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  Neutral and charged bions cancel

+ 4 like - 0 dislike

I'm reading this paper http://arxiv.org/abs/1502.06624

In the discussion surrounding equation 5, it is explained how the charged bions $M_i \bar M_j$ for monopole operators $M_{i,j}$ with $i\neq j$ and neutral bions $M_i M_i$ have the same saddle-point actions $e^{-2S_0}$ where $S_0$ is the action of a single monopole but contribute terms with *opposite* phases in the path integral. My issue with this is it seems there are N choose 2 charged bions but only $N$ neutral bions, so the overall contribution to the vacuum energy is $({N \choose 2} - N)e^{-2S_0}$, not zero as claimed.

Am I missing something dumb?

asked Mar 17, 2016 in Theoretical Physics by Ryan Thorngren (1,925 points) [ revision history ]
retagged Mar 18, 2016 by Arnold Neumaier

1 Answer

+ 5 like - 0 dislike

The bion vertex involves the root vectors$$B_{ij}\sim \vec{\alpha}_i\cdot\vec{\alpha}_j\, , $$ which implies that only "adjacent" SU(N) monopoles couple, $B_{ij}\sim  2\delta_{ij}-\delta_{ij\pm 1}$. See, for example, here.

answered Mar 18, 2016 by tmschaefer (720 points) [ no revision ]

Thanks! Makes a lot of sense now (:

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