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Is this equivalent to a photon?


hobz

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This is equivalent to saying that when you roll a fair die and a six comes up that you really don't know its a six.

 

I have to disagree with you here. This is more like rolling a ball and having it come to rest, but even that's a bad analogy.

 

That is not what quantum mechanics has to say about this. Quantum mechanics would say that the position is not determined until you measure it where at that point it becomes determined (if it wasn't already in an eigenstate of position). This phenomena is known as the collapsing of the wave function.

 

And all this talk of uncertainty only applies to quantum states which are not eigenstates of the observable. If a state is in an eigenstate of, say, energy E, then there is no uncertainty in energy in that the value that will be measured has a 100% probability of having that value.

 

Pete

 

And this is the problem — we aren't necessarily in a situation where we have superposition of states. A free electron can have any value of position and momentum. So collapsing the wave function isn't really what's going on here.

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I have to disagree with you here. This is more like rolling a ball and having it come to rest, but even that's a bad analogy.

That was an example given so as to make the physics as clear as possible. I used an observable which is quantized for clarity. The die analogy refers to quantum mechanical systems where the observable is quantized, such as spin. In such cases the uncertainty of the variable is not neccesarily zero. I think such systems are very good to think of when considering the notion of uncertainty.

And this is the problem — we aren't necessarily in a situation where we have superposition of states. A free electron can have any value of position and momentum. So collapsing the wave function isn't really what's going on here.

I'm curious. What do you think that collapsing the wave function means/refers to?

 

I was speaking on the more general situation regarding observables in QM and the general expression for the uncertainty principle.

 

An arbitrary quantum state can be expanded in terms of position eigenkets. The coefficients in such an expansion is the wave function evaluated at the relevant position. Once the position is measured the system falls into an eigenstate of position. This is what the term "collapse" actually means.

 

If the quantum state is not in a superposition then it is in an eigenstate of an observable (and can therefore be repsented in another basis as a superposition of these eigenstates and therefore whether a state is in a "superposition" is really a basis dependant phenomena). If the system is in such an eigenstate then an imediate measurement of the observable will yield the cooresponding eigenvalue with certainty.

 

If an electron is moving freely then it can be in such a superposition. If it is instead in an eigenstate of an observable of, say, position then it cannot be in an eigenstate of momentum and must therefore be in a superpostion of momentum eigenstates (when the quantum state is represented in momentum space).

 

Let me restate that for clarity - If a system is in an eigenstate of observable A then it cannot be in an eigenstate of observable B if A and B do not commute, like x and p for example.

 

Note - Any wave function is in a superposition of position eigenstates unless the wave function is a Dirac delta function.

 

Pete

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That was an example given so as to make the physics as clear as possible. I used an observable which is quantized for clarity. The die analogy refers to quantum mechanical systems where the observable is quantized, such as spin. In such cases the uncertainty of the variable is not neccesarily zero. I think such systems are very good to think of when considering the notion of uncertainty.

 

Since the values one can obtain are quantized, it's a different manifestation of the concept.

 

I'm curious. What do you think that collapsing the wave function means/refers to?

 

I was speaking on the more general situation regarding observables in QM and the general expression for the uncertainty principle.

 

An arbitrary quantum state can be expanded in terms of position eigenkets. The coefficients in such an expansion is the wave function evaluated at the relevant position. Once the position is measured the system falls into an eigenstate of position. This is what the term "collapse" actually means.

 

If the quantum state is not in a superposition then it is in an eigenstate of an observable (and can therefore be repsented in another basis as a superposition of these eigenstates and therefore whether a state is in a "superposition" is really a basis dependant phenomena). If the system is in such an eigenstate then an imediate measurement of the observable will yield the cooresponding eigenvalue with certainty.

 

If an electron is moving freely then it can be in such a superposition. If it is instead in an eigenstate of an observable of, say, position then it cannot be in an eigenstate of momentum and must therefore be in a superpostion of momentum eigenstates (when the quantum state is represented in momentum space).

 

Let me restate that for clarity - If a system is in an eigenstate of observable A then it cannot be in an eigenstate of observable B if A and B do not commute, like x and p for example.

 

Pete

 

Yes. How do you reconcile that with your earlier claim that you can simultaneously measure both variables to an arbitrary precision? And that the uncertainty was a property of the state, while now you are saying it's only a property of a measurement?

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Since the values one can obtain are quantized, it's a different manifestation of the concept.

The analogy is still valid. Just because it addresses quantized values rather than continuos ones makes no difference since what you said so far has not differentiated the two.

How do you reconcile that with your earlier claim that you can simultaneously measure both variables to an arbitrary precision?

If I explained that I'd be restating what I already did above. Its not that I really mind doing it if I believe it would be instructive to do so but I recently damaged my back and any posting I do not is at the cost of incredible amounts of agony. Sorry. I'll try to be clearer tomorrow, hoping I'll feel better.

 

If you find that my comments in this thread are inconsistent or lead to observations that are contradictory to that predicted in QM then please specify in what way. Thanks.

And that the uncertainty was a property of the state, while now you are saying it's only a property of a measurement?
I never said nor implied that.

 

In the mean time I have a question for you; Have you ever seen and/or followed the derivation of the Heisenberg-Robertson Uncertainty Relation? Analysis of the derivation will provide one with all they need to know in order to be able to interpret its meaning, application and domain of applicability.

 

 

swansont - I think I might now understand what your reservations are with what I've said so far. Perhaps you believe that since individual measurements of position and momentum can be made precisely that this would lead to a classical trajectory and thus deterministic motion in QM, contrary to what one knows of quantum theory, is that the case? If so then that what I've been saying does not lead to such a conclusion. Let me elaborate.

 

First: By simultaneous measurements of two obsevables one does not neccesarily take the measurements at exactly the same time since measurements that close to each other may be impossible in practice and it would give the wrong idea of the mearement process of incompatible observables, although not with compatible observables. What is actually meant is that one measures observable A and then, in an infinitesimally short amount of time dt later, one measures the observable B.

 

Now let's apply this to the conjugate observables X and P. If one measures the position of a particle then the wave function collapses into an eigenstate of position. The corresponding wave function is a Dirac delta function. The Fourier transform of the Dirac delta function is a constant function. Thus when one changes from position space to momentum space one gets a uniform probability density of measuring mometum. This means that [math]\Delta P[/math] is infinite. If one then measures the momentum in an infinitesimally short amount of time later one gets a specific value of mometum. The system then collapses into an eigenfunction of mometum (another Dirac delata function). The corresponding position wave function is then a constant. Thus if one measures the position and then momentum and then position again there will be a discontinuity in the value of position. Therefore there can be no classical notion of a trajectory. If one meausures x and then p then in a second run with an identical system in which one also measures x then a subsequent postion won't neccesarily be the same value of p one got before.

 

It would be much easier to temporarily place this QM analysis on hold and instead discuss probability theory and the meaning of uncertainty, expectation etc. first. We could then easily see what it means to measure an exact value of A when there is a non-zero uncertainty in A. The meaning will not change when we go back to quantum theory.

 

Does anyone here who is following along have difficulty understanding the fact that situations exist where the possibility exists (in probability theory) of measuring exact values of a variable even when the uncertainty in that variable is non-zero?

 

Pete

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I never said nor implied that.

 

Then obviously we've had a mixup earlier where you were saying that "Uncertainty is a statistical quantity which applies only to a large number of data points." (emphasis added) and objected to my contention that a photon's state could have an uncertainty associated with it.

 

When I confirmed I was not talking about measurement, but rather the state of the particle (or thought I made that clear) you continued to disagree.

 

If that's the confusion then we've probably been talking past each other.

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Then obviously we've had a mixup earlier where you were saying that "Uncertainty is a statistical quantity which applies only to a large number of data points." (emphasis added) and objected to my contention that a photon's state could have an uncertainty associated with it.

 

When I confirmed I was not talking about measurement, but rather the state of the particle (or thought I made that clear) you continued to disagree.

 

If that's the confusion then we've probably been talking past each other.When I was responding to

And that the uncertainty was a property of the state, while now you are saying it's only a property of a measurement?

I meant that I never said it's only a property of a measurement. Can you tell me what I said that led you to this conclusion?

 

Pete

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Does anyone here who is following along have difficulty understanding the fact that situations exist where the possibility exists (in probability theory) of measuring exact values of a variable even when the uncertainty in that variable is non-zero?

 

Pete

 

No, but I think the gist of your argument is in relation to QM's formalism. In that case, to be brunt, is there any relationship between this and bells theorem, could I think hidden variables could be alive in a formalism sense?

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I meant that I never said it's only a property of a measurement. Can you tell me what I said that led you to this conclusion?

 

I quoted it above.

 

"Uncertainty is a statistical quantity which applies only to a large number of data points."

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I quoted it above.

 

"Uncertainty is a statistical quantity which applies only to a large number of data points."

Yes. That is quite true. This is not merely my personal opinion but is the definition found in textbooks on quantum mechanics such as Sakurai's or the one I used in graduate school, i.e. that by Cohen-Tannoudji et al

 

Sorry for the confusion. I was having difficulties with my computer earlier which prevented me from posting correctly. What I meant was that I disagree with the following - ..while now you are saying it's only a property of a measurement?. If that is how my post was interpreted then I want to state that it was not how I meant it to be interpreted.

 

I had asked you several questions in this thread which had gone unanswered. The most important of them had to do with providing me with a precise mathematical definition of undertainty as it is used in phrasing the uncertainty principle as it is found in quantum mechanics text books, preferably at the graduate level since undergraduate level texts are sometimes vauge on this point. I also suggested that we find a derivation of the uncertainty principle that we both agree with. When both of these tasks are accompished I believe our differences will be resolved in the process.

 

Here is one that I took a brief look at - http://en.wikipedia.org/wiki/Uncertainty_principle

 

In my brief review of it, it appeared to be correct. It also mentions another synonym for uncertainty that I had forgotten, i.e. uncertainty = root-mean-square (RMS) deviation of the position from its mean.

 

Doesn't it seem reasonable to you that a wave function which has a pronounced peak at x = x0 is interpreted to mean that there is a high probability that the particle will be found in a region centered at x = x0? The smaller the region the higher the probability that it will be found near there. Thus if the wave function is spread out the region in which it is likely to be found is also spread out. According to your own interpretation of "uncertainty" (as refering to how likely the particle is to be found in a region) this also makes sense.

 

I'm also curious as to what you think about the Copenhagen Interpretation? To your knowledge what does it say and what does it mean?

 

Since the above web site is saying what I've said all aloing I recommend that you find a definition of [math]\Delta X[/math] and/or a derivation of the uncertainty principle. Preferably from a QM text that you have and respect, i.e. perhaps you have a QM text that you learned QM from or you know of/can find onlinelecture notes from a university graduate course on QM or an e-book on QM that is online that we both have access to. If you prefer I can scan the relevant pages from the text I have and then upload them to my web site so you can download them and read them. Then we'd be working from the same source(s).

 

Notice that I used an upper case X here. I've been loosey-goosey about the case until now. From now on I will use upper case letters to refer to the operator corresponding to an observable and lower case letters to refer to the corresponding eigenvalues. Some authors will use a caret over the letter to indicate that is the "observable" rather than the corresponding eigenvalue (note that in QM observables are operators).

 

Does this sound like a reasonable next step in our discussion to you? If so then I found an E-book online about quantum mechanics which is located under the physics department web site of San San Francisco State University. See

http://www.physics.sfsu.edu/~greensit/book.pdf

 

See section 4.4 Eq. (4.75). Note where the author writes

The square root of this quantity ... is referred to as the Uncertainty in the position of the particle in the quantum state [math]|\Psi>[/math].

This, of course, implies that given a different state [math]\Psi[/math], the uncertainty will be different as a result.

 

See also

http://www.lsr.ph.ic.ac.uk/~plenio/lecture.pdf

 

There is a set of graduate level quantum mechanics lecture notes at

http://farside.ph.utexas.edu/teaching/qm/389.pdf

http://quantummechanics.ucsd.edu/ph130a/130_notes/node188.html

 

See section 2.14. The author provides a derivation of Heisenberg's Uncertainty Relationship.

 

Edit - I just finished scanning and uploading the relavent portions of Quantum Mechanics, by by Claude Cohen-Tannoudji, Bernard Diu and Frank Laloe. You can download it from

http://www.geocities.com/pmb_phy/uncertainty.pdf

 

I'd like to delte that file as soon as possible so as to avoid copyright infringement problems.

 

Best wishes

 

Pete

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