How is the symmetry properties of the wave function related to the time derivative of the expectation of some observable? For example, if I have an infinite square well centered at x=0, and the particular wave function is symmetric about 0, (or antisymmetric, for that matter), what would that tell me about $\frac{d}{dt}\langle\hat{x}\rangle$?


The symmetry property of the dynamics determines something like $\frac{d\langle x\rangle}{dt}$, which is in turn determined by both the Hamiltonian and the initial state.

Generically, if the Hamiltonian $H$ commutes with an operation $U$, and the initial state is an eigenvector of $U$, then the state after evolution is still an eigenstate of $U$ with the same eigenvalue.

To see it, suppose the initial state is $|\psi\rangle$, and its eigenvalue under $U$ is $u$. That is, $U|\psi\rangle=u|\psi\rangle$. After time $t$, the state becomes $e^{-i\frac{H}{\hbar}t}|\psi\rangle$, and it is still an eigenvector of $U$ with eigenvalue $u$, because $$ Ue^{-i\frac{H}{\hbar}t}|\psi\rangle=e^{-i\frac{H}{\hbar}t}U|\psi\rangle=ue^{-i\frac{H}{\hbar}t}|\psi\rangle $$ In the first equation above, we used that $H$ and $U$ commute.

In the particular setting you give, take $U$ to be the operator that flips the coordinate (more explicitly, $U=\int dx|x\rangle\langle -x|$). The initial state is an eigenstate of $U$, and the Hamiltonian commutes with $U$, so at any time the state is an eigenstate of $U$ with the same eigenvalue. Because eigenstates of $U$ has $\langle x\rangle=\langle \psi|x|\psi\rangle=\langle\psi|U^\dagger xU|\psi\rangle=-\langle\psi|x|\psi\rangle=0$ (we used that $U|\psi\rangle=\pm|\psi\rangle$ and $U^\dagger xU=-x$), in this case $\frac{d\langle x\rangle}{dt}=0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.