Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a quantum particle in an one-dimensional infinite well of width $L$, the potential has the formal expression: $$ V(x) = \begin{cases} \infty, & x < 0 \\ 0, & 0 \le x \le L \\ \infty, & x > L \end{cases} $$

, and the "hard wall" boundary condition is imposed: $\psi(0) = \psi(L) = 0$.

However, I don't get, where does this boundary condition come from. It is explained in books like "the wavefunction has to be continuous". However, the domain of this problem is $[0, L]$, and there is plenty of continuous (in the domain $[0, L]$) solutions for the Schrodinger's equation which are not zero at the endpoints of the domain.

As I see it, probably, a better explanation would be: consider an infinite sequence of potentials: $$ V_n(x) = \begin{cases} n, & x < 0 \\ 0, & 0 \le x \le L \\ n, & x > L \end{cases} $$

Then, by looking at the solutions $\psi_m$ of the Schrodinger equation (now we have the domain $\mathbb{R}$), we will see that for any fixed energy $E$, the solutions with total energy less than $E$, tend to zero at the well boundaries: $\lim\limits_{n \to \infty} \psi_n(0) = 0$, $\lim\limits_{n \to \infty} \psi_n(L) = 0$.

So, how should I actually interpret this boundary condition?

share|cite|improve this question
If the potential energy is infinite then the probability of finding the particle there is zero. That means the modulus squared of wavefunction must be zero for $x \lt 0$ and $x \gt L$. If the wavefunction was non-zero at $x = 0$ or $x = L$ there would be a discontinuity in the wavefunction. – John Rennie May 16 '14 at 15:38
up vote 4 down vote accepted

The domain of the problem is the entire real line, not $[0,L]$. Otherwise the potential would not be specified for $x>L, x<0$. Thus, calculate the total energy of any wavefunction that does not vanish at $x = 0,L$, you will find it to be infinite. Therefore for all eigenstates that do not have infinite energy, i.e. the entire spectrum, the wavefunction vanishes at the boundaries.

share|cite|improve this answer

Consider the situation where $\psi$ is non-zero at the boundary, this would imply that the modulus square of the wave function is non-zero. Thus, there is a chance of observing the particle at the boundary. If this was true, then by the relation that $$F = \nabla V$$ this would imply infinite force, and thus infinite acceleration. Such situations are nonsense in physics, thus we must impart that $\psi$ is zero at the boundary.

share|cite|improve this answer

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.