Rotating Wave Approximation

Categorizing the full interaction Hamiltonian§

As seen in the last article

$$\hat{H}_\text{int}=\hat{H}_\text{slow} + \hat{H}_\text{fast}$$

Where

$$\begin{cases} \hat{H}_\text{slow}&=\hbar g^*\hat{b}^\dagger\hat{\sigma}^-e^{-i\Delta t}+\hbar g\hat{b}\hat{\sigma}^+e^{i\Delta t}\\ \hat{H}_\text{fast}&=\hbar g^*\hat{b}^\dagger\hat{\sigma}^+e^{i\Sigma t} + \hbar g\hat{b}\hat{\sigma}^-e^{-i\Sigma t} \end{cases}$$

Where

$$\Delta=\omega_k-\omega_{eg} \qquad \Sigma = \omega_k+\omega_{eg}$$

The Transformation§

The formula§

$$\hat{H}_\text{int}'(t)=\hat{U}(t)\hat{H}_\text{int}\hat{U}^\dagger(t)-i\hbar\hat{U}(t)\frac{\mathrm{d}}{\mathrm{d}t}\hat{U}^\dagger(t)$$

And we propose the transformation $U$ should

$$\hat{U}(t)=e^{i\hat{S}}$$

Where because of unitary $\hat{S}=\hat{S}^\dagger$.

The goal§

We what no counter rotating term after the rotation, so we should demand some calculation which we will see later.

The first term§

Since we propose the form of $U$, we can use the BCH formula, Thus, after expanding, the first term becomes

$$\hat{H}'_\text{int}(t)=\hat{H}_\text{int}+i[\hat{S},\hat{H}_\text{int}]+i^2\frac{1}{2}[\hat{S},[\hat{S},\hat{H}_\text{int}]]+\dots$$

The second term§

Expanding the gauge term with the derivative-of-exponential identity

$$e^{i\hat{S}}\frac{\mathrm{d}}{\mathrm{d}t}e^{-i\hat{S}}=-i\dot{\hat{S}}+\frac{1}{2}[\hat{S},\dot{\hat{S}}]-\dots$$

gives

$$-i\hbar\hat{U}\frac{\mathrm{d}}{\mathrm{d}t}\hat{U}^\dagger=-\hbar\dot{\hat{S}}-\frac{i\hbar}{2}[\hat{S},\dot{\hat{S}}]+\dots$$

The first piece is order $g$; the second is order $g^2$ and supplies the Bloch–Siegert term below.

The cancelation§

The leading piece $-\hbar\dot{\hat{S}}$ cancels $\hat{H}_\text{fast}$:

$$\begin{align*} \hbar\dot{\hat{S}}&=\hat{H}_\text{fast}\\ &=\hbar g^*\hat{b}^\dagger\hat{\sigma}^+e^{i\Sigma t}+\hbar g\hat{b}\hat{\sigma}^-e^{-i\Sigma t}\\ \dot{\hat{S}}&=g^*\hat{b}^\dagger\hat{\sigma}^+e^{i\Sigma t}+g\hat{b}\hat{\sigma}^-e^{-i\Sigma t} \end{align*}$$

We take the oscillatory antiderivative:

$$\hat{S}=\frac{i}{\Sigma}\left(g\hat{b}\hat{\sigma}^-e^{-i\Sigma t} -g^*\hat{b}^\dagger\hat{\sigma}^+e^{i\Sigma t}\right)$$

Collecting the order-$g^2$ terms§

The first term contributes $i[\hat{S},\hat{H}_\text{fast}]$ and the gauge term contributes $-\frac{i}{2}[\hat{S},\hat{H}_\text{fast}]$ (using $\hbar\dot{\hat{S}}=\hat{H}_\text{fast}$). Together they give

$$i[\hat{S},\hat{H}_\text{fast}]-\frac{i}{2}[\hat{S},\hat{H}_\text{fast}]=\frac{i}{2}[\hat{S},\hat{H}_\text{fast}]$$

which is the combination used below.

Evaluate the final expression§

With the given $\hat{S}$, we compute the two essential commutators.

$$\frac{i}{2}[\hat{S},\hat{H}_\text{fast}] = \frac{\hbar|g|^2}{\Sigma}[\hat{b}\hat{\sigma}^-,\hat{b}^\dagger\hat{\sigma}^+]$$

The commutator $i[\hat{S},\hat{H}_\text{slow}]$ contains only rapidly oscillating terms and is dropped. Higher order terms are $\mathcal{O}(g^3/\Sigma^2)$ and negligible when $g\ll\Sigma$.

Everything§

$$\hat{H}'_\text{int}(t)=\underbrace{\hat{H}_\text{slow}}_{\text{co-rotating}} + \frac{\hbar|g|^2}{\Sigma}\bigl(\hat{n}\hat{\sigma}_e-(\hat{n}+1)\hat{\sigma}_g\bigr)$$

Transform back to the Schrödinger frame§

Adding back the free Hamiltonian $\hat{H}_0=\hbar\omega_{eg}\hat{\sigma}_e+\hbar\omega_k\hat{b}^\dagger\hat{b}$ gives the effective Schrödinger‑picture Hamiltonian ( with the consntant energy shifted away as only energy differents is physical )

$$\hat{H}_\text{eff} = \hbar\omega_{eg}\hat{\sigma}_e + \hbar\omega_k\hat{b}^\dagger\hat{b} + \bigl(\hbar g\hat{b}\hat{\sigma}^+ + \hbar g^*\hat{b}^\dagger\hat{\sigma}^-\bigr) + \frac{\hbar|g|^2}{\omega_k+\omega_{eg}}\left[\hat{n}\hat{\sigma}_e-(\hat{n}+1)\hat{\sigma}_g\right]$$

Just for simplification, let's use the pauli z basis and shift away constnat energy

$$\hat{H}_\text{eff} = \frac{\hbar\omega_{eg}}{2}\hat{\sigma}_z + \hbar\omega_k\hat{b}^\dagger\hat{b} + \bigl(\hbar g\hat{b}\hat{\sigma}^+ + \hbar g^*\hat{b}^\dagger\hat{\sigma}^-\bigr) + \frac{\hbar|g|^2}{\omega_k+\omega_{eg}}\left[\hat{n} + \frac{1}{2}\right]\hat{\sigma}_z$$

Then to better see the physics, let's define

$$\omega' = \frac{|g|^2}{\omega_k+\omega_{eg}}$$

And group them

$$\hat{H}_\text{eff} = \hbar\left(\frac{\omega_{eg}}{2}+\omega'\left[\hat{n} + \frac{1}{2}\right]\right)\hat{\sigma}_z + \hbar\omega_k\hat{b}^\dagger\hat{b} + \bigl(\hbar g\hat{b}\hat{\sigma}^+ + \hbar g^*\hat{b}^\dagger\hat{\sigma}^-\bigr)$$

Interesting physics§

If we omit $\omega'$, which is valid when $|g|^2$ is small. We get the Jaynes–Cummings model. But if we do not omit it, we notice that a vacuum shift has happened: even without any photon, the transition frequency has shifted upward by $\omega'$; this is the vacuum Bloch–Siegert shift. When photons are present the shift grows with intensity — the AC Stark effect. All together, this is the Bloch–Siegert shift.