Why must a scalar field have a constant vacuum expectation value?

In summary, the Mandle QFT book states that if we require the vacuum states to be invariant under Lorentz transformations and under translations, then the corresponding field must be a scalar field, $\phi(x)$, and its vacuum expectation value must be constant. This is an experimental fact and can also be proven using Closed Time Path formalism in QFT.
  • #1
lalo_u
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I was reading Mandle QFT book, and it says: "If we require the vacuum states to be invariant under Lorentz transformations and under translations, then this field must be a scalar field, $\phi(x)$, and its vacuum expectation value must be constant".
Could anybody explain to me why is that?
 
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  • #2
Consider, for example, Lorentz transformations. The law of transformation of a field (of any kind) [itex]\phi(x)[/itex] is:
\begin{equation}U^\dagger(\Lambda)\phi(x)U(\Lambda)=S(\Lambda) \phi(x) \end{equation}

where [itex]U(\Lambda)[/itex] is the representation of the Lorentz group on the space of physical states. This means that if you perform a transformation [itex]\Lambda[/itex] a state [itex]|p\rangle[/itex] transform as: [itex]|p'\rangle=U(\Lambda)|p\rangle[/itex]. On the other hand of the equation, [itex]S(\Lambda)[/itex] is a representation of Lorentz group over the space of operators (fields) and which has the role to transform, for example, the field components if the field is a vector one.

When you require the invariance of the vacuum state this means to ask for:
[itex]U(\lambda)|0\rangle=|0\rangle[/itex]. If you require this and you consider the vacuum expectation value then you have:

$$\langle0|\phi(x)|0\rangle=\langle0|U^\dagger (\Lambda) \phi(x)U(\Lambda)|0\rangle$$

that is satisfied if the field transform with [itex]S(\Lambda)=1[/itex], which means that it doesn't have any components to transform, i.e. is a scalar field.

You can do the same thing for translation,s considering that, if you have a translation with a parameter [itex]a[/itex], then the law is:

\begin{equation}U^\dagger(a)\phi(x)U(a)=\phi(x+a) \end{equation}

If you apply the reasoning made previously you obtain:

$$\langle0|\phi(x)|0\rangle=\langle0|\phi(x+a)|0 \rangle$$


that is the vacuum expectation value calculated in two different points. This equation is satisfied if the expecation value itself is constant over [itex]x[/itex].

I hope I didn't make any mistakes :tongue:
 
  • #3
That´s fine to me.

Thank you very much Einj!:biggrin:
 
  • #4
I was wondering...if I have a spinor or a vector field, this means that the vacuum state cannot be invariant under a Lorentz transformation? So if I have no fermions in a reference system, I could have a fermion in another system? Thank you.
 
  • #5
Actually one usually uses a backward reasoning. The Lorentz invariance of the vacuum is always required. The point is: since the vacuum is always Lorentz invariant, what kind of field can have a non-zero vacuum expectation value? And it turns out that it must be a scalar field.
 
  • #6
Thank you so much! :)
 
  • #7
lalo_u said:
I was reading Mandle QFT book, and it says: "If we require the vacuum states to be invariant under Lorentz transformations and under translations, then this field must be a scalar field, $\phi(x)$, and its vacuum expectation value must be constant".
Could anybody explain to me why is that?

According to Wightman W0 axiom in every QFT the vacuum mut be Poincare in variant.
So the sentence "If the vacuum is poincare invariant then ..." seems bizarre.
 
  • #8
This is actually more a philosophical question. In principle the fact that the vacuum is Poincarè invariant is an experimental fact, is not a necessary condition. This is why in QFT to take it to be an axiom.
 
  • #9
One can prove the vacuum invariance without Wightman axioms, only using standard rules of QFT (scalar Lagrangian and quantization via path integrals of Hamiltonian). Then the vacuum corresponds to zero temperature and one can use Closed Time Path formalism (time is a path in complex plane). The proof is quite tricky, it is published in The European Journal of Physics 73, 2654 (2013) http://dx.doi.org/10.1140/epjc/s10052-013-2654-9 (Open Access)
 

1. What is a vacuum state invariance?

A vacuum state invariance refers to the property of a physical system remaining unchanged under the action of a specific set of transformations, known as gauge transformations. These transformations shift the zero point of the system, leaving the physical observables and laws of the system unchanged.

2. Why is vacuum state invariance important in physics?

Vacuum state invariance is important in physics because it allows for the consistency and symmetry of physical laws and theories. It also helps to explain phenomena such as conservation laws, which are a consequence of vacuum state invariance.

3. Can vacuum state invariance be broken?

Yes, vacuum state invariance can be broken in certain physical systems. This can occur when the vacuum state of a system is not the lowest energy state, or when there is spontaneous symmetry breaking. In these cases, the system is no longer invariant under gauge transformations.

4. How is vacuum state invariance related to the Higgs mechanism?

The Higgs mechanism is a proposed mechanism for spontaneous symmetry breaking, which can lead to the breaking of vacuum state invariance. This is because the Higgs field, which gives particles their mass, is responsible for the symmetry breaking and can affect the vacuum state of a system.

5. Are there any real-world applications of vacuum state invariance?

Yes, vacuum state invariance has important applications in particle physics, where it is used to explain phenomena such as the conservation of electric charge and the behavior of elementary particles. It also has applications in cosmology, as it helps to explain the symmetry of the universe and the behavior of fundamental forces.

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