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Vacuum states invariance

  1. Sep 28, 2012 #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?
     
  2. jcsd
  3. Sep 29, 2012 #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:
     
  4. Sep 30, 2012 #3

    lalo_u

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    That´s fine to me.

    Thank you very much Einj!!:biggrin:
     
  5. Mar 2, 2014 #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.
     
  6. Mar 2, 2014 #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.
     
  7. Mar 2, 2014 #6
    Thank you so much! :)
     
  8. Mar 3, 2014 #7

    naima

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    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.
     
  9. Mar 3, 2014 #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.
     
  10. Apr 29, 2014 #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)
     
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