Confusion over compatible measurements

[elemental09]

I'm still learning the formalism and structure of standard QM, so bear with me. My current confusion arises when considering the subsequent measurement of two commuting observables, for example the x-position and y-position observables of a single particle. Suppose the system is in a pure state [PLAIN][URL]http://latex.codecogs.com/gif.latex?\mid[/URL]&space;\Psi&space;\rangle .[/URL] Consider two commuting observables, [URL]http://latex.codecogs.com/gif.latex?\hat{A}[/URL] and [URL]http://latex.codecogs.com/gif.latex?\hat{B}.[/URL] Since these commute, there exists a basis of simultaneous eigenkets [URL]http://latex.codecogs.com/gif.latex?\mid[/URL] a,b\rangle , with [URL]http://latex.codecogs.com/gif.latex?\hat{A}\mid[/URL] a,b\rangle = a\mid a,b\rangle and [URL]http://latex.codecogs.com/gif.latex?\hat{B}\mid[/URL] a,b\rangle = b\mid a,b\rangle.
Now, a measurement of observable A on the system must result in an outcome of an eigenvalue, say [URL]http://latex.codecogs.com/gif.latex?a_{0}[/URL] of A. Further, the system immediately after the measurement is collapsed into the corresponding eigenstate [URL]http://latex.codecogs.com/gif.latex?\mid[/URL] a_{0},b_{0} \rangle , where [URL]http://latex.codecogs.com/gif.latex?b_{0}[/URL] is the eigenvalue of B corresponding to the eigenket the system was just collapsed into.

My question is: does this not mean that if one were to then measure observable B, the outcome would have to be [URL]http://latex.codecogs.com/gif.latex?b_{0}[/URL]? But how can this be true? After all, the two observables are compatible, and measurement of one should not affect the other. The two quantities, x-position and y-position, are independant even in QM, are they not? In other words, I should be able to measure the x-position of a particle andd obtain a definite result without altering the probability distribution of the outcomes of a subsequent y-position measurement.

Where does my confusion lie?

 

[cgk]

In a Copenhagen-y interpretation, your statements are correct up to this:

Now, a measurement of observable A on the system must result in an outcome of an eigenvalue, say [URL]http://latex.codecogs.com/gif.latex?a_{0}[/URL] of A. Further, the system immediately after the measurement is collapsed into the corresponding eigenstate [URL]http://latex.codecogs.com/gif.latex?\mid[/URL] a_{0},b_{0} \rangle, where [URL]http://latex.codecogs.com/gif.latex?b_{0}[/URL] is the eigenvalue of B corresponding to the eigenket the system was just collapsed into.

Unless you actually do a measurement of B, there is no information about the b-subsector of the state you obtained after the A measurement. In particular, there is no guarantee that what you obtained actually is a eigenstate of B (in general it is not). Just that there is a basis of simultaneous eigenstates of A and B does not imply that every eigenvector of A is a eigenvector of B.

 

[elemental09]

Thanks for the reply.
In this case, I need to clarify a point which was not made explicit in my course. I am aware that the total state vector describing a composite system, say two particles each with independent spins, consists of the direct product between two kets inhabiting two spaces, one for the first particle's spin, and one for the second. However, it seems to me that even a single spinless particle in more than one dimension requires such a factorization. That is to say, I cannot see how the formalism of QM works without each independent degree of freedom of a system having its own ket, and the total state being represented as a direct product of each. In particular, the state of a spinless particle in three dimensions should take the form
[URL]http://latex.codecogs.com/gif.latex?|\psi\rangle=|&space;x\rangle\otimes|&space;y\rangle\otimes&space;|&space;z\rangle[/URL]
With, in general, each component in its own superposition, perhaps written in terms of the position basis. For x this would be:
[URL]http://latex.codecogs.com/gif.latex?|x\rangle=\int&space;d\bar{x}C_{x}(\bar{x})|\bar{x}\rangle[/URL]

Finally, when measuring the x-position, the system immediately after measurement is in some eigenstate of the x-position operator, however the remaining y and z components of the total state are unaffected - they remain in their individual superpositions. In this sense, the individual position operators act only on their appropriate "subkets" of the total state.

Is this correct?

 

[SpectraCat]

My take on this is that it doesn't actually matter which version of your explanation is correct .. the one from post #1 or the one from post #3. The salient point is, what experiment would you propose that could tell the difference between the complete collapse you describe #1, as opposed to the partial collapse you describe in #3?

My personal view is that complete collapse after the initial measurement is more consistent with the usual interpretation of Q.M., but I don't think that the difference is really meaningful beyond the context of philosophy.

 

[Demystifier]

I am aware that the total state vector describing a composite system, say two particles each with independent spins, consists of the direct product between two kets inhabiting two spaces, one for the first particle's spin, and one for the second.

That is not true. In general, the total state may be a SUPERPOSITION of such direct products.

Consider the state
|Psi> = |a1 b1> + |a1 b2> + |a2 b3>
Before measurement, you may say that the value of B may be b1, b2, or b3.
Now suppose that you measure A and that you obtain the value a1. Now after the measurement the value of B may be either b1 or b2 (you don't know which one of them will be), but it cannot be b3.

But does it mean that you may use a measurement of A to affect the value of B? Not in any predictably useful sense, because you cannot CHOOSE to get the value a1 (when you choose to measure A). If your apparatus measured A but you don't know the value it obtained, then the predictions on B you can make are the same as before the measurement.

 

[elemental09]

SpectraCat: there is a difference - were the first case true, every time one measures observable A of the system to be in a particular eigenstate, one can also immediately predict the outcome of B - namely, the eigenvalue of the simultaneous eigenket.

However, I now see this is false, as explained by cgk. My understanding, then, of simultaneous eigenkets and compatible measurements relies on the formalism of treating different degrees of freedom for a system as sub-kets, part of product kets inhabiting a product space. I understand this now.

Demystifier: thanks for your insight. Indeed, after some reading I realize now the nature of the product space - by addition closure, it contains vectors not expressible as single product kets. This leads into entanglement and such.

Having now (hopefully) grasped the slightly more general formalism of QM, I find myself now wondering about a more practical question: if the positional co-ordinates of a particle are indeed treated as being sub-kets in product kets, as described in post #3, with the general state being an arbitrary superposition of such kets, then the formalism seemingly allows for an entangled state on a single, spinless particle: namely, an entanglement between the separate, orthogonal co-ordinates, be they the components of position or momentum. Using the latter for the convenience of countability, such a state would look like:

A measurement of one of the components of momentum would then apparently determine the remaining two. Is this really allowed in QM? Is such a state preparable? If so, can anyone imagine how?

 

[SpectraCat]

SpectraCat: there is a difference - were the first case true, every time one measures observable A of the system to be in a particular eigenstate, one can also immediately predict the outcome of B - namely, the eigenvalue of the simultaneous eigenket.

No, you couldn't. You would only know that it is in *some* eigenstate of B .. you wouldn't know which one until you made a measurement on B, so it is unpredictable. If you still think this is wrong, please do what I asked and propose and experiment that could distinguish between the two cases.

 

[elemental09]

My proposal in post #1 was based on a flawed understanding of simultaneous eigenkets and compatible observables. I was operating under the idea that a simultaneous eigenket was one single ket, not the direct product of two subkets. In that case a measurement of A would collapse it to a single such eigenket, which I though to be an eigenket of B as well, rendering the state of the system such that a measurement of B could only then return one value, which could be predicted given knowledge of the matrix elements of B. A clearer example: measuring x would then determine y and z! Definitely false.

My second post was a better, but still flawed, as pointed out by demystifier - my "general case" was not general enough.

I think our disagreement simply arises from my lack of understanding while writing post #1 - it then appeared as though I understood the true nature of simultaneous eigenkets, meaning my question looked interpretative rather than empirical.

 

[SpectraCat]

My proposal in post #1 was based on a flawed understanding of simultaneous eigenkets and compatible observables. I was operating under the idea that a simultaneous eigenket was one single ket, not the direct product of two subkets. In that case a measurement of A would collapse it to a single such eigenket, which I though to be an eigenket of B as well, rendering the state of the system such that a measurement of B could only then return one value, which could be predicted given knowledge of the matrix elements of B. A clearer example: measuring x would then determine y and z! Definitely false.

Ok .. but it still seems like you may be missing the point of what I am saying, at least partly. Kets and vector spaces and direct product are just mathematical tools we impose to help us understand and rationalize what is measured. My point is that, as far as I know, BOTH pictures you mention could be correct, so it doesn't matter how you choose to interpret it. Your example above is not "definitely false" as you claim. It may well be that measuring x *does* "determine" y and z, in the sense that the act of measurement collapses the wavefunction into a single eigenstate of each component space. At that point, measurements on y and z could only return one value, the important bit (as you perhaps realize) is that you do not know what the values of y and z are in such a case until you perform subsequent measurements on those components. After the second measurement, it is impossible to know whether it was initial x measurement, or the subsequent measurement of y or z, which collapsed the state.

I think the misunderstanding may lie with your italicized statement from the quote above. You seem to think that the act of measuring A would render the value of B predictable if the state were to collapse completely. How? Which matrix elements are you referring to? If the first measurement is on A alone, then all of the matrix elements with the B components will be zero by definition, so I don't see where any predictability would come from.

Now, in some cases, measuring one observable can unavoidably result in you getting knowledge about a commuting observable at the same time. For example, consider angular momentum .. if you measure the angular momentum of a state, and find that it is zero, the you immediately know that the projection on the z-axis must also be zero .. simply because there is no other option. But there is no contradiction there, because if you had measured a different eigenvalue for the angular momentum, say hbar, then you would know that the z-component could be -1, 0, or 1, but you would not know which one. Nothing about the angular momentum measurement in general inherently gives you information about the z-component.

 

[Demystifier]

A measurement of one of the components of momentum would then apparently determine the remaining two. Is this really allowed in QM?

Yes.