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EL equations of the SM

  1. Nov 30, 2013 #1
    I was playing around with a classical field treatment of the SM, and I came across something strange I was hoping somebody could explain to me. Basically, I found that if you plug the EL equations back into the SM lagrangian, you can essentially remove the Higgs field entirely (at least the kinetic and mass terms, which are the most important aspects). I'll briefly show my math below, just in case I screwed up somewhere:

    The electroweak lagrangian:
    [itex]L=-\dfrac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu} + W_{\mu\nu}W^{\mu\nu}) + (D_\mu\phi)^\dagger D^\mu\phi - \dfrac{\mu^2}{2v^2}(|\phi|^2-\dfrac{1}{2}v^2)^2[/itex]
    [itex]D_\mu\phi \equiv (\partial_\mu-ig_1 B_\mu - ig_2 W_\mu)\phi[/itex]
    [itex]W_{\mu\nu} \equiv \partial_\mu W_\nu - \partial_\nu W_\mu -ig_2[W_\mu,W_\nu][/itex]
    [itex]B_{\mu\nu} \equiv \partial_\mu B_\nu - \partial_\nu B_\mu[/itex]

    The "3" EL equations are:
    [itex]\partial_\nu B^{\mu\nu} = ig_1\phi^\dagger(D^\mu\phi)-ig_1(D^\mu\phi)^\dagger\phi[/itex]
    [itex][D_\nu,W^{\mu\nu}] = ig_2(D^\mu\phi)\phi^\dagger-ig_2\phi(D^\mu\phi)^\dagger[/itex]
    [itex]D_\mu D^\mu \phi = -\dfrac{\mu^2}{v^2}(|\phi|^2-\dfrac{1}{2}v^2)\phi[/itex]

    From the first equation you can derive the continuity equation:
    [itex](D_\mu\phi)^\dagger D^\mu\phi + \phi^\dagger D_\mu D^\mu \phi = 0[/itex]

    and then using the third equation you find:
    [itex](D_\mu\phi)^\dagger D^\mu\phi = \dfrac{\mu^2}{v^2}(|\phi|^2-\dfrac{1}{2}v^2)\phi[/itex]

    if you then substitute this back into L, you find the new Lagrangian:
    [itex]L=-\dfrac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu} + W_{\mu\nu}W^{\mu\nu}) + \dfrac{\mu^2}{2v^2}|\phi|^4 - \dfrac{1}{8}\mu^2 v^2[/itex]

    so that the new Lagrangian has a completely isolated Higgs field, and no spontaneously broken symmetry! Not only this but the mass term for the Higgs, and the coupling terms for the EW fields have disappeared! Now I am aware that the EL equations are only considered valid on-shell, so this suggests to me that the Higgs mechanism is entirely a quantum effect and would not occur for classical fields... Am I making some mistake here?
  2. jcsd
  3. Dec 1, 2013 #2
    Nope nvm, figured out a stupid mistake I made. The continuity expression only holds for the imaginary part, not the real. Since half of it is purely real you end up with an incredibly trivial identity that you can derive solely with the 3rd EL equation
  4. Dec 2, 2013 #3
    oops, I spoke too soon. I did make a mistake, but it turns out that it really doesn't effect my question at all, except for some minor changes:

    Replace the "continuity equation" with the identity:
    [itex]\partial_\mu(\phi^\dagger D^\mu\phi) = (D_\mu\phi)^\dagger D^\mu\phi + \phi^\dagger D_\mu D^\mu\phi[/itex]

    this then leads to the new Lagrangian:
    [itex]L=-\dfrac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu}+W_{\mu\nu}W^{\mu\nu}) + \dfrac{\mu^2}{2v^2}|\phi|^4 - \dfrac{1}{8}\mu^2v^2 + \partial_\mu X^\mu[/itex]
    where [itex]X^\mu \equiv \phi^\dagger D^\mu\phi[/itex]

    Now by dropping the total derivative term, you arrive at the same Lagrangian from my original question!

    *Note: there are two steps here I'm not very confident about, so I would bet that's where my problem lies. However, I don't see any obvious problems so if someone could help me out that would be great.

    1) The term [itex]X^\mu[/itex] must be real for consistency (every other term in L is real, and L must be real). However, it is not obvious at all that this term actually is real, and whenever I try to test it I end up in logical circles. In the end you can show that it must be real if the EL equations are satisfied, but it seems a little lacking as far as a proof goes. *edit* see below where I show that [itex]X^\mu[/itex] is NOT real, but its derivative is

    2) Usually you can drop total derivatives from the lagrangian, since they end up as surface terms in the action. However there are certain situations in which this can't be done, most notably when the term doesn't vanish at infinity (or doesn't vanish fast enough). Since we're dealing with Higgs fields here, which are scalar fields with a non-zero expectation value, I have no real confidence in my decision to drop that term. On the other hand though I can't see a good reason NOT to drop it, take a look at it expanded!
    [itex]X^\mu \equiv \phi^\dagger \partial^\mu\phi - ig_1B^\mu|\phi|^2 - ig_2\phi^\dagger W^\mu\phi[/itex]
    Surely the B/W potentials fall off rapidly at infinity, and the vacuum expectation value of the 4-derivative of the Higgs field is 0! So I'm not sure how to approach this rigorously but it appears to me that this total derivative term can be removed from the Lagrangian safely..

    On the other hand, I was able to show that:
    [itex]\partial_\mu X^\mu = \dfrac{1}{2}\partial_\mu\partial^\mu|\phi|^2[/itex]
    so that even IF you can't drop this term, the Higgs field is STILL decoupled from the EW fields! So the only issue I really have with what I've done is substituting the EL equations back into the Lagrangian
    Last edited: Dec 2, 2013
  5. Dec 2, 2013 #4


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    If you substitute all the equations of motion back into the lagrangian, you just get a constant. There is no physical or mathematical insight to be gained by just substituting back some of them.
  6. Dec 2, 2013 #5
    Hmm so you CANT plug the EL equations into the Lagrangian.. Is there a simple proof for why this is? It's far from obvious to me, but you're right: every specific example I could think of results in a lagrangian that is entirely constants and total derivatives.

    It does make some sense though, if you take the EL equations as postulates the lagrangian loses all predictive power!
  7. Dec 3, 2013 #6
    The EL equations only apply to real particles. They do not apply to virtual particles. In fact, that's the defining difference between a real particle and a virtual particle.
  8. Dec 3, 2013 #7


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    EL equations are those equations that minimizes your action variations.
    What would that imply for the Lagrangian?
    That if you put them back you will have to make it constant or maybe a total derivative, so its action would be the same thing under variation and trivially you'd get that dS=0
    I guess that's why you can't plug them in...you'll get nothing out of it

    An analogous example (of course it's only slightly the same) is like finding a solution of a differential equation, and then plugging it in the same equation expecting to find something... you'd get a trivial result...
    Last edited: Dec 3, 2013
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