I thought wavefunctions always depend on space and time. Even if it is just one energy eigenstate I do not see why there would be no time dependence. That exponent $e^{iEt/\hbar}$ should still be there, I think.

For clarification: what I call energy eigenstate is $\psi(x,t)=u(x)e^{iEt/\hbar}$ where $\hbar=h/2\pi$. My general solution for the wavefunction is $\psi(x,t)=\Sigma_n a_nu_n(x)e^{iEt/\hbar}$.

  • $\begingroup$ The eigenstates actually do have the exponent factor with time dependence. However we sometimes do not write it because we are solving time independent Schrodinger equation instead of time dependent one. $\endgroup$
    – RedGiant
    Jan 4, 2021 at 15:01
  • 4
    $\begingroup$ Hint: how do you get the probabilities from the wave function? $\endgroup$
    – kaylimekay
    Jan 4, 2021 at 15:07

1 Answer 1


You need to distinguish between the wave function and the probability density. They are not the same.

The wave function $\psi(x,t)$ is the solution of Schrödinger's equation. For an energy eigenstate you correctly wrote $$\psi(x,t)=u(x)e^{iEt/\hbar}$$

Generally the probability density is $$p(x,t)=|\psi(x,t)|^2=\psi^*(x,t)\psi(x,t)$$ For the energy eigenstate from above you get $$\begin{align} p(x,t)&=|\psi(x,t)|^2\\ &=\left| u(x)e^{iEt/\hbar} \right|^2 \\ &=|u(x)|^2 \left| e^{iEt/\hbar} \right|^2 \\ &=|u(x)|^2 \end{align}$$ You see, the time-dependency dropped out because the absolute square of the exponential is $1$.

  • $\begingroup$ Oh, now I see what I was doing wrong. Many thanks! $\endgroup$
    – Agnese
    Jan 4, 2021 at 16:03

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