# Einstein Coefficients

Einstein coefficients describe the absorption and emission of photons via electronic transitions in atoms. Suppose we have an atom with 2 energy levels with an energy difference of ${\displaystyle \Delta E=h\nu _{0}}$. Einstein coefficients describe the transition rates caused by the interaction of radiation with these discrete energy levels. There are three coefficients:

Left: Photon absorption rates are described by ${\displaystyle B_{12}}$. Center: Spontaneous photon emission rates are described by ${\displaystyle {A_{21}}}$. Right: Stimulated photon emission rates are described by ${\displaystyle B_{21}}$.

### 1 Spontaneous Emission, ${\displaystyle {A_{21}}}$

${\displaystyle {A_{21}}}$ governs decay from energy state 2 to 1. It is the transition probability per unit time for an atom, and has units of ${\displaystyle s^{-1}}$. More specifically, the probability an atom undergoes spontaneous de-excitation and releases a photon is Poisson-distributed, with mean rate ${\displaystyle {A_{21}}}$. So ${\displaystyle {A_{21}}^{-1}}$ is the mean lifetime of the excited state. As an example, ${\displaystyle H_{\alpha }}$ (${\displaystyle 3\to 2}$ transition in hydrogen) has an Einstein A coefficient of ${\displaystyle {A_{21}}\approx 10^{9}s^{-1}}$.

If ${\displaystyle n_{2}}$ describes the number density of atoms in the upper energy state, then the transition rate per volume is given by:

${\displaystyle r_{\rm {spon}}=n_{2}{A_{21}}\,\!}$

### 2 Spontaneous Absorption, ${\displaystyle {B_{12}}}$

${\displaystyle {B_{12}}}$ governs photon absorption that causes a transition from the lower to upper energy state (${\displaystyle 1\to 2}$). In contrast to the ${\displaystyle {A_{21}}}$ case, absorption requires the presence of photons, so translating ${\displaystyle {B_{12}}}$ to an excitation rate requires some knowledge of the background radiation field.

To describe the background radiation field, we define the spherically averaged specific intensity:

${\displaystyle J_{\nu }\equiv {\frac {1}{4\pi }}\int {I_{\nu }d\Omega }\,\!}$

We use ${\displaystyle J_{\nu }}$ instead of ${\displaystyle I_{\nu }}$ (the intensity) because atomic absorption does not depend on direction. However, we have to remember that there are uncertainties in the energy-level separations, which means that atoms absorb photons that are not perfectly tuned to the energy difference between electronic states. To incorporate this, we use the line profile function, ${\displaystyle \phi (\nu )}$. It describes the relative absorption probability around ${\displaystyle \nu _{0}}$ (the absorption frequency), and is subject to the requirement that: ${\displaystyle \int _{0}^{\infty }{\phi (\nu )d\nu }=1}$. We can approximate the width of ${\displaystyle \phi (\nu )}$ as an effective width ${\displaystyle \Delta \nu }$. ${\displaystyle \Delta \nu }$ is affected by many factors:

• ${\displaystyle {A_{21}}}$ (the natural, uncertainty-based broadening of at atom in isolation),
• ${\displaystyle \nu _{0}v_{\rm {therm}}/c}$ (Doppler broadening from thermal motion), and
• ${\displaystyle n_{coll}\sigma _{coll}v_{rel}}$ (collisional broadening, a.k.a. pressure broadening).

Line profile functions are of special interest for studying line emission/absorption, and is discussed later in more detail (see Line Profile Functions).

Using the line profile function, we get the transition probability per unit time associated with spontaneous absorption:

${\displaystyle t_{ex}^{-1}={B_{12}}\int _{0}^{\infty }{J_{\nu }\phi (\nu )d\nu }\approx {B_{12}}{\bar {J}}\,\!}$

### 3 Stimulated Emission, ${\displaystyle {B_{21}}}$

${\displaystyle {B_{21}}}$ governs stimulated emission. In this example, we are in energy state 2, and an incoming photon causes a transition to energy level 1 and the emission of 2 photons. The transition per unit time is ${\displaystyle {B_{21}}{\bar {J}}}$. For an important case of stimulated emission, see Masers.

## 1 Einstein Relations among coefficients

Assume we have many atoms with 2 energy states, and ${\displaystyle n_{1}}$ is the # density in state 1, ditto for ${\displaystyle n_{2}}$. Assume we are in thermal, steady-state equilibrium, so:

${\displaystyle n_{1}{B_{12}}{\bar {J}}=n_{2}{A_{21}}+n_{2}{B_{21}}{\bar {J}}\,\!}$

This is because as many atoms need to be going from energy state 1 to 2 as visa versa. A second relation is: ${\displaystyle {\bar {J}}={n_{2}{A_{21}} \over n_{1}{B_{12}}-n_{2}{B_{21}}}}$. Using the Boltzmann distribution, ${\displaystyle {n_{2} \over n_{1}}={g_{2} \over g_{1}}e^{-h\nu _{0} \over kT}}$:

${\displaystyle {\bar {J}}={{{A_{21}} \over {B_{21}}} \over {g_{1}{B_{12}} \over g_{2}{B_{21}}}e^{h\nu _{0} \over kT}-1}\,\!}$

In thermal equilibrium ${\displaystyle J_{\nu }}$ is given by the Planck Function:

{\displaystyle {\begin{aligned}{\bar {J}}&\equiv \int _{0}^{\infty }{J_{\nu }\phi (\nu )d\nu }\\&=\int _{0}^{\infty }{B_{\nu }\phi (\nu )d\nu }\\&\approx B_{\nu }(\nu _{0})\\&={2h\nu _{0}^{3} \over c^{2}(e^{h\nu _{0} \over kT}-1)}\\\end{aligned}}\,\!}

Combining this with ${\displaystyle {\bar {J}}}$ earlier, we get:

${\displaystyle {g_{1}{B_{12}}=g_{2}{B_{21}}}\,\!}$

and

${\displaystyle {{A_{21}} \over {B_{21}}}={2h\nu ^{3} \over c^{2}}\,\!}$

## 2 Rewriting ${\displaystyle j_{\nu },\alpha _{\nu }}$ in terms of Einstein coeffs

In a small volume ${\displaystyle dV}$:

{\displaystyle {\begin{aligned}j_{\nu }&\equiv {dE \over dt\,dV\,d\nu \,d\Omega }\\&={h\nu _{0}{A_{21}}n_{2}\phi (\nu ) \over 4\pi }\\\end{aligned}}\,\!}

We can express the extinction coefficient, ${\displaystyle \alpha _{\nu }}$, in terms of the Einstein coefficients. The excitation probability per time is ${\displaystyle n_{1}B_{12}{\bar {J}}}$, and the energy lost in crossing the small volume ${\displaystyle \propto n_{1}{B_{12}}{I_{\nu }d\Omega \over 4\pi }\phi (\nu )d\nu }$ (it is the probability per time per volume of going ${\displaystyle 1\to 2}$ by absorbing ${\displaystyle I_{\nu }}$ from a cone of solid angle ${\displaystyle d\Omega }$ and frequency range ${\displaystyle [\nu ,\nu +d\nu ]}$). Thus, the energy is given by:

{\displaystyle {\begin{aligned}E&=n_{1}{B_{12}}{I_{\nu }d\Omega \over 4\pi }\phi (\nu )d\nu h\nu dt\,dV\\&=\alpha _{\nu }I_{\nu }ds\,dt\,d\Omega \,dA\,d\nu \\\end{aligned}}\,\!}

Recognizing that ${\displaystyle dV=dA\,ds}$:

${\displaystyle \alpha _{\nu }={n_{1}{B_{12}}\phi (\nu ) \over 4\pi }h\nu \,\!}$

Correcting for stimulated emission, we get:

${\displaystyle {\alpha _{\nu }={(n_{1}{B_{12}}-n_{2}{B_{21}})\phi (\nu )h\nu \over 4\pi }}\,\!}$

## 3 Estimating Cross-Sections

The absorption coefficient, written in terms of Einstein constants is:

${\displaystyle \alpha _{\nu }={n_{1}{B_{12}}\phi (\nu ) \over 4\pi }h\nu =n_{1}\sigma _{12}\,\!}$

Thus, the cross-section of an atom for absorption of a photon is:

${\displaystyle \sigma _{12}={{B_{12}}\phi (\nu )h\nu \over 4\pi }\,\!}$

To estimate ${\displaystyle {B_{12}}}$, we use the fact that, ignoring g’s, ${\displaystyle {B_{12}}\sim {B_{21}}}$, and ${\displaystyle {{A_{21}} \over {B_{21}}}={2h\nu ^{3} \over c^{2}}}$. Then using the approximation that that ${\displaystyle \phi (\nu )\sim {\frac {1}{\Delta \nu }}}$, we get:

${\displaystyle \sigma _{12}\sim {{A_{21}} \over \left({2h\nu ^{3} \over c^{2}}\right)}{\frac {h\nu }{4\pi \Delta \nu }}\,\!}$
${\displaystyle {\sigma _{12}\sim {\lambda ^{2} \over 8\pi }{{A_{21}} \over \Delta \nu }}\,\!}$

In a single atom, ${\displaystyle \Delta \nu \sim {A_{21}}}$, so ${\displaystyle \sigma _{12}\sim {\lambda ^{2} \over 8\pi }}$.

## 4 Einstein coefficients: a closer look

Why do we get spontaneous decay and stimulated emission? Quantum mechanics has the answer – see below (warning: all of the following is done in SI units).

These processes can be understood using a bit of quantum mechanics. To do so, let’s consider a system with two energy eigenstates, ${\displaystyle \psi _{a}}$ and ${\displaystyle \psi _{b}}$. In general, the particle will be in some linear combination of ${\displaystyle \psi _{a}}$ and ${\displaystyle \psi _{b}}$:

${\displaystyle \Psi =c_{a}\psi _{a}+c_{b}\psi _{b}\,\!}$

The probabilities are ${\displaystyle |c_{a}|^{2}}$ and ${\displaystyle |c_{b}|^{2}}$, and they of course must sum to unity at all times.

Now, let’s suppose the electron starts off in ${\displaystyle \psi _{a}}$ at ${\displaystyle t=0}$, so that ${\displaystyle c_{a}(t=0)=1}$ and ${\displaystyle c_{b}(t=0)=0}$. To first-order, time dependent perturbation theory tells us:

$\displaystyle c_ b \approx \frac{-i}{\hbar }\int _0^ t H_{ba}e^{i\omega _0t'}dt’, \,\!$

where $\displaystyle H’_{ba} \equiv \langle \psi _ a | \hat{H'} | \psi _ b$ is the off diagonal component of the perturbing Hamiltonian $\displaystyle \hat{H}’$ and ${\displaystyle \omega _{0}\equiv (E_{b}-E_{a})/\hbar >0}$. The derivation of this result is too detailed to reproduce here, but can be found in any quantum mechanics textbook. But, note that even though the system started off entirely in ${\displaystyle \psi _{a}}$, the introduction of a perturbation means that the probabilities can change as a function of a time – eventually, it may be that ${\displaystyle c_{b}(t)=1}$ and the particle is entirely in ${\displaystyle \psi _{b}}$. This is exactly what a quantum transition is!

Now, let’s consider what happens when our perturbation is modulated by a sinusoidal time dependence – say, an electric field polarized in the ${\displaystyle z}$ direction:

$\displaystyle \hat{H}’ = V(x,y,z)\cos (\omega t) = -qE_0z\cos (\omega t) \,\!$

Then:

${\displaystyle H_{ba}=-\alpha E_{0}\cos(\omega t),\,\!}$

where we will now use:

${\displaystyle \alpha \equiv q\langle \psi _{b}|z|\psi _{a}\rangle \,\!}$

Plugging into our expression from perturbation theory and integrating we get:

${\displaystyle c_{b}(t)={\frac {-V_{ba}}{2\hbar }}\left[{\frac {e^{i(\omega _{0}+\omega )t}-1}{\omega _{0}+\omega }}+{\frac {e^{i(\omega _{0}-\omega )t}-1}{\omega _{0}-\omega }}\right].\,\!}$

Let’s now consider only perturbations oscillating at near the transition frequency – i.e., ${\displaystyle \omega _{0}\approx \omega }$. This is a reasonable thing to do since driving frequencies far from ${\displaystyle \omega _{0}}$ contribute negligibly to the above expression. Then we end up with:

${\displaystyle c_{b}(t)={\frac {i\alpha E_{0}}{\hbar }}{\frac {\sin \left({\frac {(\omega _{0}-\omega )t}{2}}\right)}{\omega _{0}-\omega }}e^{i(\omega _{0}-\omega )t/2}.\,\!}$

The probability that the particle has “transitioned” from its initial state ${\displaystyle \psi _{a}}$ to the other state ${\displaystyle \psi _{b}}$ is just given by ${\displaystyle |c_{b}|^{2}}$:

${\displaystyle P_{a\rightarrow b}(t)=-{\frac {|\alpha E_{0}|^{2}}{\hbar ^{2}}}{\frac {\sin ^{2}\left({\frac {(\omega _{0}-\omega )t}{2}}\right)}{(\omega _{0}-\omega )^{2}}}.\,\!}$

But nothing in this derivation relied on the fact that the particle started off in the lower energy state! Therefore, we must conclude that:

${\displaystyle P_{a\rightarrow b}(t)=P_{b\rightarrow a}(t).\,\!}$

Wow! These equations are full of interesting physics. They tell us that if a particle starts off in a higher energy state, and then we shine light on it then, because the light is just an oscillating electric (and magnetic) field, the resulting perturbation to the Hamiltonian can induce the particle to fall down to the lower energy state. This is exactly the phenomenon of stimulated emission. Note also that shining light on the particle can also cause it to jump up an energy state – and the probabilities of going from the lower to the upper state and from the upper to the lower state are the same! This is the reason the Einstein coefficients ${\displaystyle B_{21}}$ and ${\displaystyle B_{12}}$ are the same (up to degeneracy, but that’s a different story).

What about spontaneous decay to a lower energy state? In quantum electrodynamics, there’s no such thing as a nonzero electric field – even if you’re not shining a light, the "ground state" still has a nonzero field. And it’s this field that acts as the perturbation. So, "spontaneous" decay is really not spontaneous – it’s just stimulated decay by a different electric field!

So far we’ve considered the impact of a single plane wave polarized in the ${\displaystyle z}$ direction. In reality of course, things are a little messier. Suppose we shine light of a whole bunch of frequencies, with ${\displaystyle \rho (\omega )}$ representing the energy density in some ${\displaystyle d\omega }$. Then, the transition probability can be expressed as:

${\displaystyle P_{\rm {transition}}(t)=-{\frac {2|\alpha |^{2}}{\epsilon _{0}\hbar ^{2}}}\int _{0}^{\infty }\rho (\omega ){\frac {\sin ^{2}\left({\frac {(\omega _{0}-\omega )t}{2}}\right)}{(\omega _{0}-\omega )^{2}}}d\omega \approx {\frac {\pi |\alpha |^{2}}{\epsilon _{0}\hbar ^{2}}}\rho (\omega _{0})t\,\!}$

We’ll make one more correction, which is that we want to account for light coming in at all angles and polarizations. Therefore, we’ll just generalize:

${\displaystyle {\vec {\alpha }}=q\langle \psi _{b}|{\vec {r}}\psi _{a}\rangle ,\,\!}$

and tack on a factor of 1/3 to ${\displaystyle P_{\rm {transition}}}$ in averaging over all polarizations and directions. Then the transition probability becomes:

${\displaystyle P_{\rm {transition}}(t)={\frac {\pi |{\vec {\alpha }}|^{2}}{3\epsilon _{0}\hbar ^{2}}}\rho (\omega _{0})t\,\!}$

Finally, we can get our Einstein coefficients; for brevity, and since part of this is already covered in the video, I’ll skip a couple steps. If we demand statistical and thermal equilibrium

${\displaystyle \rho (\omega _{0})={\frac {A}{\exp[\hbar \omega _{0}/k_{B}T]B_{ab}-B_{ba}}}={\frac {\hbar }{\pi ^{2}c^{3}}}{\frac {\omega ^{3}}{\exp[\hbar \omega _{0}/k_{B}T]-1}}\,\!}$

Note that I’ve dropped the degeneracy factors here, for simplicity, but they can be easily put back in. Comparing the two expression implies that:

${\displaystyle A={\frac {\omega _{0}^{3}|{\vec {\alpha }}|^{2}}{3\pi \epsilon _{0}\hbar c^{3}}}.\,\!}$

And there we have it! The above expression allows us to calculate ${\displaystyle A}$ directly! The only thing you really need to calculate is ${\displaystyle {\vec {\alpha }}=\langle \psi _{b}|{\vec {r}}|\psi _{a}\rangle }$. For most atoms, this is a bit complicated, since it requires you to know the wavefunctions. But for hydrogen, this is completely doable – you can look up the wavefunctions of hydrogen (or solve the Schrodinger equation yourself, if you like...). Note that for a spherically symmetric Hamiltonian, the elements of ${\displaystyle {\vec {\alpha }}}$ often go to zero. If you prefer not to calculate the Einstein coefficient directly, then see Estimating Atomic Transition Strengths for a quick and dirty way.