I've been thinking about chapter 11 of Griffiths' Introduction to Elementary Particles. In section 11.7, he gives the Lagrangian density(adsbygoogle = window.adsbygoogle || []).push({});

[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.

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# Spin-half mass term with symmetry breaking

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