# Deriving Feynman rules from Lagrangian

1. Oct 10, 2006

### AlphaNumeric

I've been given a small project to ease me into life as a PhD student and part of it involves working out Feynman rules for scalar bosons coupling to gauge bosons.

I've been reading through Peskin and Schroder Chapter 9 and want to clarify if I've got my understanding right of how to do it.

The QED Lagrangian is $$L = \int d^{4}x \,\mathcal{L} = \int d^{4}x \, \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - e\psi\gamma^{\mu}\psi A_{\mu}$$

The electron propogator's Feynman diagram rule is obtained by finding the Green's function of the Dirac operator $$i\gamma^{\mu}\partial_{\mu}-m$$ because it just describes an electron moving without interaction.

Similarly, the photon propogator is found by rearranging $$-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ into the form $$A_{\mu}\Delta^{\mu\nu}A_{\nu}$$ and finding the Green's function of $$\Delta^{\mu\nu}$$.

One thing I'm a bit hazy on (it seems to work but I want to be sure) is getting the vertex Feynman rules.

Suppose $$\mathcal{L} = \mathcal{L}_{0}+\mathcal{L}_{I}$$. Suppose the interaction term involves 3 fields (like $$\psi$$, $$\bar{\psi}$$ and $$A_{\mu}$$. Is the vertex rule, in momentum space, then simply

$$\frac{\delta}{\delta \psi}\frac{\delta}{\delta \bar{\psi}}\frac{\delta}{\delta A_{\mu}}\int d^{4}x \mathcal{L}_{I}$$

That would give the QED one as $$-ie\gamma^{\mu}\int d^{4}x$$ which is right. Is this true for all interaction Lagrangians? I follow most of the "Computing the correlation function" bits and the $$Z = e^{iS...}$$ bits involving functional derivatives, but when it comes to then computing the Feynman rules I just kind of noticed that relation between the Lagrangian and the rules rather than find any step by step guide in P&S.

Am I going about it the right way or have I just hit on the right answer by (dumb) luck? Thanks :)

2. Oct 12, 2006

### Hans de Vries

The vertex function stems from the interaction term in the Lagrangian
which gets into the equation of motion as:

$$-ie\gamma^\mu A\psi$$

Where both A and psi have sinusoidal plane waves as their components.
It is because of the product of A and psi that we can express scattering

$$p_1 - p_2 = p_3$$

Because:

$$\exp{(ip_1 x_\mu)} \times \exp{(-ip_2 x_\mu)}\ =\ \exp{(ip_1 x_\mu-ip_2 x_\mu)}\ =\ \exp{(ip_3 x_\mu)}$$

Multiplication of sinusoidal plane waves becomes convolution of Dirac
functions in momentum space:

$$\delta(q-p_1) \star \delta(q+p_2)\ =\ \delta(q-p_1+p_2)$$

So the momentum preserving Dirac function is actually part of the
total vertex function but usually handled separately. Now if you just
take away the $A\psi$ part from the interaction term then you're
left with:

$$-ie\gamma^\mu$$

Thus, the remaining part of the interaction term is what gives you your
vertex function.

Regards, Hans

Last edited: Oct 12, 2006
3. Oct 12, 2006

### Norman

As a quick and dirty method, you can just erase the fields in the interaction lagrangian to get the vertex function. But then you must follow the usualy "rules" for constructing amplitudes which, as Hans already said (very clearly), you need the conservation of momentum step in the "rules".

Cheers.

4. Oct 15, 2006

### AlphaNumeric

Cheers guys. It's been getting a bit clearer over the last few days. The only bits I still trip over now and again is when an interaction term depends on the derivative of the field, such as when you're doing non-abelian Lagrangians and you end up with a 3 boson vertex which involves $$A^{a}_{\rho}A^{b}_{\sigma}\partial_{\mu}A^{c}_{\nu}$$ (a,b,c being the generator indices).

I can see how the derivative would 'pull down' factors of $$p_{\mu}$$, $$k_{\mu}$$ and $$q_{\mu}$$ (the momentum on the 3 legs of the vertex) but it's sometimes complicated to get the right signs on the terms :\

5. Oct 18, 2006

### Severian

Remember that the overall signs are just conventions. Since you need to square the matrix element, the only place signs are important is in interferences. But if you are consistant, your signs in different diagrams will match up.

I must confess, when looking at a new model, I never derive Feynman rules - I just read them off the lagrangian.

So for example $$\bar{\psi} ( i\gamma^{\mu} \partial_{\mu}-m) \psi$$ is a kinetic term (since it is bilinear) giving a propagator $$\frac{1}{\gamma^\mu p_\mu -m}$$ while $$-e\psi\gamma^{\mu}\psi A_{\mu}$$ is an interaction (since it has three fields) between an electron and a photon with Feynman rule for the interaction of $$-ie\gamma^\mu$$.

Last edited: Oct 18, 2006
6. Oct 20, 2006

### Hans de Vries

Should have mentioned that the "e" in the interaction term is not the "e" in
the vertex term.

$$\mbox{interaction term: }\ \ -ie\gamma^\mu A\psi$$
$$\mbox{vertex term: }\ \ -ie\gamma^\mu$$

The latter is the square root of the first divided by 4pi:

$$e'\ =\ \frac{e^2}{4\pi} \ \ \ \ (\ = \ \alpha)$$

Why? Actually it's a combination of three different reasons:

$$\frac{e^2}{4\pi}\ \rightarrow \ \frac{e^2}{2} \ \rightarrow \ e^2 \ \rightarrow \ e$$

The first step is because it is an interaction between plane waves and
not point charges. The factor $1/4\pi$ comes from the surface of a 3D
sphere which goes to 1/2 for a 1D "sphere" in case of a 1D propagator.

$${\cal F}\left(\frac{1}{ E^2-p^2_x }\right)\ \ \qquad \qquad \ \ \ =\ \ {\frac{1}{2}\ \cal H}(ct-|r|)\ \qquad \ \ \ \ \ \mbox{1D propagator}$$
$${\cal F}\left(\frac{1}{ E^2-p^2_x-p^2_y} \right)\ \ \ \qquad =\ \ {\frac{1}{2\pi}\ \frac{1}{\sqrt{c^2t^2-r^2}}\ \qquad \mbox{2D propagator}$$
$${\cal F}\left( \frac{1}{E^2-p^2_x-p^2_y-p^2_z} \right)\ \ =\ \ {\frac{1}{4\pi}\ \delta(ct-|r|)\ \qquad \ \mbox{3D propagator}$$

Where ${\cal H}$ is the Heaviside step function.

The propagator from the "transition current" to the virtual photon is
actually a 1D propagator. Both are plane waves and both have the same
direction.

Then there is another factor of 2 because both front and backside of the
1D propagator contribute to the photons plane wave. The factor 1/2
in the 1D propagator is just that. It goes 1/2 back- and 1/2 foreward.

The last step is the square root. This is simply because the electron-
photon-electron interaction is split into 2 steps, two vertices, each
contributing a factor e to the amplitude.

Regards, Hans.

Last edited: Oct 21, 2006
7. Oct 23, 2006

### Severian

Hans, that is just wrong or (if I am misunderstanding you) just misleading.

The 'e' in the vertex and the 'e' in the interaction term are exactly the same thing (except for renormalization). The interaction term is $$-ie \bar \psi \gamma^\mu \psi A_\mu$$ and the vertex term is the same thing with the legs amuptated (ie. feilds removed) $$-ie \gamma^\mu$$.

The reason $$\alpha = \frac{e^2}{4 \pi}$$ tends to be used in the expression for, say $$e^+e^- \to \gamma^*$$ is because the cross-section is proportional to the amplitude squared. The $$4 \pi$$ is just definition because it is convenient to move the solid from the phase space to the coupling.

8. Oct 23, 2006

### AlphaNumeric

^ Yeah, you're right. I've worked through this enough now and drawn enough Feynman diagrams that I'm sick of them to know it's the same 'e' up to the renormalisation scaling.

9. Oct 23, 2006

### Hans de Vries

You're indeed right about the e's being equal.

The factor 1/4pi shows up in the 3D propagator in space-time versus
a factor 1/2 in the 1D propagator in space-time. It is implicit however
in the propagators in momentum space. There is no need to adapt e
in momentum space.

If we go from Dirac to the non-relativistic Schroedinger equation we
find alpha in the Electrostatic interaction, but the (e)/4pi comes in
here of course because the potential field used is generaly spherical...
That's were my black-out came from. Sorry..The other e (without the
4pi) then comes from the crossterms of the square of the covariant
derivative.

$$\left( \frac{\partial}{\partial t}-ieV\right)^2\psi$$

I presume that alpha as a coupling constant came historically from the
ratio of the Coulomb radius and the electron radius. If alpha was 1 then
both would be the same. Maybe the factor 4pi slips in here because
the "1D version" of the uncertainty relation is used to obtain a (3D)

Regards, Hans

10. Oct 23, 2006

### Hans de Vries

The meaning of the "propagators" shows up if we look wat is missing from the
equations of motion. For the photon.

$$\partial^2_\mu A_\mu\ =\ \frac{4\pi}{c}\ j_\mu$$

and the inverse operation is thus:

$$\left(\partial^2_\mu\right)^{-1} j_\mu =\ \frac{c}{4\pi}\ A_\mu$$

The photon propagator translates the transition current into a photon field.
(and tosses in a factor $4\pi$......)

And similar for the electron propagator:

$$\left(\gamma^\mu \partial_\mu-m\right)\psi_\mu\ =\ ie\kern+0.25em /\kern-0.75em A_\mu \psi_\mu$$

The inverse operator becomes:

$$\left(\gamma^\mu \partial_\mu-m\right)^{-1}\ ie\kern+0.25em /\kern-0.75em A_\mu \psi_\mu \ =\ \psi_\mu$$

The electron propagator turns the interaction term into an electron field.

Regards, Hans

Last edited: Oct 23, 2006
11. Oct 23, 2006

### Severian

I teach a course where I have to get to the Feynman Rules for QED without using any QFT. It is quite a task (with a lot of hand waving). I do have a rather convoluted way of 'deriving' (rather 'justifying') the fermion propagator, but I can never resist showing them that it is a Green's function of the Dirac operator. It is just so much simpler.

12. Nov 21, 2006

### mattlorig

L_int = lambda ( phi d_mu phi d^mu phi + m^2 phi^3 )

This topic seems to be right up my ally at the moment. Hopefully somebody can strighten me out. I'm dealing with spin zero bosons

Suppose L_int = lambda ( phi d_mu phi d^mu phi + m^2 phi^3 )
my vertex is drawn with momenta k1, k2, and k3 all pointing inward
I need to find the vertex factor.

I am told the correct answer is:
-i * lambda [ (k1 k2 + k2 k3 + k3 k1) 2! - m^2 3! ]

I understand that the d_mu and d^mu each bring down a factor of (i k), so I can understand where the (k1 k2) term comes from. And (I guess) since bosons are symmetric under interchange of particles, we need to have the (k2 k3) and (k3 k1) terms as well. But I don't understand where the symmetry factor of 2! comes in.

Here's my current hypothesis. Since the vertex has three identical phi particles coming into it, it must have a symmetry factor of 3! (which is what the m^2 term carries). When we symmetrized the momenta
(i.e. k1 k2 --> ( k1 k2 + k2 k3 + k3 k1 ) / 3
this canceled a factor of 3 from the 3!, so I am left with a symmetry factor of 2!. Is that right?

so by that logic if
L_int = lambda ( phi^2 d_mu phi d^mu ph )i
then the vertx factor would be
-i * lambda (k1 k2 + k1 k3 + k1 k4 + k2 k3 + k2 k4 + k3 k4) * 4! / 6

Do I have the right idea?

13. Nov 22, 2006

### dextercioby

Yes, you do. Sometimes the symmetry factors of a diagram are absorbed in the propagators (thus rescaling the rules), so no bother with the factorials. However, a lucid analysis shows how these symmetry factors appear.

Either using Wick's theorem or computing the Green functions from their generating functional.

Daniel.

14. Nov 22, 2006

### mattlorig

Is there a nice paper on the internet, or a good book that describes (hopefully in simple terms) how to take a Lagrangian and write down:
1) vertex factors
2) the propagators
in momentum space and position space? (does anybody ever do calculations in position space? if not, forget about the position space rules)

I ask because I'm using Mark Srednicki's book, and he develops nearly everything specific to phi^3 theory. He gives feynman rules for that theory, but doesn't really explain how to take a general theory and develop feynman rules from it.

I suppose I should be able to "see" where each of the rules come from. But frankly, there are so many fourier transforms in each calculation, I get lost in all the math and forget what I'm really doing.

15. Nov 23, 2006

### dextercioby

If you're ready for the path integral approach, then Bailin & Love's intro to gauge field theory is the book to have.

Daniel.

16. Nov 26, 2006

### mattlorig

Thanks I'll check that book out.

17. Nov 26, 2006

### CarlB

Feynman came up with a method of getting a massive electron propagator from the massless one. It's mentioned as a footnote in his popular book "QED, the strange theory of matter and light". The idea is to resum a bunch of massless propagators that interact with vertices of "-im" to get an effective propagator that is massive.

I use a handed version of the same thing in section 7.6 of my incomplete book, but the chapter hasn't been proof read (and I use a different signature than most people):
http://www.brannenworks.com/dmaa.pdf

18. Nov 29, 2006

### Severian

Yes, that is exactly how I do it. I don't like it much though.

19. Nov 29, 2006

### CarlB

The basic problem is gauge invariance. To actually use Feynman's trick as a method, I had to abandon most of the methods that are used in the standard model. However, I'm convinced that this is physical, and that one can compute the lepton masses from first principles using this. See:
http://www.brannenworks.com/dmaa.pdf

Carl

Last edited: Nov 29, 2006