# Hamiltonian for Single-photon, Single-atom QED Cavity

Equation 7.71 of Nielsen and Chuang's Quantum Computation and Quantum Information gives the Hamiltonian for a two level atom and single mode photons in a cavity as:

$$H = \hbarωN + δZ + g(a^†σ_− + aσ_+)$$

where $$ω$$ is the photon frequency, $$δ = (ω_0 - ω)/2$$ with $$ω_0$$ the atomic transition frequency. $$Z$$ is the Pauli Z matrix; g is a coupling constant; $$a^†$$ and $$a$$ are creation and annihilation operators for photons, and $$σ_−$$ and $$σ_+$$ are lowering and raising operators for the atomic energy levels.

Neglecting the $$N$$ term (since it only contributes a fixed phase) and considering a single excitation in the (photon) field mode, this Hamiltonian is re-written as

$$H = -\begin{bmatrix}\delta&0&0\\0&\delta&g\\0&g&-\delta\end{bmatrix}$$

I've tried but been unable to derive this Hamiltonian from equation 7.71 above. $$H$$ in 7.71 should be a 2 x 2 matrix as Pauli matrices are 2 x 2. But here it is written as a 3 x 3 matrix. Can you please explain how this can be obtained? Thanks.

While the Pauli-$$Z$$ matrix is a 2 x 2 matrix, there are more basis states that need to be considered, namely the vacuum. The atomic basis states are $$\left|0\right>$$ and $$\left|1\right>$$, representing the number of excitations in the atom (as it's only a two-level atom, there can't be more than 1 excitation) and this is what $$Z$$ acts on. Similarly, the number of excitations (i.e. photons) in the field forms another basis, denoted by $$\left|N\right>$$.

However, we're only considering a cavity with a total of number of photons $$\leq 1$$, so the possible bases are, ordering the Hilbert space by $$\mathcal H_{\text{photons}}\otimes\mathcal H_{\text{atom}}$$, $$\left|00\right>, \,\left|10\right>$$ and $$\left|01\right>$$, as all the other possible states have $$> 1$$ photon.

The easiest way to go about figuring out what the Hamiltonian on this system looks like is to look at what each operator does to each state, where $$N=aa^\dagger$$ is ignored.

That is, $$Z$$ acts on the 'atomic' Hilbert space as $$\delta Z\left|0\right> = \delta\left|0\right>$$ and $$\delta Z\left|1\right> = -\delta\left|1\right>$$, $$\sigma_+$$ acts on the same space as $$\sigma_+\left|0\right> = \left|1\right>$$ and $$\sigma_+\left|1\right> = 0$$ and $$\sigma_-$$ acts as $$\sigma_-\left|1\right> = \left|0\right>$$ and $$\sigma_-\left|0\right> = 0$$. Similarly, on the photonic Hilbert space, $$a\left|0\right> = 0$$, $$a\left|1\right> = \left|0\right>$$, $$a^\dagger\left|0\right> = \left|1\right>$$ and $$a^\dagger\left|1\right> = \sqrt{2}\left|2\right>$$.

As a result, the term involving $$g$$ acts on the combined system as: \begin{align*}g\left(a^\dagger\sigma_- + a\sigma_+\right)\left|00\right> &= 0\\ g\left(a^\dagger\sigma_- + a\sigma_+\right)\left|10\right> &= g\left|01\right>\\ g\left(a^\dagger\sigma_- + a\sigma_+\right)\left|01\right> &= g\left|10\right>\end{align*} and the $$\delta Z$$ term similarly gives \begin{align*}\delta Z\left|00\right>&=\delta\left|00\right>\\\delta Z\left|10\right>&=\delta\left|10\right>\\\delta Z\left|01\right>&=-\delta\left|01\right>\end{align*}.

Writing this in matrix form with the basis $$\left|00\right>, \,\left|10\right>$$ and $$\left|01\right>$$ as above gives equation 7.76 as expected: $$H = \begin{bmatrix}\delta&0&0\\0&\delta&g\\0&g&-\delta\end{bmatrix}.$$

It does appear that I'm out in what I get for $$H$$ by a minus sign, but looking at equation 7.77, which has $$U=e^{-iHt} = e^{-i\delta t}\left|00\rangle\langle00\right|+...$$ suggests that this is presumably a typo (in my version, anyway), unless there's something I'm missing.

• Thanks a lot for the detailed answer. In my copy too the first delta has a negative sign. Commented Jul 3, 2019 at 23:35