In Griffiths, we are told that the expansion coefficients of the stationary states are simply complex numbers:

$$\Psi(x, \ t) \ = \ \displaystyle\sum_{n} c_n e^{-iEt/\hbar} \psi_n(x)$$

How do we know that they don't depend on time? Wouldn't it make sense that they depend on time, as $$\langle E_n | \Psi(t) \rangle \ = \ c_n,$$ and the state vector itself is changing with respect to time?

I guess the root of my question is: what is changing with respect to time, the stationary state wavefunction, or the $c_n$, where we treat the time-dependence as part of the coefficient?


This is already accounted for. In general, we have $$\Psi(x, \ t) \ = \ \displaystyle\sum_{n} b_n(t) \psi_n(x)$$ where the expansion coefficients vary in time. What Griffiths has shown is that if the $\psi_n(x)$ are energy eigenstates, then the time evolution can be written as $$b_n(t) = c_n e^{- i E_n t / \hbar}$$ where $c_n$ is a constant. This gives the equation you wrote.

  • $\begingroup$ Yes, this makes sense, except Griffiths solves for the time-dependence as the time dependent component of the stationary state wavefunctions, and then just says: "we can create linear combinations of these stationary states with complex coefficients". How do we know that there isn't some other time dependence involved when we introduce the $c_n$ coefficients? $\endgroup$ – Jack Ceroni Sep 26 '19 at 0:21
  • $\begingroup$ Oh actually, I think I understand now, $b_n(t)$ is itself the complete time-dependent component of each of the stationary state wavefunctions. $\endgroup$ – Jack Ceroni Sep 26 '19 at 0:27
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    $\begingroup$ Any additional time dependence would violate the time-dependent Schrodinger equation. $\endgroup$ – G. Smith Sep 26 '19 at 0:27

The $e^{-iE_nt/\hbar}$ is what gives the time evolution of this superposition. Note that the energy should have a subscript; in general, it can be different for each stationary state.


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