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Wilson Loop and spontaneous symmetry breaking.

  1. Jun 7, 2015 #1
    I would like to ask about the case of:

    ##SU(2)\otimes U(1) \rightarrow U(1)\otimes U(1),## spontaneous symmetry breaking.

    It is given that the Wilson Loop:
    ##W \equiv exp[ig \oint dy H T^1]= diag(−1,−1,1).##

    Where ##y## is the ##S^1/Z^2## fifth/extra dimension, ##H = \frac{1}{g R}## and ##T^1## is the first ##SU(3)## generator.
    As we can check, the Wilson Loop does not commute with ##SU(3)## generators ##T^6, T^7## but it still commutes with ##T^3, T^8##.
    The non-trivial parity in this case is ##P = diag(1,−1,−1)## and it anti-commutes only with ##T^1##.
    How does this SSB occur?
     
    Last edited: Jun 7, 2015
  2. jcsd
  3. Jun 8, 2015 #2

    fzero

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    Since I am not sure what choice you're using for generators, I will use the Gell-Mann matrices, which seem to be different from your choice. ##\lambda_8## is the most convenient choice to use for the Wilson line, since it commutes with itself and ##\lambda_{1,2,3}##. From the gauge Lagrangian, we have a potential term ##\text{Tr}[A^a_\mu,A^8_5]^2##. The Wilson line corresponds to a constant background field in the 8-component, so the gauge field components that do not commute with ##\lambda_8## get a tree-level mass proportional to the square of this background field (##m_{\text{KK}}## with the appropriate choice), whereas the massless ##A^a_\mu##, ##a=1,2,3## and ##A^8_\mu## generate an unbroken ##SU(2)\times U(1)## gauge group.

    If you want to break this further to ##U(1)\times U(1)##, a convenient choice would be to put a Wilson line for ##A^3_5## as well.
     
  4. Jun 8, 2015 #3
    Thank you very much for your answer fzero!
    First of all the matrix ##T^1## equals ##\lambda_1/2##.
    My question is mostly of algebraic nature:
    Could you please explain, in algebraic terms, how the massless ##A^a_μ, a=1,2,3## and ##A^8_μ## generate an unbroken ##SU(2)\otimes U(1)##?
    Or even better, how ##U(1) \otimes U(1)## is realised?
     
    Last edited: Jun 8, 2015
  5. Jun 8, 2015 #4

    fzero

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    These fields are massless, so must correspond to some residual gauge symmetry. You can either note that ##\lambda_{1,2,3}## project to the standard Pauli matrices or just verify that they satisfy the ##SU(2)## algebra by hand, so we get a factor of ##SU(2)##. Since ##\lambda_8## commutes with them, it is the generator of a separate ##U(1)## factor.

    If we put an additional Wilson line for ##\lambda_3##, then ##A^{1,2}_\mu## will be massive, but ##A^3_\mu## will remain massless and preserve a ##U(1)## subgroup of ##SU(2)##.
     
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