Commuting observables vs. exchanging measurements

In summary, the expectation value of the commutator of two observables does not always equal zero. If two observables have the same expectation value for a given state, they share an eigenbasis. However, the order of the measurements can affect the measurement statistics.
  • #1
greypilgrim
506
36
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?
 
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  • #3
thephystudent said:
Lol, if two operators commute, they share an eigenbasis http://math.stackexchange.com/quest...-if-they-share-a-common-basis-of-eigenvectors, so your counterexample is not applicable :)
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
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
thephystudent said:
You can, if necessary, restrict the domain of A and B to this state Ψ only, constructing new operators A',B'.
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
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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Likes vanhees71 and Nugatory

1. What are commuting observables?

Commuting observables are physical properties or quantities that can be measured simultaneously and whose order of measurement does not affect the result. In other words, their measurements can be exchanged without altering the outcome.

2. What is the difference between commuting observables and exchanging measurements?

The main difference is that commuting observables can be measured simultaneously without affecting each other's measurement results. On the other hand, exchanging measurements refers to the act of interchanging the order of measurement for non-commuting observables, which can alter the final measurement outcome.

3. Why is it important to understand commuting observables and exchanging measurements?

Understanding commuting observables and exchanging measurements is crucial in quantum mechanics and other branches of physics as it helps us determine which physical quantities can be measured simultaneously and which cannot. This knowledge is essential in accurately predicting the behavior of quantum systems.

4. Can commuting observables ever become non-commuting?

No, commuting observables will always remain commuting observables. However, it is possible for non-commuting observables to become commuting in certain circumstances, such as when they are measured in a different basis.

5. How do commuting observables and exchanging measurements relate to the uncertainty principle?

The uncertainty principle states that certain pairs of physical properties, such as position and momentum, cannot be measured simultaneously with arbitrary accuracy. This is because these properties do not commute, and exchanging their measurements can alter the results. Commuting observables, on the other hand, do not violate the uncertainty principle as their measurements can be exchanged without affecting the outcome.

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