# Spin-half mass term with symmetry breaking

1. Sep 25, 2013

### welcomeblack

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

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

and shows that the minimum occurs at

$\phi=\pm\frac{\mu}{\lambda}$

and thus that the shifted potential picks up a mass term

$m=\sqrt{2}\mu$

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

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

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

$\frac{\partial\mathcal{L}}{\partial\psi}=\mu\bar{\psi}-\lambda\bar{\psi}(\bar{\psi}\psi)$

which when set to zero gives

$\bar{\psi}\psi_{min}=\frac{\mu}{\lambda}$.

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

2. Sep 25, 2013

### Avodyne

This doesn't work in this case, because $\psi$ is an anticommuting object, not an ordinary number; $\bar\psi\psi$ cannot take on a numerical value like $\mu/\lambda$.

Another aspect of this is that the sign of $\mu$ doesn't matter; by making the field redefinition
$$\psi\to \exp(i\pi\gamma_5/2)\psi$$
$$\bar\psi\to \bar\psi\exp(i\pi\gamma_5/2)$$
we get
$$\bar\psi\psi\to -\bar\psi\psi$$
$$i\bar\psi\gamma^\nu\partial_\nu\psi\to i\bar\psi\gamma^\nu\partial_\nu\psi$$
which is equivalent to changing the sign of $\mu$.

3. Sep 25, 2013

### fzero

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

$$\psi\to \exp(i\pi\gamma_5/2)\psi,~~~\bar\psi\to \bar\psi\exp(i\pi\gamma_5/2),$$

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 $\psi_a$:

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

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. Sep 26, 2013

### andrien

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.