Problem: Before developing Fermi’s golden rule, it is necessary to first lay out the general framework of time-dependent perturbation theory, of which Fermi’s golden rule is a special case. To this effect, begin by considering the usual (Schrodinger picture) Hamiltonian decomposition \(H=H_0+V\), and obtain the “fundamental theorem of (\(1^{\text{st}}\)-order) time-dependent perturbation theory” for the transition amplitude between an initial \(H_0\)-eigenstate \(|i\rangle\) at \(t=0\) and a final (Schrodinger picture) \(H_0\)-eigenstate \(|f\rangle\) with distinct energy \(E_f\neq E_i\) as a function of the elapsed time \(t\):
\[\langle f|\mathcal T\exp\left(-\frac{i}{\hbar}\int_0^tdt’H(t’)\right)|i\rangle\sim h^{-1}V_{fi}(\omega_{fi})*_{\omega_{fi}}e^{i\omega_{fi}t/2}t\text{sinc}\frac{\omega_{fi}t}{2}\]
where \(\hbar\omega_{fi}:=E_f-E_i\), \(V_{fi}(\omega_{fi}):=\langle f|V(\omega_{fi})|i\rangle\) and the linear response Fourier transform convention is being used:
\[V(\omega):=\int_{-\infty}^{\infty}dte^{i\omega t}V(t)\]
Solution: First one has to establish the following exact decomposition lemma for the time evolution operator under the perturbed Hamiltonian \(H\):
\[\mathcal T\exp\left(-\frac{i}{\hbar}\int_0^tdt’H(t’)\right)=\exp\left(-\frac{iH_0t}{\hbar}\right)\mathcal T\exp\left(-\frac{i}{\hbar}\int_0^tdt’V_{/H_0}(t’)\right)\]
which makes sense conceptually as that is what the interaction picture is all about (to prove it, show that both sides obey the same \(1^{\text{st}}\)-order ODE \(i\hbar\dot U=HU\) with same initial condition \(U(0)=1\)). Now then, upon taking the matrix element of both sides in the \(H_0\)-eigenbasis, the free evolution piece \(\exp\left(-\frac{iH_0t}{\hbar}\right)\) simply contributes an irrelevant phase \(e^{-iE_ft/\hbar}\) when acting on the bra \(\langle f|\) since of course \(H_0\) cannot promote transitions between its own eigenstates; later when taking the mod-square of the transition amplitude to obtain a transition probability, this will drop out.
At this point, one employs the key step of (\(1^{\text{st}}\)-order) time-dependent perturbation theory, which is to Dyson-expand the time-ordered exponential to \(1^{\text{st}}\)-order (the word “order” here is being used in \(2\) different senses!):
\[\sim \langle f|1-\frac{i}{\hbar}\int_0^tdt’V_{/H_0}(t’)|i\rangle\]
Since \(E_f\neq E_i\) and \(H_0\) is Hermitian, it follows \(\langle f|i\rangle=0\) are orthogonal so the transition amplitude becomes:
\[\sim\hbar^{-1}\int_0^tdt’\langle f|V_{/H_0}(t’)|i\rangle=\hbar^{-1}\int_0^tdt’e^{i\omega_{fi}t}\langle f|V(t’)|i\rangle\]
(aside: why not just Dyson-expand the original time-evolution operator \(\mathcal T\exp\left(-\frac{i}{\hbar}\int_0^tdt’H(t’)\right)\)? Answer: because it is \(V\) that is weak, not \(H=H_0+V\)! This simple observation motivates the whole point of using the interaction picture in the first place!)
At this point, the integral is recognized as a \([0,t]\)-windowed Fourier transform of \(\langle f|V(t)|i\rangle\), so by interpreting \(\int_0^tdt’=\int_{-\infty}^{\infty}dt'[0\leq t’\leq t]\) with a top-hat filter and using the convolution theorem, one arrives at the stated result (note that with this F.T. convention, there is a factor of \(2\pi\) in the convolution theorem that makes \(\hbar\mapsto h=2\pi\hbar\)).
Thus, the key conceptual corollary of all this is that the only possible \(H_0\)-eigenstate transitions \(|i\rangle\mapsto |f\rangle\) are between those coupled by the perturbation \(V(t)\) in the sense that its spectrum \(V_{fi}(\omega_{fi})\neq 0\) has some support there.
Problem: Suppose that \(V(t)=V_0\) is time-independent. Show that the transition probability between arbitrary \(H_0\)-eigenstates \(|i\rangle,|f\rangle\) is given by:
\[P_{fi}(t)=\left(\frac{2|\langle f|V_0|i\rangle|}{\hbar\omega_{fi}}\sin\frac{\omega_{fi}t}{2}\right)^2\]
Solution: This one leads to a “Fraunhofer single-slit”:
Problem: Instead of just a single DC component at \(\omega=0\), now let the spectrum of the perturbation \(V\) be monochromatic at some frequency \(\omega\neq 0\); due to Hermiticity requirements, such a perturbation must be of the form:
\[V(t)=V_0e^{i\omega t}+V_0^{\dagger}e^{-i\omega t}\]
leading to a Hermitian “Fraunhofer double-slit” spectrum:
\[V(\omega_{fi})=2\pi(V_0\delta(\omega_{fi}+\omega)+V_0^{\dagger}\delta(\omega_{fi}-\omega))\]
In this case, what does the transition probability become?
Solution:
which correspond to the processes of stimulated absorption and emission.
Problem: Using the mathematical identity:
\[\lim_{t\to\infty}\text{sinc}^2(\omega t)=\frac{\pi}{t}\delta(\omega)\]
establish Fermi’s golden rule:
\[P_{fi}(t)\propto t\]
Solution:
Problem: What is Fermi’s useful golden rule for stimulated absorption?
Solution: Instead of a discrete final \(H_0\)-eigenstate \(|f\rangle\), there should be a continuum (or in practice a quasi-continuum) of final states described by a density of \(H_0\)-eigenstates \(g_{f,H_0}(E_f)\). The useful form of Fermi’s golden rule for the transition rate into any of these energetically compatible \(H_0\)-eigenstates is then:
\[P_{fi}(t)\approx\frac{2\pi|\langle f|V_0|i\rangle|^2}{\hbar}g_{f,H_0}(E_f=E_i+\hbar\omega)\]
where a factor of \(\hbar\) has been (optionally) absorbed to have an energy rather than angular frequency argument in the density of states. Of course, a similar useful form of Fermi’s golden rule holds for stimulated emission.
Problem: What are the \(3\) key assumptions of Fermi’s golden rule (FGR)?
Solution: The \(3\) key assumptions are:
Assumption #\(1\): Weak coupling \(V(t)\) (several ways to formalize this).
Assumption #\(2\): Continuum (or in practice quasicontinuum) of final scattering \(H_0\)-eigenstates to transition into, hence described by a density of states \(g_{f,H_0}(E_f)\).
Assumption #\(3\): Intermediate time window (this leads to the idea of the emergence and subsequent breakdown of FGR as a function of time \(t\)).
In most quantum mechanics textbooks, Assumption #\(1\) tends to be emphasized clearly by the fact that one is doing (time-dependent) perturbation theory. Assumption #\(2\) is present in Fermi’s useful golden rule, so in that sense is a bit optional. Finally, Assumption #\(3\) tends to be obscured, and sometimes omitted entirely, but is absolutely critical.
Problem: Elaborate on the importance of Assumption #\(3\).
Solution: The fact that there should be a lower bound on \(t\) makes sense, as enough time needs to pass for the \(\text{sinc}^2\) to actually look like the \(\delta\) as one of the key steps above. However, at the same time the notion of \(t\to\infty\) is a bit artificial, and for one clearly cannot be strictly correct since otherwise the transition probability \(P_{fi}(t)\propto t\) would eventually exceed \(1\).
More quantitatively, if one (optionally) defines a notion of Rabi frequency in the usual way \(\hbar\Omega_{fi}/2:=|\langle f|V_0|i\rangle|\), then it’s clear from the earlier formula for the transition probability that, taking the \(t\to 0\) limit rather than \(t\to\infty\), it instead takes off quadratically in time \(t\) as \(P_{fi}(t)=\Omega_{fi}^2t^2/4\) before the emergence of the linear-in-\(t\) FGR regime.
Meanwhile, in the large-\(t\) limit, if one drives too strongly then of course perturbation theory breaks down and instead one would get Rabi oscillations or something like that. So the exact regime of validity of FGR is a much richer and subtler topic than it seems at first. See the paper of Micklitz et al. and the paper of Chen et al. for more details.
(aside: a scattering amplitude is simply a transition amplitude between \(2\) asymptotic scattering \(H_0\)-eigenstates; thus the term “transition amplitude” is more general, referring to the probability amplitude of an \(H_0\)-measurement at time \(t\) to yield the energy \(E_f\) given the quantum system was, at time \(t=0\), in the \(H_0\)-eigenstate \(E_i\)).
Problem: A hydrogen atom in its \(s\)-wave bound ground state is ionized by light of frequency \(\omega\). Calculate the FGR transition rate to the continuum of asymptotically free scattering states (this is a simple model of the photoelectric effect).
Solution: A qualitative sketch of the calculation: first, ignore quantization of the EM field (valid for e.g. a laser) by imposing a suitable vector potential such as \(\textbf A\propto e^{i(\textbf k\cdot\textbf x-\omega_{\textbf k}t)}\) for the usual planar EM wave. The perturbation Hamiltonian \(V(t)\) is determined by expanding to first-order the usual minimally-coupled Hamiltonian \(H\) for a charge \(q=-e\) in an electromagnetic field. Since this is absorption rather than stimulated emission, compute the matrix element of the appropriate term between the hydrogen atom ground state \(|1,0,0\rangle\) and the free \(e^-\) state \(|\textbf k\rangle\) (where recall that the density of such free states is \(g_{f,H_0}(E)\propto\sqrt E\)).
Problem: (to be added, dipole approximation demonstration that rates of stimulated absorption/emission are same, and the first-order p.t. expression for that rate).
Solution: Fermi’s golden rule applied to a monochromatic perturbation (should this be treated as another assumption?), but for a more general spectrum, even transitions between discrete bound \(H_0\)-eigenstates are possible, e.g. atom bathing in a photon gas. To compute the stimulated absorption/emission rates \(\Gamma_a,\Gamma_e\) it is common to employ a so-called dipole approximation (though in this context I feel a better name would be long-wavelength approximation or even just cold approximation; note also that this is a further approximation on top of all the other approximations that have already been made before this) in which the typical wavelength \(\lambda\gg r_B\) of photons \(\gamma\) in the thermal bath is much longer than the Bohr radius \(r_B\) representing the typical length scale of atoms. Essentially what this means is that the vector potential \(\textbf A\sim e^{i(\textbf k_{\text{ext}}\cdot\textbf x-\omega_{\text{ext}} t)}\approx e^{-i\omega_{\text{ext}}t}\) is approximately spatially uniform for \(\textbf k_{\text{ext}}\cdot\textbf x\ll 1\) for \(\textbf x\) ranging over the atom, and therefore the time-dependent perturbation to the atom just has the form \(\Delta\tilde H(t)=e\textbf E_{\text{ext}}(t)\cdot\sum_{e^-}\textbf X_{e^-}\) where \(e\sum_{e^-}\textbf X_{e^-}\) is the net electric dipole moment of the atom, hence the name “dipole approximation”. Chugging this into Case #\(2\) of Fermi’s golden rules, one can compute the rates of stimulated absorption and emission to be equal and given by:
\[\Gamma_{1\to 2}=\Gamma_{2\to 1}=\frac{\pi e^2}{3\varepsilon_0\hbar^2}\left|\sum_{e^-}\langle 2|\textbf X_{e^-}|1\rangle\right|^2g(\Delta E_{12})\]
Note that non-relativistic quantum mechanics predicts that atoms can only undergo stimulated absorption or stimulated emission. In particular, non-relativistic quantum mechanics predicts that there are no such things as “spontaneous absorption” or “spontaneous emission” \(\Gamma_{1\to 2}^*=\Gamma_{2\to 1}^*=0\) where an atom can undergo a transition between \(\tilde H\)-eigenstates in the absence \(\textbf E_{\text{ext}}=\textbf B_{\text{ext}}=\textbf 0\) of external electromagnetic fields/photons (because the non-relativistic Schrodinger equation asserts that the \(t\)-evolution of \(\tilde H\)-eigenstates is to stay right where they are rather than hopping to other \(\tilde H\)-eigenstates). Indeed, it turns out there is no such thing as “spontaneous absorption” \(\Gamma_{1\to 2}^*\), but there can be spontaneous emission \(\Gamma_{2\to 1}^*\neq 0\). This phenomenon only arises once one quantizes the classical, smooth electromagnetic field \((\textbf E,\textbf B)\). When this is done, it is found that spontaneous emission is possible due to interactions between the atom and zero-point fluctuations of the quantum electromagnetic field, but to delve deeper would be the subject of quantum electrodynamics.
Nevertheless, Einstein was able to compute the rate of spontaneous emission \(\Gamma_{2\to 1}^*\) without knowing anything about the quantization of the electromagnetic field. Or more precisely, Einstein showed that if one could calculate the stimulated emission rate \(\Gamma_{2\to 1}\), then one would also get the spontaneous emission rate \(\Gamma_{2\to 1}^*\) “for free” since he found that the two were proportional to each other and what that proportionality constant was.
Einstein’s Statistical Argument for Spontaneous Emission
Consider any \(2\) energy levels \(E_1<E_2\) in an atom of degeneracies \(\Omega_1,\Omega_2\) with number of electrons \(N_1,N_2\) respectively.
Then in general, one heuristically expects probabilistic kinetics of the form (ASIDES: what happens if one also adds a “spontaneous absorption” term \(A_{12}N_1\) among the rate terms? And also, need to explicitly related the \(\Gamma\)-transition rates above in Fermi’s golden rule with these Einstein \(A,B\) coefficients):
\[\dot N_2=-\dot N_1=-B_{12}(\Delta\omega_{12})N_1\mathcal E_{\text{ext}}(\Delta\omega_{12})+B_{21}(\Delta\omega_{12})N_2\mathcal E_{\text{ext}}(\Delta\omega_{12})+A_{21}(\Delta\omega_{12})N_2\]
where the spontaneous emission term \(A_{21}(\Delta\omega_{12})N_2\) is not proportional to the spectral energy density of photons \(\mathcal E_{\text{ext}}(\Delta\omega_{12})\) with the correct frequency \(\Delta\omega_{12}=(E_2-E_1)/\hbar\) precisely because it’s spontaneous (i.e. not stimulated by the radiation and therefore independent of its “concentration”). Another way to see this is that, in the absence of any external radiation \(\mathcal E_{\text{ext}}(\Delta\omega_{12})=0\) with the correct frequency \(\Delta\omega_{12}\), the only operating mechanism is spontaneous emission in which the occupation number \(N_2(t)=N_2(0)e^{-A_{21}(\Delta\omega_{12})t}\) decays exponentially to \(N_2\to 0\) with lifetime \(\tau(\Delta\omega_{12})=1/A_{21}(\Delta\omega_{12})\) inversely correlated with the gap size \(\Delta\omega_{12}\) (since a larger gap size \(\Delta\omega_{12}\) signifies a more “unstable” state).
More generally then, when there are photons with the right frequency \(\mathcal E_{\text{ext}}(\Delta\omega_{12})\neq 0\) for stimulated absorption/emission, the transient \(t\)-dependence is tricky to model, but the steady state \(t\to\infty\) where \(\dot N_1=\dot N_2=0\) is straightforward: the rates of stimulated absorption/emission and spontaneous emission must balance:
\[-B_{12}(\Delta\omega_{12})N_1\mathcal E_{\text{ext}}(\Delta\omega_{12})+B_{21}(\Delta\omega_{12})N_2\mathcal E_{\text{ext}}(\Delta\omega_{12})+A_{21}(\Delta\omega_{12})N_2=0\]
or isolating for the spectral energy density \(\mathcal E_{\text{ext}}(\Delta\omega_{12})\) of correct-frequency photons:
\[\mathcal E_{\text{ext}}(\Delta\omega_{12})=\frac{A_{21}(\Delta\omega_{12})}{B_{12}(\Delta\omega_{12})(N_1/N_2)-B_{21}(\Delta\omega_{12})}\]
But being in the \(t\to\infty\) steady state is synonymous with the atom being in thermodynamic equilibrium with the photon gas at some temperature \(T\) in the canonical ensemble, so statistical mechanics asserts that both \(\frac{N_1}{N_1+N_2}=\Omega_1e^{-E_1/kT}/Z\) and \(\frac{N_2}{N_1+N_2}=\Omega_2e^{-E_2/kT}/Z\) will be Boltzmann distributed while the spectral energy density of the radiation \(\mathcal E_{\text{ext}}(\Delta\omega_{12})=\frac{\hbar\Delta\omega_{12}^3}{\pi^2 c^3}\frac{1}{e^{\hbar\Delta\omega_{12}/kT}-1}\) will be Planck distributed. Plugging these distributions in leads to:
\[\frac{\hbar\Delta\omega_{12}^3}{\pi^2 c^3}\frac{1}{e^{\hbar\Delta\omega_{12}/kT}-1}=\frac{A_{21}(\Delta\omega_{12})\Omega_2}{B_{12}(\Delta\omega_{12})\Omega_1}\frac{1}{e^{\hbar\Delta\omega_{12}/kT}-B_{21}(\Delta\omega_{12})\Omega_2/B_{12}(\Delta\omega_{12})\Omega_1}\]
One can then just “pattern-match” (it is an encouraging sign that this is even possible!) because the relation must hold at arbitrary temperature \(T\in\textbf R\) to obtain:
\[\frac{B_{12}(\Delta\omega_{12})}{\Omega_2}=\frac{B_{21}(\Delta\omega_{12})}{\Omega_1}\]
\[\frac{\hbar\Delta\omega_{12}^3}{\pi^2 c^3}=\frac{A_{21}(\Delta\omega_{12})\Omega_2}{B_{12}(\Delta\omega_{12})\Omega_1}\]
The first relation is just reconfirming that \(\Gamma_{12}=\Gamma_{21}\). Morally, it makes sense that these stimulated rates coincide because absorption of a photon of energy \(\pm E\) is “equivalent” to emission of a photon of energy \(\mp E\). Mathematically, the time-dependent perturbation \(\Delta\tilde H(t)=\Delta\tilde H_0e^{i\omega t}+\Delta\tilde H_0^{\dagger}e^{-i\omega t}\) is Hermitian.
Plugging the first relation into the second, one obtains a direct relation between the rates of stimulated and spontaneous emission:
\[A_{21}(\Delta\omega_{12})=\frac{\hbar\Delta\omega_{12}^3}{\pi^2 c^3}B_{21}(\Delta\omega_{12})\]
At first, even after one has seen Einstein’s statistical argument for \(\Gamma_{1\to 2}=\Gamma_{2\to 1}\) and \(A_{21}(\Delta\omega_{12})\propto\Delta\omega_{12}^3B_{21}(\Delta\omega_{12})\), it is still not clear how the argument manages to “get away” with not knowing anything about the quantization of the electromagnetic field while still obtain the correct answer. It turns out to be implicit; one has already quantized the electromagnetic field by taking the Planck distribution as an input to the derivation.
Being thermodynamic in nature, whereas transition rates are inherently kinetic, Einstein’s statistical argument of course fails to actually yield an explicit expression for the various transition rates. This “kinetics” part was of course the whole point of the time-dependent perturbation theory calculation with the dipole approximation. Combining Einstein’s thermodynamic argument with a dose of quantum kinetics, the lifetime \(\tau\) of an isolated atom in an excited state before it spontaneously emits a photon \(\gamma\) in order to decay to a lower energy is roughly:
\[\tau\approx \frac{3m_{e^-}c^3}{4\hbar\Delta\omega_{12}^3a_0}\]