How Does One Calculate the Time Evolution of a Photon in a Vacuum State?

[deepalakshmi]

TL;DR

How to evolve one photon using hamiltonian as beam splitter

During time evolution of one photon with vacuum state with hamiltonian as a^†b+b^†a, the answer is cos(t/ℏ)|0,1⟩+isin(t/ℏ)|1,0⟩. But i don't know how to do calculation to get this answer. Can someone please help me?
I tried to do this calculation:
|0⟩|1⟩(t)=e−iHtℏ|0⟩|1⟩
=(cos(tH/ℏ)−isin(tH/ℏ)) |0⟩|1⟩
=[cos(t/ℏ)−isin(t/ℏ)] H|0⟩|1⟩
=[cos(t/ℏ)−isin(t/ℏ)] [|0⟩|1⟩+i|1⟩|0⟩]

How to proceed?

 

[azntoon]

A:The Hamiltonian you are using is the Jaynes-Cummings Hamiltonian, which can be written as $$H=\hbar \omega (a^{\dagger}a+b^{\dagger}b+1/2).$$For this Hamiltonian, the time evolution of a state $|\psi(0)\rangle$ is given by $$|\psi(t)\rangle=e^{-iHt/\hbar}|\psi(0)\rangle.$$You are considering the initial state to be $|\psi(0)\rangle=|0\rangle|1\rangle$. Applying the time evolution operator to this state, we have\begin{align*}|\psi(t)\rangle&=e^{-iHt/\hbar}|0\rangle|1\rangle\\&=e^{-i\hbar \omega t/2}|0\rangle|1\rangle-ie^{-i\hbar \omega t/2}|1\rangle|0\rangle\\&=\cos(\omega t/2)|0\rangle|1\rangle+\sin(\omega t/2)|1\rangle|0\rangle\end{align*}This is the answer you are looking for. Note that I have used $\hbar \omega$ instead of $t/\hbar$ to make it consistent with the Hamiltonian.