Einstein Coefficients

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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 . Einstein coefficients describe the transition rates caused by the interaction of radiation with these discrete energy levels. There are three coefficients:

B12.png A21.png B21.png
Left: Photon absorption rates are described by . Center: Spontaneous photon emission rates are described by . Right: Stimulated photon emission rates are described by .

1 Spontaneous Emission,

governs decay from energy state 2 to 1. It is the transition probability per unit time for an atom, and has units of . More specifically, the probability an atom undergoes spontaneous de-excitation and releases a photon is Poisson-distributed, with mean rate . So is the mean lifetime of the excited state. As an example, ( transition in hydrogen) has an Einstein A coefficient of .

If describes the number density of atoms in the upper energy state, then the transition rate per volume is given by:

2 Spontaneous Absorption,

governs photon absorption that causes a transition from the lower to upper energy state (). In contrast to the case, absorption requires the presence of photons, so translating 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:

We use instead of (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, . It describes the relative absorption probability around (the absorption frequency), and is subject to the requirement that: . We can approximate the width of as an effective width . is affected by many factors:

  • (the natural, uncertainty-based broadening of at atom in isolation),
  • (Doppler broadening from thermal motion), and
  • (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:

3 Stimulated Emission,

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 . For an important case of stimulated emission, see Masers.

1 Einstein Relations among coefficients

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

This is because as many atoms need to be going from energy state 1 to 2 as visa versa. A second relation is: . Using the Boltzmann distribution, :

In thermal equilibrium is given by the Planck Function:

Combining this with earlier, we get:


2 Rewriting in terms of Einstein coeffs

In a small volume :

We can express the extinction coefficient, , in terms of the Einstein coefficients. The excitation probability per time is , and the energy lost in crossing the small volume (it is the probability per time per volume of going by absorbing from a cone of solid angle and frequency range ). Thus, the energy is given by:

Recognizing that :

Correcting for stimulated emission, we get:

3 Estimating Cross-Sections

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

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

To estimate , we use the fact that, ignoring g’s, , and . Then using the approximation that that , we get:

In a single atom, , so .

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, and . In general, the particle will be in some linear combination of and :

The probabilities are and , and they of course must sum to unity at all times.

Now, let’s suppose the electron starts off in at , so that and . To first-order, time dependent perturbation theory tells us:

Failed to parse (syntax error): {\displaystyle c_ b \approx \frac{-i}{\hbar }\int _0^ t H_{ba}e^{i\omega _0t'}dt’, \,\!}

where Failed to parse (syntax error): {\displaystyle H’_{ba} \equiv \langle \psi _ a | \hat{H'} | \psi _ b \rangle } is the off diagonal component of the perturbing Hamiltonian and . The derivation of this result is too detailed to reproduce here, but can be found in any quantum mechanics textbook. Note that even though the system started off entirely in , the introduction of a perturbation means that the probabilities can change as a function of a time – eventually, it may be that and the particle is entirely in . 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 direction:


where we will now use:

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

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

The probability that the particle has “transitioned” from its initial state to the other state is just given by :

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

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 and 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 direction. In reality of course, things are a little messier. Suppose we shine light of a whole bunch of frequencies, with representing the energy density in some . Then, the transition probability can be expressed as:

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:

and tack on a factor of 1/3 to in averaging over all polarizations and directions. Then the transition probability becomes:

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

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:

And there we have it! The above expression allows us to calculate directly! The only thing you really need to calculate is . 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 often go to zero, so this really isn’t too bad. If you prefer not to calculate the Einstein coefficient directly, then see Estimating Atomic Transition Strengths for a quick and dirty way.