path: root/content/know/concept/einstein-coefficients/index.pdc

---
title: "Einstein coefficients"
firstLetter: "E"
publishDate: 2021-07-11
categories:
- Physics
- Optics
- Quantum mechanics

date: 2021-07-11T18:22:14+02:00
draft: false
markup: pandoc
---

# Einstein coefficients

The **Einstein coefficients** quantify
the emission and absorption of photons by a solid,
and can be calculated analytically from first principles
in several useful situations.


## Qualitative description

Suppose we have a ground state with energy $E_1$ containing $N_1$ electrons,
and an excited state with energy $E_2$ containing $N_2$ electrons.
The resonance $\omega_0 \equiv (E_2 \!-\! E_1)/\hbar$
is the frequency of the photon emitted
when an electron falls from $E_2$ to $E_1$.

The first Einstein coefficient is the **spontaneous emission rate** $A_{21}$,
which gives the probability per unit time
that an excited electron falls from state 2 to 1,
so that $N_2(t)$ obeys the following equation,
which is easily solved:

$$\begin{aligned}
    \dv{N_2}{t} = - A_{21} N_2
    \quad \implies \quad
    N_2(t) = N_2(0) \exp\!(- t / \tau)
\end{aligned}$$

Where $\tau = 1 / A_{21}$ is the **natural radiative lifetime** of the excited state,
which gives the lifetime of an excited electron,
before it decays to the ground state.

The next coefficient is the **absorption rate** $B_{12}$,
which is the probability that an incoming photon excites an electron,
per unit time and per unit spectral energy density
(i.e. the rate depends on the frequency of the incoming light).
Then $N_1(t)$ obeys the following equation:

$$\begin{aligned}
    \dv{N_1}{t} = - B_{12} N_1 u(\omega_0)
\end{aligned}$$

Where $u(\omega)$ is the spectral energy density of the incoming light,
put here to express the fact that only photons with frequency $\omega_0$ are absorbed.

There is one more Einstein coefficient: the **stimulated emission rate** $B_{21}$.
An incoming photon has an associated electromagnetic field,
which can encourage an excited electron to drop to the ground state,
such that for $A_{21} = 0$:

$$\begin{aligned}
    \dv{N_2}{t} = - B_{21} N_2 u(\omega_0)
\end{aligned}$$

These three coefficients $A_{21}$, $B_{12}$ and $B_{21}$ are related to each other.
Suppose that the system is in equilibrium,
i.e. that $N_1$ and $N_2$ are constant.
We assume that the number of particles in the system is constant,
implying that $N_1'(t) = - N_2'(t) = 0$, so:

$$\begin{aligned}
    B_{12} N_1 u(\omega_0) = A_{21} N_2 + B_{21} N_2 u(\omega_0) = 0
\end{aligned}$$

Isolating this equation for $u(\omega_0)$,
gives following expression for the radiation:

$$\begin{aligned}
    u(\omega_0)
    = \frac{A_{21}}{(N_1 / N_2) B_{12} - B_{21}}
\end{aligned}$$

We assume that the system is in thermal equilibrium
with its own black-body radiation, and that there is no external light.
Then this is a [canonical ensemble](/know/concept/canonical-ensemble/),
meaning that the relative probability that an electron has $E_2$ compared to $E_1$
is given by the Boltzmann distribution:

$$\begin{aligned}
    \frac{\mathrm{Prob}(E_2)}{\mathrm{Prob}(E_1)}
    = \frac{N_2}{N_1}
    = \frac{g_2}{g_1} \exp\!(- \hbar \omega_0 \beta)
\end{aligned}$$

Where $g_2$ and $g_1$ are the degeneracies of the energy levels.
Inserting this back into the equation for the spectrum $u(\omega_0)$ yields:

$$\begin{aligned}
    u(\omega_0)
    = \frac{A_{21}}{(g_1 / g_2) B_{12} \exp\!(\hbar \omega_0 \beta) - B_{21}}
\end{aligned}$$

Since $u(\omega_0)$ represents only black-body radiation,
our result must agree with Planck's law:

$$\begin{aligned}
    u(\omega_0)
    = \frac{A_{21}}{B_{21} \big( (g_1 B_{12} / g_2 B_{21}) \exp\!(\hbar \omega_0 \beta) - 1 \big)}
    = \frac{\hbar \omega_0^3}{\pi^2 c^3} \frac{1}{\exp\!(\hbar \omega_0 \beta) - 1}
\end{aligned}$$

This gives us the following two equations relating the Einstein coefficients:

$$\begin{aligned}
    \boxed{
        A_{21} = \frac{\hbar \omega_0^3}{\pi^2 c^3} B_{21}
        \qquad \quad
        g_1 B_{12} = g_2 B_{21}
    }
\end{aligned}$$

Note that this result holds even if $E_1$ is not the ground state,
but instead some lower excited state below $E_2$,
due to the principle of *detailed balance*.
Furthermore, it turns out that these relations
also hold if the system is not in equilibrium.

A notable case is **population inversion**,
where $B_{21} N_2 > B_{12} N_1$ such that $N_2 > (g_2 / g_1) N_1$.
This situation is mandatory for lasers, where stimulated emission must dominate,
such that the light becomes stronger as it travels through the medium.


## Electric dipole approximation

In fact, we can analytically calculate the Einstein coefficients,
if we make a mild approximation.
Consider the Hamiltonian of an electron with charge $q = - e$:

$$\begin{aligned}
    \hat{H}
    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + q \phi
\end{aligned}$$

With $\vec{A}(\vec{r}, t)$ the magnetic vector potential,
and $\phi(\vec{r}, t)$ the electric scalar potential.
We reduce this by fixing the Coulomb gauge $\nabla \!\cdot\! \vec{A} = 0$,
such that $\vec{A} \cdot \vec{P} = \vec{P} \cdot \vec{A}$,
and by assuming that $\vec{A}{}^2$ is negligibly small.
This leaves us with:

$$\begin{aligned}
    \hat{H}
    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{m} \vec{P} \cdot \vec{A} + q \phi
\end{aligned}$$

The last term is the Coulomb interaction
between the electron and the nucleus.
We can interpret the second term, involving the weak $\vec{A}$, as a perturbation $\hat{H}_1$:

$$\begin{aligned}
    \hat{H}
    = \hat{H}_0 + \hat{H}_1
    \qquad \quad
    \hat{H}_0
    \equiv \frac{\vec{P}{}^2}{2 m} + q \phi
    \qquad \quad
    \hat{H}_1
    \equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
\end{aligned}$$

Suppose that $\vec{A}$ is oscillating sinusoidally in time and space as follows:

$$\begin{aligned}
    \vec{A}(\vec{r}, t) = \vec{A}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
\end{aligned}$$

The corresponding perturbative electric field $\vec{E}$ points in the same direction:

$$\begin{aligned}
    \vec{E}(\vec{r}, t)
    = - \pdv{\vec{A}}{t}
    = \vec{E}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
\end{aligned}$$

Where $\vec{E}_0 = i \omega \vec{A}_0$.
Let us restrict ourselves to visible light,
whose wavelength $2 \pi / k \approx 10^{-6} \:\mathrm{m}$.
By comparison, the size of an atomic orbital is on the order of $10^{-10} \:\mathrm{m}$,
so we can ignore the dot product $\vec{k} \cdot \vec{r}$.
This is the **electric dipole approximation**:
the radiation is treated classicaly,
while the electron is treated quantum-mechanically.

$$\begin{aligned}
    \vec{E}(\vec{r}, t)
    \approx \vec{E}_0 \exp\!(- i \omega t)
\end{aligned}$$

Next, we want to convert $\hat{H}_1$
to use the electric field $\vec{E}$ instead of the potential $\vec{A}$.
To do so, we rewrite the momemtum $\vec{P} = m \: \dv*{\vec{r}}{t}$
and evaluate this in the [Heisenberg picture](/know/concept/heisenberg-picture/):

$$\begin{aligned}
    \matrixel{2}{\dv*{\vec{r}}{t}}{1}
    &= \frac{i}{\hbar} \matrixel{2}{[\hat{H}_0, \vec{r}]}{1}
    = \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vec{r} - \vec{r} \hat{H}_0}{1}
    \\
    &= \frac{i}{\hbar} (E_2 - E_1) \matrixel{2}{\vec{r}}{1}
    = i \omega_0 \matrixel{2}{\vec{r}}{1}
\end{aligned}$$

Therefore, $\vec{P} / m = i \omega_0 \vec{r}$,
where $\omega_0 = (E_2 - E_1) / \hbar$ is the resonance frequency of the transition,
close to which we assume that $\vec{A}$ and $\vec{E}$ are oscillating.
We thus get:

$$\begin{aligned}
    \hat{H}_1(t)
    &= - \frac{q}{m} \vec{P} \cdot \vec{A}
    = - i q \omega_0 \vec{r} \cdot \vec{A}_0 \exp\!(- i \omega t)
    \\
    &= - q \vec{r} \cdot \vec{E}_0 \exp\!(- i \omega t)
    = - \vec{p} \cdot \vec{E}_0 \exp\!(- i \omega t)
\end{aligned}$$

Where $\vec{p} \equiv q \vec{r} = - e \vec{r}$ is the electric dipole moment of the electron,
hence the name *electric dipole approximation*.


## Polarized light

This form of $\hat{H}_1$ is a well-known case for
[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
which tells us that the transition probability from $\ket{a}$ to $\ket{b}$ is:

$$\begin{aligned}
    P_{ab}
    = \frac{\big|\!\matrixel{a}{H_1}{b}\!\big|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ba} - \omega) t / 2 \big)}{(\omega_{ba} - \omega)^2}
\end{aligned}$$

If the location of the nucleus of the atom has $z = 0$,
then generally $\ket{1}$ and $\ket{2}$ will be even or odd functions of $z$,
such that $\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$, leading to:

$$\begin{gathered}
    \matrixel{1}{H_1}{2} = - q E_0 V
    \qquad
    \matrixel{2}{H_1}{1} = - q E_0 V^*
    \\
    \matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
\end{gathered}$$

Where $V \equiv \matrixel{1}{z}{2}$ is a constant.
The chance of an upward jump (i.e. absorption) is:

$$\begin{aligned}
    P_{12}
    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
\end{aligned}$$

Meanwhile, the transition probability for stimulated emission is as follows,
using the fact that $P_{ab}$ is a sinc-function,
and is therefore symmetric around $\omega_{ba}$:

$$\begin{aligned}
    P_{21}
    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
\end{aligned}$$

Surprisingly, the probabilities of absorption and stimulated emission are the same!
In practice, however, the relative rates of these two processes depends heavily on
the availability of electrons and holes in both states.

In theory, we could calculate the transition rate $R_{12} = \pdv*{P_{12}}{t}$,
which would give us Einstein's absorption coefficient $B_{12}$,
for this particular case of coherent monochromatic light.
However, the result would not be constant in time $t$.

To solve this "problem", we generalize to (incoherent) polarized polychromatic light.
To do so, we note that the energy density $u$ of an electric field $E_0$ is given by:

$$\begin{aligned}
    u = \frac{1}{2} \varepsilon_0 E_0^2
    \qquad \implies \qquad
    E_0^2 = \frac{2 u}{\varepsilon_0}
\end{aligned}$$

Putting this in the previous result gives the following transition probability:

$$\begin{aligned}
    P_{12}
    = \frac{2 u q^2 |V|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
\end{aligned}$$

For a continuous light spectrum,
this $u$ turns into the spectral energy density $u(\omega)$:

$$\begin{aligned}
    P_{12}
    = \frac{2 q^2 |V|^2}{\varepsilon_0 \hbar^2}
    \int_0^\infty \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2} u(\omega) \dd{\omega}
\end{aligned}$$

From here, we the derivation is similar to that of
[Fermi's golden rule](/know/concept/fermis-golden-rule/),
despite the distinction that we are integrating over frequencies rather than states.

At sufficiently large $t$, the integrand is sharply peaked at $\omega = \omega_0$
and negligible everywhere else,
so we take $u(\omega)$ out of the integral and extend the integration limits.
Then we rewrite and look up the integral,
which turns out to be $\pi t$:

$$\begin{aligned}
    P_{12}
    = \frac{q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
\end{aligned}$$

From this, the transition rate $R_{12} = B_{12} u(\omega_0)$
is then calculated as follows:

$$\begin{aligned}
    R_{12}
    = \pdv{P_{2 \to 1}}{t}
    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
\end{aligned}$$

Using the relations from earlier with $g_1 = g_2$,
the Einstein coefficients are found to be as follows
for a polarized incoming light spectrum:

$$\begin{aligned}
    \boxed{
        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2}
        \qquad
        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{\pi \varepsilon \hbar c^3}
    }
\end{aligned}$$


## Unpolarized light

We can generalize the above result even further to unpolarized light.
Let us return to the matrix elements of the perturbation $\hat{H}_1$,
and define the polarization unit vector $\vec{n}$:

$$\begin{aligned}
    \matrixel{1}{\hat{H}_1}{2}
    = - q \matrixel{1}{\vec{r} \cdot \vec{E}_0}{2}
    = - q E_0 \matrixel{1}{\vec{r} \cdot \vec{n}}{2}
    = - q E_0 W
\end{aligned}$$

The goal is to obtain the average of $|W|^2$,
where $W \equiv \matrixel{1}{\vec{r} \cdot \vec{n}}{2}$.
In [spherical coordinates](/know/concept/spherical-coordinates/),
we integrate over all possible orientations $\vec{n}$ for fixed $\vec{r}$,
using that $\vec{r} \cdot \vec{n} = |\vec{r}| \cos\!(\theta)$:

$$\begin{aligned}
    \expval{|W|^2}
    = \frac{1}{4 \pi} \int_0^\pi \int_0^{2 \pi} |\matrixel{1}{\vec{r}}{2}|^2 \cos^2(\theta) \sin\!(\theta) \dd{\varphi} \dd{\theta}
\end{aligned}$$

Where we have divided by $4\pi$ (the surface area of a unit sphere) for normalization,
and $\theta$ is the polar angle between $\vec{n}$ and $\vec{p}$.
Evaluating the integrals yields:

$$\begin{aligned}
    \expval{|W|^2}
    = \frac{2 \pi}{4 \pi} |V|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
    = \frac{|V|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
    = \frac{|V|^2}{3}
\end{aligned}$$

With this additional constant factor $1/3$,
the transition rate $R_{12}$ is modified to:

$$\begin{aligned}
    R_{12}
    = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
\end{aligned}$$

From which it follows that the Einstein coefficients for unpolarized light are given by:

$$\begin{aligned}
    \boxed{
        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2}
        \qquad
        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{3 \pi \varepsilon \hbar c^3}
    }
\end{aligned}$$



## References
1.  M. Fox,
    *Optical properties of solids*, 2nd edition,
    Oxford.
2.  D.J. Griffiths, D.F. Schroeter,
    *Introduction to quantum mechanics*, 3rd edition,
    Cambridge.