D.41 Derivation of the Einstein B coefficients

The purpose of this note is to derive the Einstein $B$ coefficients of chapter 7.8. They determine the transition rates between the energy states of atoms. For simplicity it will be assumed that there are just two atomic energy eigenstates involved, a lower energy one $\psi_{\rm {L}}$ and an higher energy one $\psi_{\rm {H}}$. It is further assumed that the atoms are subject to incoherent ambient electromagnetic radiation. The energy in the ambient radiation is $\rho(\omega)$ per unit volume and unit frequency range. Finally it is assumed that the atoms suffer frequent collisions with other atoms. The typical time between collisions will be indicated by $t_{\rm {c}}$. It is small compared to the typical decay time of the states, but large compared to the frequency of the relevant electromagnetic field.

Unlike what you may find elsewhere, it will not be assumed that the atoms are either fully in the high or fully in the low energy state. That is a highly unsatisfactory assumption for many reasons. For one thing it assumes that the atoms know what you have selected as $z$-​axis. In the derivation below, the atoms are allowed to be in a linear combination of the states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$, with coefficients $c_{\rm {L}}$ and $c_{\rm {H}}$.

Since both the electromagnetic field and the collisions are random, a statistical rather than a determinate treatment is needed. In it, the probability that a randomly chosen atom can be found in the lower energy state $\psi_{\rm {L}}$ will be indicated by $P_{\rm {L}}$. Similarly, the probability that an atom can be found in the higher energy state $\psi_{\rm {H}}$ will be indicated by $P_{\rm {H}}$. For a single atom, these probabilities are given by the square magnitudes of the coefficients $c_{\rm {L}}$ and $c_{\rm {H}}$ of the energy states. Therefore, $P_{\rm {L}}$ and $P_{\rm {H}}$ will be defined as the averages of $\vert c_{\rm {L}}\vert^2$ respectively $\vert c_{\rm {H}}\vert^2$ over all atoms.

It is assumed that the collisions are globally elastic in the sense that they do not change the average energy picture of the atoms. In other words, they do not affect the average probabilities of the eigenfunctions $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$. However, they are assumed to leave the wave function of an individual atom immediately after a collision in some state $c_{\rm {L,0}}\psi_{\rm {L}}+c_{\rm {H,0}}\psi_{\rm {H}}$ in which $c_{\rm {L,0}}$ and $c_{\rm {H,0}}$ are quite random, especially with respect to their phase. What is now to be determined in this note is how, until the next collision, the wave function of the atom will develop under the influence of the electromagnetic field and how that changes the average probabilities $\vert c_{\rm {L}}\vert^2$ and $\vert c_{\rm {H}}\vert^2$.

The evolution equations of the coefficients $\bar{c}_{\rm {L}}$ and $\bar{c}_{\rm {H}}$, in between collisions, were given in chapter 7.7.2 (7.42). They are in terms of modified variables $\bar{c}_{\rm {L}}$ and $\bar{c}_{\rm {H}}$. However, these variables have the same square magnitudes and initial conditions as $c_{\rm {L}}$ and $c_{\rm {H}}$. So it really does not make a difference.

Further, because the equations are linear, the solution for the coefficients $\bar{c}_{\rm {L}}$ and $\bar{c}_{\rm {H}}$ can be written as a sum of two contributions, one proportional to the initial value $\bar{c}_{\rm {L,0}}$ and the other to $\bar{c}_{\rm {H,0}}$:

\begin{displaymath}
\bar c_{\rm {L}} = \bar{c}_{\rm {L,0}} \bar c_{\rm {L}}^{\...
...^{\rm {L}}
+ \bar{c}_{\rm {H,0}} \bar c_{\rm {H}}^{\rm {H}}
\end{displaymath}

Here $(\bar{c}_{\rm {L}}^{\rm {L}},\bar{c}_{\rm {H}}^{\rm {L}})$ is the solution that starts out from the lower energy state $(\bar{c}_{\rm {L}}^{\rm {L}},\bar{c}_{\rm {H}}^{\rm {L}})$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(1,0)$ while $(\bar{c}_{\rm {L}}^{\rm {H}},\bar{c}_{\rm {H}}^{\rm {H}})$ is the solution that starts out from the higher energy state $(\bar{c}_{\rm {L}}^{\rm {H}},\bar{c}_{\rm {H}}^{\rm {H}})$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(0,1)$.

Now consider what happens to the probability of an atom to be in the excited state in the time interval between collisions:

\begin{displaymath}
\vert\bar{c}_{\rm {H}}\vert^2 - \vert\bar{c}_{\rm {H,0}}\v...
... {H}}^{\rm {H}})
- \bar{c}_{\rm {H,0}}^*\bar{c}_{\rm {H,0}}
\end{displaymath}

Here $\Delta{\bar{c}_{\rm {H}}^{\rm {L}}}$ indicates the change in $\bar{c}_{\rm {H}}^{\rm {L}}$ in the time interval between collisions; in particular $\Delta{\bar{c}_{\rm {H}}^{\rm {L}}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar{c}_{\rm {H}}^{\rm {L}}$ since this solution starts from the ground state with $\bar{c}_{\rm {H}}^{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Similarly, the change $\Delta{\bar{c}_{\rm {H}}^{\rm {H}}}$ equals $\bar{c}_{\rm {H}}^{\rm {H}}-1$ since this solution starts out from the excited state with $\bar{c}_{\rm {H}}^{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

Because the typical time between collisions $t_{\rm {c}}$ is assumed small, so will be the changes $\Delta{\bar{c}_{\rm {H}}^{\rm {L}}}$ and $\Delta{\bar{c}_{\rm {H}}^{\rm {H}}}$ as given by the evolution equations (7.42). Note also that $\Delta{\bar{c}_{\rm {H}}^{\rm {H}}}$ will be quadratically small, since the corresponding solution starts out from $\bar{c}_{\rm {L}}^{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, so $\bar{c}_{\rm {L}}^{\rm {H}}$ is an additional small factor in the equation (7.42) for $\bar{c}_{\rm {H}}^{\rm {H}}$.

Therefore, if the change in probability $\vert\bar{c}_{\rm {H}}\vert^2$ above is multiplied out, ignoring terms that are cubically small or less, the result is, (remember that for a complex number $c$, $c+c^*$ is twice its real part):

\begin{displaymath}
\vert\bar{c}_{\rm {H}}\vert^2 - \vert\bar{c}_{\rm {H,0}}\v...
...}}\vert^2 2 \Re\left(\Delta\bar{c}_{\rm {H}}^{\rm {H}}\right)
\end{displaymath}

Now if this is averaged over all atoms and time intervals between collisions, the first term in the right hand side will average away. The reason is that it has a random phase angle, for one since those of $\bar{c}_{\rm {L,0}}$ and $\bar{c}_{\rm {H,0}}$ are assumed to be random after a collision. For a number with a random phase angle, the real part is just as likely to be positive as negative, so it averages away. Also, for the final term, $2\Re(\Delta\bar{c}_{\rm {H}}^{\rm {H}})$ is the approximate change in $\vert\bar{c}_{\rm {H}}^{\rm {H}}\vert^2$ in the time interval, and that equals $-\vert\Delta\bar{c}_{\rm {L}}^{\rm {H}}\vert^2$ because of the normalization condition $\vert\bar{c}_{\rm {L}}^{\rm {H}}\vert^2+\vert\bar{c}_{\rm {H}}^{\rm {H}}\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. So the relevant expression for the average change in probability becomes

\begin{displaymath}
\vert\bar{c}_{\rm {H}}\vert^2 - \vert\bar{c}_{\rm {H,0}}\v...
... {H,0}}\vert^2 \vert\Delta\bar{c}_{\rm {L}}^{\rm {H}}\vert^2
\end{displaymath}

Summing the changes in the probabilities therefore means summing the changes in the square magnitudes of $\Delta\bar{c}_{\rm {H}}^{\rm {L}}$ and $\Delta\bar{c}_{\rm {L}}^{\rm {H}}$.

If the above expression for the average change in the probability of the high energy state is compared to (7.46), it is seen that the Einstein coefficient $B_{\rm {L\to{H}}}$ is the average change $\vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}\vert^2$ per unit time. This is admittedly the same answer you would get if you assumed that the atoms are either in the low energy state or in the high energy state immediately after each collision. But as noted, that assumption is simply not reasonable.

Now the needed $\Delta\bar{c}_{\rm {H}}^{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar{c}_{\rm {H}}^{\rm {L}}$ may be found from the second evolution equation (7.42). To do so, you can consider $\bar{c}_{\rm {L}}^{\rm {L}}$ to be 1. The reason is that it starts out as 1, and it never changes much because of the assumed short evolution time $t_{\rm {c}}$ compared to the typical transition time between states. That allows $\bar{c}_{\rm {H}}^{\rm {L}}$ to be found from a simple integration. And the second term in the modified Hamiltonian coefficient (7.44) can be ignored because of the additional assumption that $t_{\rm {c}}$ is still large compared to the frequency of the electromagnetic wave. That causes the exponential in the second term to oscillate rapidly and it does not integrate to a sizable contribution.

What is left is

\begin{displaymath}
\Delta\bar{c}_{\rm {H}}^{\rm {L}} = \frac{{\cal E}_{\rm {f...
...rac{e^{-{\rm i}(\omega-\omega_0)t} - 1}{2(\omega-\omega_0)} %
\end{displaymath} (D.25)

and $\Delta\bar{c}_{\rm {L}}^{\rm {H}}$ is given by a virtually identical expression. However, since it is assumed that the atoms are subject to incoherent radiation of all wave numbers ${\vec k}$ and polarizations $p$, the complete $\Delta\bar{c}_{\rm {H}}^{\rm {L}}$ will consist of the sum of all their contributions:

\begin{displaymath}
\Delta\bar{c}_{\rm {H}}^{\rm {L}}
= \sum_{{\vec k},p} \Delta\bar{c}_{\rm {H}}^{\rm {L}}({\vec k},p)
\end{displaymath}

(This really assumes that the particles are in a very large periodic box so that the electromagnetic field is given by a Fourier series; in free space you would need to integrate over the wave numbers instead of sum over them.) The square magnitude is then

\begin{displaymath}
\vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}\vert^2 =
\sum_{{\...
...,p} \vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}({\vec k},p)\vert^2
\end{displaymath}

where the final equality comes from the assumption that the radiation is incoherent, so that the phases of different waves are uncorrelated and the corresponding products average to zero.

The bottom line is that square magnitudes must be summed together to find the total contribution of all waves. And the square magnitude of the contribution of a single wave is, according to (D.25) above,

\begin{displaymath}
\vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}({\vec k},p)\vert^2 ...
...t)}
{{\textstyle\frac{1}{2}}(\omega-\omega_0)t}
\right)^2
\end{displaymath}

Now broadband radiation is described in terms of an electromagnetic energy density $\rho(\omega)$; in particular $\rho(\omega){\,\rm d}\omega$ gives the energy per unit volume due to the electromagnetic waves in an infinitesimal frequency range ${\rm d}\omega$ around a frequency $\omega$. For a single wave, this energy equals $\frac12\epsilon_0{\cal E}_{\rm {f}}^2$, chapter 13.2 (13.11). And the square amplitudes of different waves simply add up to the total energy; that is the so-called Parseval equality of Fourier analysis. So to sum the expression above over all the frequencies $\omega$ of the broadband radiation, make the substitution ${\cal E}_{\rm {f}}^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\rho(\omega){\,\rm d}\omega$$\raisebox{.5pt}{$/$}$$\epsilon_0$ and integrate:

\begin{displaymath}
\vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}\vert^2
=
\frac{...
...}}(\omega-\omega_0)t_{\rm {c}}}
\right)^2
{\,\rm d}\omega
\end{displaymath}

If a change of integration variable is made to $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}(\omega-\omega_0)t_{\rm {c}}$, the integral becomes

\begin{displaymath}
\vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}\vert^2
=
\frac{...
...(u/t_{\rm {c}}))
\left(\frac{\sin u}{u}\right)^2 {\,\rm d}u
\end{displaymath}

Recall that a starting assumption underlying these derivations was that $\omega_0t_{\rm {c}}$ was large. So the lower limit of integration can be approximated as $\vphantom0\raisebox{1.5pt}{$-$}$$\infty$.

Note that this is essentially the same analysis as the one for Fermi’s golden rule, except for the presence of the given field strength $\rho$. However, here the mathematics can be performed more straightforwardly, using integration rather than summation.

Consider for a second the limiting process that the field strength $\rho$ goes to zero, and that the atom is kept isolated enough that the collision time $t_{\rm {c}}$ can increase correspondingly. Then the term $2u$$\raisebox{.5pt}{$/$}$$t_{\rm {c}}$ in the argument of $\rho$ will tend to zero. So only waves with the exact frequency $\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\omega_0$ will produce transitions in the limit of zero field strength. That confirms the basic claim of quantum mechanics that only the energy eigenvalues are measurable. In the absence of an electromagnetic field and other disturbances, the energy eigenvalues are purely the atomic ones. (Also recall that relativistic quantum mechanics adds that in reality, the electric field is never zero.)

In any case, while the term $2u$$\raisebox{.5pt}{$/$}$$t_{\rm {c}}$ may not be exactly zero, it is certainly small compared to $\omega_0$ because of the assumption that $\omega_0t_{\rm {c}}$ is large. So the term may be ignored anyway. Then $\rho(\omega_0)$ is a constant in the integration and can be taken out. The remaining integral is in table books, [40, 18.36], and the result is

\begin{displaymath}
\vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}\vert^2
=
\frac{...
...\rm {H}}\rangle\vert^2}{\hbar^2\epsilon_0}
\rho(\omega_0) t
\end{displaymath}

This must still be averaged over all directions of wave propagation and polarization. That gives:

\begin{displaymath}
\vert\Delta\bar{c}_{\rm {H}}^{\rm {L}}\vert^2
=
\frac{...
...e\vert^2}
{3\hbar^2\epsilon_0}
\rho(\omega_0) t_{\rm {c}}
\end{displaymath}

where

\begin{displaymath}
\vert\langle\psi_{\rm {L}}\vert e{\skew0\vec r}\vert\psi_{...
...angle\psi_{\rm {L}}\vert ez\vert\psi_{\rm {H}}\rangle\vert^2.
\end{displaymath}

To see why, consider the electromagnetic waves propagating along any axis, not just the $y$-​axis, and polarized in either of the other two axial directions. These waves will include $ex$ and $ey$ as well as $ez$ in the transition probability, making the average as shown above. And of course, waves propagating in an oblique rather than axial direction are simply axial waves when seen in a rotated coordinate system and produce the same average.

The Einstein coefficient $B_{\rm {L\to{H}}}$ is the average change per unit time, so the claimed (7.47) results from dividing by the time $t_{\rm {c}}$ between collisions. There is no need to do $B_{\rm {H\to{L}}}$ separately from $\Delta\bar{c}_{\rm {L}}^{\rm {L}}$; it follows immediately from the symmetry property mentioned at the end of chapter 7.7.2 that it is the same.