# When is a quantum state stationary?

If a quantum state is an eigenstate of the Hamiltonian, then it is stationary. But can a state be stationary if it is not an eigenstate of the Hamiltonian? If yes, how can one prove whether a state is stationary?

• Stationary states are defined to be eigenstates of the Hamiltonian. So your question is saying "If A, then A. But can we have A, if not A?" Jul 19, 2016 at 23:38
• @knzhou I thought so, but in another question I asked, someone told me "So the state is stationary if it is an eigenstate. But this is not the only possibility for stationarity." so I'm in doubt now. Jul 19, 2016 at 23:40
• That answer is wrong, in several places. Jul 19, 2016 at 23:42

A state $\Psi(x,t)$ is stationary if

$$\mid\Psi(x,t)\mid^2=\mid\Psi(x,0)\mid^2$$

This means that the time dependence of $\Psi(x,t)$ must be in the form

$$\Psi(x, t) = \Psi(x, 0) \ e^{i \phi t}$$

but you also have (Schrödinger's equation)

$$i \hbar \partial_t \ \Psi(x,t) =\hat H \ \Psi(x,t)$$

so that we obtain

$$(i \hbar)(i \phi) \Psi(x,t) =\hat H \ \Psi(x,t) \\ \to-\hbar \phi \Psi(x,t) =\hat H \ \Psi(x,t)$$

i.e. $\Psi(x,t)$ is an eigenstate of $\hat H$ with eigenvalue $-\hbar \phi$. This means, incidentally, that $\phi$ must have the dimensions of energy/action, that is to say

$$\phi=\frac E \hbar$$

So if a state is stationary, it is an eigenstate of $\hat H$. Since we know that the reverse is true, we can conclude that a state is stationary if and only if it is an eigenstate of $\hat H$.

• You're answer is good but you might want to add why $\Psi(x, t) = \Psi(x, 0) \ e^{i \phi t}$ is true. Don't assume people know that.
– Gert
Jul 20, 2016 at 2:17
• Yeah, you pretty much used circular reasoning. From $|\Psi(x,t)|^2=|\Psi(x,0)|^2$ immediately comes only that $\Psi(x,t)=e^{i\alpha(x,t)}\Psi(x,0)$. Why it should be just a $\phi t$ demands proof... I'm going to present short proof I've made up now in my answer though I'm sure there may be much more elegant way.
– OON
Jul 20, 2016 at 10:33
• Arr, no, I've stupid mistake, so I'll hide that for now
– OON
Jul 20, 2016 at 10:53