# Infinite square well bound states

For an infinite square well, beyond the walls ( including them ) are infinite potentials and the wavefunctions have to be zero when they hit the walls because of these boundary conditions. For such a case, how do you say whether the stationary states are bounded or not? In fact my confusion springs from the fact that when we solve the Schrodinger's equation we take

$$k = \sqrt{\frac{2mE}{\hbar^2}},$$

which automatically assumes $E \geq 0$ which I thought was the condition for scattering states. However it doesn't make sense to have scattering states for this case physically. So what is going on?

## 1 Answer

For the infinite square well all stationary states are bound states. Keep in mind that all wave functions $\psi(x)$ in position space are limited to the invervall $x \in (-L/2, L/2)$. They are bound to this region of space. Your condition for scattering states is only exact for potentials that that go to $0$ (or some $c \in \mathbb{R}$). for $x \rightarrow \infty$. The more general condition is therefore \begin{align} \text{Bound state: } E &< min \left(\lim_{x \to \, -\infty} V(x),\lim_{x \to \infty} V(x) \right)\\ \end{align}

The 1d harmonic oszillator with energies $E_n = \hbar \omega ( n + 1/2)$ fullfils this condition as well, as all states are bound states. This is further discussed here.

• Ohh thanks. Just wanted to verify the definition of scattering states. Makes a lot more sense. – Astronomer May 30 '18 at 14:43