- #1
Abigale
- 56
- 0
Hey,
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!
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!