# How is a half twist of a diagram equivalent to exchanging anyons?

Consider the fusion of $a$ and $b$ to form $c$, with the exchange of anyons $a$ and $b$, as per Pachos' book Introduction to Topological Quantum Computation.

I'm confused how the half twist of $c$ is equivalent to exchanging $a$ and $b$. If I naively uncross the world lines forming an exchange, isn't anyon $a$ on the right hand side rather than left as shown in figure 4.4? Alternatively, if I apply a half twist in the clockwise direction (when looking back up the world line of $c$) I can see how this unravels to $b$ on the right hand side as given, but hasn't this effectively exchanged $a$ and $b$ twice (not once) leaving them where they were? I could understand that $b$ on the left is the same as $b$ on the right if the particles were indistinguishable, but I thought $a$ and $b$ can be distinct anyons in this consideration.

Am I thinking of the crossed world lines too literally? Is the crossing symbolising that we have exchanged $a$ and $b$ without showing their positions flipped? If that was correct why not just relabel the diagram? Is the answer to this last question that then we wouldn't develop braiding intuition? I.e. are we trying to transform particle motions (exchanges, say) into equivalent braiding of world lines but with positions fixed?

• Without being able to see Fig 4.4. of Panchos' book (what book? A quick Google failed to fid anything obviously related) it's hard to comment. – John Rennie Mar 16 '16 at 7:02
• fair enough! here's a direct link to "Introduction to Topological Quantum Computation" by Panchos: researchgate.net/publication/… – physioConfusio Mar 16 '16 at 8:20
• It would help a lot if you could make your question more to the point. – Norbert Schuch Mar 16 '16 at 14:55
• The picture on the left depicts a process in which particles a and b are exchanged, then fuse to c. The picture on the right depicts a process in which particles a and b fuse to c without first being exchanged. $R_{ab}^c$ is the difference in amplitude between the two processes. There is no claim being made here about the topological equivalence of any two processes here, so I don't really understand what you're asking. – Dominic Else Mar 17 '16 at 3:41