I'll illustrate my question with a textbook picture:

enter image description here

Why can't it be discontinuous there? No obvious answer appears to me here, unlike why $\psi$ itself can't be continuous, hopefully? In my view, it's because at the region where there is a jump, it will have two probabilities at once for a single outcome (which is nonsense), but that doesn't illuminate my titular question.

  • $\begingroup$ Momentum is defined and exists, so the derivative exists, hence the function must be continuous. $\endgroup$
    – FGSUZ
    Jun 28, 2018 at 22:25
  • $\begingroup$ The action always has some kinetic terms, and physical field configurations are at the extrema of the action. Configurations in which $\psi'$ varies like crazy are not among the extremal configurations and hence can be disregarded when one solves the Schrödinger equation. This leads to the ad hoc rule you are mentioning. $\endgroup$
    – user178876
    Jun 28, 2018 at 22:30
  • $\begingroup$ One notable exception: the delta-function potential (quantummechanics.ucsd.edu/ph130a/130_notes/node154.html) $\endgroup$ Jun 28, 2018 at 22:35
  • $\begingroup$ We also break the rule for the usual first introductory system: the infinite square well. But both that cases should be thought of as approximations to very high finite potential (which has a well behaved wave-function) that is easier for students to analyze. $\endgroup$ Jun 28, 2018 at 22:48
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/19667/2451 , physics.stackexchange.com/q/149001/2451 , physics.stackexchange.com/q/262671/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 29, 2018 at 3:03

2 Answers 2


If we look at the 1D time-independent Schrödinger equation we would see,$$-{\hbar^2\over 2m}\psi''(x) + [U(x) - E]~\psi(x) = 0.$$This dependence on the second spatial derivative of $\Psi$ means that a discontinuity in $\psi'(x)$ is an infinite $\psi''(x)$ and only happens at places where $U(x)$ goes to infinity, too, so that the two infinities can compete in a way that can balance out to give zero.

One can extend this reasoning somewhat to include more dimensions of space or a dimension of time, of course; one finds for example that a discontinuous $\Psi''(x, t)$ must either coincide with an infinite $U(x)$ or else must cause $i\hbar \dot\Psi$ to go to infinity at that point, so you've got a complex number circling the origin with infinite frequency.


The quantum mechanical momentum operator is: $$p=\frac{\hbar}{i}\frac{\partial}{\partial x}.$$

This means that if you were given a wave function $\psi$ where $\psi'$ were discontinuous, you would have points where this momentum operator is not defined. If you have points where this operator is not defined, you are unable to calculate $$\langle p \rangle=\frac{\hbar}{i}\int\psi^*\frac{\partial}{\partial x}\psi dx$$...Generally we want physical systems to have well defined momenta.


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