I Commuting observables vs. exchanging measurements

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1. Feb 21, 2017

greypilgrim

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. Feb 21, 2017

thephystudent

3. Feb 21, 2017

greypilgrim

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.

4. Feb 21, 2017

thephystudent

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.

5. Feb 21, 2017

greypilgrim

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.

6. Feb 21, 2017

strangerep

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.)