What type of field can have a Lorentz invariant VEV?

[ChrisVer]

Suppose I have a field \hat{X}...
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 X 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?

 

[Vanadium 50]

But who tells me that the vacuum shouldn't be a spinor or vector?

In which direction does the background point?

 

[ChrisVer]

what is the background?

 

[Vanadium 50]

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

 

[Orodruin]

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.

 

[ChrisVer]

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?

 

[Orodruin]

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.

 

[ChrisVer]

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...

 

[Einj]

Suppose I have a field \hat{X}...
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 X 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?

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 \Lambda a generic field transforms as:
$$
U^\dagger(\Lambda)\phi(x)U(\Lambda)=S(\Lambda)\phi(\Lambda x),
$$
where U(\Lambda) belongs to the representation of the Lorentz group acting on the physical states while S(\Lambda) 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 S=1 which is true for a scalar field.