Goldstone boson state, conserved current

In summary, the first sentence is saying that the Goldstone boson is a particle that results from a broken symmetry. The second sentence is saying that the appearance of the Goldstone boson has nothing to do with whether the symmetry is completely broken or only broken down to a sub-group. The third sentence is saying that the relationship between the operator of goldstone boson creation and the operator of the particle is only valid in the long wavelength limit.
  • #1
LayMuon
149
1
I don't understand these sentences from Peskin:

"At long wavelength (*why long wavelength?*), the Goldstone bosons become infinitesimal symmetry rotations of the vacuum, ##Q^a|0 \rangle##, where ##Q^a## is the global charge associated with ##J^{\mu a}##. Thus, the operators ##J^{\mu a}## have the correct quantum numbers to create Goldstone boson state. In general, there will be a current ##J^{\mu a}## that can create or destroy this boson; we can parametrize the corresponding matrix element as $$\langle 0| J^{\mu a} | \pi_k(p) \rangle = -i p^\mu F^a{}_k e^{-ip \cdot x}$$, where ##p^{\mu}## is the on-shell momentum of the boson and ##F^a{}_k## is a matrix of constants. The elements ##F^a{}_k## vanish when ##a## deontes a generator of an unbroken symmetry.""

I don't understand every single sentence here. Can anybody explain. It is from Peskin&Schroeder 20.1 formula 20.46. Thanks.
 
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  • #2
Take for example a ferromagnet. The broken symmetry is specified by the orientation of the magnetization axis. The Goldstone boson is the magnon, an oscillation of the magnetization axis.
A long wavelength magnon corresponds to an (at least locally) uniform rotation of the magnetization axis. There is some charge (here e.g. the total spin in the z-direction## Q= S_z=\sum_i s_{zi}=\int J^0 dV ##) which induces this rotation.
The treatment is quite sloppy as the operator ##Q^a## can be shown not to exist. However you can work with an operator ##Q^a_V=\int_V J^{0 a} dV## defined for a finite region V.
 
  • #3
But in ferromagnetic, I so understand, SO(3) is broken into SO(2) so rotation around axis leaves the particles massless and has nothing to do with massless boson emerged because of the symmetry breaking. Am I wrong?

How about the second sentence?
 
  • #4
The appearance of Goldstone Bosons has nothing to to with whether the symmetry is completely broken or only broken down to a sub-group.
If you want to simplify the model consider a ferromagnet in two dimensions (or an anisotropic three dimensional one).
 
  • #5
how about the second and most importantly the third sentence?
 
  • #6
The second sentence is a rather weak statement. For a general operator we expect non-vanishing matrix elements between almost any states, including the vacuum and states with one Goldstone boson. Furthermore we know explicitly that some integral aka Q of J over V generates long wavelength GBs. So J itself certainly also has non-vanishing matrix elements between the vacuum and 1-GB states.
Again, I find analogies with non-relativistic systems helpful: E.g. rotation of all spins in a ferromagnet changes one broken symmetry state into another. Rotation of just one spin corresponds to the excitation of a superposition magnons of all wavelengths. You could also consider a crystal where translation symmetry is broken. A translation of the whole crystal transforms one broken symmetry state into another one. Translation of just one atom yields a superposition of acoustic lattice vibrations (phonons) of all wavelengths.


The third sentence is the most general expression possible with all the indices and being Poincare covariant (or only covariant with respect to the translation group).
I am sorry if this is still a bit hand-waving, but I am not so familiar with relativistic field theory.
 
  • #7
I understand these: the lagrangian has a symmetry with generators ##\hat{T}^a## and parameters say ##\alpha_a##, this leads to conserved current operators ##\hat{J}^{\mu}_a##; on the other hand we have symmetry breaking, which , according to my understanding, is that you fix the minimum of the potential and make an expansion around it and a massless boson alongside a massive one emerges. Question: how precisely mathematically is ##\hat{J}^{\mu}_a## related to operator of goldstone boson creation ##\hat{a}^\dagger##?
 
  • #8
In general such a simple relation only exists in the long wavelength limit. E.g. you won't be able to derive a complete phonon spectrum just from the broken translational symmetry.
In chapter 19.2 of "The quantum theory of fields" by Stephen Weinberg, vol. 2 might contain the answer you are looking for.
 
  • #9
I have to read previous chapters to understand it. :/
 

1. What is the Goldstone boson state?

The Goldstone boson state is a theoretical particle that was proposed by physicist Jeffrey Goldstone in the 1960s. It is a type of boson, or elementary particle, that plays a key role in the spontaneous breaking of symmetries in particle physics theories.

2. How is the Goldstone boson state related to conserved currents?

The Goldstone boson state is closely related to conserved currents in particle physics. Conserved currents are mathematical quantities that represent symmetries in a system. The Goldstone boson state can arise when these symmetries are spontaneously broken, resulting in a non-zero value for the conserved current.

3. What is meant by "spontaneous breaking of symmetries" in relation to the Goldstone boson state?

Spontaneous breaking of symmetries refers to a situation where the equations describing a physical system exhibit symmetries, but the physical system itself does not. In the case of the Goldstone boson state, the symmetry is broken and the resulting particle is no longer symmetric, but the underlying equations still exhibit the symmetry.

4. Can the Goldstone boson state be observed in experiments?

Yes, the Goldstone boson state has been observed in experiments. In 1964, physicists Peter Higgs, François Englert, and Robert Brout proposed a theory that predicted the existence of a particle, now known as the Higgs boson, which is a type of Goldstone boson state. This particle was finally observed in 2012 at the Large Hadron Collider in Switzerland.

5. What implications does the Goldstone boson state have for particle physics?

The Goldstone boson state plays a crucial role in many particle physics theories, including the Standard Model of particle physics. It helps explain why certain particles have mass and why some symmetries in nature are broken. The discovery of the Higgs boson, a type of Goldstone boson state, has further confirmed the validity of these theories.

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