Why with the same classical Lagrangian,there are the difference in symm breaking?

[ndung200790]

Please teach me this:
It seem to me that the classical Lagrangian is only ''the tip of the iceberg'' of full quantum Lagrangian which has some ''counterterms'' having ''classical parameters'' zero.With this ''counterterms'' of full Lagrangian,the vacuum expectation value of field(s) is nonzero,then the spontaneous symmetry breaking happens.So with the same classical Lagrangian,we have different theories,some of them have hidden symmetry breaking but some of them have not.By the way:Why the symmetry breaking does not affect the UV divergences?
Thank you very much in advanced.

 

[ndung200790]

Now,I have rethought that the hidden symmetry breaking is simply resulted of the shift the field by a constant.The origine Lagrangian is changed to different broken Lagrangians corresponding to different shiffted constants.But I still do not understand why the breaking does not affect on the divergences(of Feymann diagrams).

 

[ndung200790]

It seem to me that when we return from the broken Lagrangian(with the vacuum expectation value of sigma field zero) to the origine Lagrangian which is before being broken(with the vacuum expectation value of Phi field nonzero) then the loop diagrams of the returned Lagrangian is the same of the origine Lagrangian which is the Lagrangian that is not broken( with vacuum expctation value of phi field zero).But the divergences of the loops might be different,because the integral variable in the returned Lagrangian is shifted by a constant comparring with loop integral variable in the case of the Lagrangian not broken.(the constant shift in variable of divergent integral is not permissive).
Please give me any answer.Thank you very much in advanced.

 

[Parlyne]

Some computations simply won't work if you're expanding around a non-vacuum state.

Presuming, however, that the divergence you're thinking of suffers no such problems, we can think about comparing the two versions of the calculation; and, in particular, if we cut off the divergences at scale [itex]\Lambda[/itex], we can look at the behavior of the difference between the two divergence calculations as we take [itex]\Lambda[/itex] to [itex]\infty[/itex].

If we take the shift in the field to be v, we should be able to express the difference in the divergence behaviors in terms of [itex]\Lambda[/itex] and [itex]\frac{v}{\Lambda}[/itex].

The difference between the two versions of the calculation should only have to do with terms involving v. This means that we should be able to use an expansion in [itex]\frac{v}{\Lambda}[/itex] to allow us to cancel out the parts of the behavior that don't depend on the shift. But, this should mean that all remaining terms have positive powers of [itex]\frac{v}{\Lambda}[/itex], which tend to 0 as [itex]\Lambda \rightarrow \infty[/itex].

 

[Polyrhythmic]

Please teach me this:
It seem to me that the classical Lagrangian is only ''the tip of the iceberg'' of full quantum Lagrangian which has some ''counterterms'' having ''classical parameters'' zero.With this ''counterterms'' of full Lagrangian,the vacuum expectation value of field(s) is nonzero,then the spontaneous symmetry breaking happens.So with the same classical Lagrangian,we have different theories,some of them have hidden symmetry breaking but some of them have not.By the way:Why the symmetry breaking does not affect the UV divergences?
Thank you very much in advanced.

Maybe I'm getting something wrong in your argument, but spontaneous symmetry breaking refers to the situation where you chose one of several (or of continuously many, like for a [tex]\varphi^4[/tex] potential) possible vacua to do your perturbative expansion around it. In this case the Lagrangian itself remains symmetric. I don't see the point of your question, would you elaborate it again?

 

[ndung200790]

Mr.Polyrhythmic is right,but in case the vacuum expectation value of field is nonzero,then there are many possible vacuum states to appear,then symmetry is spontaneously broken,despite of the origine Lagrangian is still symmetry.

 

[ndung200790]

Thank you Mr Parlyne very much!

 

[Polyrhythmic]

Mr.Polyrhythmic is right,but in case the vacuum expectation value of field is nonzero,then there are many possible vacuum states to appear,then symmetry is spontaneously broken,despite of the origine Lagrangian is still symmetry.

That's just what I said.

 

[ndung200790]

I am trying to find out the reason for the vacuum expectation value of field nonzero!

 

[ndung200790]

I think that the reason for the vacuum expectation value of field nonzro is the ''hidden terms'' of Lagrangian,because the vacuum states are depended on the full ''quantum Lagrangian''(not on classical Lagrangian with the physical parameters nonzero).

 

[ndung200790]

Maybe my first thingking is wrong!Now I think that the ''reality Lagrangian'' that regulate the vacuum expectation value of field nonzero is the broken symmetry Lagrangian,not the origine Lagrangian.The broken Lagrangian(with sigma field) controls the vacuum expectation value of field(phi field).This field ''makes'' up the origine Lagrangian.(phi field=sigma field+<vacuum expectation value of phi field>).

 

[ndung200790]

Maybe it is nonsense to say something with the physical observations zero.But without saying it(e.g without my first thinking above)how we can argue the degenerate of vacuum states(or the appearance of vacuum expectation value of field)?

 

[ndung200790]

Sorry,now I have just understood the hidden symmetry breaking.The concrete form of the Lagrangian determines all the vacuum states.The values of field at the vacuum states are the vacuum expectation value of field calculated at those vacuum states(<vacuum/operator of field/vacuum>=value of field at position Lagrangian determines vacuum state).In general speaking,those vacuum expectation values are nonzero.When we shift the field by the vacuum expectation value of field:phi=<phi>+sigma,then the Lagrangian with sigma is broken in the symmetry.
Please teach me that correct or not.Thank you very much in advanced.

 

[ndung200790]

At quantum level,the vacuum expectation of field is determined by the minima of effective potential instead of minima of classical potential part of Lagangian.