Solutions of the optical Bloch equations

The optical Bloch equations

${\dot {\rho }}=-{\frac {i}{\hbar }}[H,\rho ]-{\frac {\Gamma }{2}}\left[{\sigma _{+}\sigma _{-}\rho -2\sigma _{-}\rho \sigma _{+}+\rho \sigma _{+}\sigma _{-}}\right]\,,$

provide a time-dependent quantum description of a spontaneously emitting atom driven by a classical electromagnetic field. Considerable insight into the physical processes involved can be gained by studying these equations in the transient excitation limit, as well as the steady-state limit, as we see in this section. We begin by considering the coherent part of the evolution, then extend this to re-visit the Bloch sphere picture of the optical Bloch equations, which provides useful visualizations of transient responses and steady state solutions.

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Eigenstates of the Jaynes-Cummings Hamiltonian

The Hamiltonian involved in the optical Bloch equations,

$H={\frac {\hbar \omega _{0}}{2}}(|e\rangle \langle e|-|g\rangle \langle g|)+{\frac {\hbar \Omega }{2}}(|g\rangle \langle e|+|e\rangle \langle g|)\,,$

describes the evolution of an atom in a classical field. We begin here by reviewing the coherent evolution under $H$.

$H$ arises from the Jaynes-Cummings interaction we have previously considered in the context of cavity QED, describing a single two-level atom interacting with a single mode of the electromagnetic field:

$H_{JC}={\frac {\hbar \omega _{0}}{2}}\left[{|e\rangle \langle e|-|g\rangle \langle g|}\right]+\hbar \omega a^{\dagger }a+{\frac {\hbar \Omega }{2}}\left[{|g\rangle \langle e|+|e\rangle \langle g|}\right]\left[{a+a^{\dagger }}\right]\,.$

The last term in this expression is $H_{I}$, the dipole interaction between atom and field. By defining $\sigma _{+}=|e\rangle \langle g|$ and $\sigma _{-}=|g\rangle \langle e|$, we may write this interaction as

$H_{I}={\frac {\hbar \Omega }{2}}\left[{\sigma _{+}+\sigma _{-}}\right]\left[{a+a^{\dagger }}\right]\,.$

In the frame of reference of the atom and field, recall that

${\begin{array}{rcl}\sigma _{+}a^{\dagger }&\rightarrow &e^{i(\omega +\omega _{0})t}\sigma _{+}a^{\dagger }\\\sigma _{-}a&\rightarrow &e^{-i(\omega +\omega _{0})t}\sigma _{-}a\\\sigma _{+}a&\rightarrow &e^{i(\omega _{0}-\omega )t}\sigma _{+}a\\\sigma _{-}a^{\dagger }&\rightarrow &e^{-i(\omega _{0}-\omega _{0})t}\sigma _{-}a^{\dagger }\,.\end{array}}$

When near resonance, $\omega \approx \omega _{0}$, and because the $\sigma _{+}a^{\dagger }$ and $\sigma _{-}a$ terms oscillate at nearly twice the frequency of $\omega$, those terms can be dropped. Doing so is known as the rotating wave approximation, and it gives us a simplified interaaction Hamiltonian

$H_{I}={\frac {\hbar \Omega }{2}}\left[{a^{\dagger }\sigma _{+}+a\sigma _{-}}\right]\,.$

Under this approximation, it is useful to note that this interaction merely exchanges one quantum of excitation from atom to field, and back, so that the total number of excitations $N=a^{\dagger }a+|e\rangle \langle e|$ is a constant of the motion. We may thus write the total Hamiltonian, in the rotating wave approximation, as

$H=\hbar \omega N+\delta \sigma _{z}+{\frac {\hbar \Omega }{2}}(a^{\dagger }\sigma _{-}+a\sigma _{+})\,,$

where we have defined $\sigma _{z}=|e\rangle \langle e|-|g\rangle \langle g|$, and $\delta =\omega _{0}-\omega$. Below, we may use $g=\hbar \Omega /2$ to simplify writing. What are the eigenstates of this Hamiltonian? It describes a two-level system coupled to a simple harmonic oscillator; when uncoupled, if $\delta =0$, then the eigenstates are simply those of $|e{\rangle }$, $|g{\rangle }$ and $|n{\rangle }$, as shown here:

When coupled, degenerate energy levels split (this is sometimes called dynamimc Stark splitting), with harmonic oscillator levels $|n{\rangle }$ and $|n+1{\rangle }$ splitting into two energy levels separated by $2g{\sqrt {n+1}}$. Since the coupling only pairs levels separated by one quantum of excitation, it is straightforward to show that the eigenstates of the Jaynes-Cummings Hamiltonian fall into well defined pairs of states, which we may label as $|\pm ,n{\rangle }$; these are

$|\pm ,n\rangle ={\frac {1}{\sqrt {2}}}\left[{|e,n\rangle \pm |g,n+1\rangle }\right]\,,$

and they have energies

$E_{\pm ,n}=\hbar \omega (n+1)\pm g{\sqrt {n+1}}\,.$

When $\delta \neq 0$, similar physics result, but with slightly more complicated expressions describing the eigenstates, as we shall see when we later return to the "dressed states" picture.

Strongly driven atom: Mollow triplet

An atom strongly coupled to a single mode electromagnetic field, or an atom driven strongly by a single mode field, will thus have an emission spectrum described by the coupled energy level diagram:

where, to good approximation, the energy level differences are $\omega$ and $\omega \pm \Omega$. These three lines which appear in the spectrum are known as the Mollow triplet:

The Mollow triplet is experimentally observed in a wide variety of systems. However, while our energy eigenstate analysis has predicted the number and frequencies of the emission lines, it fails to explain a key characteristic: the widths are not the same. If the central peak at $\omega _{0}$ has width $\Gamma /2$, the two sidebands each have a width of $3\Gamma /4$. To explain this, we need the optical Bloch equations.

Optical Bloch equation evolution on the Bloch sphere

We have previously seen that an arbitrary qubit state $|\psi \rangle =\cos \theta |0\rangle +e^{i\phi }\sin \phi |1{\rangle }$ can be represented as being a point on a unit sphere, located at $(\theta ,\phi )$ in polar coordinates. Similarly, a density matrix $\rho$ may be depicted as being a point inside or on the unit sphere, using

$\rho ={\frac {I+{\vec {r}}\cdot {\sigma }}{2}}\,,$

where ${\vec {r}}$ is the Bloch vector representation of $\rho$.

Explicitly, if we let

$\rho =\left[{\begin{array}{cc}{\rho _{ee}}&{\rho _{eg}}\\{\rho _{ge}}&{\rho _{gg}}\end{array}}\right]\,,$

then

${\begin{array}{rcl}r_{x}&=&\langle X\rangle =\rho _{ge}+\rho _{eg}\\r_{y}&=&\langle Y\rangle =(\rho _{ge}-\rho _{eg})/i\\r_{z}&=&\langle Z\rangle =\rho _{gg}-\rho _{ee}\,.\end{array}}$

Visulization of the evolution of a density matrix under the optical Bloch equations is thus helped by rewriting them in terms of a differential equation for ${\vec {r}}$. A convenient starting point for this is the optical Bloch equation

${\dot {\rho }}=-{\frac {i}{\hbar }}[H,\rho ]-{\frac {\Gamma }{2}}\left[{\sigma _{+}\sigma _{-}\rho -2\sigma _{-}\rho \sigma _{+}+\rho \sigma _{+}\sigma _{-}}\right]\,,$

using the rotating frame Hamiltonian (suppressing $\hbar$)

$H={\frac {\delta }{2}}Z+gX\,.$

This gives us the equations of motion

${\begin{array}{rcl}{\dot {r}}_{x}&=&\delta \,r_{y}-{\frac {\Gamma }{2}}r_{x}\\{\dot {r}}_{y}&=&-\delta \,r_{x}-g\,r_{z}-{\frac {\Gamma }{2}}r_{y}\\{\dot {r}}_{z}&=&g\,r_{y}-\Gamma (r_{z}-1)\,.\end{array}}$

Note how these equations of motion provide a simple set of flows on the Bloch sphere: the $\delta$ terms correspond to a rotation in the ${\hat {x}}-{\hat {y}}$ plane, $g$ corresponds to a rotation in the ${\hat {y}}-{\hat {z}}$ plane, and $\Gamma$ drives a relaxation process which shrinks ${\hat {x}}$ and ${\hat {y}}$ components of the Bloch vector, while moving the ${\hat {z}}$ component toward $r_{z}=1$.

Physically, what is the meaning of $r_{x}$, $r_{y}$, and $r_{z}$? $r_{z}$ is manifestly the population difference between the excited and ground states. The other two components may be interpreted by recognizing that the average dipole moment of the atom is

${\begin{array}{rcl}\langle d\rangle &=&d_{ge}{\rm {Tr}}(\rho (|e\rangle \langle g|+|g\rangle \langle e|))\\&=&d_{ge}(\rho _{ge}e^{i\omega t}+\rho _{eg}e^{-i\omega t})\\&=&d_{ge}(r_{x}\cos \omega t+r_{y}\sin \omega t)\,.\end{array}}$

Thus, $r_{x}$ and $r_{y}$ correspond to the phase components of the atomic dipole moment which are in-phase and in quadrature with the incident electromagnetic field.

Solutions of the optical Bloch equations

The optical Bloch equations, describing the evolution of a single atom coupled to the vacuum,

${\dot {\rho }}=-{\frac {i}{\hbar }}[H,\rho ]-{\frac {\Gamma }{2}}\left[{\sigma _{+}\sigma _{-}\rho -2\sigma _{-}\rho \sigma _{+}+\rho \sigma _{+}\sigma _{-}}\right]\,,$

are a set of three coupled differential equations,

${\begin{array}{rcl}{\dot {r}}_{x}&=&\delta \,r_{y}-{\frac {\Gamma }{2}}r_{x}\\{\dot {r}}_{y}&=&-\delta \,r_{x}-g\,r_{z}-{\frac {\Gamma }{2}}r_{y}\\{\dot {r}}_{z}&=&g\,r_{y}-\Gamma (r_{z}-1)\,,\end{array}}$

written in the rotating frame of the atom's Hamiltonian. These equations can be solved analytically, but a great deal of intuition can be obtained from just studying the solutions in several limits. Here, we consider the steady state, transient, and weak excitation limits.

Transient response of the optical Bloch equations

The optical Bloch equations allow us to study the internal state of the atom as it changes due to the external driving field, and due to spontaneous emission.

Starting from the time-independent form of the equations,

${\begin{array}{rcl}{\dot {r}}_{x}&=&\delta \,r_{y}-{\frac {\Gamma }{2}}r_{x}\\{\dot {r}}_{y}&=&-\delta \,r_{x}-g\,r_{z}-{\frac {\Gamma }{2}}r_{y}\\{\dot {r}}_{z}&=&g\,r_{y}-\Gamma (r_{z}-1)\,,\end{array}}$

we may note that when $g\rightarrow 0$ and at resonance, $\delta =0$, the Bloch vector exhibits pure damping behavior, towards $r_{z}=1$, and $r_{x}=r_{y}=0$.

When $g\gg \Gamma$, Rabi oscillations occur, represnted by rapid rotations of the Bloch vector about ${\hat {x}}$. Since the relaxation along ${\hat {y}}$ occurs at rate $\Gamma /2$, and the relaxation about ${\hat {z}}$ occurs at rate $\Gamma$, we might expect that the average relaxation rate of the rotating components under such a strong driving field would be $(\Gamma +\Gamma /2)/2=3\Gamma /4$. The remaining component $r_{x}$ does not rotate, because it sits along ${\hat {x}}$, the axis of rotation. Thus, it relaxes with rate $\Gamma /2$. Computation of the eigenvalues of the equations of motion verify this qualitative picture, and show that for $g\gg \Gamma$, and $\delta =0$, the eigenvalues of motion are $\pm ig+3\Gamma /4$ and $\Gamma /2$.

These correspond to a main peak at $\omega _{0}$ with width $\Gamma /2$, and two sidebands at $\omega _{0}\pm \Omega$, with widths $3\Gamma /4$, thus explaining the widths of the observed Mollow triplet lines.

Steady-state solution of the optical Bloch equations

The steady state solution of the optical Bloch equations are found by setting all the time derivatives to zero, giving a set of three simultaneous equations,

${\begin{array}{rcl}0&=&\delta \,r_{y}-{\frac {\Gamma }{2}}r_{x}\\0&=&-\delta \,r_{x}-g\,r_{z}-{\frac {\Gamma }{2}}r_{y}\\0&=&g\,r_{y}-\Gamma (r_{z}-1)\,.\end{array}}$

The solutions are (up to overall minus signs which can be absorbed into definitions):

${\begin{array}{rcl}r_{x}&=&(g\delta ){\frac {1}{\delta ^{2}+g^{2}/2+(\Gamma /2)^{2}}}\\r_{y}&=&\left({\frac {g\Gamma }{2}}\right){\frac {1}{\delta ^{2}+g^{2}/2+(\Gamma /2)^{2}}}\\r_{z}&=&\left(\delta ^{2}+{\frac {\Gamma ^{2}}{4}}\right){\frac {1}{\delta ^{2}+g^{2}/2+(\Gamma /2)^{2}}}\end{array}}$

Physically, these are Lorentzians; the $r_{y}$ solution (the component in quadrature with the dipole) corresponds to an absorption curve with half-width

${\sqrt {{\frac {\Gamma ^{2}}{4}}+{\frac {g^{2}}{2}}}}\,,$

and the $r_{x}$ solution (the component in-phase with the dipole) corresponds to a dispersion curve. And under a strong driving field, as $g\rightarrow \infty$, $r_{z}\rightarrow 0$, indicating that the populations in the excited and ground states are equalizing. The steady-state population in the excited state is

$\rho _{ee}={\frac {1-r_{z}}{2}}={\frac {g^{2}/4}{\delta ^{2}+g^{2}/2+\Gamma ^{2}/4}}\,,$

an important result that will later be used in studying light forces.

These solutions can be re-expressed in a simplified manner by defining the saturation parameter

$s={\frac {g^{2}/2}{\delta ^{2}+\Gamma ^{2}/4}}\,,$

in terms of which we find

${\begin{array}{rcl}r_{x}&=&{\frac {2\delta }{g}}{\frac {s}{1+s}}\\r_{y}&=&{\frac {\Gamma }{g}}{\frac {s}{1+s}}\\r_{z}&=&{\frac {1}{1+s}}\,.\end{array}}$

As $s\rightarrow \infty$, the atomic transitions become saturated, and the linewidth of the transition broadens from its natural value $\Gamma$, becoming $\Gamma '=\Gamma {\sqrt {1+s}}$ on resonance, at $\delta =0$.

Weak excitation limit solution of the optical Bloch equations

Once again, the time-independent form of the optical Bloch equations is

${\begin{aligned}{\dot {\rho }}_{ee}&=i{\frac {\Omega }{2}}(\rho _{eg}-\rho _{ge})-\Gamma \rho _{ee}\\{\dot {\rho }}_{ge}&=i(\omega _{0}-\omega _{L})\rho _{ge}-i{\frac {\Omega }{2}}(\rho _{ee}-\rho _{gg})-{\frac {\Gamma }{2}}\rho _{ge}\,,\end{aligned}}$

where the remaining two components of the density matrix are given by $\rho _{gg}=1-\rho _{ee}$, and $\rho _{eg}=\rho _{ge}^{*}$. It is insightful to study these equations in the limit of weak excitation, and for short evolution times.

It turns out that the solution of these equations to lowest order in $|\Omega |$, and in the limit $|\Omega |\ll \Gamma$, with the initial conditions $\rho _{ee}=0$ and $\rho _{ge}=0$, gives

$\rho _{ee}={\frac {{\frac {1}{4}}|\Omega |^{2}}{(\omega _{0}-\omega _{L})^{2}+\left({\frac {\Gamma }{2}}\right)^{2}}}\left[{1+e^{-\Gamma t}-2\cos[(\omega _{0}-\omega _{L})t]e^{-\Gamma t/2}}\right]\,.$

Compare this with the solution of the Jaynes-Cummings interaction we previously obtained,

$\rho _{ee}-\rho {gg}=\sum _{n}\alpha _{n}^{2}\left[{{\frac {\delta ^{2}}{\Omega _{n}^{2}}}+{\frac {4g^{2}(n+1)}{\Omega _{n}^{2}}}\cos(\Omega _{n}t)}\right]\,.$

Moreover the solution of these equations to lowest order in $|\Omega |$ in the limit $|\Omega |t\ll 1$, with the initial conditions $\rho _{ee}=0$ and $\rho _{ge}=0$, gives

$\rho _{ee}={\frac {1}{4}}|\Omega |^{2}t^{2}\,,$

irrespective of the values of $(\omega _{0}-\omega _{L})$ and $\Gamma$.

Damped vacuum Rabi oscillations

The optical Bloch equations can be generalized to describe not just an atom interacting with the vacuum, but also an atom and a single cavity mode, each interacting with its own reservoir. This is the master equation for cavity QED, and using such a master equation we can revisit the phenomenon of vacuum Rabi oscillations and see what happens in the presence of damping.

Generalization of the optical Bloch equations

The starting point for generalizing the optical Bloch equations is the Lindblad form we previously saw in the full-derivation walkthrough,

${\dot {\rho }}_{A}=-{\frac {i}{\hbar }}[H,\rho _{A}]-{\frac {\Gamma }{2}}\left[{\sigma _{+}\sigma _{-}\rho _{A}-2\sigma _{-}\rho _{A}\sigma _{+}+\rho _{A}\sigma _{+}\sigma _{-}}\right]\,.$

In this expression, $\sigma _{+}$ and $\sigma _{-}$ are "jump operators," and represent changes that occur to the atom when distinct dissipative events happen ("distinct" meaning that the environment changes between orthogonal states).

We may write the master equation for our more general scenario by replacing the atomic density matrix $\rho _{A}$ by a general density matrix $\rho$ representing the atom and cavity field, and by replacing the atomic jump operators with general jump operators $L_{k}$,

${\dot {\rho }}=-{\frac {i}{\hbar }}[H,\rho ]+\sum _{k}\left[{L_{k}\rho L_{k}^{\dagger }-{\frac {1}{2}}(L_{k}^{\dagger }L_{k}\rho +\rho L_{k}^{\dagger }L_{k})}\right]\,.$

Note that the $L_{k}$ include normalization factors which reflect their probabilities of occurrence. In other words, for the atom + vacuum model, $L={\sqrt {\Gamma }}\sigma _{-}$.

Master equation for cavity QED system

For the cavity QED model, the atom and cavity field each have possible jump operators. In general, the atom and cavity may both couple to a thermal field with $n_{th}$ average photons. In such a case, the jump operators are $L_{1}={\sqrt {\Gamma (n_{th}+1)}}\sigma _{-}$, $L_{3}={\sqrt {\Gamma n_{th}}}\sigma _{+}$, $L_{2}={\sqrt {\kappa (n_{th}+1)}}a$, and $L_{4}={\sqrt {\kappa n_{th}}}a^{\dagger }$, where $\Gamma$ parameterizes the spontaneous emission rate of the atom in free space, and $\kappa$ is parameterizes the cavity $Q$ factor.

Experimentally, typically the environment, the vacuum, is essentially at zero temperature, so $n_{th}=0$, in which case the only two relevant jump operators are $L_{1}={\sqrt {\Gamma }}\sigma _{-}$ and $L_{2}={\sqrt {\kappa }}a$.

Damped vacuum Rabi oscillations

Vacuum Rabi oscillations, in the absence of damping, involve only two states of the atom and cavity: $|e,0{\rangle }$ and $|g,1{\rangle }$. When damping is added, the $|g,0{\rangle }$ state must be included, since both the atom and cavity states can decay and loose their quanta of energy. Moreover, because only one quantum of excitation is involved in this system, we can observe the essential physics by considering the case when $\Gamma$ is zero (no spontaneous emission), but $\kappa$ is nonzero (the cavity is leaky). Let $\Omega _{0}$ be the vacuum Rabi frequency, and denote this three state space by $|1{\rangle }=|e,0{\rangle }$, $|2{\rangle }=|g,1{\rangle }$, and $|3{\rangle }=|g,0{\rangle }$. Written out explicitly in terms of the $3\times 3$ density matrix elements $\rho _{jk}$, the master equation is (for the case $\delta =0$ with the system starting in the state $|1\rangle$)

${\begin{array}{rcl}{\dot {\rho }}_{11}&=&i{\frac {\Omega _{0}}{2}}(\rho _{12}-\rho _{21})\\{\dot {\rho }}_{22}&=&-i{\frac {\Omega _{0}}{2}}(\rho _{12}-\rho _{21})-\kappa \rho _{22}\\{\dot {\rho }}_{12}&=&i{\frac {\Omega _{0}}{2}}(\rho _{11}-\rho _{22})-{\frac {\kappa }{2}}\rho _{12}\\{\dot {\rho }}_{21}&=&-i{\frac {\Omega _{0}}{2}}(\rho _{11}-\rho _{22})-{\frac {\kappa }{2}}\rho _{21}\\{\dot {\rho }}_{33}&=&\kappa \rho _{22}\,.\end{array}}$

When the cavity damping rate is small, $\kappa \ll \Omega _{0}$, then the vacuum Rabi oscillations are damped, with average damping rate $\kappa /2$ since the atoms spend roughly half their time in the excited state.

When the cavity damping rate is large, $\kappa \gg \Omega _{0}$, then the atomic excitation is irreversibly damped, and no oscillations occur. Let $P_{e}$ be the probability of being in the $|1\rangle =|e,0{\rangle }$ state. We can combine two of the above equations to write

${\begin{array}{rcl}{\dot {\rho }}_{12}-{\dot {\rho }}_{21}&=&i\Omega _{0}(\rho _{11}-\rho _{22})-{\frac {\kappa }{2}}(\rho _{12}-\rho _{21})\,.\end{array}}$

Since in this case ${\dot {\rho }}_{12}+{\dot {\rho }}_{21}\approx 0$, and $\rho _{22}\approx 0$, it follows that

$(\rho _{12}-\rho _{21})\approx {\frac {2i\Omega _{0}}{\kappa }}(\rho _{11}-\rho _{22})\,,$

Since in this regime, we slowly try to populate state $|2{\rangle }$ while it quickly decays away, we can approximate $\rho _{22}\ll \rho _{11}$ giving

$(\rho _{12}-\rho _{21})\approx {\frac {2i\Omega _{0}}{\kappa }}\rho _{11}\,,$

Finally plugging this back into the optical bloch equation for $\rho _{11}$ yields

${\dot {\rho }}_{11}=-{\frac {\Omega _{0}^{2}}{\kappa }}\rho _{11}\,,$

so $P_{e}$ decays exponentially, with rate $\Gamma _{c}=\Omega _{0}^{2}/\kappa$. Note that a large cavity damping rate slows down the decay of the atom's excited state. This is due to the fact that the stimulated transition (Rabi oscillation) to the lower state is slowed down by the cavity damping (which is a manifestation of the quantum Zeno effect).

Purcell factor: cavity enhanced spontaneous emission

How does this compare with the free space spontaneous emission rate $\Gamma$? Recall that the vacuum Rabi frequency is (for the case that the dipole matrix element vector and cavity mode polarization are aligned)

$\Omega _{0}={\frac {dE_{0}}{\hbar }}\,,$

where $d$ is the atomic dipole moment and $E_{0}$ is the electric field amplitude of a single photon at the atomic transition frequency $\omega _{0}$ (Note that this is a standing wave in a cavity so it is a factor of two larger than the field for a running wave),

$E_{0}={\sqrt {\frac {2\hbar \omega _{0}}{\epsilon _{0}V}}}\,,$

and $V$ is the cavity volume. This gives

$\Omega _{0}=d{\sqrt {\frac {2\omega _{0}}{\epsilon _{0}\hbar V}}}\,.$

Letting $Q=\omega _{0}/\kappa$, we thus find that

$\Gamma _{c}={\frac {2d^{2}}{\epsilon _{0}\hbar }}{\frac {Q}{V}}\,$

as the decay rate of the atom in the cavity.

Recall that the spontaneous emission rate of an atom in free space, as determined by Fermi's golden rule, is

$\Gamma ={\frac {d^{2}\omega _{0}^{3}}{3\pi \epsilon _{0}\hbar c^{3}}}$

The ratio of this rate to the decay rate in the cavity is

$\eta ={\frac {\Gamma _{c}}{\Gamma }}={\frac {3}{4\pi ^{2}}}{\frac {Q\lambda ^{3}}{V}}\,,$

where we take $\lambda =2\pi c/\omega _{0}$ as being the wavelength of the cavity field, which is assumed to be resonant with the atomic transition frequency $\omega _{0}$. Note that $\eta$ is independent of the atomic dipole strength, and determined solely by cavity parameters. Moreover, note that for small, high-$Q$ cavities, the decay rate of the atom in the cavity can be much larger than the free space spontaneous emission rate. This "cavity enhanced" spontaneous emission rate was predicted by Purcell (1946), an observation credited as being the starting point of cavity QED.