The symmetries of a Hamiltonian $\hat{\rm H}$ is generally reflected on the energy eigenstates of the Hamiltonian or stationary states. (For example, if the Hamiltonian is invariant under parity, each energy eigenfunction $\phi_n(\textbf{r})e^{-iE_nt}$ have definite parity in absence of degeneracy.)

What can be said about the consequence of the symmetry of the Hamiltonian for an arbitrary solution $\psi(\textbf{r},t)$ of the time-dependent Schrodinger equation $\hat{\rm H}\psi(\textbf{r},t)=i\hbar\frac{\partial}{\partial t}\psi(\textbf{r},t)$? The consequence of time-reversal invariance of the Hamiltonian on a general solution is pretty trivial: if $\psi(\textbf{r},t)$ is a solution so does $\psi^*(\textbf{r},-t)$. But what about the other invariances of the Hamiltonian such as translational or rotational symmetries? What do they tell us about the general solution $\psi(\textbf{r},t)$?

I would guess that the general solutions must also bear some reflection of the symmetry of the Hamiltonian because the Hamiltonian is after all responsible for time-development of any state. But I'm being able to justify it mathematically.

  • $\begingroup$ ? So, take a mixed-parity configuration for an oscillator and evolve it in time, as a superposition of even and odd parity eigenstates.... $\endgroup$ Commented Mar 30, 2018 at 14:48
  • $\begingroup$ Sorry, I don't understand the point. A superposition of two oscillator eigenstates, one with even and the other with odd parity, when evolved in time will, in general, do not have definite parity. Right? @CosmasZachos $\endgroup$
    – SRS
    Commented Mar 30, 2018 at 14:50
  • $\begingroup$ I'm asking something completely different. Where did I say that I expect something different? @CosmasZachos $\endgroup$
    – SRS
    Commented Mar 30, 2018 at 14:55
  • $\begingroup$ That's my very point. How do you expect configurations to respect rotational symmetry if they do not start that way? The hamiltonian will magically take them there? $\endgroup$ Commented Mar 30, 2018 at 14:58
  • $\begingroup$ I'm not saying that any state must have the same symmetry property as that of the energy eigenstates. Clearly, it's not. But there can be some consequence of the symmetry in a general state. I've edited the question a bit. Does it make sense now? And I used parity just as an example. @CosmasZachos $\endgroup$
    – SRS
    Commented Mar 30, 2018 at 15:00

1 Answer 1


A standard equation derived in most (if not all) elementary quantum mechanics textbooks is

$$ \frac{d \langle Q\rangle_t}{dt} = \frac i \hbar \langle [H,Q] \rangle_t + \langle \frac{\partial Q}{\partial t} \rangle_t.$$

Here $\langle O \rangle_t$ is the time-dependent expectation value of the operator $O$ (not necessarily hermitian) defined as $\langle \psi(t) | O | \psi(t) \rangle$, and $Q$ is a particular operator of interest.

In quantum mechanics if you have a symmetry it will manifest as a unitary or antiunitary operator $Q$ by Wigner's theorem. For simplicity I'll consider only the unitary case. In the case of a continuous symmetry we can and often do consider instead the generator of the symmetry which is hermitian but this is not necessary for the argument. If the symmetry is respected by the dynamics (so that $[H,Q] = 0$) and is itself time-independent (so that $\frac {\partial Q}{\partial t} =0$) then the right hand side is $0$, so $\langle Q \rangle$ is a conserved quantity.

This puts restrictions on how the state can evolve, but they are somewhat less practical to use because, once you impose the all the conservation laws, the subset of states that the initial state is allowed to evolve to is not a vector subspace of the Hilbert space; that is, superpositions of allowed final states will generally not be allowed. For example, if you have a spin-$\frac12$ system with $H = \omega S_z$, we have a symmetry, namely time-evolution itself, which tells you that the expected energy is conserved. If you consider the initial state $|+x\rangle$ this tells you that it can not evolve to $|+z\rangle$ since they have different expected energies, but the state can and does evolve to $|-x\rangle$ which together with $|+x\rangle$ makes a basis for the space.


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