# Beam splitter energy conservation

1. Oct 1, 2011

### McLaren Rulez

Hi,

If we describe a beam splitter as follows:

$$e^{ikx} -> \sqrt{T}e^{ikx} + \sqrt{R}e^{i\theta}e^{iky}$$
$$e^{iky} -> \sqrt{T}e^{iky} + \sqrt{R}e^{i\theta'}e^{ikx}$$

then $\theta+\theta'=\pi$ is a condition to ensure conservation of energy according to my text.

I tried working this out by taking $Ae^{ikx}+Be^{iky}$ incident on a beam splitter. The incident energy is $A^{2}+B^{2}$.

The output is

$A\sqrt{T}e^{ikx} + A\sqrt{R}e^{i\theta}e^{iky} +B\sqrt{T}e^{iky} + B\sqrt{R}e^{i\theta'}e^{ikx}$

Its energy is $A^{2} + B^{2} + AB\sqrt{TR}(e^{i\theta}+e^{-i\theta})+AB\sqrt{TR}(e^{i\theta'}+e^{-i\theta'})$

So, to preserve conservation, we must have $2cos(\theta)+2cos(\theta')=0$

That gives $\theta+\theta'=\pi$ or $\theta -\theta'=\pi$. But I never see this second result anywhere. Why is it there and how is it eliminated?

Thank you

2. Oct 2, 2011

### McLaren Rulez

Anyone? Strange thing is that I have the answer but an extra solution which should be invalid. Thank you