# 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.

## 1 Answer

1. We assume a square box, because it simplifies the argument.
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 May 30 '14 at 9:32
• @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. – Neuneck May 30 '14 at 9:34
• Oh again! I was going to write, it won't matter when you take the limit. :( – 299792458 May 30 '14 at 9:35
• I still have a question, do the results at large scale limit invariant with respect to the shapes of boundaries? Square vs rectangular seem to be obviously fine. However, The system can be sphere, cylinder, etc, as op already noticed... – user26143 May 30 '14 at 9:45
• @user26143 Yes, as long as you take the limit in such a way that your "box" covers the whole space in the end. Usually Cartesian coordinates are most convenient, though, as other geometries will lead to nasty classes of functions such as spherical Bessel functions or the like. – Neuneck May 30 '14 at 9:50