Superradiance was first described by Dicke in 1954 in this paper (official link). If you are a laser / optical physics expert, then you already know all about it. For everyone else, this post is an introductory tutorial.

Motivation: What does superradiance have to do with cold fusion? Well, I’m gearing up to discuss Peter Hagelstein’s “spin-boson model” theory of cold fusion. This theory says that the 24MeV of energy from D+D→⁴He fusion goes more-or-less directly into exciting a billion or so phonons (all of them in a single phonon mode, i.e. all at the same frequency, wavelength, etc.). Normally, this process would be extremely unlikely. However, there are two famous effects that increase the probability of transferring energy to an oscillation mode: Dicke superradiance and stimulated emission. Accordingly, the spin-boson model relies very heavily on both of these principles.

Since it’s impossible to thoroughly understand the spin-boson model without understanding superradiance, and since I couldn’t find a suitable description online, I wrote out this post. Buckle your seatbelts, let’s do some physics!

Every two-level systems is “like spin-1/2”

In introductory quantum mechanics, we learn about the quantum mechanics of spin-1/2 particles in exhaustive detail, including related concepts like spin operators and Pauli matrices. It is re-discussed in one course after another, from different points of view, until eventually most physicists grow to know it and love it and dream happy dreams about it.

The spin-1/2 particle is a 2-level system, i.e. all possible quantum states of the particle fill out a 2-dimensional complex vector space. Whenever physicists come across any other 2-level system, like an atom with a ground state and an excited state, then they often describe that system as “like a spin-1/2”. They borrow the same math and notation and terminology of spin-1/2 and apply it to this other system, even though it actually has nothing to do with spin.

In particular, I can write $|\downarrow \rangle$ as the atom’s ground state, and $|\uparrow \rangle$ as the atom’s excited state. That notation is suggestively similar to how I would write the spin-down and spin-up states for a spin-1/2 particle. Therefore, I can just copy down any formula about spin-1/2 particles and say that it also applies to atoms! For example, I can say $S_z |\downarrow \rangle = (-i \hbar/2) |\downarrow \rangle$. This formula would be part of the definition of the (so-called) spin operator S.

Later on, we will be pulling the same trick by treating certain 3-level systems as “like spin-1” and 4-level systems as “like spin-3/2” and so on.

Some facts about a spin-s system

Again, we are planning to take the math describing a spin-s system, and re-purpose it to talk about other systems like excited atoms. So maybe I should remind everyone about the math describing a spin-s system. Here are the highlights:

• s is one of the following: $0, \frac12, 1, \frac32, 2,$…..
• There are three quantum operators $S_x,S_y,S_z$, which can be lumped into a vector S, and which have certain commutation relations like $S_xS_y - S_yS_x = i \hbar S_z$.
• You can combine these into a “total spin” operator $S^2 = S_x^2 + S_y^2 + S_z^2$.
• In a spin-s system, if you measure $S^2$, you will always find that it is $\hbar^2 s (s+1)$.
• If you measure $S_z$, you can get $\hbar m$ , where m is the “spin projection quantum number”. There are only certain possibilities for m: It is s, or s-1, or s-2, … or –s. There is one quantum state for each m, which I’ll call $|m\rangle$, so the possible quantum states of the system fill out a $(2s+1)$-dimensional vector space.
• You can define raising and lowering operators $S_+ = S_x + i S_y$ and $S_- = S_x - i S_y$. When these operate on a state, it has the effect of increasing or decreasing m, according to the formulas $S_+|m\rangle = \hbar\sqrt{(s-m)(s+m+1)}|m+1\rangle,$ $S_-|m\rangle = \hbar\sqrt{(s+m)(s-m+1)}|m-1\rangle.$

The raising and lowering operator formulas are going to be critical for understanding superradiance. But we still have some work to do to figure out how they apply.

Ground-state atoms collectively absorb a photon

Now let’s say we have three atoms, pretty close together (but not so close that they directly interact with each other). All three are in the ground state: $|\downarrow \downarrow \downarrow \rangle$. A photon heads towards them. If the photon is absorbed, it is “collectively absorbed”, and the atoms enter the following state: $\frac{1}{\sqrt{3}}\left( |\uparrow \downarrow \downarrow \rangle \, + \, |\downarrow \uparrow \downarrow \rangle \, + \, |\downarrow \downarrow \uparrow \rangle \right)$

One of the atoms absorbed the photon, but it’s uncertain which one! (If the atoms are not floating in space but instead interacting with stuff, then the state would quickly decohere / dephase, and then the uncertainty would resolve and just one of the atoms, randomly chosen, would be the one that absorbed the photon, like you would intuitively expect in everyday life. But for this post, we are assuming that there is no decoherence.)

In that expression I am assuming that the distance between the atoms is much smaller than a wavelength. Otherwise the expression would be $\frac{1}{\sqrt{3}}\left( e^{i\mathbf{k}\cdot \mathbf{r}_1}|\uparrow \downarrow \downarrow \rangle \, + \, e^{i\mathbf{k}\cdot \mathbf{r}_2} |\downarrow \uparrow \downarrow \rangle \, + \, e^{i\mathbf{k}\cdot \mathbf{r}_3} |\downarrow \downarrow \uparrow \rangle \right)$

where k is the wavevector of that absorbed photon and $\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3$ are the atom positions. But let’s not overcomplicate. Forget about that. Assume that the atoms are all much closer than a wavelength, or spaced exactly one wavelength apart, or whatever, so that we can forget about those exponential factors.

Describing many 2-level atoms as one abstract “spin”

In the quantum state that I wrote out above, there are three different terms. That’s not very hard to write down. But if there were 100 atoms, and then you send in 50 photons to excite 50 of the atoms, then there would be grillions of terms in the final state, with the up and down arrows in different orders. With so many terms, it is awkward to write down equations and reason in our head about what’s going on. Luckily for us, there is a convenient mathematical framework for these kinds of states: Spin coupling!

(Remember learning about Clebsch-Gordon coefficients during introductory quantum mechanics? That’s the kind of thing I’m going to talk about next! Hey, don’t give me that look, it’s not so bad.)

We have N atoms, each with a ground and excited state, and each one has its own corresponding spin operator which we’ll call $\mathbf{S}_1, \mathbf{S}_2, \ldots, \mathbf{S}_N$. For example, the z-component of $\mathbf{S}_5$ is the operator that measures whether the 5th atom is in the ground or excited state.

Now we define a composite spin operator $\mathbf{S} = \mathbf{S}_1 + \cdots + \mathbf{S}_N$. We care about this operator because we can write down the complete interaction between photons and atoms (e.g. collective absorption and superradiance) in terms of just this S operator, rather than the individual atomic operators. For example, the raising operator $S_+ = S_x + i S_y$ describes the collective absorption process above.

By calling the operator S, I am hinting that it works like a normal spin operator, in that we can use all those spin-related math formulas from above. Is that right? Yes, but it is a bit complicated and abstract. Here’s an example with three atoms:

• The two states $(|\downarrow \downarrow \uparrow \rangle - |\downarrow \uparrow \downarrow \rangle)/\sqrt{2}$ and $(|\uparrow \downarrow \uparrow \rangle - |\uparrow \uparrow \downarrow \rangle)/\sqrt{2}$ span a 2-dimensional space of quantum states. When S operates on that space, it works just like a spin-1/2 particle.
• The two states $(|\downarrow \downarrow \uparrow \rangle + |\downarrow \uparrow \downarrow \rangle - 2 |\uparrow \downarrow \downarrow \rangle)/\sqrt{6}$ and $(|\uparrow \downarrow \uparrow \rangle + |\uparrow \uparrow \downarrow \rangle - 2 |\downarrow \uparrow \uparrow \rangle)/\sqrt{6}$ span a 2-dimensional space of quantum states. When S operates on that space, it works just like a spin-1/2 particle.
• The four states $|\downarrow \downarrow \downarrow \rangle$, and $(|\uparrow \downarrow \downarrow \rangle \, + \, |\downarrow \uparrow \downarrow \rangle \, + \, |\downarrow \downarrow \uparrow \rangle)/\sqrt{3}$, and $(|\uparrow \uparrow \downarrow \rangle \, + \, |\uparrow \downarrow \uparrow \rangle \, + \, |\downarrow \uparrow \uparrow \rangle)/\sqrt{3}$, and $|\uparrow\uparrow\uparrow\rangle$ span a 4-dimensional space of quantum states. When S operates on that space, it works just like a spin-3/2 particle.

Altogether we have accounted for 2+2+4=8 dimensions of the quantum state, which is everything. If you’re not sure how I came up with that decomposition, don’t worry, it won’t be on the test.

That means, if we have a collective quantum state of those three atoms, we can ask “what is its total spin”? The answer for three atoms is either $\frac32$ or $\frac12$, or a quantum superposition of the two. There isn’t a direct obvious relation between the state (some quantum superposition of states where different atoms are excited) and its total spin. (Once again, we are not literally talking about spin, just invoking a mathematical analogy.)

If there are N atoms, there is 1 subspace with spin N/2, and (N-1) subspaces with spin (N/2 – 1), and N(N-3)/2 subspaces with spin (N/2-2), … and more generally, $(\binom{N}{k} - \binom{N}{k-1})$ subspaces with spin (N/2 – k). (Binomial coefficient notation.)

Thanks to this new “composite spin operator” way of looking at things, we now have an alternate notation for many-atom quantum states: We can write them as $|s,m\rangle$, where s is the spin quantum number and m is the spin projection quantum number (defined above). So for example, for three atoms, if we want to describe the state $(|\uparrow \downarrow \downarrow \rangle \, + \, |\downarrow \uparrow \downarrow \rangle \, + \, |\downarrow \downarrow \uparrow \rangle)/\sqrt{3}$, we can just call it $|\frac32,-\frac12\rangle$. This $|s,m\rangle$ notation is actually ambiguous—remember, in the 3-atom example, there were two separate spin-1/2 subspaces—but it’s good enough. If it ever causes a problem, I can be more specific by adding an extra label. For example, the 3-atom system has two different states with $s=m=\frac12$; I can call one of them $|\frac12,\frac12, \text{Abdullah}\rangle$ and the other one $|\frac12,\frac12, \text{Kayla}\rangle$. But actually the ambiguous notation $|s,m\rangle$ is going to be good enough for this post.

We are finally getting to superradiance. The composite spin operator S has associated raising and lowering operators $S_\pm = S_x \pm i S_y$. The raising operator $S_+$ is associated with “collective absorption” of a photon: One atom changes from ground to excited state by absorbing a photon, but it is uncertain which atom. The lowering operator $S_-$ is associated with “collective emission” of a photon: One atom changes from the excited to ground state by emitting a photon, but it is uncertain which atom. We are going to be using the formulas mentioned above: $S_+|s,m\rangle = \hbar\sqrt{(s-m)(s+m+1)}|s,m+1\rangle,$ $S_-|s,m\rangle = \hbar\sqrt{(s+m)(s-m+1)}|s,m-1\rangle.$

Let’s consider a few examples:

Example 1: 100 excited atoms collectively emit one photon.

The initial state is $|\uparrow \uparrow \cdots\uparrow\rangle$, which in the $|s,m\rangle$ basis is $|50,50\rangle$. The effect of emitting one photon corresponds to applying the $S_-$ operator to the state. The final state is $|50,49\rangle$, or in our other notation, it’s $(|\downarrow \uparrow \uparrow \cdots \rangle + |\uparrow \downarrow \uparrow \uparrow \cdots\rangle + \cdots)/\sqrt{100}$. Using the $S_-$ formula above, we have $S_-|50,50\rangle = \hbar \sqrt{100}|50,49\rangle$

The square of the coefficient is proportional to how fast this process occurs. So the conclusion: If you have 100 excited atoms, one of them will emit a photon 100 times faster than if you had just one excited atom. This is no surprise, it’s the same thing that happens if each atom acts independently, just from statistics. This is not “superradiance” yet.

Example 2: After that has happened, this same group of atoms emit a second photon.

Now the initial state is $|50,49\rangle$ and the final state is $|50,48\rangle$: $S_-|50,49\rangle = \hbar \sqrt{198}|50,48\rangle$

The conclusion: If you have 100 excited atoms that just emitted one photon, they will emit a second photon 198 times faster than if you had just one excited atom. Aha: The first taste of superradiance! The photon emission occurs twice as fast as you would normally expect.

Example 3: This same group of atoms emits a 51st photon.

We’re now going from $|50,0\rangle$ to $|50,-1\rangle$. $S_-|50,0\rangle = \hbar \sqrt{2550}|50,-1\rangle$

The conclusion: If you have 100 excited atoms that just emitted 50 photons, they will emit a 51st photon 2550 times faster than if you had just one excited atom. This maximum speedup is proportional to the number of atoms squared. This is the heart of superradiance.

Example 4: Out of 100 atoms, 50 are excited, but you don’t know anything else about the state. They emit one photon.

Conclusion: You do not expect superradiance here, as I’ll explain. In the previous example, I gave a specific procedure to create the state: Start with 100 excited atoms, then let them collectively emit 50 photons. We wind up in the state $|50,0\rangle$. But in this example, it’s possible that the atoms are in the state $|50,0\rangle$, but it’s also possible that they are in a state $|49,0\rangle$, or $|48,0\rangle$, …, or $|0,0\rangle$. I didn’t specify! Out of these possibilities, $|50,0\rangle$ radiates super-fast, $|49,0\rangle$ a bit less fast, and so on. A state $|0,0\rangle$ cannot radiate at all (“collective subradiance”).

Now, there are ~ $10^{29}$ orthogonal quantum states that meet the specifications (100 atoms, 50 excited). Of them, just one is the special state $|50,0\rangle$. Another 49 states can be written as $|49,0\rangle$, still a negligible fraction. In fact, if we write the state as $|s,0\rangle$, then s is usually way smaller than N. I calculated: $s leq 5$ for 50% of the states, $s\leq10$ for 91% of the states, and $s\leq 25$ for 99.9999% of the states. The average photon emission rate, over all states, is exactly 50 times the single-atom emission rate, the same as if the atoms were acting totally independently.

So, superradiance only happens if the atoms are in a special kind of quantum state. That means that decoherence / dephasing is the enemy of superradiance. Superradiance requires that the state $|\uparrow \downarrow \uparrow \downarrow \cdots\rangle$ and the state $|\downarrow \uparrow \uparrow \downarrow \cdots \rangle$ and all the other possibilities have specific, prescribed amplitude and phase relationships with each other.

Postscript: Whole books have been written about superradiance. There is a lot of fun stuff I’m leaving out. But I think this covers all the material that I’ll need for later. 😀

## 2 thoughts on “Introduction to superradiance”

1. Chetan Waghela

Is synchronization that occurs in non-linear dynamics a classical analog of superradiance?

Like

1. steve Post author

Off the top of my head, I can’t see any (non-superficial) relation between synchronization in non-linear dynamics and superradiance.

Like