Beam Splitter

Beam Splitter§

By defnition, it is

$$\begin{bmatrix} \hat{a}_\text{out}\\ \hat{b}_\text{out} \end{bmatrix} = \begin{bmatrix} t & r \\ -r^* & t^* \end{bmatrix} \begin{bmatrix} \hat{a}_\text{in}\\ \hat{b}_\text{in} \end{bmatrix}$$

It is basically an output that is linear mixing of differnt input.

Effective beam splitter§

$$\hat{E}_\text{out}=t\hat{E}_\text{in} + \hat{F}$$

Where $\hat{F}$ is the noise term. But notice the $r$ have been eaten up since our goal is to find a way to proof is this system a beam splitter.

Bosoninc commutation preverving§

A true beam splitter preverve it's bosonic properties. So

$$\begin{align*} \left[\hat{E}_\text{out},\hat{E}^\dagger_\text{out}\right]&=|t|^2\left[\hat{E}_\text{in},\hat{E}^\dagger_\text{in}\right]+t\left[\hat{E}_\text{in},\hat{F}^\dagger\right]+t\left[\hat{F},\hat{E}^\dagger_\text{in}\right]+\left[\hat{F},\hat{F}^\dagger\right]\\ 1&=|t|^2+t\left[\hat{E}_\text{in},\hat{F}^\dagger\right]+t\left[\hat{F},\hat{E}^\dagger_\text{in}\right]+\left[\hat{F},\hat{F}^\dagger\right]\\ \end{align*}$$

Now, since the noise and the input should be fundelmentally unrelated. We want

$$\left[\hat{E}_\text{in}, \hat{F}^\dagger\right]=0$$

This simplifies the previous expression

$$\begin{align*} \left[\hat{E}_\text{out},\hat{E}^\dagger_\text{out}\right]&=|t|^2\left[\hat{E}_\text{in},\hat{E}^\dagger_\text{in}\right]+t\left[\hat{E}_\text{in},\hat{F}^\dagger\right]+t\left[\hat{F},\hat{E}^\dagger_\text{in}\right]+\left[\hat{F},\hat{F}^\dagger\right]\\ 1&=|t|^2+\left[\hat{F},\hat{F}^\dagger\right]\\ \left[\hat{F},\hat{F}^\dagger\right]&=1-|t|^2\\ \end{align*}$$

In conclusion, if the above condition can be proven to be true. Then that system is effectively a beam splitter.