# Quantum dimension of an anyon

In the book Introduction to Topological Quantum Computation Jiannis K. Pachos page 60 Equ.(4.16) $$\text{dim}(M_{(n)}) \propto d_a^n$$, two lines after (4.16) said that,

The dimension of $$M_{(n)}$$ is always an integer as it enumerates different fusion outcomes,while $$d_a^n$$ does not need to be an integer.

Is this correct? I think $$d_a$$ does not need to be an integer but $$d_a^n$$ is an integer.

• Just as a heads up, this book is full of small typos and errors which is often somewhat of a pain. It's good to be aware of though. Jun 3, 2023 at 7:58

The number $$n$$ counts how many anyons are present. Depending on the choice of $$n$$, there is no reason for $$d_a^n$$ to be an integer.
Maybe you're trying to say that $$d_a^n$$ is always an integer for some value of $$n$$. This is also incorrect. For example, the quantum dimension of the unique non-trivial quasiparticle type the the Fibonacci anyon theory is the golden ratio. No power of the golden ratio is an integer.
The point of all this is that $$\dim(M_n)$$ counts how many dimensions how you have to play with when $$n$$ anyons are created. The quantity $$\log_2(\dim(M_n))$$ counts how many qubits you have to work with. The formula $$\dim (M_n)\propto d_a^n$$ says that creating the $$a$$-type quasiparticle gives you asymptotically $$\log_2(d_a)$$ qubits.
When $$a$$ is abelian, you are getting $$\log_2(1)=0$$ qubits per anyon. That's why you can't do any interesting computations with abelian anyons - adding more anyons doesn't give you more power.
If $$a$$ is not abelian then, then $$d_a\geq \sqrt{2}$$. This comes from the general fact that if $$d_a<2$$ then $$d_a=2\cos(\pi/n)$$ for some $$n\geq 3$$. This means that non-abelian anyons will give you at least half a qubit per quasiparticle.