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Broken symmetry

  1. Jun 15, 2009 #1
    As an exemple of borken symmetry the teacher showed what happens with this lagrangian:

    [tex] \Lambda=\frac{1}{2}\partial_\mu \Phi^{+}\partial^{\mu}\Phi - \left( \frac{m^2}{2} \partial_\mu \Phi^{+}\partial^{\mu}\Phi +\frac{g^2}{4}(\partial_\mu \Phi^{+}\partial^{\mu}\Phi )^2 ) \right)[/tex]
    where
    [tex]\Phi=(\phi_1,\phi_2)[/tex]
    the components are two real fields and the plus sign means making the transposed conjugate of the vector (actually [tex]\Phi[/tex] should be a column vector)... of course everything depends on the space time point [tex]x[/tex]

    ok... the approach was, as far as I've understood it: let's expand [tex]\Lambda(\phi_1,\phi_2)[/tex] in one of the minima diagonalizing also the quadratic term and recognize frome the quadratic term what can be the physical meaning of the field, that is: two particle, and among them a goldstone boson... the new fields were so done:

    [tex]\sigma=\phi_1-\frac{m}{g}[/tex]

    [tex]\eta=\phi_2[/tex]

    the new lagrangian (can't make the calculation here, but it's not important) has no more an easy symmetry related to the new fields and the void is not invariant through the action of an operator implementing the symmetry... this is (as far as I've understood) the broken symmetry)

    ok, going on... one realizes that one could have chosen other minima to expand the hamiltonian.... this is my first question: choosing other minima has a physical meaning? I guess so, and the result is describing the same particles in a more difficult way: I expect the lagrangian for a single field to contain terms of self interaction that don't permit to identify the quadratic part and the relative vectors in Hilbert space as observed particles... I don't bother of the interaction terms becuase I expect them to count only when the particles are interacting...

    do you agree with the red sentence? It's an idea of mine so I wanted to have an exchange of opinions...

    Going on, my teacher says, let's try this:

    [tex]\phi_1=\rho cos(\theta)[/tex]

    [tex]\phi_2=\rho sen(\theta)[/tex]

    writing the hamiltonian this way he finds that the goldstone boson has no self interacting term (I call this way terms who contain only one field but are not quadratic)... and since the other particle has a mass he says (this is a remembering, avtually I didn't take notes on this point) that at low energy the effective lagrangian is:

    [tex]\Lambda=\frac{1}{2}\partial_\mu \alpha\partial^{\mu}\alpha [/tex]

    having relabled [tex]\theta=\alpha\frac{g}{m}[/tex]

    but now I think... why not consider also the other lagrangian the one with [tex]\phi_1,\phi_2[/tex] for low energies? if the intuitive reason above is correct, I expect the mass-particle not to count and conserve only terms with the goldstone boson, and find:

    [tex]\Lambda=\frac{1}{2}\partial_\mu \eta\partial^{\mu}\eta-\frac{g^2}{4}\eta^4+\frac{m^4}{4g^2} [/tex]

    so the second question is: at low energy we have these lagrangians, of course the second more complicated then the second one... but I expect them to describe exactly the same particles (at low energy)! and so having the same low energy "eigenstates"... and so solving the second should be in some sense equivalent to solving the first one...

    ok.... i said a lot of things... am I totally out of way with my understanding?... is there some correct idea?

    thanks to anybody who's gonna help me...
     
    Last edited: Jun 15, 2009
  2. jcsd
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