# Electric Potential Normalization

1. Sep 14, 2009

### makhoma

I was in my Electrodynamics lecture last week, still working the Laplacian and Poisson equations, when we discussed an infinite parallelpipid (infinite in the $$x$$ direction, length $$a$$ and $$b$$ in the $$y$$ and $$z$$ direction respectively) with a potential of $$\Phi=\Phi_0$$ at $$x=0$$ plane and every other face having $$\Phi=0$$. Here he said that, due to boundary conditions we should expect the potential to have the form

$$\Phi(x,y,z)=\sum_{n_{y},n_{z}}A_{n_{y}n_{z}}\eta_{n_{y}}\sin\left[\frac{n_{y}\pi y}{a}\right]\eta_{n_{z}}\sin\left[\frac{n_{z}\pi z}{b}\right]\exp\left[-\pi\sqrt{\frac{n_{y}^{2}}{a^{2}}+\frac{n_{z}^{2}}{b^{2}}}x\right]$$

where $$\eta$$ and $$A$$ are both normalization constants. I agree with this form (not sure about the normalization constants, but that's the question to come). We let the eta's equal $$\sqrt{2/a}$$ and $$\sqrt{2/b}$$ in order to coincide with QM. In looking at the case $$x=0$$, we find

$$\Phi_{0}=\sum_{n_{y}n_{z}}A_{n_{y}n_{z}}\sqrt{\frac{2}{a}}\sin\left[\frac{n_{y}\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_{z}\pi z}{b}\right]$$

Then multiply both sides by

$$\sum_{n_{y}'n_{z}'}A_{n_{y}'n_{z}'}\sqrt{\frac{2}{a}}\sin\left[\frac{n_{y}'\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_{z}^{'}\pi z}{b}\right]$$

Which after integrating over $$dy'$$ and $$dz'$$ from $$0$$ to $$a$$ and $$0$$ to $$b$$, respectively, we find that $$n_y,n_z$$ must be odd integers. So then the normalization constant is found to be

$$A_{n_{y}'n_{z}'}=\frac{8\Phi_{0}\sqrt{ab}}{n_{y}'n_{z}'\pi^{2}}$$

All of this makes sense and does work out mathematically. The trouble I find is that my professor said we could do this without having the $$\eta$$ terms in there, that $$A$$ would absorb it and we should still come out with the same answer. But following the same math as above, and removing the $$\eta$$ normalizations, I find the normalization constant to be

$$A_{n_{y}',n_{z}'}=\frac{4\Phi_{0}ab}{n_{y}'n_{z}'\pi^{2}}$$

My question is two-fold:
(1) Is my professor wrong and we should not expect the same normalization constant between the two (my answer is off by the factor $$2/\sqrt{ab}$$?
(2) Is my professor right, and I screwed up somewhere?

2. Sep 14, 2009

### makhoma

I think I found an error in what my professor did in class. We have a line that reads

$$\int_0^ady\int_0^bdz\sqrt{\frac{2}{a}}\sin\left[\frac{n_y'\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_z'\pi z}{b}\right]\Phi_0=A_{n_y'n_z'}$$

$$\int_0^ady\int_0^bdz\sqrt{\frac{2}{a}}\sin\left[\frac{n_y'\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_z'\pi z}{b}\right]\Phi_0=\frac{2}{\sqrt{ab}}A_{n_y'n_z'}$$
This makes the normalization constant $$A_{n_y'n_z'}$$ equal to what I get:
$$A_{n_y'n_z'}=\frac{4\Phi_0ab}{n_y'n_z'\pi^2}$$