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I Commuting observables vs. exchanging measurements

  1. Feb 21, 2017 #1
    Hi.
    I'm afraid I might just be discovering quite a big misunderstanding of mine concerning the meaning of the expectation value of a commutator for a given state.
    I somehow thought that if the expectation value of the commutator of two observables ##A, B## is zero for a given state ##\left|\psi\right\rangle##, ##\langle \psi|[A,B]|\psi\rangle = 0##, then the measurement statistics for both observables must be the same no matter which one is measured first.

    Obviously this cannot be true. Assume ##A## and ##B## have no common eigenstates and ##\left|\psi\right\rangle## is an eigenstate of ##A##. Then ##\sigma_A=0## but ##\sigma_B>0## and by that (assuming a situation where ##\sigma_B<\infty##)
    $$0=\sigma_A\cdot\sigma_B\geq\frac{1}{2}\cdot \left|\langle \psi|[A,B]|\psi\rangle\right|\enspace ,$$
    so ##\left\langle \psi|[A,B]|\psi\right\rangle=0##.

    But obviously if we measure ##A## first we will always get ##\left|\psi\right\rangle##, but not when we first measure ##B## and then ##A##, hence the measurement statistics depends on the order of the measurements.

    So can measurements only be exchanged if ##[A,B]=0##, i.e. they share an eigenbasis? Is the only implication of ##\langle \psi|[A,B]|\psi\rangle = 0## that ##\left|\psi\right\rangle## is an eigenstate of at least one of the observables?
     
    Last edited: Feb 21, 2017
  2. jcsd
  3. Feb 21, 2017 #2
  4. Feb 21, 2017 #3
    I'm not talking about the commutator being zero (meaning the zero operator), but its expectation value being zero for a given state
    $$\left\langle \psi|[A,B]|\psi\right\rangle=0\enspace .$$
    I might not have worded this very well in #1 though, I made some clarifications.
     
  5. Feb 21, 2017 #4
    As you say, any expectation value is taken with respect to some given state $$|\psi\rangle$$. You can, if necessary, restrict the domain of A and B to this state Ψ only, constructing new operators A',B'. Then, the old story works again: if $$\langle\psi|A'B'|\rangle=\langle\psi|B'A'|\rangle$$ they share an eigenvector, namely Ψ. And if you know you are in Ψ from the beginning, the order of measurement does not matter. Someone else may make this story a bit more rigourous.

    If the initial state is not this ψ, measurement order can play a role on the other hand.
     
  6. Feb 21, 2017 #5
    If ##\left|\psi\right\rangle## is not an eigenvector of ##B##, you cannot restrict it to a subspace spanned by ##\left|\psi\right\rangle## only (##B'## shall still be an endomorphism or square matrix). You cannot use as an assumption what you're trying to show.

    Let's make a counterexample: Assume the eigenvectors of ##A## are ##\left|0\right\rangle=\left|\psi\right\rangle## and ##\left|1\right\rangle## and the ones of ##B## are ##\frac{\left| 0\right\rangle+\left|1\right\rangle}{\sqrt{2}}## and ##\frac{\left| 0\right\rangle-\left|1\right\rangle}{\sqrt{2}}##. It's easy to show that indeed
    $$\left\langle \psi|[A,B]|\psi\right\rangle=0\enspace .$$

    Now if we measure ##A## first, we will obviously get ##\left|\psi\right\rangle##. If we measure ##B## first, ##\left|\psi\right\rangle## gets projected with probability 50 % onto either of ##B##'s eigenvectors. If we now measure ##A##, both of those states get projected with probability 50 % onto either ##\left|0\right\rangle## or ##\left|1\right\rangle##, so the order of the measurements did indeed matter.
     
  7. Feb 21, 2017 #6

    strangerep

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    Science Advisor

    It's wrong to think that an operator product AB means "first measure property B, then measure property A".

    E.g., consider a spin-1/2 particle, i.e., 2D Hilbert space and the spin observable (operators) are the Pauli matrices ##\sigma_x, \sigma_y, \sigma_z.~## Now, ##\sigma_x \sigma_y = i\sigma_z##, so measuring spin along the ##y##-axis, then measuring spin along the ##x##-axis is in no sense related to a single measurement of spin along the ##z##-axis.

    Moreover, the product ##i\sigma_z## is not hermitian, hence doesn't even qualify as an observable in the usual sense. Indeed, for 2 noncommuting hermitian observables ##A,B## we have: $$(AB)^\dagger ~=~ B^\dagger A^\dagger ~=~ BA ~\ne~ AB ~.$$ (Ballentine covers precisely this point somewhere, though I don't recall the exact section number.)
     
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