The n-Category Café

Nuclear physics forces you to imagine blobs of protons and neutrons wiggling around in a very quantum-mechanical way. Nuclei are too complicated to fully understand. We can simulate them on a computer, but simulation is not understanding, and it’s also very hard: one book I’m reading points out that one computation you might want to do requires diagonalizing a $10^{14} \times 10^{14}$ matrix. So I’d rather learn about the many simplified models of nuclei people have created, which offer partial understanding… and lots of beautiful math.

Protons minimize energy by forming pairs with opposite spin. Same for neutrons. Each pair acts like a particle in its own right. So nuclei act very differently depending on whether they have an even or odd number of protons, and an even or odd number of neutrons!

The ‘Interacting Boson Model’ is a simple approximate model of ‘even-even’ atomic nuclei: nuclei with an even number of protons and an even number of neutrons. It treats the nucleus as consisting of bosons, each boson being either a pair of nucleons — that is, either protons or neutrons — where the members of a pair have opposite spin but are the same in every other way. So, these bosons are a bit like the paired electrons responsible for superconductivity, called ‘Cooper pairs’.

However, in the Interacting Boson Model we assume our bosons all have either spin 0 (s-bosons) or spin 2 (d-bosons), and we ignore all properties of the bosons except their spin angular momentum. A spin-0 particle has 1 spin state, since the spin-0 representation of $\text{SO}(3)$ is 1-dimensional. A spin-2 particle has 5, since the spin-2 representation is 5-dimensional.

If we assume the maximum amount of symmetry among all 6 states, both s-boson and d-boson states, we get a theory with $\text{U}(6)$ symmetry! And part of why I got interested in this stuff was that it would be fun to see a rather large group like $\text{U}(6)$ showing up as symmetries — or approximate symmetries — in real world physics.

More sophisticated models recognize that not all these states behave the same, so they assume a smaller group of symmetries.

But there are some simpler questions to start with.

How do we make a spin-0 or spin-2 particle out of two nucleons? That’s easy. Two nucleons with opposite spin have total spin 0. But if they’re orbiting each other, they have orbital angular momentum too, so the pair can act like a particle with spin 0, 1, 2, 3, etc.

Why are these bosons in the Interacting Boson Model assumed to have spin 0 or spin 2, but not spin 1 or any other spin? This is a lot harder. I assume that at some level the answer is “because this model works fairly well”. But why does it work fairly well?

By now I’ve found two answers for this, and I’ll tell you the more exciting answer, which I found in this book:

• Igal Talmi, Simple Models of Complex Nuclei: the Shell Model and Interacting Boson Model, Harwood Academic Publishers, Chur, Switzerland, 1993.

In the ‘liquid drop model’ of nuclei, you think of a nucleus as a little droplet of fluid. You can think of an even-even nucleus as a roughly ellipsoidal droplet, which however can vibrate. But we need to treat it using quantum mechanics. So we need to understand quantum ellipsoids!

The space of ellipsoids in $\mathbb{R}^3$ centered at the origin is 6-dimensional, because these ellipsoids are described by equations like

$$A x^2 + B y^2 + C z^2 + D x y + E y z + F z x = 1$$

and there are 6 coefficients here. Not all nuclei are close to spherical! But perhaps it’s easiest to start by thinking about ellipsoids that are close to spherical, so that

$$(1 + a)x^2 + (1 + b)y^2 + (1 + c)z^2 + d x y + e y z + f z x = 1$$

where $a,b,c,d,e,f$ are small. If our nucleus were classical, we’d want equations that describe how these numbers change with time as our little droplet oscillates.

But the nucleus is deeply quantum mechanical. So in the Interacting Boson Model, invented by Iachello, it seems we replace $a,b,c,d,e,f$ with operators on a Hilbert space, say $q_1, \dots, q_6$, and introduce corresponding momentum operators $p_1, \dots, p_6$, obeying the usual ‘canonical commutation relations’:

$$[q_j, q_k] = [p_j, p_k] = 0, \qquad [p_j, q_k] = - i \hbar \delta_{j k}$$

As usual, we can take this Hilbert space to either be $L^2(\mathbb{R}^6)$ or ‘Fock space’ of $\mathbb{C}^6$: the Hilbert space completion of the symmetric algebra of $\mathbb{C}^6$. These are two descriptions of the same thing. The Fock space of $\mathbb{C}^6$ gets an obvious representation of the unitary group $\text{U}(6)$, since that group acts on $\mathbb{C}^6$. And $L^2(\mathbb{R}^6)$ gets an obvious representation of $\text{SO}(3)$, since rotations act on ellipsoids and thus on the tuples $(a,b,c,d,e,f) \in \mathbb{R}^6$ that we’re using to describe ellipsoids.

The latter description lets us see where the s-bosons and d-bosons are coming from! Our representation of $\text{SO}(3)$ on $\mathbb{R}^6$ splits into two summands:

• the (real) spin-0 representation, which is 1-dimensional because it takes just one number to describe the rotation-invariant aspects of the shape of an ellipsoid centered at the origin: for example, its volume. In physics jargon this number tells us the monopole moment of the mass distribution of our nucleus.

• the (real) spin-2 representation, which is 5-dimensional because it takes 5 numbers to describe all other aspects of the shape of an ellipsoid centered at the origin. You need 2 numbers to say in which direction its longest axis points, one number to say how long that axis is, 1 number to say which direction the second-longest axis point in (it’s at right angles to the longest axis), and 1 number to say how long it is. In physics jargon these 5 numbers tell us the quadrupole moment of our nucleus.

This shows us why we don’t get spin-1 bosons! We’d get them if the mass distribution of our nucleus could have a nonzero dipole moment. In other words, we’d get them if we added linear terms $G x + H y + K z$ to our equation

$$A x^2 + B y^2 + C z^2 + D x y + E y z + F z x = 1$$

But by conservation of momentum, we can assume the center of mass of our nucleus stays at the origin, and set these linear terms to zero.

As usual, we can take linear combinations of the operators $q_j$ and $p_j$ to get annihilation and creation operators for s-bosons and d-bosons. If we want, we can think of these bosons as nucleon pairs. But we don’t need that microscopic interpretation if we don’t want it: we can just say we’re studying the quantum behavior of an oscillating ellipsoid!

After we have our Hilbert space and these operators on it, we can write down a Hamiltonian for our nucleus, or various possible candidate Hamiltonians, in terms of these operators. Talmi’s book goes into a lot of detail on that. And then we can compare the oscillations these Hamiltonians predict to what we see in the lab. (Often we just see the frequencies of the standing waves, which are proportional to the eigenvalues of the Hamiltonian.)

So, from a high-level mathematical viewpoint, what we’ve done is try to define a manifold $M$ of ellipsoid shapes, and then form its cotangent bundle $T^\ast M$, and then quantize that and start studying ‘quantum ellipsoids’.

Pretty cool! And there’s a lot more to say about it. But I’m wondering if there might be a better manifold of ellipsoid shapes than just $\mathbb{R}^6$. After all, when $1+a, 1+b$ or $1+c$ become negative things go haywire: our ellipsoid can turn into a hyperboloid! The approach I’ve described is probably fine ‘perturbatively’, i.e. when $a,b,c,d,e,f$ are small. But it may not be the best when our ellipsoid oscillates so much it gets far from spherical.

I think we need a real algebraic geometer here. In both senses of the word ‘real’.