27
$\begingroup$

I am trying to understand what a term like $$ \mathcal{L}_{int} = (\partial^{\mu}A )^2 B^2 $$ with $A$ and $B$ being scalar fields for instance means. I understand how to draw an interaction term in Feynman diagrams without the derivative and how to interpret it (connecting external lines, find the correct value for the interaction coupling constant and so on).

But if I have a derivative in front of one of the fields, how do I interpret it ? Is it still two $A$ scalar particles interacting with two $B$ scalar particles ? How the derivative changes the interaction ?

I search a bit on Internet about that and found some resources : Preskill Notes (see p. 4.33) or Useful Formulae and Feynman Rules (see p. 20) but still... Don't understand.

$\endgroup$
1
  • $\begingroup$ Does the interaction term with the momentum then means, that we have a travelling particle A from the vertex to another vertex with endpoint particle B ? $\endgroup$
    – user16745
    Commented Dec 10, 2012 at 10:58

1 Answer 1

30
$\begingroup$

Here I'll try to basically connect some dots to guide you through the example of the second text you posted...

Any quantum field theory of your choice associates certain integrals to observables, which you have to compute. The Feynman diagrams are representations of these integrals. The lines correspond to propagators, which encode the different field dynamics, and the vertices are expressions which contain the coupling strengths and the right amount of indices to connect your propagators. To derive the Feynman rules, you'll expand the integrals, read off the general structure and associate certain integrands to certain pictures. Then, with the rules in your pocket, you decide on a Feynman diagram of choice you want to compute, write down all the right terms and integrate over all the loose ends.

Now, you have an expression $ \mathcal{L}_{int} = g\,(\partial^{\mu}A )^2 B^2 $, which you identify as interaction term (there are two different fields after all) and you wonder what to do with the derivative, which you only know from the kinetic term. Well, to know what the propagators of the theories are, you need the whole Lagrangian/the full dynamics of the theory anyway, so this information will certainly get incorporated there. How the vertex expression turns out (your question) is what the second paper you posted is trying to describe:

If you derive what the Feynman rules are in momentum space, where the fields $A$ and $B$ get represented in terms of their Fourier modes ($A(x)=\int\text{d}p\, \hat A(p)\text{e}^{ipx}$), then you see that a derivative $\partial^\mu$ turns into a momnetum four-vector $p^\mu$ (under the intergral).

If you had the simpler interaction structure $g\, A ^2 B^2$, then your vertex would be typically represented merely by the number $g$ and the knowledge of which propagators end up there. Now, in deriving the Feynman rule for your specific problem which involves $g\, (\partial^{\mu}A )^2 B^2 $, your integrand will also contain a function of the momentum vector, e.g. $p^2$ from $$\partial^2 A(x)=\int \text{d}p\, \hat{A}(p)\text{e}^{ipx}\cdot p^2$$ Hence your vertex term (in momentum space representation), which is essentially the integrand without the propagator expressions (some denominators which look like "$\frac{1}{p^2+m^2}$" or so) will be not only "$g$" but something like "$g \cdot p^2$".

Clearly what this means is that the higher modes (big momenta etc.) might be dangerous objects, as you want your integrals to converge - you integrate over $p$, so higher powers in $p$ under the integral are usually not your friend. Very vaguely, if the direct coupling à la $S\sim g\int\text{d}x A^2B^2$ wants to be minimized then high $A$ means low $B$. From this perspective, a term "$S\sim g\int\text{d} x \, (\partial^{\mu}A )^2 B^2 $" makes you think "Oh, so the behaviour of the field $B$ doesn't only depend on the other field amplitude of $A$, but also directly on that fields relative local dynamics". but you really have to take a look at specific theories for specific implications.

If you look for physical (but more involved) examples, you can look up the Feynman diagrams of Yang-Mills theory (scrolling down a little on the page) and try to compare the interaction structure with all the vertices containing functions of momentum (the second and the last here).

$\endgroup$
2
  • $\begingroup$ Great explanation ! Thanks, very helpful. To sum up, if I understand correctly, the derivative will modify the vertex term, but it is still interpreted as the same particle ($A$ here). By the way I'll go throughout the text again, with yours tips. :) $\endgroup$
    – AnSy
    Commented Nov 14, 2012 at 14:33
  • 2
    $\begingroup$ @Bagheera: Yes, it will "modify" it, speaking relative to the term which doesn't involve the derivative. But that's of course a totoally different theory. Roughly: differential equation -> propagator, rest -> vertex. And yes, it's the same particle, just as the classical trajectory $x(t)$ gives the position, even if your Lagrangian contains the velocity $v(t)=\dot x(t)$. You decide on what constitudes the physical degrees of freedom. What you mean by particles is labeled by the momenta. The momentum operator, your derivative, is just a (multiplicative) operator with respect to these states. $\endgroup$
    – Nikolaj-K
    Commented Nov 14, 2012 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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