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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

    [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.
     
  2. jcsd
  3. Sep 25, 2013 #2

    Avodyne

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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

    fzero

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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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