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## Justification for the matching condition for the wave function at potential jumps. Why is it both restrictive enough and sufficiently general?

Consider Schrödinger's time-independent equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi.$$ In typical examples, the potential $V(x)$ has discontinuities, called potential jumps.

Outside these discontinuities of the potential, the wave function is required to be twice differentiable in order to solve Schrödinger's equation.

In order to control what happens at the discontinuities of $V$ the following assumption seems to be standard (see, for instance, Keith Hannabus' An Introduction to Quantum Theory):

Assumption: The wave function and its derivative are continuous at a potential jump.

Questions:

1) Why is it necessary for a (physically meaningful) solution to fulfill this condition?

2) Why is it, on the other hand, okay to abandon twofold differentiability?

Edit: One thing that just became clear to me is that the above assumption garanties for a well-defined probability/particle current.

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 You may want to look at en.wikipedia.org/wiki/Weak_solution – Willie Wong Jul 26 2010 at 13:11 Thanks Willie. This provides some explanation concerning my second question. – Rasmus Jul 26 2010 at 13:40 @Downvoter: Why did you downvote? – Rasmus Aug 20 2010 at 12:37

## 3 Answers

To answer your first question:

Actually the assumption is not that the wave function and its derivative are continuous. That follows from the Schrödinger equation once you make the assumption that the probability amplitude $\langle \psi|\psi\rangle$ remains finite. That is the physical assumption. This is discussed in Chapter 1 of the first volume of Quantum mechanics by Cohen-Tannoudji, Diu and Laloe, for example. (Google books only has the second volume in English, it seems.)

More generally, you may have potentials which are distributional, in which case the wave function may still be continuous, but not even once-differentiable.

To answer your second question:

Once you deduce that the wave function is continuous, the equation itself tells you that the wave function cannot be twice differentiable, since the second derivative is given in terms of the potential, and this is not continuous.

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Your first argument is not clear to me - I'll take a look at Cohen-Tannoudji. – Rasmus Jul 26 2010 at 15:23
The idea is the following: suppose that $V$ has isolated discontinuities and let $x_0$ be the location of one such discontinuity. Replace $V$ on $[x_0-\epsilon, x_0+\epsilon]$ with another potential which is continuous and which tends to $V$ as $\epsilon\to 0$. Then you show that the wave-function which solves the Schrödinger equation for this new potential tends in the limit as $\epsilon\to0$ to the wave-function you want and that in this limit the first derivative remains continuous. This is not really proven in Cohen-Tannoudji et al. but only sketched. The details are not hard, though. – José Figueroa-O'Farrill Jul 26 2010 at 16:56
There is a very clear physical reason why the wavefunction should be continuous: it's derivative is proportional to the momentum of the particle, so discontinuities imply that the state has an infinite-momentum component. – Jess Riedel Apr 27 2011 at 3:47

Since you talk about 'jump' discontinuities, I guess you are interested in a one dimensional Schroedinger equation, i.e., $x\in\mathbb{R}$. In this situation a nice theory can be developed under the sole assumption that $V\in L^1(\mathbb{R})$ (and real valued of course). By a nice theory I mean that the operator $-d^2/dx^2+V(x)$ is selfadjoint, with continuous spectrum the positive real axis, and (possibly) a sequence of negative eigenvalues accumulating at 0. Better behaviour can be produced by requiring that $(1+|x|)^a V(x)$ be integrable (e.g. for $a=1$ the negative eigenvalues are at most finite in number). If you are interested in this point of view, a nice starting point might be the classical paper by Deift and Trubowitz on Communications Pure Appl. Math. 1979. Notice that the solutions are at least $H^1_{loc}$ (hence continuous) and even something more.

A theory for the case $V$ = Dirac delta (or combination of a finite number of deltas) was developed by Albeverio et al.; the definition of the Schroedinger operator must be tweaked a little to make sense of it. This is probably beyond your interests.

Summing up, no differentiability at all is required on the potential to solve the equation in a meaningful way. However, I suspect that this point of view is too mathematical and you are actually more interested in the physical relevance of the assumptions.

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Here is a tangential response to your first question: sometimes these discontinuities do have physical significance and are not just issues of mathematical trickery surrounding pathological cases. Wavefunctions for molecular Hamiltonians become pointy where the atomic nuclei lie, which indicate places where the 1/r Coulomb operator becomes singular. There are equations like the Kato cusp conditions (T. Kato, Comm. Pure Appl. Math. 10, 151 (1957)) that relate the magnitude of the discontinuity at the nucleus to the size of the nuclear charge. I have heard this explained as a result of requiring the energy (which is the Hamiltonian's eigenvalue) to remain finite everywhere, thus at places where the potential is singular, the kinetic energy operator must also become singular at those places. Since the kinetic energy operator also controls the curvature of the wavefunction, the wavefunction at points of discontinuity must change in a nonsmooth way.

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