Exploring Implications of Breaking U(1) in Electroweak Theory

In summary: I admit I don't know how to calculate it :)!After the answer given by Ben Niehoff I had understood that even if the gauge bosons got a mass, the global symmetry would be saved and so there would still be a noether current.Yes, the current is still the sum of the matter and the gauge field.
  • #1
jacopo23
11
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What if in the elettroweak theory we broke U(1) too?
I read sometimes that electric charge would not be conserved anymore.

Anyway, if the photon got a mass, even if U(1) gauge symmetry is broken, QED lagrangian would still be invariant under global U(1) transformation, and we should still have the noether current
[itex]\overline{\psi}\gamma^{\mu}\psi[/itex]

So why electric current shouldn't be conserved anymore?
 
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  • #2
I read sometimes that electric charge would not be conserved anymore.
This is not true. Breaking symmetry does not make conserved charges unconserved. What happens is that particles' charges get new values. I.e. electron mass and charge would be different, but still constant. From another point of view you can still say that an electron has -1 elementary charge, but its interaction force has changed.

Also, Goldstone bosons appear.

What if in the elettroweak theory we broke U(1) too?
That depends on what is left from the symmetry.

For example, superconductivity is a theory where U(1) gauge group of electromagnetism is broken.
 
  • #3
Hi, thank you for the answer. I think you are right.
Ciao.
 
  • #4
I was just wondering that if the theory was not abelian, for example SU(2), when the gauge bosons get masses the lagrangian is not invariant under the global symmetry anymore.
In fact, under a gauge transformation,
[itex] A_{\mu} \rightarrow U A_{\mu} U^{+} - i (\partial_{\mu} U) U^{+}[/itex]
and if the group is not abelian, [itex] A_{\mu} [/itex] is not invariant.
Thus, the mass term [itex] A_{\mu} A^{\mu}[/itex] is not invariant under a global transformation.

Why should still exist a conserved noether current in the case of spontaneoulsy broken non abelian gauge theories?
 
  • #5
jacopo23 said:
Why should still exist a conserved noether current in the case of spontaneoulsy broken non abelian gauge theories?
b/c the current is constructed from the fields whereas symmetry breaking is on the level of states
 
  • #6
Hi thank you for the answer but could you be more clear please :) ?
I'm not so very into the subjet. I study other things, but had to study a little of field theory for an exam.

Do you maybe mean that the mass term emerging from spontaneous symmetry breaking does not enter in the lagrangian so it is still invariant under global transformations?
I actually don't think it is right, do you say it is?

Ciao
 
  • #7
jacopo23 said:
I was just wondering that if the theory was not abelian, for example SU(2), when the gauge bosons get masses the lagrangian is not invariant under the global symmetry anymore.
In fact, under a gauge transformation,
[itex] A_{\mu} \rightarrow U A_{\mu} U^{+} - i (\partial_{\mu} U) U^{+}[/itex]
and if the group is not abelian, [itex] A_{\mu} [/itex] is not invariant.
Thus, the mass term [itex] A_{\mu} A^{\mu}[/itex] is not invariant under a global transformation.

Under a global gauge transformation, [itex]\partial_{\mu} U \equiv 0[/itex], so the mass term is still invariant.
 
  • #8
Yes, under a global transformation
[itex] \partial_{\mu} U = 0[/itex]
but since the gauge transformation is
[itex] A_{\mu} \rightarrow U A_{\mu} U^{+}-i (\partial_{\mu} U ) U^{+} [/itex],
then under a global transformation the first term gives
[itex] m^2 A_{\mu} A^{\mu} \rightarrow m^2 U A_{\mu} A^{\mu} U^{+}[/itex]
that is different from [itex] m^2 A_{\mu} A^{\mu} [/itex] if the group is not abelian.

So, since in a not abelian gauge theory a mass term spoils the global symmetry too, why should we still have a conserved noether current?
Maybe the answer is just that there is not any conserved current?
 
  • #9
Ah, I see the problem. Your mass term is not correct. It should be

[tex]m^2 \, \mathrm{tr} \, (A_\mu A^\mu)[/tex]
 
  • #10
Ah ok, now I see.
You are right, thank you :)!
 
  • #11
jacopo23 said:
What if in the elettroweak theory we broke U(1) too?
I read sometimes that electric charge would not be conserved anymore.

Anyway, if the photon got a mass, even if U(1) gauge symmetry is broken, QED lagrangian would still be invariant under global U(1) transformation, and we should still have the noether current
[itex]\overline{\psi}\gamma^{\mu}\psi[/itex]

So why electric current shouldn't be conserved anymore?

What you wrote down is not the expression for the current but for the current density. What happens if the symmetry is broken is that the integral over the current density diverges, so that the total current (or charge in case of the zero component) becomes undefined.
 
  • #12
What happens if the symmetry is broken is that the integral over the current density diverges, so that the total current (or charge in case of the zero component) becomes undefined.

Could you please explain this point?

After the answer given by Ben Niehoff I had understood that even if the gauge bosons got a mass, the global symmetry would be saved and so there would still be a noether current.
I haven't done the calculation but I expect that in general this current is the sum of the matter + gauge field contribution.
 
  • #13
As long as you are talking about global U(1) symmetry, you don't even need to consider gauge bosons. Especially it doesn't matter whether they are massive or not.
Whether symmetry is broken or not, the Lagrangian is always invariant. It is the vacuum which is no longer invariant under the symmetry operation.
 

FAQ: Exploring Implications of Breaking U(1) in Electroweak Theory

What is U(1) in Electroweak Theory?

U(1) in Electroweak Theory refers to the gauge group that describes the symmetry between electromagnetism and the weak nuclear force. It is a fundamental component of the Standard Model of particle physics.

What does it mean to break U(1) symmetry?

Breaking U(1) symmetry means that the symmetry between electromagnetism and the weak force is no longer maintained. This can happen through the Higgs mechanism, where the Higgs field acquires a non-zero vacuum expectation value and breaks the symmetry.

What implications does breaking U(1) have on the Standard Model?

Breaking U(1) has significant implications on the Standard Model as it allows for the masses of particles to be generated. It also explains the difference in strength between the electromagnetic and weak forces.

Are there any experimental observations of the implications of breaking U(1)?

Yes, there have been numerous experimental observations that support the implications of breaking U(1) in Electroweak Theory. The discovery of the Higgs boson at the Large Hadron Collider is a significant confirmation of the theory.

What further research is being done to explore the implications of breaking U(1)?

Scientists are currently conducting experiments at the Large Hadron Collider and other particle accelerators to study the properties of the Higgs boson and search for new physics beyond the Standard Model. Theoretical studies are also being done to understand the implications of breaking U(1) and to look for potential extensions to the Standard Model.

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