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For the Normalization of the Schrödinger wave equation we need the following to be true $$\begin{equation} \int_{-\infty}^{+\infty}|\psi(x,t)|^2\,\mathrm{d}x=1 \end{equation}$$ Now, if I write in variable separable form the wave function $|\psi(x,t)|=|f(x)||g(t)|$, then we have $$\int_{-\infty}^{+\infty}|f(x)|^2|g(t)|^2\,\mathrm{d}x=|g(t)|^2\int_{-\infty}^{+\infty}|f(x)|^2\,\mathrm{d}x=1$$, since for any square integratable function $\int_{-\infty}^{+\infty}|f(x)|^2\,\mathrm{d}x$ will be a finite constant, hence $|g(t)|$ will also be a constant.

Now, let us look at the probability current of the particle in a localized neighbourhood $(a<x<b)$. $$\frac{\partial}{\partial t}P(a<x<b)=\left(\int_{a}^{b}|f(x)|^2\,\mathrm{d}x\right)\frac{\partial}{\partial t}|g(t)|^2$$

Now, since I've established above, that $|g(t)|$ is a constant, hence the probability current through a predetermined neighbourhood should always be zero. Where am I going wrong, kindly help, I'm not a physicist but I'm having to do some material science right now.

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  • $\begingroup$ Looks correct, apart from the last integration boundaries. What you showed is that separable states are 'stationary'. General states are superpositions if separable states. $\endgroup$ – lalala Apr 22 '17 at 16:10
  • $\begingroup$ But, if all separable states are stationary, the general state would be a sum of the stationary states which would logically itself be another stationary state. Then Probability currents lose their significance in that all probability currents of quantum states would be zero. $\endgroup$ – ubuntu_noob Apr 22 '17 at 16:14
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    $\begingroup$ No. Add two like f1g1 and f2g2 then absolutsquare them and you see that the aforementioned reasoning does not apply anymore. $\endgroup$ – lalala Apr 22 '17 at 16:43
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There is no probability current for steady states: their probability density is time-independent. However for linear combinations of time-dependent solutions (which are not separable in $t$ and $x$) the probability current can be $\ne 0$.

Take $\Psi(x,t)= a \psi_0(x)e^{-i\omega t/2} + b\psi_1(x)e^{-i3\omega t/2} $ as linear combinations of harmonic oscillator states with any $a,b$ such that $aa^*+bb^*=1$ as an example...

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