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Poincare Invariance of vacuum 
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#1
May814, 05:43 PM

P: 762

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? 


#3
May814, 06:43 PM

P: 762

what is the background?



#4
May814, 07:17 PM

Mentor
P: 16,191

Poincare Invariance of vacuum
I'm sorry. I mean vacuum. In what direction does it point?



#5
May914, 02:54 AM

P: 270

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.



#6
May914, 05:28 AM

P: 762

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? 


#7
May914, 05:45 AM

P: 270

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.



#8
May914, 12:21 PM

P: 762

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...



#9
May914, 01:13 PM

P: 306

$$ U^\dagger(\Lambda)\phi(x)U(\Lambda)=S(\Lambda)\phi(\Lambda x), $$ 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: $$ \langle 0\phi(x)0\rangle=\langle 0U^\dagger(\Lambda)\phi(x)U(\Lambda)0\rangle=S(\Lambda)\langle 0\phi(\Lambda x)0\rangle, $$ and so it must be [itex]S=1[/itex] which is true for a scalar field. 


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