I know that the solutions to the time-dependent Schrodinger equation are always linear combinations of the form $$ \Psi(x,t)=\sum_n c_n e^{-iE_nt/\hbar} \psi_n(x) $$

If $ \Psi(x,0) $ is PURELY imaginary, $ \Psi(x,t) $ will be PURELY imaginary $ \forall t \geq 0$?

My answer should be yes because the solution can be written has a superposition of purely imaginary solutions. Is it correct?

  • 1
    $\begingroup$ Hint: $e^{iat} = \cos at + i\sin at$ $\endgroup$ Jan 22, 2019 at 15:38
  • $\begingroup$ ... hence even if $\Psi(x,0)$ is purely imaginary $\Psi(x,t)$ will not be so. $\endgroup$ Jan 22, 2019 at 15:41

1 Answer 1


Nope. The "energetic" exponents will make the solution complex at $t>0$.

  • $\begingroup$ Thanks everyone for your help. Can I also say because of the exp factor that if $ \Psi(x,0) $ is even / odd the $ \Psi(x,t) $ will NOT be even / odd or it depends from the parity of the potential? Many thanks. $\endgroup$ Jan 23, 2019 at 8:13
  • $\begingroup$ If the initial state is even/odd, it is exclusively due to $\psi_n(x)$ involved (all even or all odd, question of choice). When the time runs, this initial property will be conserved. $\endgroup$ Jan 23, 2019 at 14:48
  • $\begingroup$ Thanks for your answer but I don't understand sorry. $ \Psi(x,0) = \sum_n c_n \psi_n(x)$, right? if $ \Psi(x,0)$ is, let's say, even, this means that $ \Psi(-x,0)= \Psi(x,0) $ so of course $ \psi_n(x)$ are all even. But $\Psi(-x,-t)=\sum_n c_n e^{iE_nt/\hbar} \psi_n(-x) = \sum_n c_n e^{iE_nt/\hbar} \psi_n(x)\neq \Psi(x,t)$ no? Where is my error? Really thanks again. $\endgroup$ Jan 23, 2019 at 15:49
  • $\begingroup$ I do not understant and I did not know that you were going to replace $t$ too. I meant the usual time evolution. But even "evolving backwads" does not change the initial wave function property. There is nothing special in $\Psi(t)\ne\Psi(-t)$. It is like $\Psi(t_1)\ne\Psi(t_2)$. $\endgroup$ Jan 23, 2019 at 16:27
  • $\begingroup$ Well my full problem is the follow: consider the time-dependent Schroedinger eq. for a one-dimensional problem with Hamiltonian $H= p^2/2m + V(x)$, with $V(x)$ real and lower bounded (I mean $\min V(x) \neq \infty$). Let $\psi(x,t)$ be a solution of the Schroedinger eq.. The question is "if $V(x)$ is even / odd and $\psi(x,0)$ is even / odd, can we say that $\psi(x,t)$ will be even / odd $\forall t \geq 0$? How can the solution still be even / odd if we have an $\exp$ term? Thanks again. $\endgroup$ Jan 23, 2019 at 16:46

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