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I SO(2n) representation on n complex fields

  1. Jul 30, 2017 #1
    If I have a lagrangian which has terms of the form ##\Psi^{\dagger}_\mu \Psi^\mu## then I can decompose the n complex ##\Psi## fields into 2n real fields by ##\Psi_\mu = \eta_{2\mu+1} + i\eta_{2\mu}##. When I look at the lagrangian now it seems to have SO(2n) symmetry from mixing the 2n real fields.

    Is there any obvious choice of representing SO(2n) on the n complex ##\Psi## fields explicitly?
     
  2. jcsd
  3. Jul 31, 2017 #2

    vanhees71

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    The SU(N) is a ##N^2-1##-dimensional Lie group, while SO(2N) is ##2N(2N-1)/2=2N^2-N## dimensional. So SO(2N) is in this sense larger than SU(N).
     
  4. Jul 31, 2017 #3
    Youre not wrong, but I dont see how this helps
     
  5. Jul 31, 2017 #4

    Avodyne

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    Well we could work out the generators. The SO(2N) generator mixing ##\eta_{2\mu+1}## and ##\eta_{2\mu}## becomes a U(1) generator acting on ##\Psi_\mu##, and SO(2N) generators mixing even indexed ##\eta##'s have the same form when acting on ##\Psi##, and similarly for odd indexed ##\eta##'s. Then there are the SO(2N) generators that mix an even ##\eta## and an odd ##\eta##; I suspect will we need the charge-conjugation operator (which exchanges ##\Psi## and ##\Psi^\dagger##) to express those ...
     
  6. Aug 1, 2017 #5
    I worked on a field theory with the same enlarged symmetry a few years ago, see here: https://arxiv.org/abs/1603.05652

    For my purposes, I was largely interested in how the SU(N) irreps fit inside the larger SO(2N) irreps, see section IV (page 21). I have a mapping between the creation operators in the SU(N) and SO(2N) notation in equations 55-56, but I don't think my mapping implies a simple SO(2N) action on the complex field itself. So I don't have a good answer for you, but maybe looking at what I did will help.

    Also, chapter 23 of Georgi's textbook is entirely dedicated to [itex]SU(N) \subset SO(2N)[/itex].
     
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