Jaynes-Cummings model

Jaynes-Cummings model§

If we even drop the $|g|^2$ terms from the previous expansion, this is the Jaynes-Cummings model. And since this model conserve energy ( energy don't fluctuate ), we take

$$\begin{cases} \hat{A} = \hat{b}\hat{\sigma}^+ \\ \hat{A}^\dagger = \hat{b}^\dagger\hat{\sigma}^- \\ \end{cases}$$ $$\hat{H}_\text{JC}=\hbar\omega_{eg}\hat{\sigma}_e+\hbar\omega\hat{n} + \hbar g\hat{A}+\hbar g^*\hat{A}^\dagger$$

Notice that at some point we also have droped the summation over $k$ and $\lambda$ which is justified as resonant interaction is dominant.

Conservation§

Since this models only contain energy converating terms ( a raise on atomic level must lower the cavity photon count, or vis versa ). We expect the new number operator to be conserved. We call this the excitation number

$$\hat{N}=\hat{n}+\hat{\sigma}_e$$

We want

$$[\hat{H}_\text{JC}, \hat{N}] = 0$$

Verify term by term§

$$[\hat{\sigma}_e,\hat{N}]=[\hat{\sigma}_e,{b}^\dagger\hat{b}]+[\hat{\sigma}_e,\hat{\sigma}_e]=0$$ $$[{b}^\dagger\hat{b},\hat{N}]=[{b}^\dagger\hat{b},{b}^\dagger\hat{b}]+[{b}^\dagger\hat{b},\hat{\sigma}_e]=0$$ $$[\hat{b}\sigma^+,\hat{N}]=[\hat{b}\sigma^+,{b}^\dagger\hat{b}]+[\hat{b}\sigma^+,\hat{\sigma}_e]=0$$ $$[\hat{b}^\dagger\hat{\sigma}^-,\hat{N}]=[\hat{b}^\dagger\hat{\sigma}^-,{b}^\dagger\hat{b}]+[\hat{b}^\dagger\hat{\sigma}^-,\hat{\sigma}_e]=0$$

So indeed $\hat{N}$ is a conserved quantity

Choosing basis§

Because $\hat{N}$ is converved, we can build basis out of it, let's take

$$\ket{0}=\ket{g,n} \qquad \ket{1}=\ket{e,n-1}$$

Any arbitrary state will be $\ket{\psi}=c_g\ket{0}+c_e\ket{1}$

Rabi frequncy§

Where it is not obvious for now, but let's call The Rabi frequency at resonant

$$\Omega_R = 2|g|\sqrt{n}$$

With the general Rabi frequency, or, just Rabi frequency

$$\Omega=\sqrt{\Delta^2 + \Omega_R^2}$$

The matrix§

In order to calculate the Hamiltonian matrix, we evaluate

$$\begin{cases} \hat{H}_\text{JC}\ket{0}=\hbar\omega n\ket{0} + \hbar g\sqrt{n}\ket{1} \\ \hat{H}_\text{JC}\ket{1}=\hbar(\omega_{eg} + \omega(n-1))\ket{1} + \hbar g^*\sqrt{n}\ket{0} \\ \end{cases}$$

Thus

$$H_\text{JC}= \hbar \begin{bmatrix} n\omega & g^*\sqrt{n} \\ g\sqrt{n} & \Delta + n\omega \end{bmatrix}$$

Simplify§

Note that by adding something propotional to the Identity, we are just shifting the base energy of the system and introducing a global frequency to all state and those will not affect the physical measurement.

Let's subtract the avarage energy from the system to create symmetry.

$$E_\text{avg} = \hbar\left(\omega n + \frac{\Delta}{2}\right)$$ $$H_\text{JC}= \hbar \begin{bmatrix} -\Delta/2 & g^*\sqrt{n} \\ g\sqrt{n} & \Delta/2 \end{bmatrix}$$

Eigenstates§

We are able find it's eigenenergy trivially given its symettry, but a standard trick that makes the notation much cleaner for the eigenstates later on is

Parametrize into mixing angle§

Let

$$\tan2\theta=\frac{2|g|\sqrt{n}}{\Delta}=\frac{\Omega_R}{\Delta}$$

Thus,

$$\begin{cases} \Omega\sin2\theta=\Omega_R \\ \Omega\cos2\theta= \Delta \\ \end{cases}$$

Where

$$\Omega^2=\Delta^2 + \Omega_R^2$$

Now we rewrite the hamiltonian matrix as

$$H_\text{JC}= \frac{\Omega}{2}\hbar \begin{bmatrix} -\cos2\theta & e^{-i\phi}\sin2\theta \\ e^{i\phi}\sin2\theta & \cos2\theta \end{bmatrix}$$

Which easily we see

$$E=\pm\hbar \frac{\Omega}{2}$$

And to rewind back before we shifted the energy

$$E=\hbar\left(n\omega+\frac{\Delta}{2}\pm \frac{\Omega}{2} \right)$$

And for eigenstates, it is

$$\begin{cases} \ket{n, +}=\sin\theta\ket{0}+e^{i\phi}\cos\theta\ket{1} \\ \ket{n, -}=\cos\theta\ket{0}-e^{i\phi}\sin\theta\ket{1} \\ \end{cases}$$

This have another more specific name called dressed states.