg/p08.htm
http://www.honors.unr.edu/~fenimore/wt202/close/
http://www.mathpages.com/home/index.htm Select "Reflections on Relativity"
http://www.mtnmath.com/ Select "book"


39 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 15 april 2002 15:44

"Frank Wappler" schreef in bericht news:s4ou1qekszn.fsf@acunix1.albany.edu...
>

Nicolaas Vroom wrote:

>
> >

One of the best examples (brillant demonstrations) is at: http://www.aip.org/history/heisenberg/p08b.htm

>
> >

Problem one I have with this thought experiment is that what you measure in the microscope in each run is the position dx of the reflected/ diffracted (one) gamma ray.

>

No, AFAIU, the gamma ray is to be observed behind the microscope (in order to define a successful trial/run), and the measurement derived from observing the gamma ray e.g. on a screen, is that the electron had been in front of the aperture anywhere in regions of the order delta_x = wave_length / (2 sin( half_opening_angle ));

where further details of the observations on/by the screen might be used to decide just in which of these possibly many approximate regions the electron had been contained.

The demonstration/movie, as part of this URL, seems easy but IMO it is not as simple as shown. The movie starts with a photon which moves in a horizontal path to the right and collides with the electron. This photon has a wave length of L.

This raises immediate 3 questions.
1. How do you know that there is one photon
2. How do you know that the photon followed a straight path.
3. How do you know that the photon has a wave length of L

The only thing that you know is that there was at least one photon which was reflected by the electron (Because one pixel (CCD) of the microscope was activated) after initial set up of the experiment.
There could be more photons involved which were not detected/seen.
IMO in reality photon/electron reflections can go in any direction and the influence of the photon can be in a continuous range from 1 almost nothing, 2 a litle, 3 a litle more, 4 a lot, 5 heavy, 6 maximum Where 6 is the maximum possible disturbance of the electron.

In reality more photons could have disturbed the electron which makes the initial assumption that the electron is practically at rest very doubtfull.

Question 3 is a tricky one because the photon is generated by a photongenerator which consists of "agitated" calcium atoms ("agitated" means that the electrons are in a higher energy state) Those "agitated" electrons can jump back to a lower state by releasing a photon. You can calculate the frequency of such photons. You can also measure the frequency of such photons. But you can not measure the frequency of one particular photon specific not the frequency/wave length of the photon used in the experiment because such a measurement will "destroy" the photon.

As such as part of the experiment you use "assumed" knowledge/ information, specific you assume that the frequency of all photons generated by your photongenerator is identical, without measuring it. (I have no problems with this assumption)

> > What is not clear how you measure dp and as such that: Dpx *Dx ~ h.
>

IIUC, what's been measured about the electron's momentum is (only) that its x momentum before being illuminated must have been around zero plus/minus 2 h sin( half_opening_angle ) / wave_length

(and perhaps that afterwards it may have x momentum anywhere about within twice that range).

> >

At the end of the url is written: "The true quantum interaction, and the true uncertainty associated with it, cannot be demonstrated with any kind of picture that looks like everyday colliding objects. "

>

I'd hope that this expresses my above statement about QM in general; but then one shouldn't have to ask for ever "more true" quantum interactions "underneath" what's been described already.

> >

"To get the actual result you must work through the formal mathematics that calculates probabilities for abstract quantum states. "

>

Which you indicated being able to do, at least in principle.

That is not true. For example I do not know what abstract quantum states are nor what the outcome of those calculations is.

The tenure of the whole URL is:

We show you an (animated) movie We show you some mathematics dp*dx ~ h

but

In reality the whole issue is more "complex"

and then ..... nothing

> > I have no doubt that if a photon hits an electron that the v of the electron before and after is different,
>

_How_ would you define and measure "the v of the electron", to be so free of doubt ? (picked up below, at least for what it is worth for now.)

I have no idea (the measuring) using single photons. And that is exactly the problem I have with the uncertainty relation. (You can not measure anything twice the same)

You need millions of photons to study collissions between identical billiard balls accurately.

To study the behaviour of a single photon by using a single photon is impossible.

This is identical as to study the bahaviour of one billiard ball by using only one billiard ball once.

IMO we can not measure the behaviour of one photon (without disturbing it) but that does not mean that we can not say (that means that we can say) that a specific photon has certain properties without measuring them.

Nick


40 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 16 april 2002 22:34

Frank Wappler said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:

> Delta( A ) * Delta( B ) >= 1/2 sqrt( ((A B - B A)~(A B - B A)) ) for any two measurement operators A and B.

You mean what everone else calls an "observable"?

> Then for instance with momentum and distance operators, A == p == hbar/i d/dq(), and B == q:

A B - B A ==
hbar/i (d/dq( q _ ) - q d/dq( _ )) =
hbar/i (d/dq( q ) + q d/dq( _ ) - q d/dq( _ )) =
hbar/i d/dq( q ) =
hbar/i,

and thereby

Delta( p ) * Delta( q ) >= 1/2 hbar = 1/2 h/2Pi.

A = L^2, B = Lz => [L^2, Lz] = 0


41 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 16 april 2002 22:46

Nicolaas Vroom said some stuff about

> However I doubt if that experiment allows you to validate HUP ie dp*dx ~ h

Yes, it does. Try using a search engine. You apparently consider doing some of your own work a waste of time. I find constructing answers you won't read or attempt to figure out before snipping everything and saying "Nuh-uh", to be a lower priority.

[...]
>

I do not understand why you mention interference patterns twice.

I'm sure you don't. You seem determined to avoid it.

[...]
> IMO it is impossible to validate HUP.

You HO is wrong. I've already done as much as I can do to explain it to you without you deciding to cooperate and pick up a pen and some paper.

>

See also my comments to Frank Wappler

IMO, you should have him explain it to you so that your comments will be justified.


42 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 17 april 2002 15:59

"Bilge" schreef in bericht news:slrnabp3q0.ft0.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff about
> >

However I doubt if that experiment allows you to validate HUP ie dp*dx ~ h

>

Yes, it does. Try using a search engine.

I did http://www.google.com/search?hl=en&q=Heisenberg+Uncertainty+Principle+Experi ment

Using that one I found: http://www.aip.org/history/heisenberg/p08.htm and specific: http://www.aip.org/history/heisenberg/p08b.htm The gamma ray microscope.

See also the bottom of the (my) previous posting.

With: http://www.google.com/search?hl=en&q=aharanov-bohm I found http://fisica.udea.edu.co/~mpaez/aharanov/Bohmen.html which shows an interesting demo

Also interesting is: http://www.google.com/search?hl=en&q=aharanov-bohm+uncertainty

> You apparently consider doing some of your own work a waste of time.
No

> I find constructing answers you won't read or attempt to figure out before snipping everything and saying "Nuh-uh", to be a lower priority.

> [...]
> >

I do not understand why you mention interference patterns twice.

>

I'm sure you don't. You seem determined to avoid it.

See below

> [...]
> > IMO it is impossible to validate HUP.
>

You HO is wrong. I've already done as much as I can do to explain it to you without you deciding to cooperate and pick up a pen and some paper.

In order to validate HUP there are two strategies: By using light ie "millions" of photons "simultaneous" By using single photons.

An example of the first case are the experiments by Fresnel and Young.
In this experiment you get interference patterns. My physics book (Kronig in dutch page 389) tells me that the distance between two strong intensity bands = l * wavelength / (2a)
l = distance to screen
2a = distance between the 2 light sources ie distance between the slits.

The text near this experiment explains: "From the formula you can see that the distance between the strong intensity bands is a function of the wave length" "By measuring this distance you can calculate the wavelength of the used light. "

I have myself participated in this experiment.

However this paragraph/chapter does not mention HUP.

In the second case you need a laser and a tool to control the laser such that your "single photon generator" generates/ creates single photons one by one.

Which such a device you can do three types of experiments:
1. Use a single slit.
2. Use a double slit
3. Use an electron and gamma ray microscope.

With a single slit and performing many runs you get one intensity light band with follows a normal distribution. There is no interference pattern. It is not clear in that experiment how you can measure p of a single photon.

With a double slit and performing many runs you get many high and low intensity light bands ie there is an interference pattern. However: it is not clear in this experiment how you can measure p of a single photon. This experiment is very similar as Fresnel and Young. See also aharanov-bohm urls mentioned previous.

The third experiment is explained in detail at: http://www.aip.org/history/heisenberg/p08b.htm And in the book Physics and philosophy by Werner Heisenberg.

In this experiment a gamma ray is reflected by an electron and detected/observed in the microscope.
It is not clear how you can measure from that result the p from the electron.
Mr Heisenberg in his book writes:
"The electron may have been practically at rest before the observation."
How do you know that ? How do you know that before you observed a flash in your microscope that there was not an other photon released by your single-photon-generator which collided with the electron but which was not detected in the microscope ?

See also my new remarks to Frank Wappler

> >

See also my comments to Frank Wappler

>

IMO, you should have him explain it to you so that your comments will be justified.

HUP claims that you can measure p precisely but than you can not measure x precisely of a single particle or photon. HUP claims that you can measure x precisely but than you can not measure p precisely. HUP even "quantifys" this claim: dp * dx ~ h

IMO you can not measure p of a single photon nor p of a single electron by using one single photon

For more information see:
http://www.google.com/search?hl=en&q=Heisenberg+Uncertainty+Principle+Experiment

http://www.scienceandfaith.org/physI/heisenberg6a.htm
The following url states:
The act of measuring one magnitude of a particle, be it its mass, its velocity or its position, causes the other magnitudes to blur.
http://www.thebigview.com/spacetime/uncertainty.html
IMO you should add something like: including the one you want to measure (ie position)

http://www.honors.unr.edu/~fenimore/wt202/close/ Altijd is kort jakje ziek. Midden in de week maar zondag niet

The relation between HUP and Big Bang http://www.google.com/search?hl=en&q=Heisenberg+Uncertainty+Principle+Big+Bang&btnG=Google+Search

I do not think you can not make this "link" without first clearly understanding HUP

Nick


43 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 17 april 2002 20:15

Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>

"Bilge" schreef in bericht news:slrnabp3q0.ft0.root@radioactivex.lebesque-al.net...

[...]
>
>>

You apparently consider doing some of your own work a waste of time.

> No

I can only go by your response, which snipped everything I wrote without a single specific objection.

>>

You HO is wrong. I've already done as much as I can do to explain it to you without you deciding to cooperate and pick up a pen and some paper.

>

In order to validate HUP there are two strategies: By using light ie "millions" of photons "simultaneous" By using single photons.

Those strategies are the same. Every photon interferes with itself, therefore experiments which use "millions" of photons, differ only by having a greater rate of "single photons", thereby obtaining good statistics faster. Single photon experiments are useful to show that photons interfere with themselves, not other photons. Once that's established, both represent single photon experiments, differing only by rate.

[...]
>

The text near this experiment explains: "From the formula you can see that the distance between the strong intensity bands is a function of the wave length" "By measuring this distance you can calculate the wavelength of the used light. "

I have myself participated in this experiment.

However this paragraph/chapter does not mention HUP.

Since classical mechanics is a limiting case of quantum mechanics, there must be a limiting case which describes the uncertainty principle here. This is geometrical optics.

> In the second case you need a laser

And what I'm telling you to do is to first take a pen and some paper, do the mathematics, and work some problems involving the uncertainty principle. Many of those involve exactly these effects. If you don't make some attempt to figure out why all of these experiments are the exact same experimemt in different disguises, telling you that they are and having you repeat the same list of experiments is not going to be of much use. I'm sorry, but usenet is not a replacement for the conceptual understanding one obtains by getting stuck for a few hours (or days as the case might be) trying to work a problem and perhaps asking _specific_ questions about it.

[...]
> HUP claims that you can measure p precisely but than you can not measure x precisely of a single particle or photon. HUP claims that you can measure x precisely but than you can not measure p precisely. HUP even "quantifys" this claim: dp * dx ~ h
IMO you can not measure p of a single photon nor p of a single electron by using one single photon

I can always measure it to the precision: p = hbar/\Delta x


44 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: donderdag 18 april 2002 13:19

"Bilge" schreef in bericht news:slrnabrfb9.jgo.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
> >

HUP claims that you can measure p precisely but than you can not measure x precisely of a single particle or photon. HUP claims that you can measure x precisely but than you can not measure p precisely. HUP even "quantifys" this claim: dp * dx ~ h
IMO you can not measure p of a single photon nor p of a single electron by using one single photon

>

I can always measure it to the precision: p = hbar/\Delta x

Is that your final answer ?

The following search revealed two interesting links
http://www.google.com/search?hl=en&q=Heisenberg+Uncertainty+Principle+Schrodinger&btnG=Google+Search

Bohr Theory page 2
http://www.bcpl.net/~kdrews/bohr/bohr2.html

At this URL is written:
"Bohr Theory ran into trouble with this principle because Bohr tried to predict the movement of the electron too precisely. By restricting the electron to certain locations on the atom and having it move in paths called orbits, he violated the Heisenberg UP. " "The Modern Theory of Atomic Structure, created by Schrodinger, eventually solved the problem of Bohr's conflict with the Heisenberg UP. "

IMO Heisenberg should have put more water in the wine.

There is a recent discussion in sci.physics.research "Simple question related to basic quantum mechanics.
It discusses this issue: *** Can you prove the Schrodinger equation? The general answer is: "No"

The Uncertainty Principle (the theory)
http://library.thinkquest.org/C005775/Theory/newtheory_section2.html
This URL shows a simple demonstration of HUP. IMO it is too simple.

Heisenberg in his book at page 34 writes: "Therefore, the theoretical interpretation of an experiment requires three distinct steps:
(1) the translation of the initial experiment situation into a probability function
(2) the following up of this function in the course of time
(3) the statement of a new measurement to be made of the system, the result of which can then be calculated from the probability function."

Next he writes at page 35:
"Is the first step, the translation of the result of the observation into a probability function possible? It is possible only if the uncertainty relation is fulfilled after the observation."

Why does Mr Heisenberg writes such difficult sentences while HUP is so simple, as the above URL seems to indicate?

The following url: "Particle or wave" contains a very nice demonstration about the two slit experiment.
http://library.thinkquest.org/C005775/Theory/particle_or_wave.html
Select : See the two slit experiment.
This demonstration shows what is observed when resp.: balls, light and single electrons go through a single or double slit.
Unfortunate it does not contain a demonstration what is observed (how) when single photons go through a double slit.

IMO it should show that each single photon goes through both slits.

IMO If you want to know the position (velocity) of an electron precisely and you use electrons than you have to know the velocity of those electrons as precisely. On the other hand each electron will influence the state of the traget drastically This makes the calculation of its speed "impossible"

If you want to know the position (velocity) of an electron precisely and you use single photons (smaller particle size) than you have to know the velocity (direction) of those photons as precisely. The photons will influence the traget less. This makes the calculation of its speed more practical.

On the other hand photons are partly particle partly wave. This implies that the collision between a photon and an electron is not a clean experiment (mathematically) when you observe the results of such an event. This again makes the calculation of the speed of the electron more difficult.

IMO the best reason why you can not validate HUP is HUP itself.

Nick.


45 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 19 april 2002 7:24

Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>

"Bilge" schreef in bericht news:slrnabrfb9.jgo.root@radioactivex.lebesque-al.net...

>> Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>> >

HUP claims that you can measure p precisely but than you can not measure x precisely of a single particle or photon. HUP claims that you can measure x precisely but than you can not measure p precisely. HUP even "quantifys" this claim: dp * dx ~ h
IMO you can not measure p of a single photon nor p of a single electron by using one single photon

>>

I can always measure it to the precision: p = hbar/\Delta x

>

Is that your final answer ?

Apparently so. You don't intend to do anything on your own, and I'm certainly not going to bother wasting time explaining the same thing a thousand different ways, hoping to save you the effort of thinking. If you wan't more of an answer, work a quantum mechanics problem and ask a specific question when you get stuck and can't make any progress after a several hours.

[...]

> IMO it should show that each single photon goes through both slits.

YO is wrong, and IMO, if you spent some time and effort on some quantum mechanics, you'd feel foolish for making such statments.

> IMO If you want to know the position (velocity) of an electron precisely

YO is based upon not knowing what you are talking about.

[...]
>

IMO the best reason why you can not validate HUP is HUP itself.

YO is metaphysical mumbo jumbo. IMO, you can't learn any physics by pontificating. You have to actually pick up a pen, work problems on paper and get frustrated.


46 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 19 april 2002 10:30

"Bilge" schreef in bericht news:slrnabvava.r7f.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
> >

IMO it should show that each single photon goes through both slits.

>

YO is wrong, and IMO, if you spent some time and effort on some quantum mechanics, you'd feel foolish for making such statments.

How do you solve this problem ?
- With pen and paper ?
- By performing as many as possible ^different^ experiments ?
- By common sense ?
- ...?

Anyway what is the right answer (when one CCD is activated) ? Does the "single photon" go through:
- No hole ?
- The left hole ?
- The right hole ?
- Both holes ?
- Either the left or the right hole but not through both holes ?
- .... ?

Readers are suggested to reply by e-mail

Nick http://users.pandora.be/nicvroom/


47 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zaterdag 20 april 2002 20:13

Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>

"Bilge" schreef in bericht news:slrnabvava.r7f.root@radioactivex.lebesque-al.net...

>> Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>> >

IMO it should show that each single photon goes through both slits.

>>

YO is wrong, and IMO, if you spent some time and effort on some quantum mechanics, you'd feel foolish for making such statments.

>

How do you solve this problem ?

By first understanding it.


48 Basic question in Heisenberg's uncertainity principle

Van: "Frank Wappler"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zondag 21 april 2002 9:21

Nicolaas Vroom wrote:

> > > http://www.aip.org/history/heisenberg/p08b.htm

Frank Wappler [wrote:]
> > [the drawing don't seem to come up ...]

(I've managed to download that little animated gif separately; now we should be on the same page.)

> The movie starts with a photon which moves in a horizontal path to the right and collides with the electron. This photon has a wave length of L.

I still claim that the written page itself refers to wavelengths L, L', or L'' strictly only _after_ the "scattering" (i.e. after the electron observed the source signalling). As far as the animation shows otherwise, I consider it wrong.

> This raises immediate 3 questions. 1. How do you know that there is one photon

The entire derivation considers precisely one lightsignal being observed at the detector side of the aperture. Whether and which other lightsignals might be exchanged simply doesn't enter the presented considerations.

Perhaps, if the exchanges of sufficiently many lightsignals were being observed, if might be possible to infer the particular of the potential wrt the electron -- perhaps, for instance, the electron might turn out being actually in a tube, one end of which constitutes the aperture.

> 2. How do you know that the photon followed a straight path.

There is no requirement involved that any photon "followed a straight path"; only that the momentum associated with the lightsignal which is exchanged between the electron and the detector(s) is of order h 1/2 aperture_diameter / distance( aperture_rim, electron ) 1/L.

Also, there's no assumption expressed about any photon having "followed a straight path" except in that unfortunate animation.

> 3. How do you know that the photon has a wave length of L

There is no requirement involved for determining the value L of photon wave length as such, but only its ratio to the smallest distance at which a microscope of a given aperture_diameter can resolve patterns on or near the aperture axis, at a given distance from the aperture_rim ...

> The only thing that you know is that there was at least one photon which was reflected by the electron (Because one pixel (CCD) of the microscope was activated) after initial set up of the experiment. There could be more photons involved which were not detected/seen.

Exactly.

> IMO in reality photon/electron reflections can go in any direction and the influence of the photon can be in a continuous range from 1 almost nothing, 2 a [little], 3 a little more, 4 a lot, 5 heavy, 6 maximum Where 6 is the maximum possible disturbance of the electron.

Yes; essentially your description corresponds to the range of interactions that are called "Compton scattering". But note that the calculation presented on the referenced webpage is concerned only with the momentum associated with the lightsignal which is exchanged between the electron and the detector(s), and _not_ at all with the momentum associated with the lightsignal(s) which is exchanged between the source(s) and the electron.

> In reality more photons could have disturbed the electron which makes the initial assumption that the electron is practically at rest very doubtfull.

I cannot find the word "rest" on the referenced website. AFAIU a possibly related phrase which actually appears there, namely that the

. electron sits directly beneath the center of the microscope's lens

it doesn't particularly seem to restrict the momentum and/or possible "disturbedness" of the electron.

> Question 3 is a tricky one because the photon is generated by a photongenerator which consists of "agitated" calcium atoms ("agitated" means that the electrons are in a higher energy state) Those "agitated" electrons can jump back to a lower state by releasing a photon. You can calculate the frequency of such photons.

How would one know the (most probable) potential of the various electron "within" some particular calcium atom to begin with ?

> [...] You need millions of photons to study [potentials]

Yes, measuring potentials is tricky; but, as indicated above, that's not really required for following the presentation on the referenced website.

> You can also measure the frequency of such photons. But you can not measure the frequency of one particular photon specific not the frequency/wave length of the photon used in the experiment because such a measurement will "destroy" the photon.

If one calcium atom having been agitated as you described were observed by some other atom (which were thus in turn agitated in some manner that can at least statistically be defined specificly) -- then has not one photon been exchanged, and has its frequency not been determined/calculated/measure as you indicated above?

> [...] The tenure [tenor?] of the whole URL is: We show you an (animated) movie We show you some mathematics dp*dx ~ h but In reality the whole issue is more "complex" and then ..... nothing

Well, yes, the uncertainty principle can be expressed and proven quite generally and abstractly. The website gives a concrete and quite precise example, but in order to recognize this, of course you'll have to concentrate on the example itself, not its context or incorrectly suggestive rendering in some animation.

> To study the behaviour of a single photon by using a single photon is impossible. [...]

But studying the geometric relations of specific charges based on their mutual observations does appear feasible.

Regards, Frank W ~@) R

p.s.

Recently (in this, and another newsgroup) unfortunately I mangled Gerard 't Hooft's name to "t'Hooft".

Could you please spell out what's abbreviated by "'t" in Dutch names, and whether (and for what) "t'" might a valid abbreviation, too?


49 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 22 april 2002 12:33

"Frank Wappler" schreef in bericht news:s4od6wtbjp6.fsf@acunix1.albany.edu...
>

Nicolaas Vroom wrote:

> > > >

http://www.aip.org/history/heisenberg/p08b.htm

> >

The movie starts with a photon which moves in a horizontal path to the right and collides with the electron. This photon has a wave length of L.

>

I still claim that the written page itself refers to wavelengths L, L', or L'' strictly only _after_ the "scattering" (i.e. after the electron observed the source signalling). As far as the animation shows otherwise, I consider it wrong.

The movie starts with a drawing of a photon with the shape of a wave. Below this wave is written photon. Above the wave is written the letter L ie the wavelength before the scattering.

L' and L'' are after the scattering

> > This raises immediate 3 questions. 1. How do you know that there is one photon
>

The entire derivation considers precisely one lightsignal being observed at the detector side of the aperture. Whether and which other lightsignals might be exchanged simply doesn't enter the presented considerations.

>
> >

The only thing that you know is that there was at least one photon which was reflected by the electron (Because one pixel (CCD) of the microscope was activated) after initial set up of the experiment. There could be more photons involved which were not detected/seen.

>

Exactly.

Let me propose a slightly different experiment.

Instead of a microscope at a distance r from the electron we build a device with a sphere of CCD's with radius r. The small area at the Northpole of radius dx/2 is our original microscope. The advantage of this device is that we can detect all photons involved. The single photon generator is at the West position. This photon generator generates at average 60 photons per min.

First there is no electron in the centre. What will happen in one minute ? IMO we will get 60 hits at the CCD's in the East position. There will be a normal distribution. Most of the hits are at the CCD opposite photon generator.

Next we place one electron in the centre. What will happen in one minute ? IMO the first photon will collide with the electron and can hit almost any CCD. next we get 59 hits at the CCD's in the East position. There is a small change that the photon will hit the small area identified as microscope.

We can repeat this whole experiment 1000 times.

The question is now as a result of those experiments Can we say something about the position of the electron ? Can we say something about the momemtum of the electron ?

HUP is very clear dx*dp ~ h.

> > IMO in reality photon/electron reflections can go in any direction and the influence of the photon can be in a continuous range from 1 almost nothing, 2 a [little], 3 a little more, 4 a lot, 5 heavy, 6 maximum Where 6 is the maximum possible disturbance of the electron.
>

Yes; essentially your description corresponds to the range of interactions that are called "Compton scattering". But note that the calculation presented on the referenced webpage is concerned only with the momentum associated with the lightsignal which is exchanged between the electron and the detector(s), and _not_ at all with the momentum associated with the lightsignal(s) which is exchanged between the source(s) and the electron.

> >

In reality more photons could have disturbed the electron which makes the initial assumption that the electron is practically at rest very doubtfull.

>

I cannot find the word "rest" on the referenced website. AFAIU a possibly related phrase which actually appears there, namely that the . electron sits directly beneath the center of the microscope's lens

Heisenberg writes: "The position of the electron will be known with an accuracy given by the wave length of the g-ray. The electron may have been practically at rest before the observation. But in the act of observation at least one light quantum of the g-ray must have passed the microscope and must first have been deflected by the electron. Therefore the electron has been pushed by the light quantum it has changed its momemtum and its velocity and one can show that the uncertainty of this change is just big enough to guarantee the validity of the uncertainty relations. "

This change in the electron can be from almost zero (by one photon) to its maximum which means that the electron absorps the photon. (See also below video by prof. Veltman)

IMO one of the weak points is the initial setup: How do you place a photon (at rest) in the centre. IMO this smells to a priory information.

The mathematical derivation as shown in the url states that you can use the following:

"If A is small, then the wavelengths are approximately the same, L' ~ L" ~ L. So we have etc"

How do we know that ? Is this allowed for all photons detected. ?

IMO when there is a collision/reflection there is a transfer of energy between the photon(source) and the electron(target)

(The maximum when there is absorption....)

> it doesn't particularly seem to restrict the momentum and/or possible "disturbedness" of the electron.
>
> >

You can also measure the frequency of such photons. But you can not measure the frequency of one particular photon specific not the frequency/wave length of the photon used in the experiment because such a measurement will "destroy" the photon.

>

If one calcium atom having been agitated as you described were observed by some other atom (which were thus in turn agitated in some manner that can at least statistically be defined specificly) -- then has not one photon been exchanged, and has its frequency not been determined/calculated/measure as you indicated above?

> >

[...] The tenure [tenor?] of the whole URL is: We show you an (animated) movie We show you some mathematics dp*dx ~ h but In reality the whole issue is more "complex" and then ..... nothing

>

Well, yes, the uncertainty principle can be expressed and proven quite generally and abstractly. The website gives a concrete and quite precise example, but in order to recognize this, of course you'll have to concentrate on the example itself,

I have done above

> not its context or incorrectly suggestive rendering in some animation.
I take the animation with "a grain of salt" if you understand that expression.

> > To study the behaviour of a single photon by using a single photon is impossible. [...]
>

But studying the geometric relations of specific charges based on their mutual observations does appear feasible.

The more I discuss it, the least I understand HUP

> Regards, Frank W ~@) R

I'am very gratefull for your effort

Nick.

> p.s.

Recently (in this, and another newsgroup) unfortunately I mangled Gerard 't Hooft's name to "t'Hooft".

Could you please spell out what's abbreviated by "'t" in Dutch names, and whether (and for what) "t'" might a valid abbreviation, too?

The two physicists who won the 1999 are professor Gerardus 't Hooft and professor Martinus J.G. Veltman

please go to http://www.nobel.se/physics/laureates/1999/ and study first the video of professor Veltman (46 min) and then the video of Professor 't Hooft. (48 min)

the 't in his name is part of his full name.

Specific listen about the absorption and emission of single photons from electrons in the speech by prof. Veltman. (after roughly 23 minutes)

Nick


50 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 22 april 2002 15:12

"Nicolaas Vroom" schreef in bericht news:9GRw8.31547$Ze.5934@afrodite.telenet-ops.be...
>

Let me propose a slightly different experiment.

Instead of a microscope at a distance r from the electron we build a device with a sphere of CCD's with radius r. The small area at the Northpole of radius dx/2 is our original microscope. The advantage of this device is that we can detect all photons involved. The single photon generator is at the West position. This photon generator generates at average 60 photons per min.

First there is no electron in the centre. What will happen in one minute ? IMO we will get 60 hits at the CCD's in the East position. There will be a normal distribution. Most of the hits are at the CCD opposite photon generator.

Next we place one electron in the centre. What will happen in one minute ? IMO the first photon will collide with the electron and can hit almost any CCD. next we get 59 hits at the CCD's in the East position. There is a small change that the photon will hit the small area identified as microscope.

We can repeat this whole experiment 1000 times.

The question is now as a result of those experiments Can we say something about the position of the electron ? Can we say something about the momemtum of the electron ?

HUP is very clear dx*dp ~ h.

The purpose of this device is also to study what happens if you make the radius twice as large. That means you need 4 times as much CCD's However, You keep the distance of photon generator to the centre the same and also the size of the microscope.

Each experiment of 1 minute will be identical. The first photon will be reflected and the next 59 photon's will hit the CCD's opposite the generator. The area will be larger.

If you repeat this experiment 1000 times less photons (4 times) will hit the microscope.

Does that mean that you know now the position of the electron more precise ?

Does that mean that you know now the momentum of the electron less precise ?

Or does that mean that of the electron you know the position more precise and the momentum less precise ?

(and that this in agreement with dx*dp ~ h ?)

I have doubts

Nick


51 Basic question in Heisenberg's uncertainity principle

Van: "Frank Wappler"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: maandag 22 april 2002 23:07

Nicolaas Vroom wrote:

> http://www.aip.org/history/heisenberg/p08b.htm

> The movie starts with a drawing of a photon with the shape of a wave. Below this wave is written photon. Above the wave is written the letter L ie the wavelength before the scattering.

That's what I recognize there as well. But let me quote and compare that with what I recognize on that website in writing (notation slightly tightened - FW):

. In the extreme case of diffraction of the gamma ray to the right edge of the lens, the total momentum in the x direction would be the sum of the electron's momentum p'x in the x direction and the gamma ray's momentum in the x direction:

. p'x + (h sinA ) / L',

. where L' is the wavelength of the deflected gamma ray.

. In the other extreme, the observed gamma ray recoils backward, just hitting the left edge of the lens. In this case, the total momentum in the x direction is:

. p"x - (h sinA ) / L".

... Btw., note the opposite sign of momentum "to the left" vs. momentum "to the right" ...

. The final x momentum in each case must equal the initial x momentum, since momentum is never lost (it is conserved).

... and, I should add, since in this thought experiment we imagine having all observations accounted for, i.e. for the experimental setup being "closed".

. Therefore, the final x momenta are equal to each other:

. p'x + (h sinA ) / L' = p"x - (h sinA ) / L"

... because: either side of this equation must have been equal to the initial (prescattering) momentum component in the x direction, which is explicitly (but without allowing to assign particular values):

p_x_source_after_signalling - p_x_source_before_signalling + p_x_electron_before_scattering.

Now consider the following crucial statement:

> . If A is small, then the wavelengths are approximately the same, . L' ~ L" ~ L.

> [...] How do we know that ?

If you recall and accept the Compton relation:

lambda_final = lambda_initial + h/(c m_electron) (1 - cos( angle )),

and thereby

L' = lambda_initial + h/(c m_electron) (1 - cos( angle_to_the_left ))

L" = lambda_initial + h/(c m_electron) (1 - cos( angle_to_the_right )),

consequently if "A is small" (in comparison to? ... the distance rim to electron, I presume) can be taken to mean that

-2Pi << angle_to_the_left - angle_to_the_right << 2Pi

then

cos( angle_to_the_left ) =~= cos( angle_to_the_right ) =~= cos( angle )

and thus

L' =~= L".

However, h/(c m_electron) (1 - cos( angle )) does _not_ need to be small in comparison e.g. to L' at all.

L' ?=~=? lambda_initial is neither in general satisfied nor required;

and therefore L' ?=~=? L is sensible only if ``L'' here abbreviates any approximate wavelength _after_ scattering, such as L', L" or any value "inbetween" (as indexed by angle).

As far as the drawing/animation indicates otherwise (which I believe it does), I choose to follow the writing (which I believe to understand), and I call the drawing/animation wrong.

The website

http://www.aip.org/history/heisenberg/p01.htm

contains e_mail addresses and a phone number for contacting those who claim the copyright and/or were apparently responsible for the disign of this site and the drawing/animation in particular.

Whoever considers such websites useful references might wish to consult these contacts about the consistency of their usage of the label ``L''.

> Instead of a microscope at a distance r from the electron we build a device with a sphere of CCD's with radius r. The small area at the Northpole of radius dx/2 is our original microscope. The advantage of this device is that we can detect all photons involved.

At least we consider having better coverage in making this assumption.

> The single photon generator

... what's that? ...

> is at the West position. This photon generator generates at average 60 photons per min.

Fine, we can surely consider that an average of 60 CCD elements per minute observe the "photon generator" having stating a signal.

> First there is no electron in the centre. What will happen in one minute ? IMO we will get 60 hits at the CCD's in the East position. There will be a normal distribution. Most of the hits are at the CCD opposite photon generator.

Well, at least, given that the geometry of the CCD's wrt each other has been and continues to be measured, the geometric relations wrt those CCD's who observed the generator signalling (and thus exchanged light signals with the generator) may give clues about the most probable geometry of various generator constituents (and/or of whatever setup elements don't count as CCDs and "the sphere").

> Next we place one electron in the centre. What will happen in one minute ? IMO the first photon will collide with the electron and can hit almost any CCD.

By such a drastic change in the measured most probable geometry of the setup (especially if it is repeated with statistical significance) we'd know that "an electron had been placed in (the center of) the sphere in the first place.

> next we get 59 hits at the CCD's in the East position. There is a small [chance] that the photon will hit the small area identified as microscope.

> We can repeat this whole experiment 1000 times.

... with the outcomes being repeated, on average; yes.

> The question is now as a result of those experiments Can we say something about the position of the electron ?

Yes: the electron (or at least: an electron) has been "in the sphere". (Though not necessarily always in "the center of" the sphere, i.e. in "the middle between" any pair of CCD with maximal/diameter distance wrt each other, as defined and to be measured through Einstein's calibration procedure.)

> Can we say something about the momentum of the electron ?

Per definition of "momentum" as invariant under translations, and by the generality of the uncertaintly relations which are thereby implied, yes:

The average momentum of the electron equals the momentum of the sphere, +/- hbar/2 1/sphere_diameter, at least.

> Heisenberg [wrote]:

... can you please give the reference ...

> . "The position of the electron will be known with an accuracy given by the wave length of the g-ray. The electron may have been practically at rest before the observation. But in the act of observation at least one light quantum of the g-ray must have passed the microscope and must first have been deflected by the electron.
Therefore the electron has been pushed by the light quantum it has changed its [momentum] and its velocity and one can show that the uncertainty of this change is just big enough to guarantee the validity of the uncertainty relations. "

... perhaps that's a version for which

. Bohr pointed out some flaws [...].

> This change in the electron can be from almost zero (by one photon) to its maximum which means that the electron [absorbs] the photon.

> (See also below video by prof. Veltman) please go to http://www.nobel.se/physics/laureates/1999/ and study first the video of professor Veltman (46 min) and then the video of Professor 't Hooft. (48 min)

Without having having had the opportunity to see these video's yet, I'd find it very unlikely that either of these two Nobel laureats had made a statement as you seem to have understood it. But yes, light scattering will change the momentum of an electron more or less (depending on the wavelength, and the potential).

As I have tried to point out, the uncertainty relation is concerned with the definiteness of momentum values _besides and on top of_ any such change. In the example considered in the referenced website it is derived entirely by considering possible assignments of momenta _after_ the scattering.

> IMO one of the weak points is the initial setup: How do you place a photon (at rest) in the centre.

I can answer how you'd _know whether_ an electron E had been (placed) in the center of four observers (such as CCD elements, A, B, C and D):

by Einstein's calibration procedure -- essentially, if in every trial

- the electron observes each lightsignal roundtrip interval to A to be the same as a lightsignal roundtrip interval to B, as well as to C, as well as to D,

- A observes each lightsignal roundtrip interval to B to be the same as two lightsignal roundtrip intervals to E,

- B observes each lightsignal roundtrip interval to A to be the same as two lightsignal roundtrip intervals to E,

- C observes each lightsignal roundtrip interval to D to be the same as two lightsignal roundtrip intervals to E,

- D observes each lightsignal roundtrip interval to C to be the same as two lightsignal roundtrip intervals to E,

- A observes each lightsignal roundtrip interval to C to be the same as a lightsignal roundtrip interval to D,

- B observes each lightsignal roundtrip interval to C to be the same as a lightsignal roundtrip interval to D,

- C observes each lightsignal roundtrip interval to A to be the same as a lightsignal roundtrip interval to B, and

- D observes each lightsignal roundtrip interval to A to be the same as a lightsignal roundtrip interval to B,

then E had been (placed) in the center between A, B, C and D; and otherwise not.

Two notes: obviously it is more difficult to identify condition based on which one would _expect that_ some particular electron might be thus found to be the center of any particular four observers (but just as obviously, such conditions couldn't be identified at all without having defined the sought/expected result in the first place).

And: obviously, if indeed some E were found to have been the center of given A, B, C, and D, throughout an experiment, then the momentum of E wrt. any and all of these four were _undefined_ and thus maximally uncertain (if translation is zero, then it cannot be evaluated what may or may not stay invariant under translation).

> I'am very gratefull for your effort

You're very welcome.

Regards, Frank W ~@) R

p.s.

> > I mangled [...] Could you please spell out what's abbreviated by "'t" in Dutch names, and whether (and for what) "t'" might [be] a valid abbreviation, too?

> The two physicists who won the 1999 are professor Gerardus 't Hooft and professor Martinus J.G. Veltman

> [...] the 't in his name is part of his full name.

Fine, thanks; but would you please consider my questions, which are here stated more explicitly again:

(1)

- does "'t" abbreviate a separate Dutch word (such as for instance "can't" abbreviates the English word "cannot"), and if so, which Dutch word ?

- or _is_ "'t" a separate Dutch word, and if so, how is it to be translated into English ?

- and/or is "'t Hooft" as a whole a meaningful Dutch phrase, and if so, how is it to be translated into English ?

(2)

- does "t'" abbreviate a separate Dutch word, and if so, which ?

- or _is_ "t'" a separate Dutch word, and if so, how is it to be translated into English ?

- and/or is "t'Hooft" as a whole a meaningful Dutch phrase, and if so, how is it to be translated into English ?

Thanks again, Frank W.


52 Basic question in Heisenberg's uncertainity principle

Van: "Frank Wappler"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 23 april 2002 22:27

Frank Wappler wrote:

> [...] the final x momenta are equal to each other:

> . p'x + (h sinA ) / L' = p"x - (h sinA ) / L"

> [...] if "A is small" (in comparison to? ...

... since A is an angle, therefore surely in comparison to 2Pi. IOW: the diameter of the aperture being small in comparison to ...

> ... the distance rim to electron, I presume)

> can be taken to mean that

> -2Pi << angle_to_the_left - angle_to_the_right << 2Pi

... where angle_to_the_left - angle_to_the_right == 2A ...

> then [...] L' =~= L".


53 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: woensdag 24 april 2002 22:36

"Frank Wappler" schreef in bericht news:s4oadrvqw6k.fsf@acunix1.albany.edu...
>

Nicolaas Vroom wrote:

> >

http://www.aip.org/history/heisenberg/p08b.htm

>
> >

The movie starts with a drawing of a photon with the shape of a wave. Below this wave is written photon. Above the wave is written the letter L ie the wavelength before the scattering.

> . The final x momentum in each case must equal the initial x momentum, since momentum is never lost (it is conserved).

... and, I should add, since in this thought experiment we imagine having all observations accounted for, i.e. for the experimental setup being "closed".

. Therefore, the final x momenta are equal to each other:

. p'x + (h sinA ) / L' = p"x - (h sinA ) / L"

However you have to be very carefull here. What "you" are doing is that "you" are comparing the results of two experiments.

In the first experiment you have a photon with a momenta ph1x in the x direction After the collision the electron has momenta p'x and the photon (h sinA ) / L' resulting in the equation ph1x = p'x + (h sinA ) / L'

In the second experiment you have a photon with a momenta ph2x in the x direction resulting in the equation ph2x = p''x - (h sinA ) / L''

Now "you" say both photons initial are the same ie phx1=phx2 resulting in p'x + (h sinA ) / L' = p"x - (h sinA ) / L"

> If you recall and accept the Compton relation:

> and thus

L' =~= L".

I have no problems that all photons which are detected at the same place on my CCD sphere have the same wavelength and or frequency and or energy This is ofcourse also true for the microscope.

as such L' = L"

> However, h/(c m_electron) (1 - cos( angle )) does _not_ need to be small in comparison e.g. to L' at all.

L' ?=~=? lambda_initial is neither in general satisfied nor required;

and therefore L' ?=~=? L is sensible only if ``L'' here abbreviates any approximate wavelength _after_ scattering, such as L', L" or any value "inbetween" (as indexed by angle).

I have a great difficult to understand that L' = L (IMO it is not true, but that is not important)

What is happening is an energy transfer between the photon and the electron. The energy of the electron increases this is reflected that the speed of the electron increases. The energy of the photon decreases this is reflected in a change in f and wavelength of the photon. (all: IMO)

> As far as the drawing/animation indicates otherwise (which I believe it does), I choose to follow the writing (which I believe to understand), and I call the drawing/animation wrong.

The website

http://www.aip.org/history/heisenberg/p01.htm

contains e_mail addresses and a phone number for contacting those who claim the copyright and/or were apparently responsible for the disign of this site and the drawing/animation in particular.

Whoever considers such websites useful references might wish to consult these contacts about the consistency of their usage of the label ``L''.

IMO what the url not demonstrates is that HUP is correct. ( I am not saying that HUP is wrong)

As such I wish that someone points me to an url which demonstrates HUP in an unambiguous fashion.

> > Instead of a microscope at a distance r from the electron we build a device with a sphere of CCD's with radius r. The small area at the Northpole of radius dx/2 is our original microscope. The advantage of this device is that we can detect all photons involved.
>

At least we consider having better coverage in making this assumption.

> >

The single photon generator

>

... what's that? ...

It is a laser which emits photons one by one.

> > is at the West position. This photon generator generates at average 60 photons per min.
>

Fine, we can surely consider that an average of 60 CCD elements per minute observe the "photon generator" having stating a signal.

> >

First there is no electron in the centre. What will happen in one minute ? IMO we will get 60 hits at the CCD's in the East position. There will be a normal distribution. Most of the hits are at the CCD opposite photon generator.

>

Well, at least, given that the geometry of the CCD's wrt each other has been and continues to be measured, the geometric relations wrt those CCD's who observed the generator signalling (and thus exchanged light signals with the generator) may give clues about the most probable geometry of various generator constituents (and/or of whatever setup elements don't count as CCDs and "the sphere").

> >

Next we place one electron in the centre. What will happen in one minute ? IMO the first photon will collide with the electron and can hit almost any CCD.

>

By such a drastic change in the measured most probable geometry of the setup (especially if it is repeated with statistical significance) we'd know that "an electron had been placed in (the center of) the sphere in the first place.

I do not like this whole concept (a priory information) but it is the start point of the derivation of the formula: p'x + (h sinA ) / L' = p"x - (h sinA ) / L" ie there should be an electron straight below the center of microscope.

> > next we get 59 hits at the CCD's in the East position. There is a small [chance] that the photon will hit the small area identified as microscope.
>
> >

We can repeat this whole experiment 1000 times.

>

... with the outcomes being repeated, on average; yes.

> > Can we say something about the momentum of the electron ?
>

Per definition of "momentum" as invariant under translations, and by the generality of the uncertaintly relations which are thereby implied, yes:

The average momentum of the electron equals the momentum of the sphere, +/- hbar/2 1/sphere_diameter, at least.

I can not agree on this statement solely based on the information in this url. because the derivation of the uncertainty relations is not clear.

> > Heisenberg [wrote]:
>

... can you please give the reference ...

Heisenberg: "Physics and Philosophy" ISBN 0 14 014660-1
>
> >

. "The position of the electron will be known with an accuracy given by the wave length of the g-ray. The electron may have been practically at rest before the observation. But in the act of observation at least one light quantum of the g-ray must have passed the microscope and must first have been deflected by the electron. Therefore the electron has been pushed by the light quantum it has changed its [momentum] and its velocity and one can show that the uncertainty of this change is just big enough to guarantee the validity of the uncertainty relations. "

>

... perhaps that's a version for which

. Bohr pointed out some flaws [...].

I wish to known which flaws Bohr identified.

The most critically sentence is: "The electron may have been practically at rest before the observation." That means v initial electron = 0 That means the electron after observation (reflection) has v>0 That means after the observation the electron has energy 0.5*m*v*v How much is v ? How much is m*v ? This energy must come from the photon etc see above.

My next question would be based on "one can show" please do, please give the details. Heisenberg in this book does not.

The book also is written:(page 36)

"Therefore, the first light quantum is sufficient to knock the electron out of the atom and one can never observe more than one point in the orbit of the electron; therefore there is no orbit in the ordinary sense."

IMO there are many outcomes possible making IMO it impossible to quantize anything.

"The next observation will show the electron on its path from the atom" IMO there is no next observation, with one photon, because you do not know where the electron is. ie it is impossible to measure v and p of one electron (and to demonstrate HUP)

> > IMO one of the weak points is the initial setup: How do you place a photon (at rest) in the centre.
>

I can answer how you'd _know whether_ an electron E had been (placed) in the center of four observers (such as CCD elements, A, B, C and D):

by Einstein's calibration procedure -- essentially, if in every trial

- the electron observes each lightsignal roundtrip interval to A to be the same as a lightsignal roundtrip interval to B, as well as to C, as well as to D,

- A observes each lightsignal roundtrip interval to B to be the same as two lightsignal roundtrip intervals to E,

etc

> then E had been (placed) in the center between A, B, C and D; and otherwise not.

I think you can only do that for the atoms in a substrate but not for the electrons which move around the nucleus.

IMO a practical experiment which demonstrates HUP is impossible.

> Two notes: obviously it is more difficult to identify condition based on which one would _expect that_ some particular electron might be thus found to be the center of any particular four observers (but just as obviously, such conditions couldn't be identified at all without having defined the sought/expected result in the first place).

And: obviously, if indeed some E were found to have been the center of given A, B, C, and D, throughout an experiment, then the momentum of E wrt. any and all of these four were _undefined_ and thus maximally uncertain (if translation is zero, then it cannot be evaluated what may or may not stay invariant under translation).

> p.s.
> > >

I mangled [...] Could you please spell out what's abbreviated by "'t" in Dutch names, and whether (and for what) "t'" might [be] a valid abbreviation, too?

Answered via e-mail

Nick


54 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: vrijdag 26 april 2002 16:26

"Frank Wappler" schreef in bericht news:s4oadrvqw6k.fsf@acunix1.albany.edu...
>

Nicolaas Vroom wrote:

> >

http://www.aip.org/history/heisenberg/p08b.htm

>
> >

The movie starts with a drawing of a photon with the shape of a wave. Below this wave is written photon. Above the wave is written the letter L ie the wavelength before the scattering.

>

> Now consider the following crucial statement:
> >

. If A is small, then the wavelengths are approximately the same, . L' ~ L" ~ L.

>
> >

[...] How do we know that ?

>

If you recall and accept the Compton relation:

See: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html

The whole concept of Compton scattering is based around the idea that lambda_intial <> lambda_final ie that L <> L'

> lambda_final = lambda_initial + h/(c m_electron) (1 - cos( angle )),

and thereby

L' = lambda_initial + h/(c m_electron) (1 - cos( angle_to_the_left ))

L" = lambda_initial + h/(c m_electron) (1 - cos( angle_to_the_right )),

consequently if "A is small" (in comparison to? ... the distance rim to electron, I presume) can be taken to mean that

-2Pi << angle_to_the_left - angle_to_the_right << 2Pi

then

cos( angle_to_the_left ) =~= cos( angle_to_the_right ) =~= cos( angle )

and thus

L' =~= L".

However, h/(c m_electron) (1 - cos( angle )) does _not_ need to be small in comparison e.g. to L' at all.

L' ?=~=? lambda_initial is neither in general satisfied nor required;

and therefore L' ?=~=? L is sensible only if ``L'' here abbreviates any approximate wavelength _after_ scattering, such as L', L" or any value "inbetween" (as indexed by angle).

As far as the drawing/animation indicates otherwise (which I believe it does), I choose to follow the writing (which I believe to understand), and I call the drawing/animation wrong.

IMO the mathematics used is wrong

> The website

http://www.aip.org/history/heisenberg/p01.htm

contains e_mail addresses and a phone number for contacting those who claim the copyright and/or were apparently responsible for the disign of this site and the drawing/animation in particular.

Whoever considers such websites useful references might wish to consult these contacts about the consistency of their usage of the label ``L''.

Maybe the following gives better information ? http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html#c1

Anyway the compton photon/electron scattering experiment and thought experiment by Heisenberg with the microscope are almost identical but the implications are so different. (Both start from the assumption that the electron initial is at rest)

The first one gives exact predictions/results. The second one gives the suggestion that those exact/precise results are "impossible".

Compton scattering seems mathematical rather straight forward. I expect that precise experimental validation of the compton scattering angle is more difficult.

Nick.


55 Basic question in Heisenberg's uncertainity principle

Van: "Nicolaas Vroom"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: zaterdag 27 april 2002 17:22

"Bilge" schreef in bericht news:slrnabrfb9.jgo.root@radioactivex.lebesque-al.net...
> Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
> > HUP claims that you can measure p precisely but than you can not measure x precisely of a single particle or photon. HUP claims that you can measure x precisely but than you can not measure p precisely. HUP even "quantifys" this claim: dp * dx ~ h IMO you can not measure p of a single photon nor p of a single electron by using one single photon
>

I can always measure it to the precision: p = hbar/\Delta x

The question is how do you that ? (IMO this is a method to calculate Delta p, but not to measure p or if you do that 1000 times to calculate Delta p based on measurements of p)

At 13/04/2002 you replied on my question:

Quote
> Is that the one hole (slit) experiment with single photons?

That's a good example. In principle, you could move the source of those single photons as far back as you wish so that, the angle formed by the location of the source and the edges of the aperature -> 0. Since p is the total momentum, it's a vector sum, p = p_x + p_y, so that the vector, p = |p|[cos(A)i + sin(A)j], i,j are unit vectors in x,y direction. The vector r from the source to the screen is given similarly by: r = |r|[cos(A)i + sin(A)j].

         |
         |      p_y
           p . |
       . 'A    +--p_x
S : '__________
    ' .   A
         '  .
         |
         |
As the angle A->0, you would expect classically to preferentially select only those photons that have p_y very small, so that the photons follow a straight line through the aperature and strike the screen, just like a bullet would do. But that is not what happens. The aperature constitutes _both_ a position measurement and a momentum measurement. If you choose to view it as a position measurement, then the momentum is completely undetermined, so that it points just like you would expect from a huygens construction. You add up the phases of all the rays to a particlular point on the screen. If you view it as a momentum measurement, i.e., since p = hv/c, you are trying to force p to point along a line from the source to the screen, p == p_x, then the position across the aperature is left completely undetermined, and again you take the rays from every point. However, I'm not going to do all of your work for you.

End of quote
(Unfortunate your drawing is slightly scrambled)
The part specific after: If you view it as a momentum measurement is important: "you take the rays from every point" apparently more photons are involved but still for me it is not clear how you can measure v (or p) of a single photon. (IMO as claimed previously it is impossible because you need two position measurements)

Suppose I wrote:
> IMO you can not measure x of a single photon nor x of a single electron by using one single photon

Would you answer than: "I can always measure it to the precision: x = hbar/\Delta p"

IF x and p are canonically conjugate variables (See 3/04/2002) than I should expect that, because both variables should be interchangable. (commutateble)

Nick http://users.pandora.be/nicvroom/


56 Basic question in Heisenberg's uncertainity principle

Van: "Bilge"
Onderwerp: Re: Basic question in Heisenberg's uncertainity principle
Datum: dinsdag 30 april 2002 9:49

Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>

"Bilge" schreef in bericht news:slrnabrfb9.jgo.root@radioactivex.lebesque-al.net...

>> Nicolaas Vroom said some stuff about Re: Basic question in Heisenberg's uncertainity principle to usenet:
>> > HUP claims that you can measure p precisely but than you can not measure x precisely of a single particle or photon. HUP claims that you can measure x precisely but than you can not measure p precisely. HUP even "quantifys" this claim: dp * dx ~ h IMO you can not measure p of a single photon nor p of a single electron by using one single photon
>>

I can always measure it to the precision: p = hbar/\Delta x

>

The question is how do you that ? (IMO this is a method to calculate Delta p, but not to measure p

I don't care. I suggested trying to work some quantum mechanics problems involving the uncertainty principle. Did you do that? I didn't think so, since you just repeated the same question.


Created: 27 April 2002

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