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The Wigner-Weisskopf Approximation (also known as the Born-Markov Approximation) is an approximation used in quantum optics to calculate the rate of spontaneous emission of an atom. Victor Weisskopf and Eugene Wigner constructed the theory when studying the natural line width of atomic transition and published it in 1930. The assumption is that $$ \Gamma <<\omega_0 $$, where $$\Gamma$$ is the rate of change of the excited state amplitude and $$ \omega_0 $$ is the atomic transition frequency.

Derivation of Rate of Spontaneous Emission with WWA
The theory begins with a two-level atom in vacuum initially in the excited state. The atom couples to the quantum electromagnetic vacuum and is effectively an open quantum system so the excited state is not an eigenstate. Therefore, it evolves and after some time it ends up in the ground state as it spontaneously emits a photon to some mode of the electromagnetic vacuum. The rate at which this occurs is called the rate of spontaneous emission or the “Einstein A coefficient”. The Hamiltonian of this open quantum system after the Rotating Wave Approximation (RWA) (dominating modes are those with frequencies close to the atom transition frequency) and the dipole approximation (the wavelength of the field is much smaller than the characteristic length scale), the Hamiltonian in the Schrödinger picture is:



\hat{H} = \hbar\omega_{0}\hat{\sigma_z} + \sum_{\vec{k},\mu} \hbar \omega_k \hat{a}^{\dagger}_{\vec{k},\mu} \hat{a}_{\vec{k},\mu} + \sum_{\vec{k},\mu} (\hbar g_{\vec{k},\mu} \hat{a}_{\vec{k},\mu} \hat{\sigma_{+}} + \hbar g^{*}_{\vec{k},\mu} \hat{a}^{\dagger}_{\vec{k},\mu} \hat{\sigma_{-}}) $$

The first term in the Hamiltonian is the atom term, the second is the field term and the third is the atom-field interaction Hamiltonian. The coupling coefficient $$g_{\vec{k},\mu}$$ is defined by:


 * $$ \hbar g_{\vec{k},\mu} = - \langle e | \hat{\vec{d}} | g \rangle \cdot \vec{\epsilon}_{\vec{k},\mu} \sqrt{ \frac{2\pi \hbar \omega_{\vec{k}}} {V_{\vec{k}}} } e^{i\vec{k}\cdot \vec{R}} $$

Like in the Jaynes-Cummings model, the Hilbert Space is spanned by the product states $$ \{ |e,0\rangle, | g, 1_{\vec{k},\mu} \rangle \} $$ (atom in excited state and field in vacuum state, ground state with photon emitted in some mode $$(\vec{k},\mu)$$ ) since the number of excitations is conserved. So, the state at any time can be written as a superposition of the excited state and the ground states with coefficients $$ c_{e,0}(t) $$ and $$ c_{g,1_{\vec{k},\mu}}(t) $$ respectively.

Thus, according to the Schrödinger Equation, the probability amplitudes satisfy the following differential equations.



\begin{align} & \dot{c}_{e,0} = -\sum_{\vec{k},\mu} |g_{\vec{k},\mu}|^2 \int^{t}_0 dt' \text{exp}(-i(\omega_k - \omega_0)(t-t')) c_{e,0}(t') \\ & \dot{c}_{g,1_{\vec{k},\mu}} = -ig^{*}_{\vec{k},\mu}\int^{t}_0 dt' \text{exp}((\omega_k - \omega_0)t')c_{e,0}(t') \end{align} $$

Since the electromagnetic modes are a continuous set when the quantization volume becomes infinite, then the discrete sum in the first equation becomes an integral of the density of states $$ D(\vec{k}) $$ over all space. When the quantization volume was finite, the discrete modes were $$ \vec{k} = (n_x\pi/L_x, n_y\pi/L_y, n_z\pi/L_z ) $$ so the density of states can be written as $$ D(\vec{k}) = V/(2\pi)^3 $$.



\sum_{\vec{k},\mu} \longrightarrow \sum^2_{\mu = 1}\int d^3\vec{k} D(\vec{k}) = \sum^2_{\mu = 1} \frac{V}{(2\pi)^3} \int d\Omega_k k^2 dk = \sum^2_{\mu = 1} \int d\Omega_k d\omega_k \frac{\omega^2_k}{c^3} \frac{V}{(2\pi)^3} $$

The density of the frequency modes is $$ D(\omega_k) = \frac{\omega^2_k}{c^3} \frac{V}{(2\pi)^3}$$. Note that the sum over the magnitude squared of the coupling strength is the calculation for the angular distribution of the dipole emission which is proportional to $$ \sum^2_{\mu = 1} |\vec{d}_{eg} \cdot \vec{\epsilon}_{\vec{k},\mu}|^2 = |\vec{d}_{eg}|^2 \sin^2\theta $$. Thus, the integral over this is



\int d \Omega_k \sum^2_{\mu = 1} |g_{\vec{k},\mu}|^2 = \frac{2\pi\omega_k}{\hbar V}\int^{\pi}_0 d\theta \sin^3\theta |\vec{d}_{eg}|^2 \int^{2\pi}_0 d\phi = \frac{16\pi^2\omega_k}{3\hbar V} |\vec{d}_{eg}|^2 = \overline{g^2{\omega_k}} $$

And the differential equation for the probability amplitude of the excited state is:



\dot{c}_{e,0} = -\int^{\infty}_0 d \omega_k D(\omega_k) \overline{g^2{\omega_k}} \int^{t}_0 dt' \text{exp}(-i(\omega_k - \omega_0)(t-t')) c_{e,0}(t') $$

Now, to calculate the time integral, we use the Wigner-Weisskopf Approximation. We use that the time scale of $$c_{e,0(t)}$$ is $$\gamma << \omega_0 $$ so the exponential term will oscillate very fast compared to $$c_{e,0(t)}$$ and the significant contribution will be for $$ t \sim t' $$. So, we approximate $$c_{e,0}(t') = c_{e,0(t)}$$ and we factor it out of the integral. This action is very important since we are essentially taking away the atom's memory. By not integrating, and only keeping the state of the atom at a certain time the reversible dynamics of the atom are removed from our calculations. For the remaining part of the integral we can extend the upper limit to infinity since we assume that the the integrand vanishes for $$ t >> t' $$. This integral doesn't converge but we can use Cauchy's Principal Part to assign a value to it.



\begin{align} & \zeta (\omega_k - \omega_0) = \int^{t}_0 d\tau \text{exp}(-i(\omega_k - \omega_0)\tau) \\ & \Longrightarrow \zeta (\omega) = \lim_{\epsilon \to 0^+} \left( \frac{\epsilon}{\epsilon^2 + \omega^2} - i\frac{\omega}{\epsilon^2 + \omega^2} \right) = \pi \delta(\omega) - i P\left(\frac{1}{\omega}\right) \end{align} $$

Where $$P$$ is Cauchy's Principal Part. Finally, we have the final expression for the differential equation:



\dot{c}_{e,0} = - \left( \frac{\Gamma}{2} + i\delta \right) c_{e,0} \quad \Longrightarrow |c_{e,0}(t)|^2 = e^{-\Gamma t}|c_{e,0}(0)|^2 $$

As expected, the probability density to be in the excited state decays as a result of spontaneous emission. The imaginary part of the differential equation leads to a frequency shift which is actually a contribution to the Lamb shift. The variables in the solution are:



\begin{align} & \Gamma = 2\pi \overline{g^2{\omega_k}}D(\omega_0) = \frac{4\omega_0^3}{3\hbar c^3}|\vec{d}_{eg}|^2 \\ & \delta = P \int^{\infty}_0 d\omega_k D(\omega_k) \frac{\overline{|g^2{\omega_k}|}}{\omega_0 - \omega_k} \end{align} $$

This result is consistent with Fermi's Golden Rule since the Einstein A Coefficient here is proportional to the density of states and the magnitude squared of the off-diagonal matrix element of the dipole operator. However, this approximation also gave insight on the energy level shifts resulting from the atom coupling to the quantum electromagnetic vacuum unlike Fermi's Golden Rule. It is important to note that the differential equation demonstrates an irreversible process. This was not the case in the Jaynes Cummings Model because now the atom was coupled to a continuum of modes instead of a single mode (the cavity mode). Because of this coupling, there was a separation of time scales that allowed for the WWA to be done and remove the reversibility of the atom's evolution.

Vacuum Fluctutations vs. Radiation Reaction
An interesting point of view is the Heisenberg picture. The evolution of the annihilation and creation operators is composed of two terms. One of which can be seen as being due to vacuum fluctuations and another due to radiation reaction.



\begin{align} & \frac{d}{dt}\hat{a}_{k,\mu} = \frac{-i}{\hbar} [\hat{a}_{k,\mu},\hat{H}] = -i\omega_k \hat{a}_{k,\mu} - ig^*_{k,\mu}\hat{\sigma}_{-} \\ & \Longrightarrow \hat{a}_{k,\mu}(t) = \hat{a}_{k,\mu}(0) e^{-I\omega_k t} - ig^*_{k,\mu}\int^{t}_0 dt' e^{-1\omega_k(t-t')}\hat{\sigma}_{-}(t') \end{align} $$

The first term doesn't involve the atom at all so it exists even when there is no atom. The second one, however, includes the atom-field coupling term and $$\sigma_{-}$$. It turns out that the vacuum field plays no part in the evolution of the expectation values of $$ \{ \sigma_{+}, \sigma_{z} \}$$ when calculated in normal order. This means that we can explain spontaneous emission mostly with radiation reaction and vacuum fluctuations play a small part. One can note that if the initial state is in the excited state, there is no radiating dipole generating a radiation field. To get the radiation field in the first place, the vacuum field "kicks" the atom so that it starts oscillating on the Bloch sphere. Then, because of radiation reaction the state evolves from the excited state to the ground state. A key part of this interpretation of spontaneous emission is that calculations were done in normal order. Had they been done in "anti-normal" order, then spontaneous emission could have been entirely explained with vacuum fluctuations. It is in "symmetric" order that both effects can be seen simultaneously.

Purcell Effect
From the Purcell Effect, it is known that the rate of spontaneous emission of an atom in a cavity is affected by boundary conditions. It is clear to see this from Fermi's Golden Rule; the quantization volume affects the density of states and therefore changes the rate of spontaneous emission. In the study of the rate of spontaneous emission of a two-level atom in a cavity, the rate of spontaneous emission is:



\Gamma_{can} = \Gamma_{free} F_p\frac{\kappa^2/4}{(\omega_0 - \omega_c)^2 + \kappa^2/4}\frac{|\vec{u}(\vec{R})\cdot\vec{d}_{eg}|^2}{|\vec{d}_{eg}|^2} $$

Where $$ \kappa $$ is the decay rate of the cavity mode, $$ \vec{u}(\vec{R}) $$ is the normalized cavity mode with volume $$ V $$, and $$ F_p = \frac{3\omega_c \lambda_{eg}}{4\pi^2 \kappa V}$$ is the Purcell factor. Therefore, when the cavity is tuned right on resonance, the rate is enhanced significantly. Else, if it is tuned far off resonance, the emission is suppressed. The same result can be obtained by following a similar approach to the one above except the Hamiltonian is non-Hermitian and includes the decay of the cavity mode ($$\kappa$$) and of the atom into all modes not in the cavity ($$\Gamma'$$). The result is that the probability amplitude of being in the excited state satisfies:



\dot{c}_{e,0} = -\left[ \frac{\Gamma'}{2} + \frac{\Gamma_{cav}}{2} -i\frac{\delta E_{cav}}{\hbar} \right] $$