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

$$\begin{array}{rl}{\displaystyle [{\widehat{E}}_{\text{out}},{\widehat{E}}_{\text{out}}^{†}]}& {\displaystyle =\mathrm{\mid }t{\mathrm{\mid }}^2[{\widehat{E}}_{\text{in}},{\widehat{E}}_{\text{in}}^{†}]+t[{\widehat{E}}_{\text{in}},{\widehat{F}}^{†}]+t[\widehat{F},{\widehat{E}}_{\text{in}}^{†}]+[\widehat{F},{\widehat{F}}^{†}]}\\ {\displaystyle 1}& {\displaystyle =\mathrm{\mid }t{\mathrm{\mid }}^2+t[{\widehat{E}}_{\text{in}},{\widehat{F}}^{†}]+t[\widehat{F},{\widehat{E}}_{\text{in}}^{†}]+[\widehat{F},{\stackrel{}{}}^{}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle [{\widehat{E}}_{\text{out}},{\widehat{E}}_{\text{out}}^{†}]}& {\displaystyle =\mathrm{\mid }t{\mathrm{\mid }}^2[{\widehat{E}}_{\text{in}},{\widehat{E}}_{\text{in}}^{†}]+t[{\widehat{E}}_{\text{in}},{\widehat{F}}^{†}]+t[\widehat{F},{\widehat{E}}_{\text{in}}^{†}]+[\widehat{F},{\widehat{F}}^{†}]}\\ {\displaystyle 1}& {\displaystyle =\mathrm{\mid }t{\mathrm{\mid }}^2+[\widehat{F},{\widehat{F}}^{†}]}\\ {\displaystyle [\widehat{F},{\widehat{F}}^{†}]}& {\displaystyle =1-\mathrm{\mid }t{\mathrm{\mid }}^{}}\end{array}$$

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