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It is well known that wave functions that are discontinuous in the space variable cannot be solutions of the Schrödinger equation, because the Schrödinger equation is a second-order differential equation in the spatial variable, and the second-order spatial derivative wouldn't exist if $\psi(x)$ or its first derivative in the space variable $\frac{\delta \psi(x)}{x}$ were discontinuous, except in the case of an infinite potential well.

The question is, can a wave function $\psi(x,t)$ that is continuous in the space variable and discontinuous in the time variable be a solution of the Schrödinger equation, and most importantly, can you show why?

I am interested in keeping everything within the accepted theory i.e. the Schrödinger equation as it is with no adjustments.

It may be a bit concerning that this question is very similar to many previous questions, although it specifically asks for discontinuity in the time variable with continuity in the coordinate variable.

saturn
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In principle, why not? Just make sure your Hamiltonian has a term proportional to $\delta(t)$. As an example think of a particle that receives an instant kick at $t=0$. At $t<0$ it was in the state $|k=0\rangle$; at $t>0$ it is in $|k\neq0\rangle$. The wavefunction $\psi(k) = \langle \psi | k\rangle$ is discontinuous at $t=0$ as you see.

EDIT: Since the original question explicitly mentions continuity is space I am adding these extra few lines. Consider the instant kick at $t=0$ mentioned above, and think that it is such that it adds certain momentum $q$. Then any pure state (in momentum representation) $|k\rangle \rightarrow |k+q\rangle$. If the state $|\psi\rangle$ at $t<0$ was such that $\psi(k) = \langle\psi|k\rangle$ is a $continuous$ function of $k$, then after $t=0$ $|\psi \rangle \rightarrow |\phi \rangle$ where $\phi(k) = \langle\phi|k\rangle = \psi(k-q)$. Hence the total wavefunction is by construction continuous in variable $k$ but discontinuous in time. Here momentum was chosen just for the merit of it providing a more intuitive picture, but can be reproduced for any complete basis, including space representation $\{ |x \rangle\}$.

John
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