# 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.)