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in my notes I have calculated the Eigenenergie of the Hamiltonian:

[itex]

H= \hbar \omega (n+\frac{1}{2}) \cdot

\begin{pmatrix}

1 & 0 \\

0 & 1

\end{pmatrix}

+\hbar

\begin{pmatrix}

\frac{\Omega_0 -\omega_0}{2} & g \sqrt{n+1} \\

g \sqrt{n+1}& -\frac{\Omega_0 -\omega_0}{2}

\end{pmatrix}

[/itex]

We have just calculated:

$$det\begin{pmatrix}

\frac{\Omega_0 -\omega_0}{2} - \lambda & g \sqrt{n+1} \\

g \sqrt{n+1}& -\frac{\Omega_0 -\omega_0}{2} - \lambda

\end{pmatrix} => \lambda = \pm \sqrt{

\frac{\Omega_0 -\omega_0}{4}+g^2(n+1)

}$$

And then we said the Eigenenergie is:

[itex]E=\hbar \omega(n+\frac{1}{2}) \pm \hbar\sqrt{

\frac{\Omega_0 -\omega_0}{4}+g^2(n+1)

}[/itex]

Why can I add [itex]\hbar \omega(n+\frac{1}{2})[/itex] to the result of the determinant?

Or why is it possible to neglect the first term of the hamiltonian in the determinant?

I also know that:

[itex] [N,H]=[a^\dagger a + c_{1}^\dagger c_1 , H] =0[/itex]

a is a photon annihilation operator and c a fermionic annihilation operator.

The Hamiltonian in an other notation is [itex]H = \hbar \omega a^\dagger a + \frac{1}{2} \hbar \Omega(c^\dagger _1 c_1 -c^\dagger _0 c_0) +\hbar g(a c^\dagger _1 c _0 + a^\dagger c^\dagger _0 c_1)[/itex].

I regard interaction of photons and fermions. So the states looks like [itex]|01>|n>[/itex] or [itex]|10>|n+1>[/itex] where n is the number of photons.

Thank you very much!

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# Photon in cavity

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