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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 $\frac{d}{dt}\langle\hat{x}\rangle$?

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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 $\frac{d\langle x\rangle}{dt}=0$.

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