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Spontaneous symmetry breaking in the SM

  1. Mar 7, 2006 #1
    [tex]\mathcal{L} = \frac{1}{2}(\partial_{\mu}\underline{\phi}).(\partial^{\mu}\underline{\phi}) + \frac{1}{2}\mu^{2}\underline{\phi}.\underline{\phi} - \frac{\lambda}{4}(\underline{\phi}.\underline{\phi})^{2} + \bar{\psi}(i\gamma . \partial )\phi - g\bar{\psi}(\phi_{1}+i\gamma^{5}\phi_{2})\psi [/tex]

    where [tex]\underline{\phi} = \left( \begin{array}{c} \phi_{1} \\ \phi_{2} \end{array} \right) [/tex]

    I've shown this Lagrangian is invariant under [tex]\phi_{1} \to \cos \alpha \phi_{1} - \sin \alpha \phi_{2}[/tex] [tex]\phi_{2} \to \sin \alpha \phi_{1} + \cos \alpha \phi_{2}[/tex] [tex]\psi \to \exp\left( -\frac{i \alpha \gamma^{5}}{2} \right)\psi[/tex]

    The question then asks to show that the solution to the classical equations of motion with minimal energy lead to a vacuum which breaks the symmetry spontaneously. Then, to pick a suitable vacuum solution, and use it to show the fermion field acquires a mass proportional to g.

    If someone could give me pointers in the right direction I've be very grateful. I've tried mucking about with the equations of motion for the phi's and psi's, but seem to going round in circles. Thanks :smile:
  2. jcsd
  3. Mar 7, 2006 #2

    Physics Monkey

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    Hi AlphaNumeric,

    Since you are asked to find the classical lowest energy field configuration it makes sense to construct the Hamiltonian for your system. This step is essentially trivial since your Lagrangian is nice and simple in its time derivatives (note that I think you have a typo in the kinetic term for the dirac field, that [tex] \phi [/tex] should be a [tex] \psi [/tex]). I won't spoil it for you, but you will find that some terms in the Hamiltonian should obviously be zero in the lowest energy configuration. Once you set these terms to zero in your classical field equations, everything becomes simple and the solution will pop right out.
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