- #1

lalo_u

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Could anybody explain to me why is that?

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- Thread starter lalo_u
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- #1

lalo_u

Gold Member

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Could anybody explain to me why is that?

- #2

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\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

lalo_u

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That´s fine to me.

Thank you very much Einj!!

Thank you very much Einj!!

- #4

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- #5

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- #6

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Thank you so much! :)

- #7

naima

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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

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- #9

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