# Unitary Transformation of an Interfering Beam Splitter

I was reading this research paper Quantum interference enables constant time quantum information processing and was confused by one particular expression involving the Hamiltonian of a beam splitter.

We consider two interfering modes $$a$$ and $$b$$ on a beam splitter device. The hamiltonian of the beam splitter is then

$$\hat{H}_{B S}=\frac{i \hbar}{2}\left(\hat{a}^{\dagger} \hat{b} e^{-i \varphi}-\hat{a} \hat{b}^{\dagger} e^{i \varphi}\right)$$

where $$\hat{a}$$, and $$\hat b$$ are respective annihilation operators of states $$a$$ and $$b$$.

Then the unitary transformation generated by the beam splitter is given by $$U_{bs}=e^{-i \theta H_{B S} / \hbar}$$ where $$\sin\left(\frac{\theta}{2}\right)=\sqrt{r}$$ (the reflection coefficient).

My question given this how is it possible to derive the expression given on page 32 for the annihilation operatorrs for output states $$a_{r}$$ and $$a_{t}$$

$$\begin{array}{l}{a_{r}=U_{B S}^{\dagger} a U_{B S}=a \cos \frac{\theta}{2}+b e^{-i \varphi} \sin \frac{\theta}{2}} \\ {a_{t}=U_{B S}^{\dagger} b U_{B S}=-a e^{i \varphi} \sin \frac{\theta}{2}+b \cos \frac{\theta}{2}}\end{array}$$

How can we derive this result using only the information given above? I tried evaluating the matrix exponential in Mathematica for some low dimensional cases and got several contradictions (mostly sign errors) which just made me more confused.

## 1 Answer

1. Write the evolution equation in the Heisenberg picture for $$a$$ (and the same for $$b$$): $$\dot a = i[H,a]\ .$$
2. Simplify the expression for the given Hamiltonian (omitting $$\hbar$$): \begin{aligned} \dot a &= -\tfrac12[a^\dagger b e^{-i\phi},a]+\tfrac12[ab^\dagger e^{i\phi},a] \\ & = -\tfrac12[a^\dagger,a] b e^{-i\phi} \\ & = \tfrac12 b e^{-i\phi}\ . \end{aligned}
3. Do the same for $$b$$: $$\dot b = -\tfrac12 a e^{i\phi}\ .$$
4. Integrate the two coupled differential equations $$\begin{pmatrix}\dot a\\ \dot b\end{pmatrix} = \frac12\begin{pmatrix} & e^{-i\phi}\\-e^{i\phi} \end{pmatrix} \begin{pmatrix} a\\ b\end{pmatrix}\ .$$