# Pure imaginary Schroedinger wave function

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?

• Hint: $e^{iat} = \cos at + i\sin at$ – GodotMisogi Jan 22 '19 at 15:38
• ... hence even if $\Psi(x,0)$ is purely imaginary $\Psi(x,t)$ will not be so. – ZeroTheHero Jan 22 '19 at 15:41

Nope. The "energetic" exponents will make the solution complex at $$t>0$$.
• 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. – user3796805 Jan 23 '19 at 8:13
• 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. – Vladimir Kalitvianski Jan 23 '19 at 14:48
• 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. – user3796805 Jan 23 '19 at 15:49
• 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 $\Psi(t_1)\ne\Psi(t_2)$. – Vladimir Kalitvianski Jan 23 '19 at 16:27
• 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. – user3796805 Jan 23 '19 at 16:46