The n-Category Café

Let’s consider an example: $G = SU(2)$. In physics this group is important because its representations describe the ways a particle can transform under rotations. There is one irreducible representation for each ‘spin’ $s = 0, \frac{1}{2}, 1, \dots$ When $s$ is an integer, this representation describes the angular momentum states of a boson; when $s$ is a half-integer (meaning an integer plus $\frac{1}{2}$) it describes the angular momentum states of a fermion.

For example, take any spin-$\frac{1}{2}$ particle, and consider only its rotational symmetries. Then we can describe this particle using the obvious unitary representation of $SU(2)$ on $\mathbb{C}^2$. This is called the spin-$\frac{1}{2}$ representation, and it’s quaternionic.

Why? Because we can think of a pair of complex numbers as a single quaternion, and $SU(2)$ as the group of unit quaternions. The obvious representation of $\SU(2)$ on $\mathbb{C}^2$ then turns out to be the action of unit quaternions on $\mathbb{H}$ via left multiplication. And since this commutes with right multiplication by quaternions, we’ve got a quaternionic representation!

In slogan form: qubits are not just quantum—they are also quaternionic!

More generally, all particles of half-integer spin are quaternionic, while particles of integer spin are real — as long as we consider them only as representations of $SU(2)$.

We shall see why this is true shortly, but I can’t resist making a little table that summarizes the pattern:

$$\begin{aligned} integer \; spin & bosonic & \quad & real & J^2 = 1 \\ half-integer \; spin & fermionic & & quaternionic & J^2 = -1 \end{aligned}$$

Why does it work this way?

We’ve seen it’s true for the spin-$\frac{1}{2}$ representation. But we can build all the irreducible unitary representations of $SU(2)$ as symmetrized tensor powers of the spin-$\frac{1}{2}$ representation. The $n$th symmetrized tensor power, $S^n(\mathbb{C}^2)$, is the spin-$s$ representation with $s = n/2$. At this point a well-known general result will help:

Theorem: Suppose $G$ is a Lie group and $H, H' \in Rep(G)$. Then:

• $H$ and $H'$ are real $\implies$ $H \otimes H'$ is real.
• $H$ is real and $H'$ is quaternionic $\implies$ $H \otimes H'$ is quaternionic.
• $H'$ is quaternionic and $H'$ is real $\implies$ $H \otimes H'$ is quaternionic.
• $H$ and $H'$ are quaternionic $\implies$ $H \otimes H'$ is real.

To see this, just notice that $H$ is real (resp. quaternionic) if and only if it can be equipped with an invariant antiunitary operator $J: H \to H$ with $J^2 = 1$ (resp. $J^2 = -1$). So, pick such an antiunitary $J$ for $H$ and also one $J'$ for $H'$. Then $J \otimes J'$ is an invariant antiunitary operator on $H \otimes H'$, and $$(J \otimes J')^2 = J^2 \otimes {J'}^2 .$$ This makes the result obvious.

In short, the ‘multiplication table’ for tensoring real and quaternionic representations is just like the multiplication table for the numbers $1$ and $-1$… and also, not coincidentally, just like the usual rule for combining bosons and fermions!

Next, note that any subrepresentation of a real representation is real, and any subrepresentation of a quaternionic representation is quaternionic. It follows that since the spin-$\frac{1}{2}$ representation of $\SU(2)$ is quaternionic, its $n$th tensor power is real or quaternionic depending on whether $n$ is even or odd, and similarly for the subrepresentation $S^n(\mathbb{C}^2)$. This is the spin-$s$ representation for $s = n/2$. So, the spin-$s$ representation is real or quaternionic depending on whether $s$ is an integer or half-integer.

But what is the physical meaning of the antiunitary operator $J$ on the spin-$s$ representation? In fact, it describes time reversal.

To see this, let’s call our representation $\rho : \SU(2) \to \U(H)$, where $H$ is the Hilbert space $S^n(\mathbb{C}^2)$. Choose any element $X$ in the Lie algebra $su(2)$ and let $S = d\rho(X)$. Then $$J \, \exp(t S) = \exp(t S) \, J$$ for all $t \in \mathbb{R}$, so differentiating gives $$J S = S J$$

But the operator $S$ is skew-adjoint. As discussed earlier, in quantum mechanics observables are self adjoint. To turn $S$ into an observable, say $A$, we should write $S = i A$. This observable $A$ tells us the particle’s angular momentum along some axis.

Since $J$ anticommutes with $i$, we have $$J A = -A J$$ So, for every state of our spin-$s$ particle, say $v \in H$, we have a new state $J v$ where the expected value of the observable $A$ is exactly opposite: $$\langle J v, A J v \rangle = - \langle J V, J A v \rangle = - \langle A v, v \rangle = - \langle v, A v \rangle$$ So, the antiunitary operator $J$ reverses angular momentum! Since the time-reversed version of a spinning particle is a particle spinning the opposite way, physicists call $J$ time reversal.

It seems natural that time reversal should obey $J^2 = 1$, as it does in the bosonic case; it may seem strange to have $J^2 = -1$, as we do for fermions. Shouldn’t applying time reversal twice get us back where we started? It actually does, in a crucial sense: the expectation values of all observables are unchanged when we multiply a unit vector $v \in H$ by $-1$. Still, this minus sign should remind you of the equally curious minus sign that a fermion picks up when you rotate it a full turn. Are these signs related? Is there a nice way to see the connection?

It seems so, and we’ll talk about that next time…