Fermi's Golden Rule

Derivation

Fermi's Golden Rule

D-227 Home PU-301 Threads energy · chance · light Depends on Time-Dependent Perturbation Theory

Statement

For a system prepared in an eigenstate \(|i\rangle\) of an unperturbed Hamiltonian \(\hat{H}_0\) and subjected at \(t=0\) to a weak, time-independent perturbation \(\hat{V}\), first-order time-dependent perturbation theory gives, in the long-time limit, a constant transition rate into a dense (continuum) group of final states \(|f\rangle\): \[ \Gamma_{i\to f} \;=\; \frac{2\pi}{\hbar}\,\bigl|\langle f|\hat{V}|i\rangle\bigr|^{2}\,\rho(E_f)\Big|_{E_f=E_i}, \] where \(\rho(E_f)\) is the density of final states per unit energy evaluated at the initial energy. This is Fermi's Golden Rule.

Why it matters

The Golden Rule is the workhorse of every process in which a quantum system decays or scatters into a continuum: spontaneous and stimulated emission of photons, photoionisation and the photoelectric effect, radioactive \(\alpha\)- and \(\beta\)-decay, non-radiative recombination in semiconductors, and scattering cross-sections in nuclear and particle physics. It converts an oscillating first-order amplitude into a steady rate, which is exactly what makes lifetimes, linewidths and cross-sections well-defined observables.

It is also the bridge between the microscopic matrix element \(\langle f|\hat{V}|i\rangle\) — which encodes the physics of the coupling — and macroscopic, measurable quantities like the mean lifetime \(\tau = 1/\Gamma\). Almost all selection rules follow from the vanishing of that matrix element.

Assumptions

The perturbation is weak.Without \(|\langle f|\hat V|i\rangle|\ll\) the relevant level spacing, first-order theory is invalid: the initial-state amplitude depletes, higher orders and back-transitions matter, and the linear-in-\(t\) probability breaks down.

The final states form a continuum (or a dense quasi-continuum).If the final states are discrete, the "sum over states" does not become an integral, no \(\rho(E_f)\) exists, and the probability keeps oscillating (Rabi flopping) rather than growing linearly — there is no constant rate.

The observation time is long compared to \(\hbar/\Delta E\) yet short compared to the decay time.Too short and the \(\mathrm{sinc}^2\) function has not collapsed to a \(\delta\)-function (energy not yet sharply conserved); too long and first-order perturbation theory over-counts because \(P\) would exceed unity.

The matrix element \(\langle f|\hat V|i\rangle\) is smooth across the width of the \(\mathrm{sinc}^2\) peak.If \(\hat V\) or \(\rho\) varies sharply within a few \(\hbar/t\) of \(E_i\), it cannot be pulled outside the energy integral and the simple product form fails; a convolution must be kept.

The perturbation is switched on effectively suddenly and held constant.For a harmonic perturbation \(\hat V e^{-i\omega t}\) the same derivation gives \(\delta(E_f-E_i\mp\hbar\omega)\); dropping the constant/adiabatic assumption changes which resonance condition appears.

Derivation

\[ c_f^{(1)}(t) = \frac{1}{i\hbar}\int_0^{t} \langle f|\hat{V}|i\rangle\, e^{\,i\omega_{fi}t'}\,\mathrm{d}t', \qquad \omega_{fi}\equiv\frac{E_f-E_i}{\hbar} \]

First-order transition amplitude from time-dependent perturbation theory (prior result), for \(\hat V\) turned on at \(t=0\). A

\[ c_f^{(1)}(t) = \frac{\langle f|\hat{V}|i\rangle}{i\hbar}\int_0^{t} e^{\,i\omega_{fi}t'}\,\mathrm{d}t' = \frac{\langle f|\hat{V}|i\rangle}{i\hbar}\,\frac{e^{\,i\omega_{fi}t}-1}{i\omega_{fi}} \]

\(\hat V\) is time-independent, so the matrix element leaves the integral; elementary integration of the exponential. A

\[ \frac{e^{\,i\omega_{fi}t}-1}{i\omega_{fi}} = e^{\,i\omega_{fi}t/2}\,\frac{e^{\,i\omega_{fi}t/2}-e^{-i\omega_{fi}t/2}}{i\omega_{fi}} = e^{\,i\omega_{fi}t/2}\,\frac{2\sin(\omega_{fi}t/2)}{\omega_{fi}} \]

Factor out the half-angle phase and use \(2i\sin\theta = e^{i\theta}-e^{-i\theta}\); this isolates a pure phase times a real amplitude. A

\[ P_{i\to f}(t) = \bigl|c_f^{(1)}(t)\bigr|^{2} = \frac{\bigl|\langle f|\hat{V}|i\rangle\bigr|^{2}}{\hbar^{2}}\, \frac{4\sin^{2}(\omega_{fi}t/2)}{\omega_{fi}^{2}} \]

Modulus squared; the pure phase \(e^{i\omega_{fi}t/2}\) and the \(1/i\) drop out, leaving a real, positive probability. A

\[ P_{i\to f}(t) = \frac{\bigl|\langle f|\hat{V}|i\rangle\bigr|^{2}}{\hbar^{2}}\; t^{2}\, \operatorname{sinc}^{2}\!\left(\frac{\omega_{fi}t}{2}\right), \qquad \operatorname{sinc}(x)\equiv\frac{\sin x}{x} \]

Rewrite \(4\sin^2(\omega_{fi}t/2)/\omega_{fi}^2 = t^2\,\operatorname{sinc}^2(\omega_{fi}t/2)\) to expose the peak height \(\propto t^2\) and width \(\propto 1/t\). B

\[ \Gamma_{i\to f} = \sum_{f\in\text{cont.}}\frac{P_{i\to f}(t)}{t} \;\longrightarrow\; \frac{1}{t}\int P_{i\to f}(t)\,\rho(E_f)\,\mathrm{d}E_f \]

Because the states are dense, replace the sum over final states by an energy integral weighted by the density of states \(\rho(E_f)\); the rate is the total probability divided by \(t\). B

\[ \Gamma_{i\to f} = \frac{1}{t}\int \frac{\bigl|\langle f|\hat{V}|i\rangle\bigr|^{2}}{\hbar^{2}}\, t^{2}\,\operatorname{sinc}^{2}\!\left(\frac{(E_f-E_i)t}{2\hbar}\right)\rho(E_f)\,\mathrm{d}E_f \]

Substitute \(P_{i\to f}(t)\) and \(\omega_{fi}=(E_f-E_i)/\hbar\). The \(\operatorname{sinc}^2\) is sharply peaked at \(E_f=E_i\) with width \(\sim\hbar/t\). B

\[ \int_{-\infty}^{\infty} \operatorname{sinc}^{2}\!\left(\frac{(E_f-E_i)t}{2\hbar}\right)\mathrm{d}E_f = \frac{2\hbar}{t}\int_{-\infty}^{\infty}\operatorname{sinc}^2(x)\,\mathrm{d}x = \frac{2\hbar}{t}\cdot\pi = \frac{2\pi\hbar}{t} \]

For large \(t\) the peak is narrow, so pull the smooth factors \(|\langle f|\hat V|i\rangle|^2\rho(E_f)\) out at \(E_f=E_i\) and use \(\int_{-\infty}^{\infty}\operatorname{sinc}^2 x\,\mathrm{d}x=\pi\) with \(x=(E_f-E_i)t/2\hbar\). C

\[ \lim_{t\to\infty}\frac{t}{2\pi\hbar}\,\operatorname{sinc}^{2}\!\left(\frac{(E_f-E_i)t}{2\hbar}\right) = \delta(E_f-E_i) \]

The normalised \(\operatorname{sinc}^2\) is a nascent delta function: unit area (Step 8) and vanishing width. This is the precise statement of energy conservation emerging from the long-time limit. C

\[ \Gamma_{i\to f} = \frac{2\pi}{\hbar}\int \bigl|\langle f|\hat{V}|i\rangle\bigr|^{2}\,\delta(E_f-E_i)\,\rho(E_f)\,\mathrm{d}E_f = \frac{2\pi}{\hbar}\bigl|\langle f|\hat{V}|i\rangle\bigr|^{2}\rho(E_i) \]

Insert the delta function of Step 9 into Step 7 (the factors of \(t\) cancel), then collapse the integral. C

Result

\[ \boxed{\;\Gamma_{i\to f} \;=\; \frac{2\pi}{\hbar}\,\bigl|\langle f|\hat{V}|i\rangle\bigr|^{2}\,\rho(E_f)\Big|_{E_f=E_i}\;} \]

Reading. The transition rate is constant in time (hence a well-defined lifetime \(\tau=1/\Gamma\)). It is proportional to the squared coupling strength \(|\langle f|\hat V|i\rangle|^2\) — so a weak perturbation gives a slow rate — and to how many final states are available at the conserved energy, \(\rho(E_i)\). Energy is conserved between initial and final states, enforced by the delta function that emerged in the long-time limit.

Units check. \([\hat V]=\mathrm{J}\), so \(|\langle f|\hat V|i\rangle|^2=\mathrm{J}^2\). The density of states \([\rho]=\mathrm{J^{-1}}\) (states per joule). Then \(\frac{1}{[\hbar]}\,\mathrm{J^2\cdot J^{-1}} = \frac{\mathrm{J}}{\mathrm{J\cdot s}} = \mathrm{s^{-1}}\). A rate, as required.

Limiting cases

Vanishing coupling \(\langle f|\hat V|i\rangle\to 0\): \(\Gamma\to 0\), the state is stable — this is the origin of selection rules (forbidden transitions).
No available final states \(\rho(E_i)\to 0\): \(\Gamma\to 0\) even with strong coupling — e.g. photoemission below threshold, or a decay blocked by a band gap or Pauli blocking.
Harmonic perturbation \(\hat V(t)=\hat V\,e^{-i\omega t}\): the delta function becomes \(\delta(E_f-E_i-\hbar\omega)\), the resonant absorption condition — the constant-\(\hat V\) result is the \(\omega\to0\) case.
Short times \(t\ll\hbar/\Delta E\): the \(\operatorname{sinc}^2\) is broad, \(P\propto t^2\) (coherent, quadratic growth), and no constant rate exists yet.
Discrete two-level limit \(\rho\to\) single state: recovers Rabi oscillations, \(P\propto\sin^2(\Omega t/2)\), not a monotonic rate.

Breaks when

Strong coupling / long times. When \(\Gamma t \gtrsim 1\) the initial state is appreciably depleted, so \(|c_i|\approx1\) fails and \(P_{i\to f}\) would exceed unity. The exponential decay \(|c_i(t)|^2=e^{-\Gamma t}\) (Weisskopf–Wigner) must replace linear-in-\(t\) growth; the constant-rate result survives only as the leading exponent.
Genuinely discrete final spectrum. With no continuum there is no \(\rho(E_f)\); the \(\operatorname{sinc}^2\) never collapses to a delta function and the probability oscillates forever (Rabi flopping) instead of settling to a rate.
Sharply structured continuum. If \(|\langle f|\hat V|i\rangle|^2\rho(E_f)\) varies on the scale \(\hbar/t\) (e.g. a narrow resonance in the final-state density, or a photonic band edge), it cannot be pulled out of the integral — non-Markovian, non-exponential dynamics appear (e.g. Fano lineshapes, fractional decay).
Very short times (Zeno regime). For \(t\to0\), \(P\propto t^2\) not \(\propto t\); repeated fast measurement can suppress the transition entirely (quantum Zeno effect), which the linear rate cannot describe.

Failure modes

Forgetting the density of states. Writing \(\Gamma=\frac{2\pi}{\hbar}|\langle f|\hat V|i\rangle|^2\) with no \(\rho\) — the answer then has units of \(\mathrm{J/(J\cdot s)}\times\mathrm{J}\), wrong. The \(\rho\) is not optional; it is what makes the continuum sum finite.
Using \(\rho\) at the wrong energy. \(\rho\) must be evaluated at \(E_f=E_i\) (or \(E_i+\hbar\omega\) for a harmonic drive), the energy the delta function selects — not at some average or at \(E_f=0\).
Squaring after summing. One squares the amplitude for each final state and then sums (\(\sum_f|c_f|^2\)); squaring a coherent sum \(|\sum_f c_f|^2\) is wrong because distinct final states are orthogonal, distinguishable outcomes.
Dropping a factor of 2 in the width. The argument is \(\omega_{fi}t/2\), not \(\omega_{fi}t\); mishandling the half-angle gives the wrong normalisation and a spurious factor in \(\int\operatorname{sinc}^2\).
Double counting spin/polarisation. If \(\rho\) already includes a spin or polarisation degeneracy, summing over those quantum numbers again over-counts the rate.
Applying it at \(t=0^+\). Quoting a "rate" before the \(\operatorname{sinc}^2\) has narrowed; in that regime the growth is quadratic and the notion of a rate is meaningless.

Discussion

The physical heart of the derivation is the transmutation of an oscillating amplitude into a steady rate. The single-state probability \(P_{i\to f}(t)\propto t^2\operatorname{sinc}^2(\omega_{fi}t/2)\) always oscillates and never grows without bound. What produces monotone growth is the sum over a continuum: as \(t\) increases the central peak of the \(\operatorname{sinc}^2\) narrows like \(1/t\) but its height grows like \(t^2\), so its area grows like \(t\). Dividing by \(t\) leaves a constant. The Golden Rule is therefore a statement about interference between many final states, not about any single transition.

The emergent delta function is an energy–time uncertainty relation in action. At finite \(t\) energy is conserved only to \(\Delta E\sim\hbar/t\); only in the strict long-time limit does the transition become sharply energy-conserving. This is precisely why unstable states have a finite linewidth: a state that lives for time \(\tau\) has an energy width \(\Gamma_E=\hbar/\tau\), and the Golden Rule rate \(\Gamma\) is exactly \(\Gamma_E/\hbar\). Lifetime and linewidth are the same physics seen in the time and energy domains.

The rule connects deep threads. Applied to the dipole coupling \(\hat V=-\hat{\vec d}\cdot\vec E\) with the photon density of states it yields spontaneous emission and the Einstein \(A\) coefficient; applied to a scattering potential it yields the Born-approximation cross-section; applied to the weak interaction it yields \(\beta\)-decay rates and the Sargent rule. In every case the matrix element carries the specific physics and \(\rho\) carries the phase-space counting, and their product is a measurable rate.

Rigorously, the Golden Rule is the first term of a controlled expansion whose resummation is the resolvent (self-energy) formalism. The exact decay is governed by the complex self-energy \(\Sigma(E)\); its imaginary part evaluated on shell reproduces \(\Gamma=\frac{2\pi}{\hbar}|\langle f|\hat V|i\rangle|^2\rho(E_i)\) (the optical theorem), while its real part gives the Lamb-type level shift the naive rule omits. The Markovian, exponential decay of Weisskopf–Wigner theory then follows from assuming \(\Sigma(E)\) is slowly varying — exactly the smooth-continuum assumption of Step 8 — which is why non-exponential decay at very short and very long times is not a violation of quantum mechanics but a breakdown of that assumption.

Common misconceptions. The Golden Rule does not say the transition happens instantaneously or that energy is conserved at every instant — energy conservation is an asymptotic (\(t\to\infty\)) property. It is not exact even at first order for all times; it is the long-time slope of \(P(t)\), valid only in the window \(\hbar/\Delta E\ll t\ll1/\Gamma\). And the "\(2\pi\)" is not a fudge factor: it is \(\int_{-\infty}^{\infty}\operatorname{sinc}^2 x\,\mathrm{d}x=\pi\) doubled by the change of variables — pure geometry of the \(\operatorname{sinc}^2\) peak.

Worked examples

Example 1 — Rate into a free-particle (box-normalised) continuum.

\[ \rho(E)=\frac{\mathrm{d}N}{\mathrm{d}E},\qquad N(E)=\frac{V}{(2\pi)^3}\cdot\frac{4}{3}\pi k^3,\qquad E=\frac{\hbar^2k^2}{2m} \]

Count plane-wave states \(\mathbf k\) inside a sphere of radius \(k\) in a box of volume \(V\); each occupies \((2\pi/L)^3\) in \(k\)-space. A

\[ \rho(E)=\frac{V}{(2\pi)^3}\,4\pi k^2\,\frac{\mathrm{d}k}{\mathrm{d}E} =\frac{V}{(2\pi)^3}\,4\pi k^2\,\frac{m}{\hbar^2 k} =\frac{V\,m\,k}{2\pi^2\hbar^2} \]

Chain rule with \(\mathrm{d}E/\mathrm{d}k=\hbar^2 k/m\); this is the free-particle density of states (per spin). B

\[ \Gamma=\frac{2\pi}{\hbar}|V_{fi}|^2\,\frac{V m k}{2\pi^2\hbar^2} =\frac{|V_{fi}|^2\,V m k}{\pi\hbar^3} \]

Insert \(\rho\) into the Golden Rule. Take a final electron with \(E=1.0\,\mathrm{eV}\): \(k=\sqrt{2mE}/\hbar\). B

\[ k=\frac{\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})}}{1.055\times10^{-34}} =5.12\times10^{9}\ \mathrm{m^{-1}} \]

Numbers: \(m_e=9.11\times10^{-31}\,\mathrm{kg}\), \(1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J}\). A

\[ \rho=\frac{V m k}{2\pi^2\hbar^2} =\frac{(1.0\times10^{-27})(9.11\times10^{-31})(5.12\times10^{9})}{2\pi^2(1.055\times10^{-34})^2} =2.12\times10^{18}\ \mathrm{J^{-1}} \]

Use a box \(V=(10^{-9}\,\mathrm{m})^3=10^{-27}\,\mathrm{m}^3\); \(\rho\) scales with \(V\) but the physical rate will not (a real matrix element \(\propto1/\sqrt V\)). B

\[ \Gamma=\frac{2\pi}{\hbar}|V_{fi}|^2\rho =\frac{2\pi}{1.055\times10^{-34}}\,(1.60\times10^{-22})^2\,(2.12\times10^{18}) \]

Take a coupling \(|V_{fi}|=10^{-3}\,\mathrm{eV}=1.60\times10^{-22}\,\mathrm{J}\). A

\[ \Gamma \approx 3.2\times10^{12}\ \mathrm{s^{-1}} \qquad(\tau=1/\Gamma\approx0.31\ \mathrm{ps}) \]

Reading. A milli-electron-volt coupling into a free-electron continuum gives sub-picosecond decay — typical of fast electronic relaxation. Units check. \(\mathrm{s^{-1}\cdot J^2\cdot J^{-1}/(J\cdot s)}=\mathrm{s^{-1}}\).

Example 2 — Spontaneous emission lifetime of hydrogen \(2p\to1s\).

\[ \Gamma=\frac{2\pi}{\hbar}\sum_{\text{pol}}\int |\langle 1s,\mathbf k\lambda|\hat V|2p\rangle|^2\,\rho_{\text{ph}}(E)\,\mathrm{d}\Omega =\frac{\omega^3\,e^2\,|\langle 1s|\mathbf r|2p\rangle|^2}{3\pi\varepsilon_0\hbar c^3} \]

Golden Rule with the dipole interaction \(\hat V=-e\,\mathbf r\!\cdot\!\mathbf E\) and the photon density of states, summed over two polarisations and integrated over emission directions (standard result). C

\[ |\langle 1s|\mathbf r|2p\rangle|^2=\left(\frac{2^{7}}{3^{5}}\right)^{2}a_0^2 =(0.7449\,a_0)^2,\qquad a_0=5.29\times10^{-11}\,\mathrm{m} \]

Exact hydrogenic radial matrix element for \(2p\to1s\); \(2^7/3^5=0.7449\). B

\[ \hbar\omega=E_{2}-E_{1}=13.6\left(1-\tfrac14\right)\mathrm{eV}=10.2\,\mathrm{eV} \;\Rightarrow\;\omega=1.55\times10^{16}\ \mathrm{s^{-1}} \]

Lyman-\(\alpha\) transition energy divided by \(\hbar\). A

\[ |\langle 1s|\mathbf r|2p\rangle|^2=(0.7449\times5.29\times10^{-11})^2=1.55\times10^{-21}\ \mathrm{m^2} \]

Multiply out; radial length squared. A

\[ \Gamma=\frac{(1.55\times10^{16})^3(1.60\times10^{-19})^2(1.55\times10^{-21})} {3\pi(8.85\times10^{-12})(1.055\times10^{-34})(3.0\times10^{8})^3} \]

Insert \(\omega\), \(e\), \(\varepsilon_0\), \(\hbar\), \(c\) and the matrix element into the boxed formula. B

\[ \Gamma \approx 6.3\times10^{8}\ \mathrm{s^{-1}} \qquad \tau=\frac{1}{\Gamma}\approx1.6\ \mathrm{ns} \]

Reading. The \(2p\) state of hydrogen lives about \(1.6\,\mathrm{ns}\), the textbook Lyman-\(\alpha\) lifetime — a direct triumph of the Golden Rule. Units check. \(\frac{\mathrm{s^{-3}\cdot C^2\cdot m^2}}{(\mathrm{C^2\,J^{-1}m^{-1}})(\mathrm{J\,s})(\mathrm{m^3 s^{-3}})}=\mathrm{s^{-1}}\).

Problems

(A) A state has a Golden-Rule decay rate \(\Gamma=2.0\times10^{9}\ \mathrm{s^{-1}}\). Find its mean lifetime and its natural energy linewidth in \(\mathrm{eV}\).
Solution
\(\tau=1/\Gamma=5.0\times10^{-10}\,\mathrm{s}=0.50\,\mathrm{ns}\). Linewidth \(\Gamma_E=\hbar\Gamma=(1.055\times10^{-34})(2.0\times10^{9})=2.11\times10^{-25}\,\mathrm{J}=1.3\times10^{-6}\,\mathrm{eV}=1.3\ \mu\mathrm{eV}\).
(A) By dimensional analysis, verify that \(\frac{2\pi}{\hbar}|V_{fi}|^2\rho\) has units of inverse seconds, given \([V_{fi}]=\mathrm{J}\) and \([\rho]=\mathrm{J^{-1}}\). What would go wrong dimensionally if you omitted \(\rho\)?
Solution
\(\frac{1}{\mathrm{J\,s}}\cdot\mathrm{J}^2\cdot\mathrm{J}^{-1}=\frac{\mathrm{J}}{\mathrm{J\,s}}=\mathrm{s}^{-1}\). Omitting \(\rho\) leaves \(\frac{1}{\mathrm{J\,s}}\cdot\mathrm{J}^2=\mathrm{J/s}=\mathrm{W}\), a power, not a rate — the density of states is what supplies the missing \(\mathrm{J^{-1}}\) and turns the single-state (dimensionally wrong) expression into a genuine rate summed over the continuum.
(B) Show that the single-state probability \(P(t)=\dfrac{|V_{fi}|^2}{\hbar^2}t^2\operatorname{sinc}^2(\omega_{fi}t/2)\) reduces on resonance (\(\omega_{fi}=0\)) to \(P=|V_{fi}|^2t^2/\hbar^2\), and explain why this cannot be a valid "rate times \(t\)".
Solution
At \(\omega_{fi}=0\), \(\operatorname{sinc}(0)=1\), so \(P=|V_{fi}|^2t^2/\hbar^2\). This grows quadratically, not linearly, so \(P/t\propto t\) is not constant — there is no time-independent rate for a single resonant state. A steady rate appears only after summing the \(\operatorname{sinc}^2\) over a continuum of \(\omega_{fi}\), which converts the \(t^2\) peak of shrinking width into an area \(\propto t\).
(B) A perturbation couples \(|i\rangle\) to a continuum with constant density \(\rho=1.0\times10^{19}\ \mathrm{J^{-1}}\) and matrix element \(|V_{fi}|=5.0\times10^{-23}\ \mathrm{J}\). Find \(\Gamma\) and \(\tau\). Then find the matrix element needed to halve the lifetime.
Solution
\(\Gamma=\frac{2\pi}{\hbar}|V_{fi}|^2\rho=\frac{2\pi}{1.055\times10^{-34}}(5.0\times10^{-23})^2(1.0\times10^{19})=\frac{2\pi(2.5\times10^{-45})(1.0\times10^{19})}{1.055\times10^{-34}}=1.49\times10^{9}\,\mathrm{s^{-1}}\). \(\tau=1/\Gamma=6.7\times10^{-10}\,\mathrm{s}\). Halving \(\tau\) means doubling \(\Gamma\), and since \(\Gamma\propto|V_{fi}|^2\), the matrix element must increase by \(\sqrt2\): \(|V_{fi}|'=\sqrt2\times5.0\times10^{-23}=7.1\times10^{-23}\,\mathrm{J}\).
(C) For a harmonic perturbation \(\hat V(t)=\hat V_0\cos\omega t\), sketch the derivation showing the Golden Rule becomes \(\Gamma=\frac{2\pi}{\hbar}\cdot\frac14|\langle f|\hat V_0|i\rangle|^2\,[\rho(E_i+\hbar\omega)+\rho(E_i-\hbar\omega)]\), and identify the two physical processes.
Solution
Write \(\cos\omega t=\tfrac12(e^{i\omega t}+e^{-i\omega t})\). The first-order amplitude \(c_f^{(1)}=\frac{1}{i\hbar}\int_0^t\langle f|\hat V_0|i\rangle\tfrac12(e^{i\omega t'}+e^{-i\omega t'})e^{i\omega_{fi}t'}\mathrm{d}t'\) has two terms whose denominators are \(\omega_{fi}\pm\omega\). Each produces its own \(\operatorname{sinc}^2\) peaked where \(\omega_{fi}=\mp\omega\), i.e. \(E_f=E_i\mp\hbar\omega\). Cross terms oscillate and average away for \(\omega\neq0\) (rotating-wave/long-time). Squaring and taking the continuum limit gives, with the factor \((\tfrac12)^2=\tfrac14\), \(\Gamma=\frac{\pi}{2\hbar}|\langle f|\hat V_0|i\rangle|^2[\rho(E_i+\hbar\omega)+\rho(E_i-\hbar\omega)]\). The term at \(E_i+\hbar\omega\) is absorption (system climbs by one quantum \(\hbar\omega\)); the term at \(E_i-\hbar\omega\) is stimulated emission (system drops by \(\hbar\omega\)). Equality of the two matrix elements is the microscopic root of \(B_{12}=B_{21}\).