Do Lagrangian and hidden symm broken Lagr describe the same physics?

[ndung200790]

Please teach me this:
Do the Lagrangian(before broken in symmetry) and corresponding hidden symmetry broken Lagrangian describe the same physics or not?Because the field in the Lagrangian before broken is shifted by a constant in comparision with the field in the broken Lagrangian,but at quantum level we must consider the field operators.Then I do not understand whether the two Lagrangian are equivalent or not.
Thank you very much in advanced.

 

[Demystifier]

The two Lagrangians are equivalent.

What is different, is a "natural" definition of the vacuum state for two field operators. The "vacuum" is something that, in general, is not well defined in quantum field theory. A physical intuition is needed in order to find an appropriate definition of the vacuum.

 

[ndung200790]

Please pleasure elaborate more detail for me.I do know that the constant is the vacuum expectation value of field,then the theory is self-consistent for the one-point correlation function(the tadpole diagrams).But how about the other diagrams?

 

[Demystifier]

Please pleasure elaborate more detail for me.I do know that the constant is the vacuum expectation value of field,then the theory is self-consistent for the one-point correlation function(the tadpole diagrams).But how about the other diagrams?

I am not saying that the theory is not self-consistent, I am saying that it is not unique.

More precisely, as long as you calculate only the expectation values of products of fields in a given state, the theory is unique. But the problem appears when you try to say that some particular state is the "vacuum", that another particular state is a "1-particle state", etc. It cannot be done in a unique straightforward way, without using some intuition not derivable from first principles.