Poincare Invariance of vacuum

  1. May 8, 2014 #1

    ChrisVer

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    Suppose I have a field [itex]\hat{X}[/itex]...
    What kind of operator should it be in order to develop a vev which doesn't break the Poincare invariance?
    I am sure that a scalar field doesn't break the poincare invariance, because it doesn't transform.
    However I don't know how to write it down mathematically or prove it...
    Also, because I don't know how to "prove" it, I am not sure if there can exist some other [itex]X[/itex] field/operator which would keep the poincare invariance untouched after getting a vev...So why couldn't it be a fermion? or a vector field?

    Is it the same as looking at the Lorentz group? So that you have the scalar in (0,0)repr, while the fermions can be in (1/2,0) or (0,1/2) and vectors in (1/2,1/2)?
    But who tells me that the vacuum shouldn't be a spinor or vector?
     
    Last edited: May 8, 2014
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  3. May 8, 2014 #2

    Vanadium 50

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    In which direction does the background point?
     
  4. May 8, 2014 #3

    ChrisVer

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    what is the background?
     
  5. May 8, 2014 #4

    Vanadium 50

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    I'm sorry. I mean vacuum. In what direction does it point?
     
  6. May 9, 2014 #5

    Orodruin

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    Apart from scalar fields, rank 2 tensor fields may develop a vev proportional to the metric tensor without breaking Lorentz invariance since the metric by definition is invariant.
     
    Last edited: May 9, 2014
  7. May 9, 2014 #6

    ChrisVer

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    I thought that a general metric breaks poincare invariance (and brings instead general coord transfs)?q
    For minkowski metric, doesn't it transform like a 2nd rank tensor?
     
  8. May 9, 2014 #7

    Orodruin

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    Exactly, and since the form of the metric is preserved, it looks the same in all frames. Thus, a vev proportional to the metric does not break Lorentz invariance.
     
  9. May 9, 2014 #8

    ChrisVer

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    is there any source dealing with such a thing (vev of the minkowski metric)? I am not even sure how the metric would act on the vacuum...
     
  10. May 9, 2014 #9
    A VEV coming from a scalar field is Poincarè invariant for the following reason (I'm excluding the case of the VEV being proportional to the metric since I'm not very familiar with it). Under a Lorentz transformation [itex]\Lambda[/itex] a generic field transforms as:
    $$
    U^\dagger(\Lambda)\phi(x)U(\Lambda)=S(\Lambda)\phi(\Lambda x),
    $$
    where U([itex]\Lambda[/itex]) belongs to the representation of the Lorentz group acting on the physical states while [itex]S(\Lambda)[/itex] belongs to the representation acting on the operators.

    The vacuum is clearly Lorentz invariant. If you want for your VEV to be Lorentz invariant it must be:
    $$
    \langle 0|\phi(x)|0\rangle=\langle 0|\phi(\Lambda x)|0\rangle,
    $$
    however, because of the invariance of the vacuum:
    $$
    \langle 0|\phi(x)|0\rangle=\langle 0|U^\dagger(\Lambda)\phi(x)U(\Lambda)|0\rangle=S(\Lambda)\langle 0|\phi(\Lambda x)|0\rangle,
    $$
    and so it must be [itex]S=1[/itex] which is true for a scalar field.
     
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