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Effective field theories, eliminating fields using equations of motion

  1. Jul 26, 2012 #1
    First year grad student here, I've taken two terms QFT.

    I'm studying some effective field theories, and one of the techniques I've seen used for writting down the effective lagrangian is identifying some fields or components of fields that are "small" and removing them from the lagrangian by plugging in their equations of motion. So for example if you had something like

    [tex] \frac{1}{2}\partial_{\mu}f\partial^{\mu}f + \frac{1}{2}\partial_{\mu}g\partial^{\mu}g + fg\phi(x)[/tex]

    you would as have equations of motion

    [tex] \partial^2f = g\phi(x),\quad\quad \partial^2g = f\phi(x)[/tex]

    so if you wanted to eliminate g you would substitute

    [tex] \frac{1}{2}\partial_{\mu}f\partial^{\mu}f + \frac{1}{2}\partial_{\mu}(\partial^2f/\phi(x))\partial^{\mu}(\partial^2f/\phi(x)) + f\partial^2f[/tex]

    My problem is I'm having a hard time wrapping my head around what exactly I'm doing when I do this. If this were just a regular classical functional minimization problem where I was minimizing something like volume or cost rather than an action, I think I haven't changed anything, I've just hidden information about g. The solutions to f's (new) equations of motion should give me the correct mimization of the original problem, and then I could for example use g's equations of motion to obtain g from f.

    However, in field theory I'm not really sure what I've done. A source I'm reading says that I've "removed g at tree-level". I can understand this in the sense that as I guessed above I should still have the right answer classically, so I haven't mess up tree-level. But where exactly along the line did I break the quantum corrections? Does this have something to do with the fact that off-shell particles don't solve their equations of motion, so the assumption that g will always be on shell is wrong?

    Any comments about what's going on here would be extrememly helpful, thanks!
     
    Last edited: Jul 26, 2012
  2. jcsd
  3. Jul 26, 2012 #2
    I'm not sure I've got this entirely correct, but:

    Consider the path integral formulation. The partition function contains integrals over both f and g. We want to eliminate g by doing the g integral, leaving an effective theory of the f field.

    However, this is too hard. Let's do a semiclassical approximation. We are integrating the expression exp(iS). Most of the contribution to the integral comes from the regions near where the action S is stationary with respect to the fields: elsewhere it mostly cancels out. But recall that the condition that S be stationary with respect to g is exactly what gives the classical equation of motion for g. We can always expand the action functional around the classical trajectory: it looks like

    [tex]S[g_{classical}+\Delta g] = S[g_{classical}] + \frac{1}{2}\frac{\delta^2 S}{\delta g^2}|_{g=g_{classical}}(\Delta g)^2 + O((\Delta g)^3)[/tex]

    Doing this in the path integral (and remembering that g_classical actually depends on f), we get

    [tex]Z = \int Df \int Dg \exp(iS[f, g]) = \int Df \int D(\Delta g) \exp\{iS[f, g_{classical}(f)] + O((\Delta g)^2)\}[/tex]

    If we throw away the higher order terms in delta-g, the integral over delta-g becomes trivial, because delta-g no longer appears in the integrand. So we get

    [tex]Z \propto \int Df \exp\{iS[f, g_{classical}(f)]\}[/tex]

    that is, we have a quantum field theory of one field, f. Its action is the action you obtain by substituting for g in the original action using the classical equation of motion. But clearly this theory is only an approximation to the original theory, since we threw away a bunch of terms above.
     
  4. Jul 26, 2012 #3

    fzero

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    The_Duck's answer is basically correct. I thought I'd illustrate all of the ideas in a slightly more physical example and it turned out to be rather long.

    Suppose we have a massless scalar field ##\phi## and a massive scalar ##\Phi## coupled through an interaction ##\lambda \phi^2 \Phi##. The Lagrangian is

    $$ L [\phi,\Phi]= \frac{1}{2} (\partial \phi)^2 + \frac{1}{2} (\partial \Phi)^2 -\frac{M^2}{2} \Phi^2 + \lambda \phi^2 \Phi.$$

    Suppose that we consider the process where we scatter ##\phi## particles of momentum ##p_1,p_2##. For nonzero angle, at order ##\lambda## we can create a ##\Phi## particle as long as the center of mass energy is larger than the ##\Phi## mass, ##\sqrt{s}\geq M##. Here we are using the Mandelstam variable ##s = (p_1+ p_2)^2##.

    At low energies, ##\sqrt{s}< M##, there is not enough energy available to create a ##\Phi## particle in the final state. Nevertheless, we know from quantum scattering theory that off-shell ##\Phi## particles on internal lines of Feynman diagrams induce scattering between the ##\phi## states. The lowest-order non-trivial diagrams are at order ##\lambda^2## and are ##2\phi \rightarrow 2\phi## processes. There are 3 diagrams consisting of ##s,t## and ##u## channels. It is a simple exercise to apply the Feynman rules to these tree-level diagrams and, neglecting the overall numerical factor, the total matrix element is

    $$ A \sim \lambda^2 \left( \frac{1}{M^2-s} + \frac{1}{M^2-t} + \frac{1}{M^2-u} \right) \sim \frac{\lambda^2}{M^2} + \cdots . $$

    where in the last part we have given the leading contribution in the energy expansion (##s,t,u\ll M^2##). Here we have a factor of ##\lambda## for each vertex and a factor of ##1/M^2## for the single ##\Phi## propagator.

    Now the idea of effective field theory is the following. At low enough energies, where we cannot create the heavy particle ##\Phi## in the final state, is there an approximate Lagrangian that can reproduce ##\phi## physics? This would be a Lagrangian involving only ##\phi##, but with new interaction vertices that encode the physics of exchanging the heavy particle ##\Phi##.

    One way to arrive at such an effective Lagrangian would be to simply write down all possible terms consistent with the symmetries of the original Lagrangian. The original Lagrangian is symmetric under ##\phi\rightarrow -\phi##, so all such terms involve even numbers of ##\phi##. The original Lagrangian was parity-invariant, so all terms in the effective Lagrangian must involve even numbers of derivatives. So we can have terms like ##\phi^4, \phi^2(\partial\phi)^2, (\partial \phi)^4,\ldots##. We can write these terms in such a way that their coefficients are dimensionless numbers by including a factors of ##M## for each extra derivative appearing in the interaction terms:

    $$L_\mathrm{eff} [\phi] = \frac{1}{2} (\partial \phi)^2 + c_1 \phi^4 + \frac{c_2}{M^2} \phi^2(\partial\phi)^2 + \frac{c_3}{M^4} (\partial \phi)^4 + \cdots.$$

    We can approach the question of what the coefficients ##c_i## should be in a few ways. First we might just consider this as a phenomenological theory and leave their determination to a comparison with experiment. However, we have a high energy theory in mind, so we can get a theoretical understanding of the coefficients for this case.

    Now, we can proceed formally by using generating functions and path integrals. We couple a source ##j## to ##\phi## and a source ##J## to ##\Phi##. Then the generating function for connected graphs is ##W[j,J]##, where

    $$\exp \left( i W[j,J] \right) = \int \mathcal{D}\phi \mathcal{D}\Phi \exp \left\{ i \int d^x \left( L[\phi,\Phi] + j\phi + J\Phi \right)\right\} , $$

    while the generating function for 1-particle irreducible graphs is

    $$\Gamma[\phi,\Phi] = W[ j[\phi],J[\phi]] - \int d^4x \left( j\phi + J \Phi\right).$$

    Given ##\Gamma##, we can solve for the source ##J## using

    $$ \frac{\delta \Gamma[\phi,\Phi]}{\delta \Phi} + J =0.$$

    Now the effective field theory computes diagrams with no external ##\Phi## lines, which corresponds to setting the source ##J=0##. This corresponds to computing ##\Gamma[\phi,\Phi]## at a stationary point with respect to the heavy field ##\Phi##. However, in the loop expansion,

    $$\Gamma[\phi,\Phi] = S[\phi,\Phi] + \Gamma_1[\phi,\Phi] + (\mathrm{2+loops}),$$

    i.e., tree-level graphs are generated by the classical action. So at tree-level, evaluating the effective action consists of evaluating ##\Phi## on its equation of motion.


    In the model at hand, the equation of motion for ##\Phi## is

    $$ (\partial^2 -M^2) \Phi = - \lambda \phi^2.$$

    We can formally invert this as

    $$ \Phi = \frac{\lambda}{ M^2 - \partial^2 } \phi^2 = \frac{\lambda}{ M^2 } \left( 1 + \frac{\partial^2}{M^2} +\cdots \right) \phi^2 ,$$

    where on the RHS we've written down a formal power series in derivatives. Since each derivative acting on ##\phi## is like bringing out a factor of the average energy (a Fourier transform would make this more precise), when ##E \ll M##, we can neglect higher order terms in the derivative expansion.

    If we use the expansion for ##\Phi## in our original Lagrangian, we find

    $$L_\mathrm{eff}[\phi] = \frac{1}{2} (\partial \phi)^2 +\frac{\lambda}{2M^2} \phi^4 + \cdots,$$

    where I've haven't bothered to write down the higher derivative terms. There is also a correction of order ##\lambda^2/M^4## to the kinetic term for ##\phi## that I've neglected.

    As a small test of this effective Lagrangian, we can note that, to lowest order in the energy expansion, tree-level ##2\phi\rightarrow 2\phi## scattering is given by the coefficient of the quartic term in the effective Lagrangian. This term precisely matches the leading term of the amplitude we calculated in the full theory above. A more interesting test would be to try to reproduce the numerical factors in both results, as well as the subleading terms in that expansion by including the higher-derivative terms in the effective Lagrangian.
     
  5. Jul 27, 2012 #4
    Thank you both, that will be a big help!
     
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