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

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    $\begingroup$ $\psi(x,t)$ 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

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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 $t$.

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$\Psi(x,y,z)$ 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.

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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.

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    $\begingroup$ 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
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    $\begingroup$ My second point still OK. that's a main point. $\endgroup$ Commented Feb 25, 2017 at 15:52

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