# Box normalization

Whenever we study free fields, the solutions of these fields (or particles, whatever feels most comfortable) are always given by plane waves. The dispersion-relation $\omega=\omega(k)$ will of course depend on your kind of system (but let's ignore this for the moment).

In fact i'm going to look at the entire thing non-relativistic (not that this is necessary).

If we want to solve our equation (let me denote the equation by an operator $L$):$$L\psi(\vec{r},t)=0,$$where $\psi(\vec{r},t)$ is the quantum-field and $L$ is some kind of wave-equation for the free field. Then we are able to solve the above eqation by filling in the Fourier-expansion of the free field: $$\psi(\vec{r},t)=\sum_\vec{k}\psi(\vec{k})\exp\left(i(\vec{k}\cdot\vec{r}-\omega(\vec{k})t)\right),$$ with $\psi(\vec{k})$ the different Fourier-coëfficiënts.

Since we want our quantum-mechanical wavefunction to be normalized (easier for perturbaton-theory), we impose the normalization as: $$1=\int\text{d}^3\vec{r}\left(\psi^*(\vec{r},t)\psi(\vec{r},t)\right).$$ If we look at free space (so an infinite vacuum), the plane waves are non-normalizable. Which is of course a problem.

In order to be able to normalize the plane wave we confine our system to a box with a finite volume $V$ which we take to be infinity at the end of the calculations (or usually one since $V$ tends to drop out everywhere). This box is chosen so that it's square and has side $L$. Now simply imposing a box won't do, we of course need boundary conditions. In order to preserve momentum we impose periodic boundary conditions on this box $$\psi(x+L,y,z,t)=\psi(x,y,z,t),$$ $$\psi(x,y+L,z,t)=\psi(x,y,z,t),$$ $$\psi(x,y,z+L,t)=\psi(x,y,z,t),$$ which leads to a quantization in the momenta.

With this box-normalization we are able to normalize the wave-function and continue our calculations. Now my question is the following:

Questions:

1. Why do we always assume a square box when imposing box-normalization?
2. Does this yield the same results as a rectangular box where the different sides $L_1$, $L_2$ and $L_3$ are not equal?
3. Or do we impose that all the sides $L_1$, $L_2$ and $L_3$ are equal for the sake of homogeneity and isotropy of the free space ?

Note: Let's stick to Cartesian coordinates for the sake of simplicity. The fact that the box might become a sphere in spherical coordinates and a cylinder in cylindrical coordinates is clear to me.

2. Yes, in the limit of $L_1, L_2, L_3 \to \infty$ this is equivalent to a square box in the limit $L \to \infty$ (we can't measure the difference between infinities). Also, in the limit $L \to \infty$ the quantized momenta will eventually cover all of momentum space, making the distinction unneccesary. (For a case from solid-state physics, where this limit can not be taken, see http://en.wikipedia.org/wiki/Landau_quantization).
3. No, the introduction of a box breaks homogeneity and isotropy anyways. This is not bad though, since we always take the limit $L \to \infty$ which restores these principles. This can also be seen from the fact that the results are only dependent on the volume of the box.
• @Neuneck For 2. (which was actually my biggest concern), we can't measure the difference between infinities. But the quantization in the momenta is also different is $L_1$, $L_2$ and $L_3$ differ. Won't that lead to any problems? And New_new_newbie, héhé been there :p.
• @Nick Of course the intermediate steps will depend on your box. But in the limit $L \to \infty$ the quantized momenta become continuous, eventually covering all of momentum space. At that point previous differences in $L_1, L_2, L_3$ do not matter anymore. May 30 '14 at 9:34