# Normalizing the free particle wave function

One way to normalize the free particle wave function

"is to replace the the boundary condition $\psi(\pm{\frac{a}{2}}) = 0$ [for the infinite well] by periodic boundary conditions expressed in the form $\psi(x)=\psi(x+a)$" -- Quantum Physics, S. Gasiorowicz

How does this work? What does this mean physically? Or more precisely, why does this approximation suffice?

I understand that this makes the wavefunction square-integrable (when integrated from $x=0$ to $x=a$) hence normalizable.

Thanks.

## 2 Answers

The physical idea is that you'll let $a$ go to infinity for a truly free particle, and if you take this limit, then the specific details of the boundary conditions should be irrelevant, because the boundaries are so far away anyway.

Therefore, you are welcome to choose convenient boundary conditions, and the periodic ones are convenient, because then you have just plain waves $e^{ikx}$, with the admitted $k$-values determined by $e^{ika} = 1$, so $ka = 2\pi n$, and $n \in \mathbb{Z}$.

• The crucial convenience here seems to be an easy way to get propagating (non-zero linear momentum) states in one shot. Otherwise, if I stick to Dirichlet boundary conditions, I need to construct superpositons of eigenstates -- wave packets -- to get the particle moving. – Slaviks Aug 24 '11 at 5:54

You want boundary conditions that conserve the total probability within your finite box. The probability current is proportional to, roughly, $J \sim \psi \partial \psi$. Setting $\psi=0$ at the boundary forces $J=0$ at the boundary (particles get reflected) so probability is conserved (once can see that Dirichlet boundary condition $\partial\psi = 0$ does the same). Setting $\psi(x+a)=\psi(x)$ forces $J(x+a) = J(x)$ so the current going out of your box on the left is equal to that coming in on the right (i.e. your box is really a torus), and probability is again conserved.