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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?

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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.

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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.

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