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O(2) symmetry breaking trouble

  1. May 15, 2012 #1
    Hi, I'm going to quote a lot of a book so that I can get some help, brace yourselves...

    First, [itex]\phi_{a}[/itex] is my field with [itex]a=0,1[/itex] as internal components and my lagrangian is
    [tex]
    L=\frac{1}{2}\partial_{\mu}\phi_a \partial^\mu \phi_a +\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2
    [/tex]
    with [itex]U=\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2 [/itex]
    now in Jan Smit's book, "Introduction to Quantum Fields on a Lattice" the page has the following:
    My first question: Is that [itex]v[/itex] in the symmetry breaking part the same [itex]v[/itex] we defined above?

    Now we go on to expand around [itex]\phi_{g}^a[/itex], and the results in the book are
    with [itex]\pi=\phi^1[/itex] and [itex]\phi^0=v+\sigma[/itex]
    and
    Now the reason I asked the above question is, substituting the first definition of [itex]v^2[/itex] into these mass definitions, [itex]m_{\pi}[/itex] equals zero...

    So then, what is [itex]v^2[/itex] in the symmetry breaking equations, why use the same letter IF it's different, and not say so?

    Thanks,
     
  2. jcsd
  3. May 15, 2012 #2

    fzero

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    The vacuum expectation value, [itex]v[/itex], is defined by [itex]\phi^a_g = v \delta_{a0}[/itex]. Its value is set by the classical potential to be a root of [itex](\mu^2+ \lambda v^2)v=\epsilon[/itex]. When [itex]\epsilon =0[/itex], it takes the value [itex]v^2 = - \mu^2/\lambda[/itex], but not in general. It should be obvious why the same letter is used when viewed in this way.
     
  4. May 15, 2012 #3
    That helps, thanks.
     
  5. May 16, 2012 #4
    geez, I still can't get that second set of quoted equations. Here are my steps. I am starting with the EOM
    [tex]
    (-\partial^2 +\mu^2 +\lambda \phi^2)\phi^a=\epsilon \delta_{a,0}
    [/tex]
    and I am expanding around [itex]v[/itex] as [itex]\phi^0 =v+\sigma[/itex] and [itex]\phi^1 =\pi[/itex]. That gives me
    [tex]
    (-\partial^2 +\mu^2 +\lambda (\phi^0 \phi^0 + \phi^1 \phi^1))\phi^a =\epsilon \delta_{a,0}
    [/tex]
    [tex]
    (-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))\phi^a =\epsilon \delta_{a,0}
    [/tex]
    [tex]
    (-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))(v+\sigma)=\epsilon ,\quad (-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))\pi=0
    [/tex]
    The first equation turns to, using [itex](\mu^2+\lambda v^2)v=\epsilon [/itex]
    [tex]
    -\partial^2 (v+\sigma )+\mu^2 \sigma +\lambda \pi^2 v +\lambda \pi^2 \sigma +3\lambda v^2 \sigma =0
    [/tex]
    and the second
    [tex]
    -\partial^2 \pi +\mu^2 \pi +\lambda v^2 \pi +2\lambda v \sigma \pi =0
    [/tex]
    Both of these have many of the terms I need, plus a number of them that don't show up in the book. Where am I going wrong?

    Thanks,
     
  6. May 16, 2012 #5

    fzero

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    The terms which are of higher order in the fields are interaction terms. To find the mass we want the terms in the e.o.m. that are linear , so write those as

    [tex]
    -\partial^2 \sigma+(\mu^2 +3\lambda v^2 )\sigma = - \lambda v \pi^2 -\lambda \pi^2 \sigma
    [/tex]
    [tex]
    -\partial^2 \pi + (\mu^2 +\lambda v^2) \pi = - 2\lambda v \sigma \pi .
    [/tex]

    The terms on the RHS are just [itex]\partial V/\partial \sigma[/itex] and [itex]\partial V/\partial \pi[/itex], where [itex]V(\sigma,\pi)[/itex] is the part of the scalar potential without the mass terms.
     
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