The model of interest is a 2-level system (e.g. an atomic transition) inside a damped single-mode cavity.

The purcell effect states that the atomic decay rate inside the cavity $\Gamma_{cav}$, on resonance, is enhanced compared to the free-space decay rate $\Gamma_{free}$ by the following factor: $\frac{\Gamma_{cav}}{\Gamma_{free}} = \frac{\lambda ^3}{V} \frac{Q}{8 \pi}$ (up to a constant) .

For cavities with large volume, this ratio goes to zero. Even though the atom is resonant with the cavity mode. One could now place an atom inside a macroscopic cavity and prepare an excited state for a very long time, even though the transition to the groundstate is dipole allowed. This is not observed in practice(?) So my question is now, where does the model assume a volume "that is not too large".

In the next paragraphs I present parts of a derivation of the purcell factor. A heuristic argument can be made with Fermis Golden rule, which gives an expression for the decay rate $\Gamma = 2\pi \frac{|\langle f| e\vec{r} \cdot \vec{E}|i \rangle|^2}{\hbar^2}D(w)$.

$\Gamma_{free}$: Because the coupling matrix element scales like 1/V and the density of states scales as $D_{free} $ ~ $ V$, the free space decay rate is independent of the volume. $\Gamma = \frac{8 \pi^2 \epsilon_{0} \vec{d}^2}{3\hbar\epsilon_{0}\lambda^3}$ with $\vec{d} = -\langle f|e\vec{r}|i \rangle$. [Derived more rigorously with a Wigner-Weisskopf treatment in most Quantum Optics textbooks]

An expression for $\Gamma_{cav}$ can also be derived by evaluating Fermis Golden rule. The density of states inside the cavity is volume independent: $D_{cav}(\omega) = \frac{1}{\pi}\frac{\omega_{c} /2Q}{(\omega - \omega_{c})^2 + (\omega_{c} /2Q)^2}$ and integrates to 1 (single mode cavity). Where $Q = \frac{\omega_{c}}{\delta\omega_{c}} $ is the quality factor describing the width of the lorentzian, which is centered around the frequency $\omega_{c}$. If the cavity is tuned near the atomic frequency, then the density of states reads: $D_{cav}(\omega ) \approx 2Q/\pi \omega_{c}$.

For the single mode cavity, the diple interaction term reduces to a Jaynes-Cummings Hamiltonian: $H_{AC} = -e\vec{r}\vec{E} = ( \hat{a}^\dagger + \hat{a})(g\hat{\sigma_{+}} + g^{*}\hat{\sigma_{-}})$, where the coupling is $ g = \vec{d}\frac{ \vec{E} }{|E|} \sqrt{\frac{\hbar \omega_{c}}{2 \epsilon_{0} V}}f(r) $. The factor $f(r)$ is introduced in order to describe spatial variatons of the electric field, it is obtained by solving the classical maxwells equations with appropriate boundary conditions. The quantization volume here is then not the volume of the cavity, but rather the integral $\int_{}^{} f(\vec{r}) \,d\vec{r} $.

A real world cavity has non-zero transmission, which is modelled by a phenomenological Hamiltonian coupling the cavity mode $\hat{a}$ to a reservoir of modes outside the cavity. This quasimode treatment is a good approximation to the real situation under certain conditions, for example the requirement of a high Q-factor [Source: Theory of pseudomodes in quantum optical processes; Dalton, B. J. ; Barnett, Stephen M. ; Garraway, B. M.]. For low Q-cavities, the lorentzian lineshape is broad, and the single mode model with phenomenological coupling to the outside breaks down. My guess is that, similarly for large quantization volume, the energy difference between different discrete modes is very small and so the atom really couples to many modes, even in a very high Q cavity.


1 Answer 1


These are some good observations and a good question! I will give a few hints which may help piecing things together.

  • The mode volume formulation for the Purcell factor is indeed not without approximations as you point out, and some of them are somewhat non-trivial. However, the notion of a mode volume can be made precise, as for example discussed in https://arxiv.org/abs/1312.5769.
  • That said, the Purcell factor can also be calculated without using mode volumes. The precise formula is (see e.g. https://arxiv.org/abs/0902.3586) $$\Gamma = −\mu_0 \omega^2_a \mathrm{Im}[\mathbf{d}^* ·\mathbf{G}(\mathbf{r}_a, \mathbf{r}_a, \omega_a)·\mathbf{d}]\,,$$ where $\omega_a$ is the atom's transition frequency, $\mathbf{d}$ the transition dipole moment, $\mathbf{r}_a$ is the atom's position and $\mathbf{G}(\mathbf{r}, \mathbf{r}, \omega)$ is the Green's function of the cavity. This formula essentially only assume the dipole + rotating wave approximation and weak coupling. In a sense, it can therefore be considered the most precise formula for the Purcell effect, since without these approximations, the atom in the cavity does not only decay (instead you get Rabi oscillations or other fancy stuff). Most importantly, the above formula makes no approximations on the cavity geometry, cavity quality or single mode character.
  • So what happens for large cavities? Your argument is essentially correct, except that the quality factor $Q$ is typically not independent of $V$. Practically, this is what leads to reasonable behavior, since taking the "high $V$-limit" on its own is not meaningful.
  • That said $Q$ and $V$ are not strictly locked to each other. Indeed, different types of cavities realise different regimes. The usually Fabry-Perot cavities with highly reflecting mirrors typically feature high $Q$ and high $V$. Plasmonic cavities on the other hand feature high confinement of the electric field (low $V$) and large losses (low $Q$).
  • Plasmonic cavities sometimes feature another regime where the standard Purcell formula may break down: when the mode volume becomes on the order of the resonant wavelenght.
  • To really understand the high $V$ limit in the sense of physical cavity volume, I would like to suggest an exercise: try to compute the above formula for a simple 1D Fabry-Perot cavity. The Green's function is computable analytically in this case. Then you can play with the cavity parameters at will.
  • $\begingroup$ @DLosc1 thanks for the accept! Fyi, there is a python package that makes it much easier to try the task I suggested in the last point: pypi.org/project/pyrot $\endgroup$ Aug 4, 2022 at 9:48
  • $\begingroup$ This is great. How would you model a similar situation, but this time there is a broadband solid state emitter instead of a simple 2-level system? I posed another question few days ago: "physics.stackexchange.com/questions/721389/…". Maybe you have a good idea. Thanks. $\endgroup$
    – DLosc1
    Aug 8, 2022 at 13:26

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