Goldstone boson without symmetry?

In summary, the conversation discusses the U(1)_A problem and how the Lagrangian exhibits a U(1)_A symmetry in the limit of vanishing quark masses but due to the chiral anomaly, the current J_5^{\mu} is not conserved. The G\tilde{G} term is itself the divergence of the non-gauge invariant current K^{\mu}. It is mentioned that although there isn't actually a U(1)_A symmetry in the theory, a Goldstone boson still appears due to the conserved charge \tilde{Q}_5. However, Goldstone bosons only arise from spontaneously broken symmetries. The concept of Goldstone Dipoles is also
  • #1
QuantumCosmo
29
0
Hi,

I was wondering about the U(1)_A problem. The Lagrangian exhibits a (in the limit of vanishing quark masses) U(1)_A symmetry but due to the chiral anomaly, the current [tex]J_5^{\mu}[/tex] is not conserved:

[tex]\partial_{\mu}J_5^{\mu} = G\tilde{G} + 2i\bar{u}\gamma_5 u +...[/tex]

The [tex]G\tilde{G}[/tex] term is itself the divergence of the (not gauge invariant) current [tex]K^{\mu}[/tex].
(I have left out constant factors etc)

So in the limit of vanishing quark masses, the current [tex]\tilde{J}_5^{\mu} = J_5^{\mu} - K^{\mu}[/tex] is conserved and so is the charge

[tex] \tilde{Q}_5 = \int \tilde{J}_5^{\mu} \d^3[/tex]

Now it seems that although there actually isn't a U(1)_A symmetry in my theory, I still get a Goldstone boson because [tex] \tilde{Q}_5[/tex] is conserved.

But I thought Goldstone bosons occurred because of spontenously broken continuous symmetries and not because of conserved charges?

Can anyone help me with that?

Thank you very much,
Quantum
 
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  • #2
AH! But this charge is not gauge invariant! Goldstone's Theorem as you normally see it requires gauge invariance.

In fact, you DO get poles in NON-Gauge invariant correlation functions, which Coleman refers to as "Goldstone Dipoles" in his "Aspects of Symmetry". But since these aren't gauge invariant, they don't contribute to physics (which is only described by gauge-invariant quantities).
 
  • #3
QuantumCosmo said:
But I thought Goldstone bosons occurred because of spontenously broken continuous symmetries and not because of conserved charges?

Can anyone help me with that?

I just noticed this last sentence!

Of course, Goldstone bosons only come from spontaneously broken symmetries. So if you don't have that, then there are no Goldstone bosons at all.

This stuff about the Goldstone Dipole I mentioned earlier is relevant for the chiral Lagrangian, for example, to explain why there is no singlet after chiral symmetry breaking.

Is that what you were asking about?
 
  • #4
Yeah... because I don't understand this Kogut Susskind solution to the U(1) problem at all...
 

1. What is a Goldstone boson without symmetry?

A Goldstone boson without symmetry is a type of particle predicted by the Goldstone theorem in theoretical physics. It is a type of boson, or force-carrying particle, that arises when there is a spontaneous breaking of symmetry in a physical system.

2. How is a Goldstone boson without symmetry different from a regular Goldstone boson?

A regular Goldstone boson is associated with a broken continuous symmetry, while a Goldstone boson without symmetry is associated with a broken discrete symmetry. This means that a regular Goldstone boson arises when a continuous property, such as rotation or translation, is no longer conserved in a physical system, while a Goldstone boson without symmetry arises when a discrete property, such as parity or charge conjugation, is no longer conserved.

3. What are some examples of systems that exhibit a Goldstone boson without symmetry?

One example is the Higgs boson, which is associated with the spontaneous breaking of electroweak symmetry in the Standard Model of particle physics. Another example is the pion, which arises from the spontaneous breaking of chiral symmetry in quantum chromodynamics (QCD).

4. What are the implications of a Goldstone boson without symmetry?

The presence of a Goldstone boson without symmetry in a physical system can have significant implications for the behavior and properties of the system. For example, it can lead to the existence of massless particles and long-range interactions, and can also play a role in phase transitions and the formation of topological defects.

5. How are Goldstone bosons without symmetry being studied in current research?

There is ongoing research to better understand the properties and behavior of Goldstone bosons without symmetry in various physical systems. This includes experimental studies at particle accelerators, as well as theoretical studies using mathematical tools such as effective field theory and lattice simulations. Additionally, there is interest in finding potential applications for these particles in fields such as condensed matter physics and cosmology.

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