Phy5646/Einstein coefficients example

(Submitted by Team 1)

This example was taken from "Theory and Problems of Quantum Mechanics", Yoaf Peleg, et al, p. 297-298.

Problem: A two-level system with eigenvalues $E_{2}>E_{1}$ is in the thermodynamics equilibrium with a heat reservoir at absolute temperature T. The system undergoes the following transitions: (i) Absorption $1\rightarrow 2$, (ii) induced emission $2\rightarrow 1$, and (iii) spontaneous emission $2\rightarrow 1$. The transition rates for each of these processes are given by:

$W_{21}^{abs}=P_{1}B_{21}u(w_{21})$

$W_{12}^{ind}=P_{2}B_{12}u(w_{21})$

$W_{12}^{spon}=P_{2}A_{12}$

where $u(w_{21})$ is the energy distribution of the radiation field, $P_{j}$ is the probability of finding the system in level j of degeneracy $g_{j}$ (j=1,2), and $A_{12}$ and $B_{12}$ are the Einstein coefficients for spontaneous and induced emission, respectively. (a) Calculate the probabilities $P_{1}$ and $P_{2}$ under equilibrium conditions. (b) Use the rates together with Planck's formula for black body radiation to show that

$g_{1}B_{21}=g_{2}B_{12}$

$A_{12}={\frac {\hbar w_{21}^{3}}{\pi ^{2}c^{3}}}B_{12}$

Solution (a):

Under thermal equilibrium at absolute temperature T, the probability of finding the system in one of its stationary states |i> with and eigenvalue $\epsilon _{i}$ is proportional to the Boltzmann factor $e^{\frac {-\epsilon _{i}}{kT}}$. In this problem $\epsilon _{i}$ assumes the value $E_{1}$, $E_{2}$ with respective degenerecies $g_{i}=g_{1},g_{2}$ (a two-level system). Therefore,

$P_{1}=Cg_{1}e^{\frac {-E_{1}}{kT}}$

$P_{2}=Cg_{2}e^{\frac {-E_{2}}{kT}}$

where C is the normalization constant. Since $P_{1}+P_{2}=1$, we immediately find that

$C^{-1}=g_{1}e^{\frac {-E_{1}}{kT}}+g_{2}e^{\frac {-E_{2}}{kT}}$

Since $E_{2}-E_{1}=\hbar w_{21}$, we have

${\frac {P_{1}}{P_{2}}}={\frac {g_{1}}{g_{2}}}e^{\frac {\hbar w_{21}}{kT}}$

Solution (b):

Suppose that a larger number of systems, such as in part (a), form a closed cavity that is kept in equilibrium with its own thermal radiation at constant temperature T. In this case,

$W_{21}^{abs}=W_{12}^{ind}+W_{12}^{spon}$

Then from the transitions rates, we obtain

$P_{1}B_{21}u(w_{21})=P_{2}B_{12}u(w_{21})+P_{2}A_{12}$

or from the last result in part (a)

$g_{1}B_{21}(e^{\frac {\hbar w_{21}}{kT}}-{\frac {g_{2}B_{12}}{g_{1}B_{21}}}){\frac {\hbar w_{21}^{3}}{\pi ^{2}c^{3}}}=g_{2}A_{12}(e^{\frac {\hbar w_{21}}{kT}}-1)$

Hence,

$g_{1}B_{21}=g_{2}B_{12}$

$A_{12}={\frac {\hbar w_{21}^{3}}{\pi ^{2}c^{3}}}B_{12}$