Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Consider the Schrödinger equation for a particle in one dimension, where we have at least one boundary in the system (say the boundary is at $x=0$ and we are solving for $x>0$). Sometimes we want to impose a boundary condition in which the wavefunction vanishes (Dirichlet boundary condition).

We can indirectly impose this boundary condition through the physical assumptions by using an infinite potential outside the relevant region (like in the "particle in a box" model): $$ V(x<0)=\infty ~~~~\Longrightarrow ~~~~\psi(x=0)=0 $$ What if we want to impose a boundary condition in which the derivative of the wavefunction vanishes (Neumann boundary condition)? $$ ? ~~~~\Longrightarrow ~~~~ \left. \frac{\partial \psi}{\partial x} \right|_{x=0} = 0 $$ Is there a way to choose the potential, or maybe change something else in the Hamiltonian, in order to indirectly impose this boundary condition?

P.S. This question is not of great practical importance, it is more of a curiosity.

share|cite|improve this question

2 Answers 2

By mirroring $V(x)$ about $x = 0$, i.e., by setting $V(-x) = V(x)$, the wavefunction can be taken to be even or odd. The even solution satisfies the Neumann boundary condition since the derivative of an even function is odd and thus zero at $x = 0$.

share|cite|improve this answer
It is true that the even solutions satisfy the desired boundary condition, but unfortunately there will always be odd solutions as well, which do not satisfy it. I'm looking for an approach that imposes the boundary condition for all solutions. – Joe Jun 19 '12 at 17:48
@Joe that is impossible, such material would behave as a perfect reflector with zero London depth. Not even superconductors have zero London depth – lurscher Jun 19 '12 at 17:59
@lurscher - That's an interesting point. The infinite potential that is used for a particle in a box is also impossible. In reality there will always be some penetration depth, and the wavefunction will completely vanish only when we take the (unphysical) limit of an infinite potential. In an analogous manner, there's no reason there can't be some parameter that when a certain (unphysical) limit is taken the material turns into a perfect reflector, even though in reality there will always be some finite London length. – Joe Jun 19 '12 at 19:47

It's not really a physical condition, but when one is doing R-matrix theory for scattering (which is arguably not for the faint of heart) the condition does come up. One resource I saw recently is a lecture by Hugo van der Hart (go to the slide titled Basic Applications).

share|cite|improve this answer
This is an old answer, but it's linked elsewhere so I won't change it. The linked resource is probably not where you want to go learn about R-matrix theory. It's also not clear what the role of the Neumann condition is in this development. – Emilio Pisanty Aug 12 at 20:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.