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Spin-half mass term with symmetry breaking

  1. Sep 25, 2013 #1
    I've been thinking about chapter 11 of Griffiths' Introduction to Elementary Particles. In section 11.7, he gives the Lagrangian density


    and shows that the minimum occurs at


    and thus that the shifted potential picks up a mass term


    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


    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


    which when set to zero gives


    That's as far as I can really get without confusing myself entirely. Any insight would be much appreciated.
  2. jcsd
  3. Sep 25, 2013 #2


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    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].
  4. Sep 25, 2013 #3


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    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.
  5. Sep 26, 2013 #4
    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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