In the derivation of Spontaneous Emission in two level systems in Quantum Optics (be it Wigner Weisskopf or a different approach, such as density operators to find the master equation), one makes (several) assumptions. One of the more prominent ones here is the Markov approximation, which I think is most easily described in the density operator context. Here one assumes that to find the state of the atomic system $S$ at time $t$, one does not have to integrate $\rho_s(t')$ from $0$ to $t$, but instead can just take it to be $\rho_s(t)$. The way I see it, this means that the system does not have a memory of what has happened to it before time $t$. But what I do not completely understand is what is required for this to hold.

What I can find from other sources is that it requires 'a broad range of frequencies' to be present. Why is this the case? I read one hypothetical scenario in which one would construct a photonic crystal such that the density of modes would be 0 up to $\omega_{eg}$ (the transition of the two level system), and constant afterwards. The writer then claimed that the Markov approximation would not hold, because we have a real difference between lower and higher than the emitter frequency, meaning that the emitter has a real memory of what has happened to it. I personally do not understand this line of reasoning, but I assume it to be true. Could someone explain why this is so, and perhaps also elaborate on the conditions the Markov approximation requires?

Edit: Perhaps another example to illustrate what I don't understand: another source writes that if you were to put your emitter in a bandgap, so that the only radiation it could couple to would be in the range $\omega_{eg}-\gamma,\omega_{eg}+\gamma$, spontaneous emission would also not occur. I really do not see why not; why do we need to have it couple to so many different modes of the field?

  • 2
    $\begingroup$ Let me test the waters here: if I say that the atom is coupled to exactly one electromagnetic mode (e.g. because it is inside a Fabrey-Perot cavity) then an excitation initially in the atom oscillates between the atom and cavity in the so-called "vacuum Rabi oscillation". There's no decay because the excitation comes back to the atom after one oscillation period. Does that answer your question or should I go on? $\endgroup$
    – DanielSank
    Feb 2, 2015 at 16:45
  • 1
    $\begingroup$ Okay, so we have a vacuum Rabi oscillation with the field in the vacuum, and Rabi flopping with frequency $g_c \sqrt{n+1}$ with n photons in the mode $c$. Then if you have a lot of different modes, they will all cause oscillations with different frequencies, which destructively interfere, hence causing collapse. Is that what you mean? I can see how that causes the spontaneous emission to happen. However, I don't see how that for example excludes the case I talked about above, where you only have modes above the atomic transition frequency? $\endgroup$
    – user129412
    Feb 2, 2015 at 16:54
  • 2
    $\begingroup$ @DanielSank I believe there are two related but distinct points to be made here. The first is that a continuum of modes is required in order to prevent persistent oscillations, i.e. one needs an infinite number of modes so that the Poincare recurrence time is infinite. The second is that the spectral density of these modes must vary slowly over the relevant frequency range so that there are no oscillations at all on the coarse-grained timescale. $\endgroup$ Feb 2, 2015 at 17:00
  • $\begingroup$ @MarkMitchison: Yes, you are absolutely correct. I just wanted to see if user129412 and I were on the same page. Apparently we are, and the only remaining question is the second one you mention in your comment. $\endgroup$
    – DanielSank
    Feb 2, 2015 at 17:02
  • 1
    $\begingroup$ @DanielSank :) Well I was planning on writing an answer up today, but I have to get some actual research done first! $\endgroup$ Feb 2, 2015 at 17:05

1 Answer 1


A short, mathematical answer to the question is found in the properties of Fourier transforms. The temporal response of the environment to a perturbation is given by the Fourier transform of its frequency response to the same perturbation. Therefore, if a broad range of frequencies in the bath are perturbed, the response occurs over a narrow range of times. Let me try to briefly explain how this mathematical structure arises from the physics.

Spontaneous emission can be understood from the following hand-wavy arguments. The electron in an excited state produces an electric field. This field fluctuates over time; these fluctuations drive the transitions in the electronic state. The spontaneous emission therefore arises from the effect of the electron on its environment, which in turn produces a back-action that affects the electron.

The response of the electromagnetic field to a perturbation $\mathbf{E} = E\hat{\mathbf{z}}$ (I have arbitrarily chosen polarisation in the $z$ direction) is captured by the response function: $\Gamma(t) = \langle E(t) E(0)\rangle,$ where $E(t)$ denotes the Heisenberg-picture operator. This function is central to the theory of linear response to a small perturbation. For example, if one introduces a classical electric dipole that oscillates with a time-dependent dipole moment $d(t)$, the resulting electric field at that point is given by the convolution $$ \delta\langle E(t) \rangle \approx -i\int_0^t\mathrm{d}s\; d(t-s) \Gamma(s).$$

The previous paragraphs serve merely to motivate the appearance of the response function $\Gamma(t)$. In a physically realistic case where we have a quantum dipole (e.g. an atom) with two states separated by a frequency $\epsilon$, the response function determines the rate of spontaneous emission, which is proportional to the quantity: $$ \gamma(t) \sim \int_0^t\mathrm{d}s\; e^{i\epsilon s} \Gamma(s). $$ Assuming that $\Gamma(s)$ decays much more rapidly than $1/\epsilon$, for times $t\gg 1/\epsilon$ we can make the Markov approximation $$ \gamma(t) \to \gamma = \int_0^{\infty}\mathrm{d}s\; e^{i\epsilon s} \Gamma(s), $$ so that we effectively have a constant spontaneous emission rate over time, leading to pure exponential decay.

When does $\Gamma(s)$ decay rapidly enough so that we can make the Markov approximation? The electric field $E(t)$ contains many components (normal modes) which oscillate at different frequencies. If we make this decomposition we get a Fourier representation like $$\Gamma(t) = \int_0^\infty\mathrm{d}\omega\; e^{-i\omega t} J(\omega),$$ where the spectral density $J(\omega)$ quantifies the degree to which the field at frequency $\omega$ is perturbed by a dipole. For an atom interacting with the electromagnetic field in free space, you will normally get something like $$J(\omega) \sim \lambda \frac{\omega^3}{\omega_c^2}e^{-\omega/\omega_c}.$$ Here $\lambda$ is a small dimensionless coupling parameter, and $\omega_c$ is a large frequency cutoff on the order of $c/a_0$, where $a_0$ is the Bohr radius. From purely dimensional arguments you can see that $$ \Gamma(t) \sim \frac{\lambda \omega_c^2}{(\omega_c t)^4}. $$ This tells you that $\Gamma(t)$ vanishes after times much bigger than $\tau = 1/\omega_c$. This time $\tau$ is called the memory time. Since here $\omega_c \approx 10^{18} \text{Hz}$, while the typical optical frequencies are $\epsilon \approx 10^{14} \text{Hz}$, the Markov approximation is well justified.

The extreme example of Markovian noise (white noise) corresponds to $J(\omega) = \text{const.}$, in which case $\Gamma(t) = \delta(t)$, i.e. the bath memory time is infinitesimally small. The opposite extreme is something like a photonic crystal, where the environment has a sharp band edge at frequency $\Omega$ where $J(\omega)$ goes to zero. In that case the response function ends up something like $$ \Gamma(t) = \int_0^\Omega e^{-i\omega t} J(\omega) \sim f(t) e^{-i\Omega t}$$ where $f(t)$ is some function of time. Now if $\Omega$ is comparable to $\epsilon$, you can imagine that there will be resonance effects, and there will be no smooth irreversible transfer of energy into the environment. Rather $\gamma(t)$ becomes a complicated function of time and you will see non-Markovian dynamics. If the frequency $\epsilon$ lies deep within the band-gap then there is no spontaneous emission at all, since there are simply no electromagnetic field modes to couple to, i.e. effectively $J(\omega) = 0$ in the relevant frequency range.

Hopefully these examples should convince you that the frequency scale which sets the bandwidth of perturbed modes in the environment ($\omega_c$, $\Omega$) is on the order of the inverse memory time. Thus, large bandwidths correspond to shorter memory times, i.e. more Markovian environments.

DISCLAIMER: All equations given here are based purely on memory and minimal back-of-the-envelope consistency checks. The proportionality factors and various terms which I arbitrarily deemed irrelevant are almost certainly missing.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.