# Poincare Invariance of vacuum

1. May 8, 2014

### 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?

Last edited: May 8, 2014
2. May 8, 2014

### Vanadium 50

Staff Emeritus
In which direction does the background point?

3. May 8, 2014

### ChrisVer

what is the background?

4. May 8, 2014

### Vanadium 50

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

5. May 9, 2014

### Orodruin

Staff Emeritus
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
6. May 9, 2014

### 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?

7. May 9, 2014

### Orodruin

Staff Emeritus
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. May 9, 2014

### 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...

9. May 9, 2014

### Einj

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.

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