Spontaneous Emission (Wigner-Weisskopf theory)

Model§

Known that a single mode won't create spontatnous emission. It, could only be rabi oscilation or lamb shift, or stark effect. We instead, use a multimode of field. Thus

$$\hat{H}=\hbar\omega_{eg}\hat{\sigma}_e+\sum_{k,\lambda}\hbar\omega_{k,\lambda}\hat{n}_{k,\lambda}+\sum_{k,\lambda}\hbar g \hat{A}_{k,\lambda} + \hbar g^* \hat{A}_{k,\lambda}^\dagger$$

With the general state

$$\ket{\psi}=c_e\ket{e}+\sum_{k,\lambda}c_{k,\lambda}\ket{g_{k,\lambda}}$$

Where $\ket{e} \equiv \ket{e, 0}$, and $\ket{g_{k,\lambda}} \equiv \ket{g, 1_{k,\lambda}}$

This is not a JC-model anymore by defniiton, but rather Wigner-Weisskopf theory

Initial condition§

For spontanuous emission to happen of course it should have something to emmit form, with total excitation of only 1 in our model, initial condition must ve

$$\ket{\psi}=\ket{e}$$

Thus

$$\begin{cases} c_e(t=0) = 1 \\ c_{k,\lambda}(t=0) = 0 \end{cases}$$

Solving the Schordinger equation§

$$\begin{cases} \bra{e}i\hbar\partial_t\ket{\psi}=i\hbar\dot{c}_e\\ \bra{g_{k,\lambda}}i\hbar\partial_t\ket{\psi}=i\hbar\dot{c}_{k,\lambda}\\ \end{cases}$$ $$\begin{cases} \bra{e}\hat{H}\ket{\psi} = \hbar\omega_{eg}c_e + \displaystyle\sum_{k,\lambda} \hbar g c_{k,\lambda}\\ \bra{g_{k,\lambda}}\hat{H}\ket{\psi} = \hbar\omega_{k,\lambda}c_{k,\lambda} + \hbar g^* c_e \\ \end{cases}$$

Solving

$$\hat{H}\ket{\psi}=i\hbar\ket{\psi}$$

We get

$$\begin{cases} i\dot{c}_e = \omega_{eg}c_e + \displaystyle\sum_{k,\lambda}gc_{k,\lambda} \\ i\dot{c}_{k,\lambda} = \omega_{k,\lambda}c_{k,\lambda} + g^*c_e \end{cases}$$

Thus

$$c_{k,\lambda}(t) = -i g_{k,\lambda} \int_{0}^{t} dt' c_e(t') e^{-i\omega_k (t - t')}\\ \dot{c}_e(t) = -i\omega_{eg}c_e(t) - \sum_{\mathbf{k},\lambda} |g_{\mathbf{k},\lambda}|^2 \int_0^t dt' c_e(t') e^{-i\omega_k (t - t')}$$

Spectral Density§

An easily intepreted quantity is Spectral Density, which my definition is

$$J(\omega) = \sum_{k,\lambda} |g_{k,\lambda}|^2 \delta(\omega - \omega_k)$$

where $g$ in here is the coupling strenght ok that $\omega$, and $\int J = \sum g$.

$$\begin{align*} \dot{c}_e(t) &= -i\omega_{eg}c_e(t) - \sum_{\mathbf{k},\lambda} |g_{\mathbf{k},\lambda}|^2 \int_0^t dt' c_e(t') e^{-i\omega_k (t - t')}\\ &= -i\omega_{eg}c_e(t) - \int_0^t dt' c_e(t') K(t - t')\\ \end{align*}$$

Where the kernal $K$ is

$$K(t - t')=\sum_{\mathbf{k},\lambda}|g_{\mathbf{k},\lambda}|^2 e^{-i\omega_k (t - t')}=\int_0^\infty d\omega\,J(\omega)e^{-i\omega (t-t')}$$

The Markov approximation§

As we only cares about a timescale about the inverse of decay rate, which experimentally gives fentosecond scales.

$$(t-t')\propto \frac{1}{\gamma} = \text{fento second scale}$$

And also, by the Rimann-Lebegue lemma, a smooth forier integral with large time should goes to 0. Thus

$$K(\tau \to \infty) \to 0$$

gives

$$\begin{align*} \dot{c}_e(t) &= -i\omega_{eg}c_e(t) - \int_0^t dt' c_e(t') K(t - t')\\ &= -i\omega_{eg}c_e(t) - \int_0^\infty dt' c_e(t') K(t - t')\\ \end{align*}$$

To solve this, we need to handle the fast oscillation of $c_e(t')$ inside the integral. Since $c_e(t')$ mostly oscillates at $\omega_{eg}$, we approximate $c_e(t') \approx c_e(t) e^{-i\omega_{eg}(t-t')}$ in the integral. Let $\tau = t - t'$, the equation becomes

$$\begin{align*} \dot{c}_e(t) &= -i\omega_{eg}c_e(t) - c_e(t) \int_0^\infty d\tau \, K(\tau) e^{-i\omega_{eg}\tau} \\ &= -i\omega_{eg}c_e(t) - c_e(t) \int_0^\infty d\tau \int_0^\infty d\omega \, J(\omega) e^{-i(\omega - \omega_{eg})\tau} \end{align*}$$

Evaluating the integral§

We need to solve the time integral $\int_0^\infty d\tau \, e^{-i(\omega - \omega_{eg})\tau}$. By the Sokhotski-Plemelj theorem, this is a standard identity

$$\int_0^\infty d\tau \, e^{-i(\omega - \omega_{eg})\tau} = \pi \delta(\omega - \omega_{eg}) - i \, \mathcal{P}\frac{1}{\omega - \omega_{eg}}$$

where $\mathcal{P}$ is the Cauchy principal value. Plugging this back in gives

$$\dot{c}_e(t) = -i\omega_{eg}c_e(t) - c_e(t) \left[ \pi J(\omega_{eg}) - i \int_0^\infty d\omega \, J(\omega) \mathcal{P}\frac{1}{\omega - \omega_{eg}} \right]$$

Decay rate and Lamb shift§

We can now define two physical quantities from the real and imaginary parts of the bracket.

The decay rate $\gamma$ comes from the delta function picking out the spectral density at the atom's frequency:

$$\gamma \equiv 2\pi J(\omega_{eg})$$

The Lamb shift $\Delta_{LS}$ comes from the principal value integral, representing a small energy shift due to the coupling to the vacuum:

$$\Delta_{LS} \equiv \int_0^\infty d\omega \, J(\omega) \mathcal{P}\frac{1}{\omega - \omega_{eg}}$$

Substituting these back, we get a very simple ordinary differential equation

$$\dot{c}_e(t) = - \left( \frac{\gamma}{2} + i(\omega_{eg} + \Delta_{LS}) \right) c_e(t)$$

Final Solution§

This is trivial to solve. With the initial condition $c_e(0) = 1$

$$c_e(t) = \exp\left[ - \left( \frac{\gamma}{2} + i(\omega_{eg} + \Delta_{LS}) \right) t \right]$$

The probability of the atom still being in the excited state is the modulus squared

$$P_e(t) = |c_e(t)|^2 = e^{-\gamma t}$$

Thus we derived the exponential decay of spontatnous emission, where the decay rate is strictly dictated by the spectral density of the vacuum at the transition frequency. Note that the Lamb shift although is mathematically hard to solve ( it even direrges at uv ), it is irrelavent in physics as it is a constnat shift which doesnt appear upon measurement ( energy differents ).

And from here, if we solve for $c_{k,\lambda}$ Which is trivial and direct, we get

$$c_{k,\lambda}(\infty) = -i g_{k,\lambda} \int_0^\infty dt \, e^{-\gamma t/2} e^{-i(\omega_k - \omega_{eg}) t} = -i g_{k,\lambda} \frac{1}{\frac{\gamma}{2} + i(\omega_k - \omega_{eg})}$$ $$|c_{\mathbf{k},\lambda}(\infty)|^2 = \frac{|g_{\mathbf{k},\lambda}|^2}{(\omega_k - \omega_{eg})^2 + (\gamma/2)^2}$$

Is Lorentzian.

Single mode Cavity§

The exact same derivation can be done for Cavity Which is with the Hamiltonian

$$H=\hbar\omega_{c}a^{\dagger}a+\sum_{k}\hbar\omega_{k}b_{k}^{\dagger}b_{k}+\sum_{k}\hbar(\xi_{k}a^{\dagger}b_{k}+\xi_{k}^{*}b_{k}^{\dagger}a)$$

Thus we will have a cavity decay $\kappa$ which is analogous to $\gamma$ with also,

$$\kappa \equiv 2\pi J_\text{cav}(\omega_c) = 2\pi \xi^2(\omega_c)\rho(\omega_c)$$

Now, if we calculate the total decay rate, using the Lindblad master equation, we can find that, the decay rate for a poor cavity ( high $\kappa$ ), gives total decay of

$$\gamma + \frac{4g^2}{\kappa}$$

And for a real good cavity ( low $\kappa$ but then the allowed mode is limited ), the total decay is

$$\frac{\gamma}{2}$$