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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.

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  • $\begingroup$ 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. $\endgroup$ Jun 3, 2023 at 7:58

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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.

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