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U(1) breaking

  1. Oct 21, 2011 #1
    What if in the elettroweak theory we broke U(1) too?
    I read sometimes that electric charge would not be conserved anymore.

    Anyway, if the photon got a mass, even if U(1) gauge simmetry is broken, QED lagrangian would still be invariant under global U(1) transformation, and we should still have the noether current
    [itex]\overline{\psi}\gamma^{\mu}\psi[/itex]

    So why electric current shouldn't be conserved anymore?
     
  2. jcsd
  3. Oct 21, 2011 #2
    This is not true. Breaking symmetry does not make conserved charges unconserved. What happens is that particles' charges get new values. I.e. electron mass and charge would be different, but still constant. From another point of view you can still say that an electron has -1 elementary charge, but its interaction force has changed.

    Also, Goldstone bosons appear.

    That depends on what is left from the symmetry.

    For example, superconductivity is a theory where U(1) gauge group of electromagnetism is broken.
     
  4. Oct 21, 2011 #3
    Hi, thank you for the answer. I think you are right.
    Ciao.
     
  5. Oct 30, 2011 #4
    I was just wondering that if the theory was not abelian, for example SU(2), when the gauge bosons get masses the lagrangian is not invariant under the global simmetry anymore.
    In fact, under a gauge transformation,
    [itex] A_{\mu} \rightarrow U A_{\mu} U^{+} - i (\partial_{\mu} U) U^{+}[/itex]
    and if the group is not abelian, [itex] A_{\mu} [/itex] is not invariant.
    Thus, the mass term [itex] A_{\mu} A^{\mu}[/itex] is not invariant under a global transformation.

    Why should still exist a conserved noether current in the case of spontaneoulsy broken non abelian gauge theories?
     
  6. Oct 30, 2011 #5

    tom.stoer

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    b/c the current is constructed from the fields whereas symmetry breaking is on the level of states
     
  7. Oct 31, 2011 #6
    Hi thank you for the answer but could you be more clear please :) ?
    I'm not so very into the subjet. I study other things, but had to study a little of field theory for an exam.

    Do you maybe mean that the mass term emerging from spontaneous symmetry breaking does not enter in the lagrangian so it is still invariant under global transformations?
    I actually don't think it is right, do you say it is?

    Ciao
     
  8. Oct 31, 2011 #7

    Ben Niehoff

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    Under a global gauge transformation, [itex]\partial_{\mu} U \equiv 0[/itex], so the mass term is still invariant.
     
  9. Nov 1, 2011 #8
    Yes, under a global transformation
    [itex] \partial_{\mu} U = 0[/itex]
    but since the gauge transformation is
    [itex] A_{\mu} \rightarrow U A_{\mu} U^{+}-i (\partial_{\mu} U ) U^{+} [/itex],
    then under a global transformation the first term gives
    [itex] m^2 A_{\mu} A^{\mu} \rightarrow m^2 U A_{\mu} A^{\mu} U^{+}[/itex]
    that is different from [itex] m^2 A_{\mu} A^{\mu} [/itex] if the group is not abelian.

    So, since in a not abelian gauge theory a mass term spoils the global symmetry too, why should we still have a conserved noether current?
    Maybe the answer is just that there is not any conserved current?
     
  10. Nov 1, 2011 #9

    Ben Niehoff

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    Ah, I see the problem. Your mass term is not correct. It should be

    [tex]m^2 \, \mathrm{tr} \, (A_\mu A^\mu)[/tex]
     
  11. Nov 1, 2011 #10
    Ah ok, now I see.
    You are right, thank you :)!
     
  12. Nov 2, 2011 #11

    DrDu

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    What you wrote down is not the expression for the current but for the current density. What happens if the symmetry is broken is that the integral over the current density diverges, so that the total current (or charge in case of the zero component) becomes undefined.
     
  13. Nov 2, 2011 #12
    Could you please explain this point?

    After the answer given by Ben Niehoff I had understood that even if the gauge bosons got a mass, the global simmetry would be saved and so there would still be a noether current.
    I haven't done the calculation but I expect that in general this current is the sum of the matter + gauge field contribution.
     
  14. Nov 2, 2011 #13

    DrDu

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    As long as you are talking about global U(1) symmetry, you don't even need to consider gauge bosons. Especially it doesn't matter whether they are massive or not.
    Whether symmetry is broken or not, the Lagrangian is always invariant. It is the vacuum which is no longer invariant under the symmetry operation.
     
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