Spin-half mass term with symmetry breaking

  • #1
welcomeblack
13
0
I've been thinking about chapter 11 of Griffiths' Introduction to Elementary Particles. In section 11.7, he gives the Lagrangian density

[itex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi)+\frac{1}{2}\mu^{2}\phi^{2}-\frac{1}{4}\lambda^{2}\phi^{4}[/itex]

and shows that the minimum occurs at

[itex]\phi=\pm\frac{\mu}{\lambda}[/itex]

and thus that the shifted potential picks up a mass term

[itex]m=\sqrt{2}\mu[/itex]

in natural units. Now, I'm wondering if you can do the same thing with a spin-half particle. For example, if we have the spinor Lagrangian density

[itex]\mathcal{L}=i\bar{\psi}\gamma^{\nu}\partial_{\nu}\psi+\mu\bar{\psi}\psi-\frac{1}{2}(\bar{\psi}\psi)^{2}[/itex]

can we shift the spinors in such a manner that we can read off the mass term? Even the simple task of minimizing the potential is confusing. If we take the partial w.r.t. ψ, then

[itex]\frac{\partial\mathcal{L}}{\partial\psi}=\mu\bar{\psi}-\lambda\bar{\psi}(\bar{\psi}\psi)[/itex]

which when set to zero gives

[itex]\bar{\psi}\psi_{min}=\frac{\mu}{\lambda}[/itex].

That's as far as I can really get without confusing myself entirely. Any insight would be much appreciated.
 

Answers and Replies

  • #2
Avodyne
Science Advisor
1,396
90
This doesn't work in this case, because [itex]\psi[/itex] is an anticommuting object, not an ordinary number; [itex]\bar\psi\psi[/itex] cannot take on a numerical value like [itex]\mu/\lambda[/itex].

Another aspect of this is that the sign of [itex]\mu[/itex] doesn't matter; by making the field redefinition
[tex]\psi\to \exp(i\pi\gamma_5/2)\psi[/tex]
[tex]\bar\psi\to \bar\psi\exp(i\pi\gamma_5/2)[/tex]
we get
[tex]\bar\psi\psi\to -\bar\psi\psi[/tex]
[tex]i\bar\psi\gamma^\nu\partial_\nu\psi\to i\bar\psi\gamma^\nu\partial_\nu\psi[/tex]
which is equivalent to changing the sign of [itex]\mu[/itex].
 
  • #3
fzero
Science Advisor
Homework Helper
Gold Member
3,119
290
Now, I'm wondering if you can do the same thing with a spin-half particle. For example, if we have the spinor Lagrangian density

[itex]\mathcal{L}=i\bar{\psi}\gamma^{\nu}\partial_{\nu}\psi+\mu\bar{\psi}\psi-\frac{1}{2}(\bar{\psi}\psi)^{2}[/itex]

can we shift the spinors in such a manner that we can read off the mass term? Even the simple task of minimizing the potential is confusing. If we take the partial w.r.t. ψ, then

[itex]\frac{\partial\mathcal{L}}{\partial\psi}=\mu\bar{\psi}-\lambda\bar{\psi}(\bar{\psi}\psi)[/itex]

which when set to zero gives

[itex]\bar{\psi}\psi_{min}=\frac{\mu}{\lambda}[/itex].

That's as far as I can really get without confusing myself entirely. Any insight would be much appreciated.

This is a very interesting question. In 4d, the quartic term is nonrenormalizable (##\lambda## has dimensions of ##[\mathrm{mass}]^{-2}##)and doesn't affect the vacuum state. In 2d, the 4-fermi term is marginal (has the same mass dimension as the kinetic term and ##\lambda## is dimensionless) and can bind fermions (if the sign is chosen appropriately). In 2d, the model with ##\mu=0## is known as the Gross-Neveu model (a detailed solution can be found here).

The study of the vacuum state in the Gross-Neveu model is different than that of the massive ##\phi^4## theory, since the symmetry breaking is due to quantum effects, whereas in the massive ##\phi^4## theory it is already present at the classical level.

The interest in studying the model with ##m=0## is that it is only in that case that the classical model has the chiral symmetry

[tex]\psi\to \exp(i\pi\gamma_5/2)\psi,~~~\bar\psi\to \bar\psi\exp(i\pi\gamma_5/2),[/tex]

that Avodyne mentions. For nonzero ##\mu##, the chiral symmetry is broken. Instead of concluding that the sign of ##\mu## doesn't matter, one should instead conclude that instead it must be the case that ##\mu=0## in order for the Lagrangian to be invariant under chiral symmetry.

Gross and Neveu studied the quantum theory of the Lagrangian with ##N## fermions ##\psi_a##:

[itex]\mathcal{L}=i\sum_a \bar{\psi}_a\gamma^{\nu}\partial_{\nu}\psi_a +\frac{g_0}{2}\left(\sum_a \bar{\psi}_a\psi_a\right)^{2}.[/itex]

The + sign of the 4-fermi term corresponds to the attractive force. They found that quantum effects cause the expectation value of

$$ \sigma = - g_0^2 \sum_a \bar{\psi}_a\psi_a$$

to take a nonzero value. Therefore, even in the absence of a mass, the vacuum state breaks the chiral symmetry (that would have taken ##\sigma \rightarrow - \sigma##).

The Gross-Neveu model is interesting because it is considered a toy model of how chiral symmetry that acts on the quarks in QCD is broken by the strong interactions in the QCD vacuum.
 
  • #4
andrien
1,024
32
there is a reason why he chose the scalar field for example.It is in this way the higgs acquire a vacuum expectation value.Anyway if you want to give masses to fermions,you can generate masses via their coupling to a scalar field(higgs field so far).The coupling is the yukawa coupling between fermions and this higgs field.The coupling strength however is free parameter and determined by measured fermion mass.
 

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