# Significance when expectation of a commutator is zero

It is clear to me what it means when the commutator of two operator $$[A, B]$$ is zero and what it implies. However, is there any significance when the expectation of the commutator $$\langle[A, B]\rangle$$ is zero?

I know $$\langle[H, O]\rangle = 0$$, where $$H$$ is the hamiltonian and $$O$$ the observable implies $$\langle O \rangle$$ is stationary but not sure about the general case for two operators.

I don't think we can conclude anything if $$\langle[A,B]\rangle=0$$, if not the condition on the uncertainties of previous answer: but even for that, the fact that on a particular state the product of the uncertainties has to be greater or equal to zero, doesn't imply that this particular state will realize the condition with the equal sign.

About the operators or the state, this condition can come from very etherogeneous situations, so unless further details are provided we can't draw many conclusions:

1) $$[A,B]=0$$: we know what it means, $$A$$ and $$B$$ are compatible observables (I assumed they're Hermitian, am I wrong?) and a common eigenbasis can be found; on the other hand, $$|\psi\rangle$$ can be whatever;

2) $$[A,B]\neq0$$, $$[A,B]|\psi\rangle=0$$: this, for example, might happen in those peculiar cases where $$A$$ and $$B$$ don't commute but $$|\psi\rangle$$ is still a simultaneous eigenstate of both of them;

(example: the eigenstate of $$\vec L^2$$, $$L_z$$ for $$l=0$$, $$|l,l_z\rangle=|0,0\rangle$$, by the rotational symmetry of the situation, is a simultaneous eigenstate also of $$L_x$$ and $$L_y$$ with eigenvalue 0, although obviously $$[L_z,L_{x/y}]\neq0$$)

3) $$[A,B]\neq0$$, $$[A,B]|\psi\rangle\neq0$$: in this case the only possibility to have $$\langle[A,B]\rangle=0$$ is that the vector $$[A,B]|\psi\rangle$$ is orthogonal to the vector $$|\psi\rangle$$.

This in turn means that $$|\psi\rangle$$ is not eigenstate of $$(AB-BA)$$, so is not simultaneous eigenstate of $$A$$ and $$B$$.

(1) The Heisenberg equation of motion gives tells us that the time evolution of the expectation value of an operator $$A$$ is given by

$${\frac {d}{dt}} \langle A(t) \rangle={\frac {i}{\hbar }} \langle [H,A(t)] \rangle.$$

where I have assumed that there is no explicit time dependence on the operator. As you can see, if $$A$$ commutes with the Hamiltonian, this tells us that

$$\frac {d}{dt} \langle A(t) \rangle = 0$$

This means that our observable $$A$$ corresponds to a conserved quantity.

(2) The generalized Heisenberg uncertainty relation states that

$${\displaystyle \sigma _{A}\sigma _{B}\geq \left|{\frac {1}{2i}}\langle [{\hat {A}},{\hat {B}}]\rangle \right|={\frac {1}{2}}\left|\langle [{\hat {A}},{\hat {B}}]\rangle \right|}$$

So if $$B$$ is, say$the hamiltonian we have that $${\displaystyle \sigma _{O}\sigma _{H}\geq \left|{\frac {1}{2i}}\langle [{\hat {O}},{\hat {H}}]\rangle \right| }$$ Now, if this is zero this means that there exists a common eigenbasis in which both $$\hat{O}$$ and $$\hat{H}$$ are diagonal. That is, they are commuting observables and can both simultaneously be measured to arbitrary precision. So I guess the moral of the story is that Heisenberg really knew what he was doing with expectation values of commutators. • actually this isn’t quite right in the sense that you stictly need$[A,B]=0$(not just “on average”) for a common basis of eigenvectors. – ZeroTheHero Jan 8 at 0:07 • lol I published a similar answer, but then I deleted it as I'm afraid it's wrong: a state$|\psi\rangle$could be simultaneous eigenstate of noncommuting$A,B$but we should have at least$[A,B]|\psi\rangle$. The condition$\langle\psi|[A,B]|\psi\rangle=0$is less strict so it's not enough by itself to conclude that$|\psi\rangle$is eigenstate of$A,B$, because it could be simply that$|\phi\rangle \equiv [A,B]|\psi\rangle\neq 0$is orthogonal to$|\psi\rangle\$. I'm writing an answer on this so you're welcome to bring your comments – Francesco Bernardini Jan 8 at 1:05