Talk:Measurement in quantum mechanics

Archive

This talk page hasn't received any comments since May 11, 2007. It's been since June 26 2006 since an actual conversation took place. I've archived the old talk page to start fresh on cleaning up this article. I'll put the appropriate tags on the main page. Here's hoping we can all work together to make this article better. JFlav 02:33, 22 September 2007 (UTC)

Cleanup Effort

This article needs to be revised or rewritten. It is heavy on derivations and examples and light on any explanation of concepts. In fact, it reads as if it was copied out of someone's quantum mechanics notes. JFlav 02:52, 22 September 2007 (UTC)

you wanna volunteer? :-) Mct mht 03:13, 22 September 2007 (UTC)

Yeah, I'll do my best. But I don't know how much I qualify as an "expert." JFlav 16:34, 22 September 2007 (UTC)

You'll notice I did a thorough cleanup on this article recently. Does it address the problems? In what ways could it be improved? --Steve 00:00, 1 December 2007 (UTC)

Erroneous Example

The example can't be salvaged. Wavefunctions can't collapse to single position eigenkets without violating the uncertainty relation. This can be seen as follows: if a measurement is made which collapses the wavefunction to the position ket $|S\rang$, we can project this into position space as $\lang x | S \rang = \delta(x-S)$, and the position standard deviation for this wavefunction would be zero. The assumption that the function has collapsed to this nonphysical state leads to an error in the derived probability of measuring the nth energy value immediately after the position measurement. Suppose, for example, that a particle is measured at L/2 with such accuracy that it is left in the state $|\tfrac{L}{2}\rang$. Then: $\Pr(E_n) = |\lang \psi_n | \tfrac{L}{2} \rang|^2 = \frac{2}{L}~{\rm sin}^2\left(\frac{n \pi \tfrac{L}{2}}{L}\right)=\frac{2}{L}~{\rm sin}^2\left(\frac{n \pi}{2}\right)=\frac{2}{L}$ for every odd n. This is, of course, impossible as these probabilities must add to 1. Auspex1729 (talk) 17:56, 2 August 2009 (UTC)

First, since (non-relativistic) quantum mechanics imposes no limitation on the accuracy with which a parameter can be measured there is no problem with collapsing onto eigenkets.

Second, the formula:$\Pr(E_n) = |\lang \psi_n | \tfrac{L}{2} \rang|^2$ is incomplete. You have assumed a discrete orthonormal basis. More generally it should be:

$\Pr(E_n) = \frac{|\lang \psi_n | \frac{L}{2} \rang|^2 }{ |\lang \psi_n | \psi_n \rang||\lang \tfrac{L}{2} | \tfrac{L}{2} \rang|}$

Since $|\lang \tfrac{L}{2} | \tfrac{L}{2} \rang| = \infty$ each individual probability vanishes, but they still sum to one.--Michael C. Price ^talk 19:21, 3 August 2009 (UTC)

You have assumed a discrete orthonormal basis. Because the infinite square well has a discrete orthonormal basis! You've assumed, to the contrary, that they aren't (fine, it's your life) and thrown some orthonormalization terms into the probability. You should get the same thing as someone who started with an orthonormal basis, but you don't. You don't because you're using unbelievably sloppy math here, and drawing incorrect conclusions as a result. Then you threw some redundant notation into the position kets in the example... why? And doesn't your conclusion that "each individual $\Pr(E_n)$ vanishes" disturb you, especially in an example which was supposed to demonstrate the probability of collapse to the energy eigenstates after a position measurement? The position has a continuous spectrum, and a position measurement will cause the wavefunction to collapse onto a continuous superposition of position states about the measured value. If you don't believe it because you're getting unphysical results from this example, if you don't believe it because delta functions such as $\lang x | S \rang = \delta(x-S)$ aren't normalizable, then at least sit down and use some uncertainty estimates or the DeBroglie relation to see the energy requirements of making a measurement which would justify writing the state of the system as a single position ket (Spoiler: They're infinite). If that doesn't make you give this up, the error is explained in Cohen-Tannoudji Vol. 1 on p. 278, again on Griffiths p. 106 in a footnote, and again in Sakurai.Auspex1729 (talk) 18:42, 4 August 2009 (UTC)

Obviously the energy eigenbasis is orthonormal and discrete, but the position basis is not. I should have thought that it was obvious which eigenbasis I was referring to. I suggest you digest that and then perhaps calm down. And, no, the fact that each individual probability vanishes does not disturb me; since the position measurement was infinitely precise the associated disturbance has been infinite (as a consequence of the very uncertainty relations which you maintain have been violated by this example -- I don't think so....). I was being polite in my first response, but I will be blunter now; your statement: Wavefunctions can't collapse to single position eigenkets without violating the uncertainty relation. reveals that you really don't have a clue about the physics. --Michael C. Price ^talk 18:58, 4 August 2009 (UTC)

The formalism of the problem was based on the energy basis. The only thing you could obviously have been referring to was the set of energy states. ...reveals that you really don't have a clue about the physics. Based on the little gems you've peppered this thread with, I don't doubt for a second that you believe there's a $\sigma_p$ satisfying $0\cdot\sigma_p\ge\frac{\hbar}{2}$. Maybe you should review a text on elementary quantum mechanics before you cram this article full of your egregiously wrong original research.Auspex1729 (talk) 19:40, 4 August 2009 (UTC)

$\sigma_p$ can be infinite, and is in this case. But Auspex1729 has a bit of a point...if we're going to put in an example, it might as well be one that's at least somewhat physically plausible. (No energy expectation values being infinity, etc.) It's enough already to confront some poor reader with their first example of doing a QM measurement calculation...it's too much if, at the same time, we confront them with infinities and delta-functions and all the other stuff that we're used to but is hard to swallow the first time you see it. Can we come up with a better/simpler example? Maybe measuring the spin of a spin-1/2 particle along the x-axis then y-axis then x-axis? Or something else? Do we even really want or need an example? This is sorta moving into textbook territory out of wikipedia territory if we're not careful. :-) --Steve (talk) 20:02, 4 August 2009 (UTC)

Examples are not out of place on wikipedia, especially when we have a lot of abstract concepts floating around. The example is only illustrating the disturbance induced by the measurement of one parameter on another non-commutting parameter. The actual values of the probabilities involved aren't that important, so this is all a bit of a red herring. The example illustrates what it is intended to illustrate quite well.

The trouble with an example involving spin is that this isn't something a non-physics reader can really relate to. But of course, no reason why we can't have another example as well.

BTW there are neither delta functions nor infinities in the example (only on this talk page), so that is another red herring we don't have to worry about. Please don't delete this example, based on some superious irrelevant objections -- at least until we have some equally/more acessible examples to replace it with. --Michael C. Price ^talk 21:12, 4 August 2009 (UTC)

The only thing spurious here is your handwaving upthread. I have cited two standard textbooks on this, the footnote in Griffiths p. 106 and the example in Cohen-Tannoudji. Incidentally, I was wrong when I said the example couldn't be salvaged. Cohen-Tannoudji does write the system as prepared in a state represented by an integral over a small interval of position eigenkets (as I suggested it should be upthread), and obtains solutions and probabilities which, you know, actually work.Auspex1729 (talk) 05:05, 5 August 2009 (UTC)

Salvage it then, if you feel so strongly. However, as I said, the explanation currently does not mention infinities, so why are we bothered? Why complicate a simple example?--Michael C. Price ^talk 06:04, 5 August 2009 (UTC)

It seems that the edit by IP address 86.73.72.40 : reference is not relevant to the encyclopedic nature of wikipedia. It may even be self promotion. The IP address 86.73.72.40 has gone about adding this reference and information rtelated to this reference at several Physics pages. Kanwarpreet Grewal 09:31, 2 April 2010 (UTC) —Preceding unsigned comment added by Kp grewal (talk • contribs)

small query

In the section on observables it mentions the properties of observable operators. Amongst these it says that the operators are Hermitian, and they have real eigenvalues. But ALL Hermitian opertaors have real eigenvalues, so it might be better to put these in the same point. Currently it reads like having real eigenvalues is specific to operators corresponding to observables. —Preceding unsigned comment added by 82.6.96.22 (talk) 21:44, 24 May 2011 (UTC)

minor changes

I'll be adding a sentence here and there, or inserting a few words, to hopefully improve comprehension. I'll start with the lead which is not very informative. I'm somewhat familiar with the discussion on this topic among experts, but wish to make things more accessible to non-experts.

regards, Jonathan JonathanD (talk) 03:57, 12 June 2011 (UTC)

I've reverted the changes, since the measurement problem is not largely the result of Quantum indeterminacy - the cause is a matter of debate that relates to whatever interpretation of quantum mechanics one holds. Jonathan, I would advise you not continue making large additions/changes to the lead of articles on this subject (or any subject for that matter). The lead should only mention material that is discussed in more detail within the body of the main article. -- cheers, Michael C. Price ^talk 06:19, 12 June 2011 (UTC)