What criteria should I consider when determining the non-relativistic limit of a Lagrangian density? For example, how would I take the non-relativistic limit of the following Lagrangian density:

$$\mathcal{L}=-\frac{1}{4}F_{\mu \nu }F^{\mu \nu }+\left( D_{\mu }\phi \right) \left( D^{\mu }\phi \right) -U\left( \left\vert \phi \right\vert \right) $$

A general method of approaching this problem would also be much appreciated.

  • 1
    $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$ – Qmechanic Mar 26 '14 at 18:01
  • 1
    $\begingroup$ Thak you vary much! $\endgroup$ – lucenalex Mar 26 '14 at 18:38

QED is an innately relativistic theory since it has massless particles (photons). Thus to take a non-relativistic limit you must remove the photon degree of freedom. This can be done by treating the electromagnetic field as a classical field, i.e. one that isn't quantized. Practically this amounts to only having real (on-shell) photons.

The scalar QED Lagrangian is explicitly \begin{equation} {\cal L} = \partial _\mu \phi ^\ast \partial ^\mu \phi - V \left( \phi \right) + \left( i e A _\mu \phi ^\ast \partial ^\mu \phi + h.c. \right) + e ^2 A _\mu A ^\mu \left| \phi \right| ^2 \end{equation}

The procedure for finding the Feynman rules for this theory are the same as for when you quantize $ A _\mu $. The only difference is instead of writing $ A _\mu \sim e ^{ - i k x } a _k \epsilon _\mu + e ^{ i k x} a _k ^\dagger \epsilon _\mu $ you just leave it as a function.

To calculate the Feynman rule associated with the first interaction you can calculate the two point function to first order in $e$ (depending on how comfortable you are with derivative couplings you may just be able to guess the answer),

\begin{equation} \left\langle 0 \right| T \phi _1 \phi _2 \int \,d^4y ( i e A _\mu \phi ^\ast \partial ^\mu \phi + h.c. ) \left| 0 \right\rangle \end{equation}

This leads the Feynman rule,

$\hspace{2cm}$enter image description here

where the $ \times $ denotes the effect of an external field. Note that the external field cannot appear as an intermediate particle since it was not quantized. The $ \tilde{A} _\mu $ is just the Fourier transform of the vector potential of the system. One can analogously calculate the Feynman rule for the second interaction.

Once you have the Feynman rules for scalar QED you go ahead and in principle go ahead derive all of classical electromagnetism (at least acting on spinless particles).

EDIT: Thus far what we have taken QED and reduced it to having a background electromagnetic field instead of single photon interactions. As mentioned in the comments by @LoveLearning this approach isn't complete as we are still allowing relativistic scalars to propagate. However, in the non-relativistic limit these are suppressed. To see why this must be true consider the diagram with two interaction vertices,

$\hspace{5cm}$enter image description here

This diagram is given by, \begin{equation} ( 2 i e ) ^2 ( p _1 - p _2 ) _\mu \tilde{A} ^\mu ( p _2 - p _3 ) _\nu \tilde{A} ^\nu \frac{ i }{ p _1 ^\alpha \gamma _\alpha - m } \end{equation} But in the non-relativistic limit, $ m \gg p _i $. The propagator suppresses the diagram by a factor of $ 1/ m $. This suppression occurs for every diagram with an off-shell particle. Thus in the non-relativistic limit we can just omit any diagrams with off-shell particles.

  • $\begingroup$ "Practically this amounts to only having real (on-shell) photons."Hi @JeffDror, is it not true that all particle (and not only the photons) be on shell in the classical limit? $\endgroup$ – Your Majesty Mar 28 '14 at 8:09
  • 1
    $\begingroup$ @LoveLearning: Good point. I updated the answer $\endgroup$ – JeffDror Mar 28 '14 at 11:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.