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If a potential $V(x,t)$ exhibits a finite discontinuity in space, the wavefunction $\phi(x,t)$ and its spatial derivative will be continuous.

If a potential exhibits a finite discontinuity in time, the wavefunction $\phi(x,t)$ is continuous but the time derivative will be discontinuous.

Can anyone explain why this is so?

Qmechanic
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user1887919
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  • Comment to the question (v1): Temporal continuity of the wave-function follows from a bootstrap argument applied to the integral TDSE: $\psi(x,t_f)-\psi(x,t_i)=\frac{i}{\hbar}\int_{t_i}^{t_f} !dt~\hat{H}(t)\psi(x,t) $. The analogous question for the TISE is e.g. discussed here and links therein. – Qmechanic Apr 08 '15 at 18:50

1 Answers1

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This is because Schrödinger equation is second order in space and first order in time, $$ i \hbar \partial_t \psi = \frac{1}{2m} \nabla^2 \psi + V \psi \, .$$ Imagine a wave function that is constant in time. Then, its second derivative is proportional to the discontinuous potential. The zero and first derivatives can still be continuous. On the other hand, if the wave function is constant in space, it is the first time derivative that is proportional to the discontinuous potential.

Steven Mathey
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