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A question about ghost number

  1. Feb 17, 2014 #1
    In BRST symmetry the action is

    Inew=I0[[itex]\phi[/itex]]+s[itex]\psi[/itex][[itex]\phi[/itex],[itex]\omega[/itex],[itex]\varpi[/itex],h].
    Where ω and [itex]\varpi[/itex] is ghost and anti ghost.
    If we expand I in series of terms In that satisfy sIn=0(s is BRST operator(Slavnov operator)).
    We introduce Hodge operator t=[itex]\varpi[/itex][itex]\delta[/itex]/[itex]\delta[/itex]h.Then
    ψ=-∑tIn/n.
    They conclude that ψ has ghost number is -1,but I do not understand why?
     
  2. jcsd
  3. Feb 18, 2014 #2

    Avodyne

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    It follows from your first line. Ghost number is conserved (by construction), the action has ghost number zero, the BRST operator s has ghost number +1, and so ψ must have ghost number -1.
     
  4. Feb 18, 2014 #3
    I do not understand why s operator has ghost number 1?Is it(s operator) being the ''charge'' of BRST symmetry?
     
  5. Feb 18, 2014 #4
    Or is s operator being a ''ladder'' operator of ghost number?
     
  6. Feb 18, 2014 #5

    Avodyne

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    s represents (anti)commutation with the BRST charge, which carries ghost number +1. So acting with s raises ghost number by +1.

    Conservation of ghost number is not a fundamental principle, and you could image using a ψ that includes terms of different ghost number (thus breaking ghost number conservation). It's just that it's more convenient to have ghost number conservation as an additional tool to restrict possible counterterms.
     
  7. Feb 18, 2014 #6
    Is it the relation:

    [Q,[itex]\Phi[/itex]][itex]_{\pm}[/itex]=is[itex]\Phi[/itex]?Then how can we see s being ''ladder'' operator of ghost number?
     
  8. Feb 20, 2014 #7

    Avodyne

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    Let N = ghost number operator.

    [N,Q]=+Q
    [N,ψ]=-ψ

    Compare the harmonic oscillator, N=a+a:

    [N,a+]=+a+
    [N,a]=-a
     
  9. Feb 21, 2014 #8
    Why we know [N,Q]=+Q?And why by this we know ghost number of s =+1?
     
  10. Feb 21, 2014 #9

    Avodyne

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    Sorry, I can't help any more. I suggest taking a look at Srednicki's text.
     
  11. Feb 22, 2014 #10
    Thank very much for your kind helping.
    At the moment,I have a new look: the full BRST transformation is θs,where θ is fermionic variable,I think the ghost number of θ is -1(?).Because of BRST symmetry θs has ghost number 0(?) then s has ghost number +1.
    Please consider my question again.
     
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