Why must the Higgs' gauge symmetry be broken?

[epsilon]

The part I understand:

I understand that the spontaneous symmetry breaking of the Higgs produces the 'Mexican hat' potential, with two non-zero stable equilibria.

I understand that as the Higgs is a complex field, there exists a phase component of the field. Under gauge transformations of this Higgs potential (in particular the rotation: $\phi \rightarrow \phi_0 e^{i \theta}$), you are simply moving between the degenerate ground state of this potential, where this motion simply generates the massless Goldstone bosons, and hence the potential is gauge invariant.

The part I do not understand:

You must now fix the gauge by using a condensate, such that the rotations are gauge variant as "you want the Higgs to take a specific value". I don't understand what you must fix the gauge. As the ground states are all degenerate, surely the Higgs already has a specific value? Please do not use the Lagrangian to explain it! Thank you in advance!

 

[Orodruin]

You must now fix the gauge by using a condensate, such that the rotations are gauge variant as "you want the Higgs to take a specific value". I don't understand what you must fix the gauge. As the ground states are all degenerate, surely the Higgs already has a specific value?

It is not that the potential becomes gauge variant, it is that the ground state is degenerate and one particular choice will be implemented in nature. This choice is not going to respect gauge symmetry and therefore the ground state is not gauge invariant. This is why it is called spontaneous symmetry breaking - the theory itself displays the invariance but the ground state of the theory breaks it.

 

[mfb]

I understand that the spontaneous symmetry breaking of the Higgs produces the 'Mexican hat' potential, with two non-zero stable equilibria.

Spontaneous symmetry breaking happens due to the shape of the potential, not the other way round.

 

[epsilon]

It is not that the potential becomes gauge variant, it is that the ground state is degenerate and one particular choice will be implemented in nature. This choice is not going to respect gauge symmetry and therefore the ground state is not gauge invariant. This is why it is called spontaneous symmetry breaking - the theory itself displays the invariance but the ground state of the theory breaks it.

Thank you for your answer, you cleared up the misunderstanding that I had!

 

[epsilon]

Spontaneous symmetry breaking happens due to the shape of the potential, not the other way round.

Thank you! This is why it wasn't making sense.

 

[nikkkom]

The part I understand:

I understand that the spontaneous symmetry breaking of the Higgs produces the 'Mexican hat' potential, with two non-zero stable equilibria.

The potential has that shape from the beginning; and there aren't two non-zero minima - there is a continuum of them. The "mexican hat" is IIUC five-dimensional - the "trough" is not one-dimensional as it would be in 3D-case, but has three directions along which potential stays at minimum, their existence creates three Goldstone bosons. The fourth direction is the Higgs particle.