Jaynes-Cummings model dynamics special example

Initial condition§

Setup§

Known that

$$\ket{\psi} = e^{-i\omega nt}e^{-i\frac{\Delta}{2}t}\left((c_+ e^{-\frac{i}{2}\Omega t}\sin\theta + c_- e^{\frac{i}{2}\Omega t}\cos\theta)\ket{0} + (c_+ e^{-\frac{i}{2}\Omega t}e^{i\phi}\cos\theta - c_- e^{\frac{i}{2}\Omega t}e^{i\phi}\sin\theta)\ket{1}\right)$$

We can notate

$$\begin{cases} c_0 = e^{-i\omega nt}e^{-i\frac{\Delta}{2}t}(c_+ e^{-\frac{i}{2}\Omega t}\sin\theta + c_- e^{\frac{i}{2}\Omega t}\cos\theta) \\ c_1 = e^{-i\omega nt}e^{-i\frac{\Delta}{2}t}(c_+ e^{-\frac{i}{2}\Omega t}e^{i\phi}\cos\theta - c_- e^{\frac{i}{2}\Omega t}e^{i\phi}\sin\theta) \end{cases}$$

Making

$$\ket{\psi}=c_0\ket{0} + c_1\ket{1}$$

Excited state§

Want at $t=0$ $c_0 = 0$, $c_1 = 1$, thus

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

We will get

$$\begin{cases} c_+ = e^{-i\phi}\cos\theta \\ c_- = -e^{-i\phi}\sin\theta \end{cases}$$

Letting the relative phase between $c_+$ and $c_-$ be $\delta$, we find $\delta = \pi$ since $c_+c_-^*$ is a negative real number.

With this initial condition, we will get

$$\ket{\psi} = e^{-i\omega nt}e^{-i\frac{\Delta}{2}t}\left(-\frac{i\Omega_R}{\Omega}e^{-i\phi}\sin\frac{\Omega}{2}t\ket{0} + \left( \cos\frac{\Omega}{2}t - i\frac{\Delta}{\Omega}\sin\frac{\Omega}{2}t\right)\ket{1}\right)$$

Thus,

$$P_e(t)=1-\frac{\Omega_R^2}{\Omega^2}\sin^2\left(\frac{\Omega}{2} t\right)$$

On resonant, it is

$$\ket{\psi} = e^{-i\omega nt}\left(-ie^{-i\phi}\sin\frac{\Omega_R}{2}t\ket{0} + \cos\frac{\Omega_R}{2}t \ket{1}\right)$$ $$P_e(t)=\cos^2\left(g\sqrt{n+1}t\right)$$

Note that the Rabi frequency is directly related to $g$ and $\sqrt{n+1}$, which implies that the oscillation is driven entirely by the coupling constant $g$ ( and the number of photon ); the stronger the coupling, the faster the oscillation. Note, however, that this is strictly true only on perfect resonance. Also, even if $n=0$, the term still oscillates, at frequency $2g$ — this is the vacuum Rabi oscillation, the atom exchanging its excitation with the empty cavity mode purely because the field is quantized. Note that the fact that it is $\sqrt{n+1}$, and not simply $n$, is what hints at quantization: the atom couples to the field even with zero photons present.

On very off resonant

$$\Omega = \sqrt{\Delta^2 + \Omega_R^2} \approx |\Delta|\left(1 + \frac{1}{2}\frac{\Omega_R^2}{\Delta^2}\right) \approx |\Delta|$$

Thus,

$$\ket{\psi} = e^{-i\omega nt}e^{-i\frac{\Delta}{2}t}\left(-\frac{i\Omega_R}{\Delta}e^{-i\phi}\sin\frac{\Delta}{2}t\ket{0} + \left( \cos\frac{\Delta}{2}t - i\frac{\Delta}{\Omega}\sin\frac{\Delta}{2}t\right)\ket{1}\right)$$ $$P_e(t) \approx 1- \frac{\Omega_R^2}{\Delta^2}\sin^2\left(\frac{\Delta}{2} t\right)$$

Note that here, the coupling strength no longer controls the oscillation frequency, as it is already far detuned such that the photon almost can't interact with the atom. Also, since this is far detuned ($\Omega_R \ll \Delta$), the amplitude of this oscillation can approach 0, further verifying that the atom can only interact well with resonant photons.

Ground state§

Want at $t=0$ $c_0 = 1$, $c_1 = 0$, thus

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

We will get

$$\begin{cases} c_+ = \sin\theta \\ c_- = \cos\theta \end{cases}$$

Similarly, letting the relative phase be $\delta$, we find $\delta = 0$ since $c_+c_-^*$ is a positive real number.

With this initial condition, we will get

$$\ket{\psi} = e^{-i\omega nt}e^{-i\frac{\Delta}{2}t}\left(\left( \cos\frac{\Omega}{2}t + i\frac{\Delta}{\Omega}\sin\frac{\Omega}{2}t\right)\ket{0} -\frac{i\Omega_R}{\Omega}e^{i\phi}\sin\frac{\Omega}{2}t\ket{1}\right)$$

Thus,

$$P_e(t)=\frac{\Omega_R^2}{\Omega^2}\sin^2\left(\frac{\Omega}{2} t\right)$$

On resonant, it is

$$P_e(t)=\sin^2\left(g\sqrt{n+1} t\right)$$

On very off resonant

$$\Omega = \sqrt{\Delta^2 + \Omega_R^2} \approx |\Delta|\left(1 + \frac{1}{2}\frac{\Omega_R^2}{\Delta^2}\right) \approx |\Delta|$$

Thus,

$$P_e(t) \approx \frac{\Omega_R^2}{\Delta^2}\sin^2\left(\frac{\Delta}{2} t\right)$$

After comparing both cases, we see that for a far-detuned photon, the atom mostly stays close to its initial state and oscillates slightly away due to the weakly interacting photon.

Coherent state§

As seen before, a coherent state, but with the current notation is

$$\ket{\psi(0)}=e^{-\frac{1}{2}|\alpha|^2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}\ket{1}_n$$

Where since we are fixed to 2-level atom, the subscript n indicate the number of photons in the system. And each $N=n+1$ as we assume the atom is already in excited state initially.

Then easily, the wavefunction will be

$$\ket{\psi}=e^{-\frac{1}{2}|\alpha|^2}\sum_{m=0}^\infty \frac{\alpha^m}{\sqrt{m!}}e^{-i\omega mt}e^{-i\frac{\Delta}{2}t}\left(-\frac{i\Omega_R}{\Omega}e^{-i\phi}\sin\frac{\Omega}{2}t\ket{0}_m + \left( \cos\frac{\Omega}{2}t - i\frac{\Delta}{\Omega}\sin\frac{\Omega}{2}t\right)\ket{1}_m\right)$$

This will give,

$$P_e(t)=e^{-|\alpha|^2}\sum_{m=0}^\infty\frac{|\alpha|^{2m}}{m!}\left(1-\frac{\Omega_R^2}{\Omega^2}\sin^2\left(\frac{\Omega}{2} t\right)\right)$$

Which on resonant, since the coupling between $\ket{e,m}$ and $\ket{g,m+1}$ gives $\Omega_R(m)=2g\sqrt{m+1}$, it is

$$P_e(t)=e^{-\braket{n}}\sum_{m=0}^\infty\frac{\braket{n}^{m}}{m!}\cos^2\left(g\sqrt{m+1} t\right)$$

For large $\braket{n}$, the poisson distribution can be approximated using gaussian distribution with $m=\braket{n}$ as the peak, so that we can easily see, different $m$ have different frequency, but their distribution density will be symmetric due to the approximation. Collapse happens when these different frequencies drift out of phase with each other — i.e. when the phase difference between the two edges of the distribution ($m=\braket{n}\pm k$, with $k=\sqrt{\braket{n}}$ the Poisson width) reaches $\pi$:

$$\begin{align*} (\Omega_{\braket{n}+k}-\Omega_{\braket{n}-k})\,t&=\pi\\ 2g\left(\sqrt{\braket{n}+k}-\sqrt{\braket{n}-k}\right)t&=\pi \\ 2g\sqrt{\braket{n}}\left[\left(1+\frac{k}{2\braket{n}}\right)-\left(1-\frac{k}{2\braket{n}}\right)\right]t&=\pi \\ 2g\sqrt{\braket{n}}\cdot\frac{k}{\braket{n}}\,t&=\pi \end{align*}$$

With $k=\sqrt{\braket{n}}$, the $\braket{n}$-dependence cancels:

$$2g\,t=\pi \quad\Rightarrow\quad t_c=\frac{\pi}{\Omega_R(n=0)}$$

Note that the collapse time comes out independent of $\braket{n}$ — this makes sense, since collapse is set by the spread of Rabi frequencies across the photon distribution, not by their mean.

Revival happens instead when neighbouring Fock components ($\Delta m=1$) rephase, i.e. when their phase difference reaches $2\pi$:

$$(\Omega_{m+1}-\Omega_m)\,t_r=2\pi$$

Using $\Omega_m=2g\sqrt{m+1}$ and expanding around $m=\braket{n}\gg1$:

$$\Omega_{m+1}-\Omega_m\approx\left.\frac{d\Omega_m}{dm}\right|_{\braket{n}}=\frac{g}{\sqrt{\braket{n}+1}}\approx\frac{g}{\sqrt{\braket{n}}}$$

so that

$$t_r=\frac{2\pi\hbar\sqrt{\braket{n}}}{g}$$

which grows with $\sqrt{\braket{n}}$, as expected — the more photons in the coherent state, the longer it takes for the discrete Fock components to rephase.

On very off resonant,

$$\begin{align*} P_e(t)&=\frac{4g^2}{\Delta^2}e^{-\braket{n}}\sin^2\left(\frac{\Delta}{2} t\right)\sum_{m=0}^\infty \frac{\braket{n}^{m}}{m!}(m+1)\\ &=\frac{4g^2(\braket{n}+1)}{\Delta^2}\sin^2\left(\frac{\Delta}{2} t\right)\\ &=\frac{\braket{\Omega_R}^2}{\Delta^2}\sin^2\left(\frac{\Delta}{2} t\right)\\ \end{align*}$$

So, the collapse and revival only happen on resonant.