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Poincare Invariance of vacuum

by ChrisVer
Tags: invariance, poincare, vacuum
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ChrisVer
#1
May8-14, 05:43 PM
P: 736
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?
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Vanadium 50
#2
May8-14, 06:15 PM
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Quote Quote by ChrisVer View Post
But who tells me that the vacuum shouldn't be a spinor or vector?
In which direction does the background point?
ChrisVer
#3
May8-14, 06:43 PM
P: 736
what is the background?

Vanadium 50
#4
May8-14, 07:17 PM
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Poincare Invariance of vacuum

I'm sorry. I mean vacuum. In what direction does it point?
Orodruin
#5
May9-14, 02:54 AM
P: 196
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
#6
May9-14, 05:28 AM
P: 736
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
#7
May9-14, 05:45 AM
P: 196
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
#8
May9-14, 12:21 PM
P: 736
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
#9
May9-14, 01:13 PM
P: 305
Quote Quote by ChrisVer View Post
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?
A VEV coming from a scalar field is Poincarč 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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