Einstein A and B Coefficients

Derivation

Einstein A and B Coefficients

D-352 Home PU-401 Threads light · chance · energy Depends on Semiclassical Radiation and Dipole Selection Rules, planck-radiation-law

Statement

For a two-level atom with a lower level of degeneracy \(g_1\) and an upper level of degeneracy \(g_2\) separated by energy \(\hbar\omega = E_2 - E_1\), the rates of spontaneous emission (\(A_{21}\)), stimulated emission (\(B_{21}\)), and absorption (\(B_{12}\)) are not independent. Demanding that the level populations reach thermal equilibrium in contact with blackbody radiation forces the two exact relations \(g_1 B_{12} = g_2 B_{21}\) and \(A_{21}/B_{21} = \hbar\omega^3/(\pi^2 c^3)\), where \(B_{21}\) multiplies the spectral energy density per unit angular frequency.

Why it matters

These ratios are the thermodynamic backbone of all light–matter interaction. They show that spontaneous emission is not an optional add-on but is fixed by the same matrix element that controls absorption; an atom that can absorb must also radiate, at a rate that grows as \(\omega^3\). This is why radio transitions are metastable while optical transitions decay in nanoseconds.

The relations were derived by Einstein in 1917, a decade before quantum electrodynamics could compute \(A_{21}\) from first principles. They remain the practical link between a measured lifetime and an absorption cross-section, and the condition \(B_{21}=B_{12}\) (for equal degeneracies) is precisely what makes population inversion — and therefore the laser — possible.

Assumptions

Only two levels participate.If other levels feed or drain \(N_1,N_2\), the steady-state balance acquires extra terms and the naive population ratio no longer equals the Boltzmann factor, so the extracted ratios are contaminated. The radiation field is isotropic, unpolarised blackbody radiation at temperature \(T\).If the field is directional or non-thermal (e.g. a laser beam), the detailed-balance argument against the Planck spectrum does not apply and the coefficients cannot be read off this way. In equilibrium the populations obey the Boltzmann distribution \(N_2/N_1 = (g_2/g_1)e^{-\hbar\omega/k_BT}\).This is the load-bearing thermodynamic input; drop it and the whole argument collapses, since it is the mismatch between Boltzmann and Planck that pins the ratios. The coefficients \(A_{21}, B_{21}, B_{12}\) are intrinsic atomic constants, independent of \(T\) and of the field.If they depended on temperature the identification with the Planck denominator would only hold at one \(T\); Einstein's insight is precisely that they cannot, because they encode the same atomic matrix element that governs the semiclassical selection rules.

Derivation

\[ \frac{dN_2}{dt} = \underbrace{N_1 B_{12}\,\rho(\omega)}_{\text{absorption}} - \underbrace{N_2 B_{21}\,\rho(\omega)}_{\text{stimulated}} - \underbrace{N_2 A_{21}}_{\text{spontaneous}} \]

Write the rate of change of the upper population as a sum of the three elementary processes. Absorption and stimulated emission are proportional to the spectral energy density \(\rho(\omega)\); spontaneous emission is not. A

\[ N_1 B_{12}\,\rho(\omega) = N_2\big(A_{21} + B_{21}\,\rho(\omega)\big) \]

Impose steady state, \(dN_2/dt = 0\). In thermal equilibrium every process is balanced, so this is detailed balance between the levels. A

\[ \rho(\omega) = \frac{A_{21}}{\dfrac{N_1}{N_2}B_{12} - B_{21}} \]

Solve algebraically for the field density \(\rho(\omega)\) that the atoms would themselves be in equilibrium with. Only ratios of coefficients appear. A

\[ \frac{N_1}{N_2} = \frac{g_1}{g_2}\,e^{+\hbar\omega/k_BT} \]

In equilibrium the populations are Boltzmann-distributed over the two levels, weighted by their degeneracies. This is the external thermodynamic input. A

\[ \rho(\omega) = \frac{A_{21}}{B_{21}}\;\frac{1}{\dfrac{g_1 B_{12}}{g_2 B_{21}}\,e^{\hbar\omega/k_BT} - 1} \]

Substitute the Boltzmann ratio into Step 3 and divide numerator and denominator by \(B_{21}\). This already has the structure of a Planck spectrum. B

\[ \rho(\omega) = \frac{\hbar\omega^3}{\pi^2 c^3}\;\frac{1}{e^{\hbar\omega/k_BT} - 1} \]

Quote the assumed prior result: the equilibrium field is the Planck spectral energy density per unit angular frequency. The atomic result of Step 5 must equal this for every \(T\) and \(\omega\). A

\[ \frac{g_1 B_{12}}{g_2 B_{21}} = 1 \qquad\Longrightarrow\qquad \boxed{\,g_1 B_{12} = g_2 B_{21}\,} \]

Match the denominators. The exponential prefactor in Step 5 must equal the bare \(e^{\hbar\omega/k_BT}\) of Planck at all temperatures; since it is \(T\)-independent it must be exactly unity. This is legal precisely because the \(B\)'s are intrinsic constants. C

\[ \frac{A_{21}}{B_{21}} = \frac{\hbar\omega^3}{\pi^2 c^3} \]

With the denominators matched, equate the surviving prefactors of Step 5 and Step 6. The absorption/stimulated symmetry and the spontaneous/stimulated ratio are now both fixed. B

Result

\[ g_1 B_{12} = g_2 B_{21}, \qquad \frac{A_{21}}{B_{21}} = \frac{\hbar\,\omega^3}{\pi^2 c^3} \]

Reading. Absorption and stimulated emission are the same process run in opposite directions: their rate constants are equal up to the degeneracy weighting. Spontaneous emission is tied to stimulated emission by a universal factor that scales as the cube of the transition frequency — so a single atomic matrix element controls all three, and high-frequency transitions radiate spontaneously far faster than low-frequency ones.

Units check. \(\rho(\omega)\) has units \(\mathrm{J\,s\,m^{-3}}\) (energy density per unit angular frequency), so \(B_{21}\rho\) must be a rate \(\mathrm{s^{-1}}\), giving \([B_{21}] = \mathrm{m^3\,J^{-1}\,s^{-2}}\). Then \(A_{21}/B_{21}\) has units \(\mathrm{s^{-1}}\cdot\mathrm{J\,s^2\,m^{-3}} = \mathrm{J\,s\,m^{-3}}\). On the right, \(\hbar\omega^3/c^3\) has units \(\mathrm{(J\,s)(s^{-3})/(m^3 s^{-3})} = \mathrm{J\,s\,m^{-3}}\). They agree.

Limiting cases

Low frequency / high temperature (\(\hbar\omega \ll k_BT\)): stimulated emission dominates, \(B_{21}\rho \gg A_{21}\); the field and matter exchange energy reversibly (the classical, Rayleigh–Jeans regime).
High frequency / low temperature (\(\hbar\omega \gg k_BT\)): the ratio of stimulated to spontaneous emission is \(B_{21}\rho/A_{21} = 1/(e^{\hbar\omega/k_BT}-1)\to 0\); spontaneous emission wins and radiation is essentially one-way (the Wien, "shot-noise" regime).
Equal degeneracies (\(g_1=g_2\)): the coefficients collapse to \(B_{12}=B_{21}\), the symmetric form used in laser rate equations.
Frequency scaling: at fixed matrix element \(A_{21}\propto\omega^3 B_{21}\), so radio transitions (\(\omega\) small) are metastable while X-ray transitions decay almost instantly.

Breaks when

Non-thermal or directional fields. The derivation assumed the equilibrium field is isotropic Planck radiation. In a single-mode laser cavity or a collimated beam \(\rho(\omega)\) is not the blackbody density, so while the coefficients themselves stay valid (they are atomic constants), the balance equation that defines them here does not describe the dynamics; one must use the mode-resolved form.
Multi-level or optically thick systems. When collisional transfer, cascades, or radiation trapping connect the two levels to others, \(N_2/N_1\) is no longer the two-level Boltzmann ratio and the detailed-balance step (Step 2/4) is invalid.
Degenerate broadening comparable to \(\omega\). The identification \(A/B=\hbar\omega^3/\pi^2c^3\) assumes a sharp transition frequency; for very broad lines the coefficients must be defined as integrals over the lineshape and the simple ratio holds only line-centre-averaged.

Failure modes

Frequency-density confusion. Using \(\rho(\nu)\) (per unit ordinary frequency, \(A/B = 8\pi h\nu^3/c^3\)) with a \(B\) defined against \(\rho(\omega)\). The factor \(2\pi\) mismatch changes \(A/B\) by \((2\pi)^3\)-ish factors. Fix the convention first.
Dropping degeneracies. Writing \(B_{12}=B_{21}\) unconditionally; it is only true when \(g_1=g_2\). For an \(S\to P\) transition \(g_2/g_1 = 3\) or more.
Thinking spontaneous emission is "small correction". Setting \(A_{21}=0\) at optical frequencies; in fact \(A_{21}\) dominates at room temperature because \(\hbar\omega\gg k_BT\).
Using energy density vs specific intensity. Confusing \(\rho\) (\(\mathrm{J\,s\,m^{-3}}\)) with intensity \(I\) (\(\mathrm{W\,m^{-2}}\)); they differ by a factor \(c\) and a solid-angle integration.
Claiming the result needs QED. It does not — it is pure thermodynamics; QED is only needed to compute the value of \(A_{21}\), not the ratio.

Discussion

The deepest point is that a purely thermodynamic argument — equilibrium against a blackbody bath — forces the existence of spontaneous emission. If only absorption and stimulated emission existed, Step 3 would give \(\rho\to\infty\) at equilibrium: the atoms could not stay in balance with a finite field. Einstein was compelled to invent the \(A_{21}\) term to rescue thermal equilibrium, years before it could be computed. The \(\omega^3\) factor is the same one that appears in the density of electromagnetic modes, which is no coincidence: \(A_{21}\) is stimulated emission driven by the vacuum's zero-point field, and the mode density is \(\propto\omega^2\) with an extra \(\omega\) from the photon energy.

The relation \(g_1B_{12}=g_2B_{21}\) is the microscopic statement of reciprocity between absorption and emission — a special case of the fluctuation–dissipation theorem. Because \(B_{12}\) and \(B_{21}\) share the same squared matrix element \(|\langle 2|\hat{\mathbf{d}}|1\rangle|^2\) inherited from the semiclassical selection rules, a transition that is dipole-forbidden in absorption is equally forbidden in emission, and its \(A\) coefficient is suppressed accordingly.

A subtle consequence: since \(A_{21}/B_{21}\propto\omega^3\), the ratio of spontaneous to stimulated emission into a single mode is exactly the mode's occupation-number reciprocal. Writing \(\bar{n}=(e^{\hbar\omega/k_BT}-1)^{-1}\), the total emission rate is \(A_{21}(1+\bar n)\): the "1" is spontaneous, the "\(\bar n\)" is stimulated. This is the origin of the bosonic \((1+\bar n)\) enhancement factor, and it foreshadows the second-quantised result in which spontaneous emission is stimulated emission by vacuum fluctuations. The same algebra underlies the Purcell effect, where engineering the mode density modifies \(A_{21}\) even though the atom is unchanged.

Common misconceptions. Students often think stimulated and spontaneous emission produce different kinds of photon; they do not — both emit a photon of energy \(\hbar\omega\). The difference is coherence: stimulated photons are emitted into the driving mode (same phase, direction, polarisation), spontaneous photons into a random mode. Also, "stimulated absorption" is redundant: all absorption is stimulated, which is why there is no "\(A_{12}\)".

Worked examples

Sodium D-line: at what temperature does stimulated emission match spontaneous emission?

The two rates are equal when \(B_{21}\rho = A_{21}\), i.e. when the mode occupation \(\bar n = 1\). A

\[ \frac{B_{21}\rho}{A_{21}} = \frac{1}{e^{\hbar\omega/k_BT}-1} = 1 \;\Rightarrow\; e^{\hbar\omega/k_BT}=2 \;\Rightarrow\; T = \frac{\hbar\omega}{k_B\ln 2} \]

Set the ratio to unity and solve for \(T\). Symbols first. A

\[ \hbar\omega = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34})(3.00\times10^{8})}{589\times10^{-9}} = 3.37\times10^{-19}\,\mathrm{J} = 2.11\,\mathrm{eV} \]

Insert \(\lambda = 589\,\mathrm{nm}\). A

\[ T = \frac{3.37\times10^{-19}}{(1.381\times10^{-23})(0.693)} \approx 3.5\times10^{4}\,\mathrm{K} \]

Numeric substitution. A

\[ T \approx 3.5\times10^{4}\ \mathrm{K} \]

Reading. Even at the surface temperature of a hot O-star (\(\sim3\times10^4\,\mathrm{K}\)) stimulated emission only just competes with spontaneous emission on the Na D line; at room temperature spontaneous emission dominates by a factor \(e^{\hbar\omega/k_BT}\sim 10^{35}\). This is why ordinary sources are incoherent.

Units check. \(\mathrm{J}/(\mathrm{J\,K^{-1}}) = \mathrm{K}\). Correct.

Compare \(A/B\) for an optical line and the 21 cm hydrogen line, given \(A/B_{21}=\hbar\omega^3/\pi^2c^3\).

The ratio depends only on \(\omega^3\), so the comparison is the cube of the frequency ratio. A

\[ \frac{(A/B)_{\text{opt}}}{(A/B)_{21\,\text{cm}}} = \left(\frac{\omega_{\text{opt}}}{\omega_{21}}\right)^3 = \left(\frac{\lambda_{21}}{\lambda_{\text{opt}}}\right)^3 \]

Since \(\omega = 2\pi c/\lambda\), the \(2\pi c\) factors cancel in the ratio. Symbols before numbers. B

\[ \frac{\lambda_{21}}{\lambda_{\text{opt}}} = \frac{0.211\,\mathrm{m}}{589\times10^{-9}\,\mathrm{m}} = 3.58\times10^{5} \]

Insert \(\lambda_{21}=21.1\,\mathrm{cm}\), \(\lambda_{\text{opt}}=589\,\mathrm{nm}\). A

\[ \left(3.58\times10^{5}\right)^3 \approx 4.6\times10^{16} \]

Cube the ratio. A

\[ \frac{(A/B)_{\text{opt}}}{(A/B)_{21\,\text{cm}}} \approx 5\times10^{16} \]

Reading. For the same \(B\) (same matrix element), spontaneous emission is roughly \(10^{17}\) times more effective at optical than at 21 cm frequencies. This is exactly why the 21 cm line has an emission lifetime of \(\sim10^7\) years and is observable only because interstellar hydrogen is so abundant, whereas optical lines de-excite in nanoseconds.

Units check. A dimensionless ratio; both sides are pure numbers. Correct.

Problems

Show explicitly that if spontaneous emission were absent (\(A_{21}=0\)), no finite radiation density can keep a two-level system in thermal equilibrium at any \(T>0\).
Solution
With \(A_{21}=0\), Step 2 becomes \(N_1B_{12}\rho = N_2B_{21}\rho\), so either \(\rho=0\) or \(N_1B_{12}=N_2B_{21}\). Using \(g_1B_{12}=g_2B_{21}\) the latter requires \(N_1/g_1 = N_2/g_2\), i.e. equal occupation per state, which is the \(T\to\infty\) Boltzmann limit. For any finite \(T\), \(N_2/N_1=(g_2/g_1)e^{-\hbar\omega/k_BT}<g_2/g_1\), so balance is impossible for finite \(\rho\); Step 3 gives \(\rho\to\infty\). Spontaneous emission is required.
A transition has \(A_{21}=6.16\times10^{7}\,\mathrm{s^{-1}}\) at \(\lambda=589\,\mathrm{nm}\). Compute \(B_{21}\) (per unit angular frequency energy density).
Solution
\(\omega = 2\pi c/\lambda = 2\pi(3.00\times10^8)/(589\times10^{-9}) = 3.20\times10^{15}\,\mathrm{rad\,s^{-1}}\). Then \(A/B_{21}=\hbar\omega^3/\pi^2c^3\). Compute \(\omega^3 = 3.27\times10^{46}\); \(\hbar\omega^3 = (1.055\times10^{-34})(3.27\times10^{46}) = 3.45\times10^{12}\); \(\pi^2c^3 = 9.87\times(2.70\times10^{25}) = 2.66\times10^{26}\). So \(A/B_{21}=1.30\times10^{-14}\,\mathrm{J\,s\,m^{-3}}\). Thus \(B_{21}=A_{21}/(1.30\times10^{-14}) = 6.16\times10^{7}/1.30\times10^{-14} = 4.7\times10^{21}\,\mathrm{m^3\,J^{-1}\,s^{-2}}\).
An \(S\to P\) transition has \(g_1=1\) (S state, \(J=0\)) and \(g_2=3\) (P state, \(J=1\)). If \(B_{21}=2.0\times10^{21}\,\mathrm{m^3\,J^{-1}\,s^{-2}}\), find \(B_{12}\).
Solution
\(g_1B_{12}=g_2B_{21}\Rightarrow B_{12} = (g_2/g_1)B_{21} = 3\times2.0\times10^{21} = 6.0\times10^{21}\,\mathrm{m^3\,J^{-1}\,s^{-2}}\). Absorption is three times stronger per atom than stimulated emission because there are three upper substates to reach.
At \(T=6000\,\mathrm{K}\) (solar photosphere), find the ratio of stimulated to spontaneous emission for the Na D line (\(\hbar\omega=2.11\,\mathrm{eV}\)).
Solution
\(\hbar\omega/k_BT = (2.11\,\mathrm{eV})/(8.617\times10^{-5}\,\mathrm{eV\,K^{-1}}\times6000\,\mathrm{K}) = 2.11/0.517 = 4.08\). Ratio \(= 1/(e^{4.08}-1) = 1/(59.1-1) = 1/58.1 = 0.017\). Stimulated emission contributes under 2%; spontaneous emission dominates even in the Sun.
Derive the general emission rate into a single mode as \(A_{21}(1+\bar n)\) with \(\bar n = (e^{\hbar\omega/k_BT}-1)^{-1}\), and identify each term.
Solution
Total downward radiative rate per upper atom is \(A_{21}+B_{21}\rho\). Using the results, \(B_{21}\rho = B_{21}\cdot\frac{A_{21}}{B_{21}}\cdot\frac{1}{e^{\hbar\omega/k_BT}-1} = A_{21}\bar n\). Hence total \(= A_{21}(1+\bar n)\). The "\(1\)" is spontaneous emission (present even at \(T=0\), \(\bar n=0\)); the "\(\bar n\)" is stimulated emission, proportional to the mode occupation. This is the bosonic enhancement factor and shows spontaneous emission as the \(\bar n\to\bar n+1\) "vacuum" term.