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Spontaneous symmetry breaking in the standard model

  1. Feb 25, 2014 #1
    In the standard model, the Lagrangian contains scalar and spinor and vector fields. But when we consider spontaneous symmetry breaking, we only account for the terms contain only scalar fields, " the scalar potential", in the Lagrangian. And if the scalar fields have vacuum expectation value, then we recognize the symmetry is broken spontaneously. Why don't we have to contain spinors and vectors? I think it's related to Lorentz invariance.
     
  2. jcsd
  3. Feb 25, 2014 #2

    Bill_K

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    If a spinor or vector had a nonzero expectation value, it would single out a preferred direction. The vacuum state must be rotationally invariant (Lorentz invariant too) so this is not possible.
     
  4. Feb 25, 2014 #3
    Thank you, Bill_K.
    Could you explan more formally or mathematically by using language of quantum theory?
    I would like to proof [itex] \langle 0|\psi_\alpha|0\rangle=0[/itex] from Lorentz invariance of the vacuum. Your explanation seems that the vacuum expectation vales is supposed to be Lorentz invariant but I think Lorentz invariance means [itex] U^\dagger(\Lambda)|0\rangle=|0\rangle[/itex].

    And another question occurred. Your explanation seems that the propagator, the vacuum expectation value of two point function also vanishes. Why propagators don't vanish even if they have Lorentz indices or spinor indices?
     
  5. Feb 25, 2014 #4

    samalkhaiat

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    See post#35 in
    www.physicsforums.com/showthread.php?t=172461

    Sam
     
  6. Feb 25, 2014 #5

    Bill_K

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