In the time-dependent Schrödinger equation we have $\Psi(x,y,z,t)=\Psi(\boldsymbol{r},t)$, is it actually an abbreviation for $\Psi(x(t),y(t),z(t),t)=\Psi(\boldsymbol{r}(t),t)$?

  • 1
    $\begingroup$ $\psi(x,t)$ represents the wavefuntion at position $x$ and time $t$ ,what will $\psi(x(t),t)$ represent? $\endgroup$
    – Paul
    Commented Feb 25, 2017 at 15:16

3 Answers 3


No, wave function is not a function of just time — it's a function of spacetime, just like displacement of a vibrating membrane, for example, is a function of position $(x,y)$ on the membrane and time $t$.


$\Psi(x,y,z)$ obeys the time-independent Schrodinger equation,$$\hat H\Psi_n =E_n \Psi_n.$$ We call it the stationary state.

$\Psi(x,y,z,t)$ obeys the time-dependent Schrodinger equation, $$\hat H\Psi =-i \hbar \frac{\partial}{\partial t} \Psi.$$

Both of them can be related by $$\Psi(x,y,z,t)=\sum_n\; e^{-i\,E_n t/\hbar}\Psi_n(x,y,z).$$

We do not write $\Psi(x(t),y(t),z(t))$ because our coordinate does not change with time.


I may write it as $\Psi(x,y,z,t)=\Psi_x(x,t)\Psi_y(y,t)\Psi_z(z,t)\;.$ The expression $\Psi(x(t),y(t),z(t),t)=\Psi(\boldsymbol{r}(t),t)$ maybe look likes the wave function of a particle with path or trajectory $\{x(t),y(t),z(t)\}$ which is not allowed in the standard Quantum Mechanics.

  • 5
    $\begingroup$ You cannot in general write it as a product of independent coordinate wavefunctions. In fact, it's more often non-separable than separable. $\endgroup$
    – Ruslan
    Commented Feb 25, 2017 at 15:11
  • 4
    $\begingroup$ My second point still OK. that's a main point. $\endgroup$ Commented Feb 25, 2017 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.