1 "Akira" |
A Question about Uncertianty Principle | dinsdag 26 maart 2002 9:47 |
2 "franz heymann" |
Re: A Question about Uncertianty Principle | dinsdag 26 maart 2002 10:40 |
3 "Pmb" |
Re: A Question about Uncertianty Principle | dinsdag 26 maart 2002 19:40 |
4 "Robert Hamilton" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 5:25 |
5 "Mike Hanson" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 11:49 |
6 "Nicolaas Vroom" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 13:05 |
7 "Robert Hamilton" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 15:09 |
8 "franz heymann" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 23:47 |
9 "Nicolaas Vroom" |
Re: A Question about Uncertianty Principle | donderdag 28 maart 2002 19:02 |
10 "Mike Hanson" |
Re: A Question about Uncertianty Principle | vrijdag 29 maart 2002 18:13 |
11 "franz heymann" |
Re: A Question about Uncertianty Principle | vrijdag 29 maart 2002 20:47 |
12 "Nicolaas Vroom" |
Re: A Question about Uncertianty Principle | maandag 1 april 2002 10:49 |
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its position is at X = a. According to the hypothesis,we have delta(x) is zero. and delta(p) is infinite!!
How do we explain this phenomenon? just said we can determine the delta(p) ? But back to the problem, we have a rest body then its momentum is ZERO. How can I say its delta(p) is infinite?
Akira
> |
It is a question confuses me long time. That is the delta(x) * delta(p) >= h / (4*pi) If the problem is following... There is a rest body with its mass M which its position is defined, and its position is at X = a. According to the hypothesis,we have delta(x) is zero. and delta(p) is infinite!! |
That is correct. If you know the position of a particle precisely, you can say nothing at all about its momentum
> |
How do we explain this phenomenon? |
What do you mean by "explain"? The HUP is derivable from the postulstes of quantum mechanics. These postulates have never made a prediction which has been contradicted by experiment.
> | just said we can determine the delta(p) ? But back to the problem, we have a rest body then its momentum is |
It sounds weird, but no actual observation has ever given results which are in disagreement with the predictions of the HUP.
> | How can I say its delta(p) is infinite? |
It simply means that you know nothing at all about the momentum under circumstances when you have determined the position to indefinitely small presision. Roughly speaking, if you want to see with precision where a particle is, you have to illuminate it with light which has a wavelength smaller than the precision you wish to achieve. Such light consists of photons which will impart a large momentum to the particle when a photon hits it. So you have disturbed its original momentum by a correspondingly large amount.
Franz Heymann
> |
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi) If the problem is following... There is a rest body with its mass M which its position is defined, and its position is at X = a. According to the hypothesis,we have delta(x) is zero. and delta(p) is infinite!! How do we explain this phenomenon? just said we can determine the delta(p) ? But back to the problem, we have a rest body then its momentum is ZERO. How can I say its delta(p) is infinite? |
Well first off let's clear up what the "delta" means. It refers to a statistical quantity. It has to do, not with one single measurement, but with many measurements of a system in a given (i.e. known) state. so when you measure "x=a" it means that you go and measure the position of the particle and the result of the measurement was x=a. Now you measure the momentum of the particle at the same time and you get P=b. you repeat this many many many times. You calculate the quanties delta(x) and delta(p) and you'll find that
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I always have to look that up).
Given the definition of dx and dp and the popstulates of quantum mechanics you can actually derive the expression dx*dp >= hbar/2
Pmb
On 26 Mar 2002 09:40:57 -0800, Pmb
> |
"Akira" |
>> |
It is a question confuses me long time. That is the delta(x) * delta(p) >= h / (4*pi) If the problem is following... There is a rest body with its mass M which its position is defined, and its position is at X = a. According to the hypothesis,we have delta(x) is zero. and delta(p) is infinite!! How do we explain this phenomenon? just said we can determine the delta(p) ? But back to the problem, we have a rest body then its momentum is ZERO. How can I say its delta(p) is infinite? |
> |
Well first off let's clear up what the "delta" means. It refers to a statistical quantity. It has to do, not with one single measurement, but with many measurements of a system in a given (i.e. known) state. |
This is not true
> |
so when you measure "x=a" it means that you go and measure the
position of the particle and the result of the measurement was x=a.
Now you measure the momentum of the particle at the same time and you
get P=b. you repeat this many many many times. You calculate the
quanties delta(x) and delta(p) and you'll find that
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I always have to look that up). Given the definition of dx and dp and the popstulates of quantum mechanics you can actually derive the expression dx*dp >= hbar/2 |
This is true.
> |
Pmb |
"franz heymann"
> | That is correct. If you know the position of a particle precisely, you can say nothing at all about its momentum |
Absolutely. However:
> | Roughly speaking, if you want to see with precision where a particle is, you have to illuminate it with light which has a wavelength smaller than the precision you wish to achieve. Such light consists of photons which will impart a large momentum to the particle when a photon hits it. So you have disturbed its original momentum by a correspondingly large amount. |
Absolutely not! Or rather, although this would certainly happen if that's what you did, it has nothing whatsoever to do with the uncertainty principle.
I would imagine that you yourself know better than to see HUP as 'unavoidable disturbance during measurement'. And I know that such an explanation is commonly given for the sake of simplicity (even Asimov explains it in these terms). But that does not justify giving out misleading information which subsequently needs to be corrected once a higher level of understanding is achieved.
If this is the way that HUP is introduced to people, then it takes an awful lot of effort for them to unlearn this misconception and fully appreciate the implications of uncertainty.
Mike.
"Robert Hamilton"
> |
On 26 Mar 2002 09:40:57 -0800, Pmb |
> > |
"Akira" |
> >> |
It is a question confuses me long time. That is the delta(x) * delta(p) >= h / (4*pi) If the problem is following... There is a rest body with its mass M which its position is defined, and its position is at X = a. According to the hypothesis,we have delta(x) is zero. and delta(p) is infinite!! How do we explain this phenomenon? just said we can determine the delta(p) ? But back to the problem, we have a rest body then its momentum is ZERO. How can I say its delta(p) is infinite? |
> > |
Well first off let's clear up what the "delta" means. It refers to a statistical quantity. It has to do, not with one single measurement, but with many measurements of a system in a given (i.e. known) state. |
> |
This is not true |
> > |
so when you measure "x=a" it means that you go and measure the position of the particle and the result of the measurement was x=a. Now you measure the momentum of the particle at the same time and you get P=b. you repeat this many many many times. |
How do you measure momentum p = mv ? In ONE measurement ? and what is than delta p (if you do this many times)
Is there some one who can give an example ?
(I think this is all very very difficult besides I do not think there IS any true uncertainty in nature except OUR limitations to measure those quantities )
> > |
You calculate the
quanties delta(x) and delta(p) and you'll find that
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I always have to look that up). Given the definition of dx and dp and the popstulates of quantum mechanics you can actually derive the expression dx*dp >= hbar/2 |
> |
This is true. |
> > |
Pmb |
> |
On Wed, 27 Mar 2002 11:05:09 GMT, Nicolaas Vroom
"Robert Hamilton"
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its
position is at X = a.
According to the hypothesis,we have delta(x) is zero. and delta(p) is
infinite!!
How do we explain this phenomenon? just said we can determine the delta(p) ?
But back to the problem, we have a rest body then its momentum is ZERO.
How can I say its delta(p) is infinite?
Well first off let's clear up what the "delta" means. It refers to a
statistical quantity. It has to do, not with one single measurement,
but with many measurements of a system in a given (i.e. known) state.
This is not true
so when you measure "x=a" it means that you go and measure the
position of the particle and the result of the measurement was x=a.
Now you measure the momentum of the particle at the same time and you
get P=b. you repeat this many many many times.
How do you measure momentum p = mv ?
In ONE measurement ?
and what is than delta p (if you do this many times)
Is there some one who can give an example ?
(I think this is all very very difficult
besides I do not think there IS any true uncertainty in nature
except OUR limitations to measure those quantities )
You calculate the
quanties delta(x) and delta(p) and you'll find that
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I
always have to look that up).
Given the definition of dx and dp and the popstulates of quantum
mechanics you can actually derive the expression dx*dp >= hbar/2
This is true.
Pmb
>
>>
On 26 Mar 2002 09:40:57 -0800, Pmb
>> >
"Akira"
>> >>
>> >
>>
>> >
>
>> >
>>
>> >
>>
Not objecting to the measurment theorem. The theorem itself is correct as
far as it goes. But it's a mistake to think of the uncertainty principle
in only in terms of a statistical sampling, because it is more fundamental
than that.
>
The objection: when presented this way, one is led to think of the measurement as a coupling of the state to some external system which performs the physical act of measurement. Moreover, that kind of measurement is correctly described using mixed states. I think that something as fundamental to quantum mechanics as is the uncertainty relation should be applicable to the states themselves and not to state+measuring device, and moreover to individual states or superpositions rather than to statistical mixtures. That naturally leads one to look for a more fundamental statement of the uncertainty principle in those terms. If such exists it is surely more appropriate.
Well, in fact one does exist in the noncommutativity of the operators p and x. Many (myself included) think of this relation, [x,p]=i as the most fundamental statement of the uncertainty principle, and the delta p delta x formula becomes a necessary consequence of it. The operator form tells us, among other things, that a state cannot have both a definite momentum and a definite position. This is a condition that has to be satisfied by individual states of individual particles so it describes an intrinsic property of the state itself, not just the act of measurement of the state.
Mike Hanson
> |
"franz heymann" |
> > |
That is correct. If you know the position of a particle precisely, you can say nothing at all about its momentum |
> |
Absolutely. However: |
> > |
Roughly speaking, if you want to see with precision where a particle is, you have to illuminate it with light which has a wavelength smaller than the precision you wish to achieve. Such light consists of photons which will impart a large momentum to the particle when a photon hits it. So you have disturbed its original momentum by a correspondingly large amount. |
> |
Absolutely not! Or rather, although this would certainly happen if that's what you did, it has nothing whatsoever to do with the uncertainty principle. I would imagine that you yourself know better than to see HUP as 'unavoidable disturbance during measurement'. And I know that such an explanation is commonly given for the sake of simplicity (even Asimov explains it in these terms). But that does not justify giving out misleading information which subsequently needs to be corrected once a higher level of understanding is achieved. If this is the way that HUP is introduced to people, then it takes an awful lot of effort for them to unlearn this misconception and fully appreciate the implications of uncertainty. |
You are indisputably correct. I appologise for falling into this trap. Regrettably, that was how I was taught seventy years ago, and my thoughts still veer in that direction if I don't keep them in check.
Franz Heymann
"franz heymann"
> |
Mike Hanson |
> > |
"franz heymann" |
> > > |
That is correct. If you know the position of a particle precisely, you can say nothing at all about its momentum |
> > |
Absolutely. However: |
I expect that also the reverse is true. If you know its momentum precisely etc. It is interesting that the sentence starts with IF
> > > | Roughly speaking, if you want to see with precision where a particle is, you have to illuminate it with light which has a wavelength smaller than the precision you wish to achieve. Such light consists of photons which will impart a large momentum to the particle when a photon hits it. So you have disturbed its original momentum by a correspondingly large amount. |
> > |
Absolutely not! Or rather, although this would certainly happen if that's what you did, it has nothing whatsoever to do with the uncertainty principle. |
Okay - but not agreed.
> > | I would imagine that you yourself know better than to see HUP as 'unavoidable disturbance during measurement'. And I know that such an explanation is commonly given for the sake of simplicity (even Asimov explains it in these terms). But that does not justify giving out misleading information which subsequently needs to be corrected once a higher level of understanding is achieved. |
If the above is not the way than how do you measure
p (and x) precisely ?
Do you use light ?
I think it is not so easy.
> > | If this is the way that HUP is introduced to people, then it takes an awful lot of effort for them to unlearn this misconception and fully appreciate the implications of uncertainty. |
Please explain those implications. (I have great doubts)
> | You are indisputably correct. I appologise for falling into this trap. |
IMO you do not have to appologise.
> | Regrettably, that was how I was taught seventy years ago, and my thoughts still veer in that direction if I don't keep them in check. |
If this was taught to you 70 years ago than your youth must be at the craddle of GR and Quantum Mechanics. Assuming you where than 15 years old than you must now be ? I still have 25 years to go to reach that age.
> | Franz Heymann |
"Nicolaas Vroom"
Mike Hanson
That is correct. If you know the position
of a particle precisely,
you can say nothing at all about its momentum
Absolutely.
I expect that also the reverse is true.
If you know its momentum precisely etc.
>
"franz heymann"
> >
> > >
"franz heymann"
> > > >
> > >
>
Yes, that's the crux of HUP. Or rather, it's the most extreme case. The less you know about one of them, the more you can predict about the other. But please note that it's not just position and momentum that are related in this way ('conjugate attributes'). There are all sorts of attributes one can measure of a particle, and each has its conjugate attribute, such that one can never predict both of a conjugate pair beyond a certain combined precision.
> | It is interesting that the sentence starts with IF |
You can measure any attribute to any arbitrary precision, dependant only upon how good your measuring equipment is.
> > > > | Roughly speaking, if you want to see with precision where a particle is, you have to illuminate it with light which has a wavelength smaller than the precision you wish to achieve. Such light consists of photons which will impart a large momentum to the particle when a photon hits it. So you have disturbed its original momentum by a correspondingly large amount. |
> > > |
Absolutely not! Or rather, although this would certainly happen if that's what you did, it has nothing whatsoever to do with the uncertainty principle. |
> |
Okay - but not agreed. |
Well, as I went on to say:
> > > | If this is the way that HUP is introduced to people, then it takes an awful lot of effort for them to unlearn this misconception and fully appreciate the implications of uncertainty. |
Having had HUP introduced to you in terms of disturbance, I should imagine you are extremely reluctant to let go of this idea. After all, it makes such intuitive sense, doesn't it?
Unfortunately, it isn't correct. And I stress: this is not a point of view. It is a fact. Even realist, hidden-variable models (which seek to get rid of the random element that Einstein so disliked) do not see HUP as a disturbance during measurement. Of course, it has everything to do with measurement, but it is an intrinsic property of the particle itself, rather than something that is 'done to it' from the outside.
> |
If the above is not the way than how do you measure
p (and x) precisely ? Do you use light ? I think it is not so easy. |
When an electron smacks into the phosphor screen of your monitor, you know where it is (or where it was when it hit) because the screen emits a flash of light at that position. The flash of light, however, is not emitted immediately but only after a (very small) interval of time. That interval is not the same every time round, and it cannot be predicted beyond a certain accuracy. Yep, it's HUP.
Another way to determine position to arbitrary accuracy is to shoot particles through a small hole that you can make smaller. The precision of the 'sideways position' of the particle as it passes through the hole is then constrained to be simply the width of the hole. Making the hole smaller increases precision in this regard, but it will result in an increased unpredictability of the particle's sideways momentum.
Measuring momentum is easy if the particle is charged: pass it through a magnetic field (going 'across' the path of the particle) and the path of the particle will curve. You know how massive the particle is, so you can work out the momentum from the curvature.
There are various clever ways of working out the particle's momentum if it isn't charged, generally involving collisions with other particles whose momentum you can measure.
> > > | the implications of uncertainty. |
> |
Please explain those implications. (I have great doubts) |
Doubtless you have doubts: you have a 'disturbance picture' in your mind as you read this. As a result, what I am about to write is going to seem like groundless mysticism to you. But I assure you that this is only because you cannot see the grounds for such apparent weirdness, and it is therefore unacceptable to you. The first time you do see them, it's quite a shock, and it does force a paradigm shift on you.
Here goes. You will have heard it said that, prior to measurement, a particle cannot really be said to have a position at all. This is commonly misinterpreted as meaning "the particle's position is fuzzy". But whilst that would certainly be weird and hard to picture, it is still a naive oversimplification; the truth is even worse.
What is actually meant is that the attribute 'position' does not exist in any real sense at all until, by the act of making a position measurement, you help to create it. Likewise for all other attributes - momentum, spin, polarisation, etc: they are all figments of the measurement process, and have no meaningful existence outside of that process. (N.B. this is not the same as saying the particle isn't real: the particle *is* real, but its attributes are not, until the act of measurement calls them into existence. Attributes are a joint creation of the particle and of the measurement process.)
There are only two ways of escaping this conclusion. One is to adopt the 'Many Worlds' interpretation ("If you flip a 'quantum coin', the universe splits into two copies, identical in every respect save that in one the coin came up heads, and in the other it came up tails."). This gives you real attributes, but the price you have to pay for it is a burgeoning number of almost identical copies of yourself reading articles on USENET.
The other escape is to go for a realist, 'hidden variable' model. Very few physicists subscribe to these, though, because they create even greater absurdities than the ones they are trying to get rid of.
And those are your three choices. Disturbance during measurement doesn't cut it as an explanation, I'm afraid.
Mike.
Nicolaas Vroom
> |
"franz heymann" |
> > |
Mike Hanson |
> > > |
"franz heymann" |
> > > > |
That is correct. If you know the position of a particle precisely, you can say nothing at all about its momentum |
> > > |
Absolutely. However: |
> |
I expect that also the reverse is true. If you know its momentum precisely etc. It is interesting that the sentence starts with IF |
> > > > |
Roughly speaking, if you want to see with precision where a particle is, you have to illuminate it with light which has a wavelength smaller than the precision you wish to achieve. Such light consists of photons which will impart a large momentum to the particle when a photon hits it. So you have disturbed its original momentum by a correspondingly large amount. |
> > > |
Absolutely not! Or rather, although this would certainly happen if that's what you did, it has nothing whatsoever to do with the uncertainty principle. |
> |
Okay - but not agreed. |
> > > |
I would imagine that you yourself know better than to see HUP as 'unavoidable disturbance during measurement'. And I know that such an explanation is commonly given for the sake of simplicity (even Asimov explains it in these terms). But that does not justify giving out misleading information which subsequently needs to be corrected once a higher level of understanding is achieved. |
> |
If the above is not the way than how do you measure
p (and x) precisely ? |
> > > |
If this is the way that HUP is introduced to people, then it takes an awful lot of effort for them to unlearn this misconception and fully appreciate the implications of uncertainty. |
> |
Please explain those implications. (I have great doubts) |
> > |
You are indisputably correct. I appologise for falling into this trap. |
> |
IMO you do not have to appologise. |
Yes, It was correct of me to apologise, since the example I gave was a bad one. The real background to the HUP may be illustrated by the following example:
A particle with a perfectly well known momentum has a wave function of the form exp(ikx). This wave is smeared out over all x. To make a wave which has an amplitude which is large only in a restricted region of x, you have to add together (integrate) a range of waves with a range of values of k. (Consider the Fourier spectrum of a pulse extending over a short range of x only). If you look at the relationship between a pulse and its Fourier transform, you will find that (delta x)/(delta k) just gives the usual HUP relationship between position and momentum.
> |
> > |
Regrettably, that was how I was taught seventy years ago, and my thoughts still veer in that direction if I don't keep them in check. |
> |
If this was taught to you 70 years ago than your youth must be at the craddle of GR and Quantum Mechanics. Assuming you where than 15 years old than you must now be ? I still have 25 years to go to reach that age. |
Slip of the tongue! 60 years ago, not 70. (Actually 58 years ago, to be precise). It just shows what creeping senility is doing to me. :-)
Franz Heymann
"Mike Hanson"
> |
"Nicolaas Vroom" |
> > |
"franz heymann" |
> > > |
> > |
I expect that also the reverse is true. If you know its momentum precisely etc. |
> |
Yes, that's the crux of HUP. Or rather, it's the most extreme case. The less you know about one of them, the more you can predict about the other. But please note that it's not just position and momentum that are related in this way ('conjugate attributes'). There are all sorts of attributes one can measure of a particle, and each has its conjugate attribute, such that one can never predict both of a conjugate pair beyond a certain combined precision. |
> > |
It is interesting that the sentence starts with IF |
> |
You can measure any attribute to any arbitrary precision, dependant only upon how good your measuring equipment is. |
> |
> > > > |
If this is the way that HUP is introduced to people, then it takes an awful lot of effort for them to unlearn this misconception and fully appreciate the implications of uncertainty. |
> |
Having had HUP introduced to you in terms of disturbance, I should imagine you are extremely reluctant to let go of this idea. After all, it makes such intuitive sense, doesn't it? Unfortunately, it isn't correct. And I stress: this is not a point of view. It is a fact. Even realist, hidden-variable models (which seek to get rid of the random element that Einstein so disliked) do not see HUP as a disturbance during measurement. Of course, it has everything to do with measurement, but it is an intrinsic property of the particle itself, rather than something that is 'done to it' from the outside. |
Starting point is first: measurements Those measurements IMO will disturb the particle involved. General speaking the more accurate you want to measure the more you will disturb the particle. Using those measurements you can come up with a model of the particle and or with intrinsic properties of the particle.
> > |
If the above is not the way than how do you measure
p (and x) precisely ? Do you use light ? I think it is not so easy. |
> |
When an electron smacks into the phosphor screen of your monitor, you know where it is (or where it was when it hit) because the screen emits a flash of light at that position. The flash of light, however, is not emitted immediately but only after a (very small) interval of time. That interval is not the same every time round, and it cannot be predicted beyond a certain accuracy. Yep, it's HUP. Another way to determine position to arbitrary accuracy is to shoot particles through a small hole that you can make smaller. The precision of the 'sideways position' of the particle as it passes through the hole is then constrained to be simply the width of the hole. Making the hole smaller increases precision in this regard, but it will result in an increased unpredictability of the particle's sideways momentum. |
If the hole is larger you will not observe any change ie no x and you will not measure x If the hole is small and the particle can go through the whole what you claim is that than you can measure x ie is a function of the size of the hole. If the hole is very small the particle will not pass and again you will not measure x ie. you can not measure x as precise as you want.
> | Measuring momentum is easy if the particle is charged: pass it through a magnetic field (going 'across' the path of the particle) and the path of the particle will curve. You know how massive the particle is, so you can work out the momentum from the curvature. |
Do you call this a precise or a not precise measurement ?
If it is precise
does this mean that you can measure p of such a particle
twice precisely ?
It seems that for such particles HUP does not apply.
> | There are various clever ways of working out the particle's momentum if it isn't charged, generally involving collisions with other particles whose momentum you can measure. |
I think this is very very tricky and the uncertainty principle
prevents you from doing this.
IMO if you want to measure p precisely of particle_b you need
a particle_c with p precisely but than you do not know the x of
that particle.
Suppose I start with particle_b and particle_c and
I let the two collide.
Of particle_b I know p precise before the collision.
The original p of particle_c is unknown.
After the collision I must measure the p of particle_b and the p of particle_c precisily in order to calculate the original p of particle_c But how do I do that ?
I need collisions with other particles with other particles you can measure "precisely?"
IMO you will not succeed.
> > > > | the implications of uncertainty. |
> > |
Please explain those implications. (I have great doubts) |
> |
Doubtless you have doubts: you have a 'disturbance picture' in your mind as you read this. As a result, what I am about to write is going to seem like groundless mysticism to you. But I assure you that this is only because you cannot see the grounds for such apparent weirdness, and it is therefore unacceptable to you. |
> |
The first time you
do see them, it's quite a shock, and it does force a paradigm shift on
you.
Here goes. |
> |
You will have heard it said that, prior to measurement, a
particle cannot really be said to have a position at all. This is
commonly misinterpreted as meaning "the particle's position is fuzzy".
But whilst that would certainly be weird and hard to picture, it is
still a naive oversimplification; the truth is even worse.
What is actually meant is that the attribute 'position' does not exist in any real sense at all until, by the act of making a position measurement, you help to create it. |
Each measurement is a small process an action by using a measurement tool. As a result of this process or action both the tool and the object will change. The amount of change of both give you information ie you get more information about the object. You learn something from this experiment.
It is difficult to accept that as such that you create something.
(IMO parameters of objects do not exist and they cannot be created. Only the particle ansich exists)
> | Likewise for all other attributes - momentum, spin, polarisation, etc: they are all figments of the measurement process, and have no meaningful existence outside of that process. (N.B. this is not the same as saying the particle isn't real: the particle *is* real, but its attributes are not, until the act of measurement calls them into existence. Attributes are a joint creation of the particle and of the measurement process.) |
IMO general speaking it is for me difficult to accept that a measurement creates the spin of a particle and that before the measurement it did not have a spin.
You can also repeat this measurement n times. Does that mean that each time you create this spin ?
(or position or momentum ?)
> |
Mike. |
Nick
http://users.pandora.be/nicvroom/
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