Checkpoint: Fusion and Braiding

This checkpoint ties together fusion rules, the fusion space, and qubit encoding for the non-abelian model closest to experiment: Ising anyons, the anyon type carried by Majorana zero modes. You will count fusion outcomes to size a topological qubit register.

The Ising fusion algebra

Ising anyons have three sectors $\{1, \sigma, \psi\}$ with

Only $\sigma\times\sigma$ has more than one channel, which is what makes $\sigma$ non-abelian, with quantum dimension $d_\sigma = \sqrt2$.

Counting the fusion space of $n$ $\sigma$'s

Fuse $n$ $\sigma$ anyons sequentially and track the running total charge. Because $\psi\times\sigma=\sigma$ and $1\times\sigma=\sigma$ are single-channel while $\sigma\times\sigma$ splits, the number of fusion trees of total charge $1$ (vacuum) for even $n$ works out to

matching the $d_\sigma^{\,n} = 2^{\,n/2}$ growth split between the two total-charge sectors $\{1,\psi\}$. Since each factor of $2$ is one logical qubit, $n$ $\sigma$ anyons encode $n/2 - 1$ qubits — the standard "four $\sigma$'s per qubit" rule of Ising/Majorana TQC, with $n=4$ giving exactly one qubit.

Braiding on this space

Within a fixed-charge fusion space, exchanging neighboring $\sigma$'s applies the $R$-matrix unitaries of the braiding lesson. For Ising anyons these braids generate the single- and two-qubit Clifford group — protected and exact, but (by Gottesman–Knill) not universal, so a non-topological magic ingredient is needed for universality. Counting the dimension first tells you how many qubits those braids act on.

Try it

Implement the Ising fusion recursion and return the dimension of the space of $n=6$ $\sigma$ anyons that fuse to the vacuum (total charge $0$). Track the 3-vector $[\,\#\to 1,\ \#\to\sigma,\ \#\to\psi\,]$, start it at $[0,1,0]$ (one $\sigma$), and apply the fusion map $n-1$ times. The answer is $2^{\,6/2-1}=4$ — a two-qubit Ising register.

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