- #1

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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

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?

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?

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