Uncovering Feynman Rules from Paper

[latentcorpse]

Consider question 1 in this paper:

http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2002/Paper65.pdf

How do we "read off" the Feynman rules? I can see that the interaction terms give contributions $$\lambda_3 , \lambda_4 , \lambda_6$$ to the relevant 3,4 and 6 point vertices respectively.

How do we get the propagator out of this without doing any calculations?

Or indeed, if we were asked to do a calculation to find it, what would it be?

Am I correct in assuming that the only Feynman rules we can deduce from this is the vertex and propagator terms? Or are there other things we could write? Such as "integrate over undetermined loop momenta, etc..."

Thanks.

 

[Avodyne]

You can only "read off" the Feynman rules by knowing the general procedure, which is discussed in any QFT textbook. These problems are apparently intended for someone who knows QFT, and are mostly "regurgitation" problems.

 

[sgd37]

the propagator is given by the free part of the theory from the generating functional the higher order terms give the interactions. Look back through our notes at the generating functionals with sources

 

[latentcorpse]

the propagator is given by the free part of the theory from the generating functional the higher order terms give the interactions. Look back through our notes at the generating functionals with sources

Sure. So take for example the xAx+bx example. There we have introduced a source b and then when we complete the square we we have to take of a A^{-1} term.

Therefore, when we take the derivatives, the A^{-1} drops down as the propagator.

So what we want to do is write the x's in the interaction terms as derivatives wrt the source right? But I don't see how we can manipulate this into the form of something we can complete the square of?

Thanks.

 

[sgd37]

you only want to express the potential terms in terms of derivatives wrt the source the quadratic terms you keep and use them to complete the square

e.g for

$$S = \int dx \partial^{\mu} \phi \partial_{ \mu} \phi + m^2 \phi^2 + V( \phi)$$

$$e^{i S[ \phi] + J \phi} = e^{i \int dx V( \frac{\delta}{\delta J})} e^{i \int dx \phi \partial^{\mu} \partial_{ \mu} \phi + m^2 \phi^2 + J \phi}$$

which you can complete the square for

 

[latentcorpse]

you only want to express the potential terms in terms of derivatives wrt the source the quadratic terms you keep and use them to complete the square

e.g for

$$S = \int dx \partial^{\mu} \phi \partial_{ \mu} \phi + m^2 \phi^2 + V( \phi)$$

$$e^{i S[ \phi] + J \phi} = e^{i \int dx V( \frac{\delta}{\delta J})} e^{i \int dx \phi \partial^{\mu} \partial_{ \mu} \phi + m^2 \phi^2 + J \phi}$$

which you can complete the square for

Right. It seems that I am able to do this for the case in the notes but I'm unsure here.

So we'd get $$\int [d \phi] e^{-i\int d^dx \lambda_3 (\frac{\delta}{\delta J_x})^3}e^{-i\int d^dx \lambda_4 (\frac{\delta}{\delta J_x})^4}e^{-i\int d^dx \lambda_6 (\frac{\delta}{\delta J_x})^6} e^{i\int d^dx -\frac{1}{2} \phi \Delta \phi +J \phi}|{J=0}$$
where $$\Delta=-\partial_\mu \partial^\mu +m^2$$
So then I complete the square right?
This gives
$$\int d^dx -1/2 \phi \Delta \phi + J \phi = \int d^dx - 1/2 \tilde{\phi} \Delta \tilde{\phi} + 1/2 J \int d^dx \Delta^{-1} J$$
where $$\tilde{\phi} = \phi - \int d^dx \Delta^{-1} J$$
I find

$$\frac{Z[J]}{Z_0[0]}= \int [d \phi] e^{-i \int d^dx \lambda_3 ( \frac{\delta}{\delta J})^3}e^{-i \int d^dx \lambda_4 ( \frac{\delta}{\delta J})^4}e^{-i \int d^dx \lambda_6 ( \frac{\delta}{\delta J})^6} e^{\frac{i}{2} \int d^dx \int d^dy J(x) \Delta^{-1} J(y)}|_{J=0}$$

Now usually what happens is we do the derivatives and THEN set the source to zero.
However, usually there is only one interaction term. Should I take the derivatives individually? Will this produce a different propagator depending on what interaction (vertex) it's associated to?

Thanks a lot!

 

[sgd37]

you already have the propagator i.e. $$\Delta ^{-1}$$ all the derivatives do is to connect different points with the propagator (here we only have vertices connected to each other).

Remember that $$( \frac{ \delta}{ \delta J})^3 = \frac{ \delta^3}{ \delta J_1 \delta J_2 \delta J_3}$$ and that $$\frac{ \delta J_x}{ \delta J_y} = \delta ^{(d)} (x - y )$$

 

[latentcorpse]

you already have the propagator i.e. $$\Delta ^{-1}$$ all the derivatives do is to connect different points with the propagator (here we only have vertices connected to each other).

Remember that $$( \frac{ \delta}{ \delta J})^3 = \frac{ \delta^3}{ \delta J_1 \delta J_2 \delta J_3}$$ and that $$\frac{ \delta J_x}{ \delta J_y} = \delta ^{(d)} (x - y )$$

but consider the cubed term

we'd have $$e^{\frac{\delta^3}{\delta J_1 \delta J_2 \delta J_3} e^{\int d^dxd^dy J(x) \Delta^{-1}(x,y) J(y)}|_{J=0}$$ ignoring everything else.

now we have 3 derivatives wrt j and only 2 j's attached to the propagator (in the 2nd exponential). won't the first j derivative remove one of those j's then the 2nd j derivative will remove the 2nd j and then the third one will have no j's to act on?

 

[sgd37]

well what is $$\frac{d^3}{dx^3} e^{ax^{2}}$$ remembering that $$\frac{d}{dx} e^{x} = e^{x}$$