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Hi,

I have a couple of questions on this (mostly based on Srednicki ch14).

If I understand things correctly the difference between the exact propagator (always shown in bold in Srednicki) and the "usual propagator" from earlier chapter (i.e. the one corresponding to lines in the Feynman graphs of Chapter 9 say) is that propagator was for the free theory and this exact propagator is for the interacting theory.

I understand that by the definition [tex] \frac{1}{i} {\bf \Delta}(x_1-x_2) =\delta_1 \delta_2 iW(J) \mid_{J=0} [/tex], that this means sum over all connected diagrams which originally had two sources (2 external lines), with the sources removed(by the [tex] \delta_i [/tex]'s), since all other diagrams are either killed by the differentiation (i.e. those diagrams with less than two J's) or are killed when we put J=0 at the end, because they still contain J's. But J=2 to start with survives.

It is obvious that the simplest and lowest order in g diagram is just the barbell type diagram. Since this has two external lines to start with, and has no vertices (so it is of order ~1), after the [tex] \delta_i [/tex]'s act, we are left with just the free propagator.

To get the next order of g, which will be O(g^2), since to have a connected diagram with two external sources, this means at least 2 propagators hanging around with the end not connected to the source, connected to a vertex. Considering if we just had one vertex it would have the two source propagators connected but then just a loose line if we didn't have another vertex, so next contributing term must be of order O(g^2).

So far so good, however I don't understand now why my 2nd, on left, diagram isn't included by Srednicki as a O(g^2) correction? Is this something to do with the fact we don't need to include diagrams with tadpoles (diagrams that when a single line is cut falls into two diagrams one of which has no sources)?

Now we also have counterterms at this order, from chapter 9 we have:

[tex] Z(J)=exp[-\frac{i}{2} \int d^4x (\frac{1}{i}\frac{\delta}{\delta J(x)}) (-A\partial^{2}_x+Bm^2)(\frac{\delta}{\delta J(x)})] Z_Y(J) [/tex]

This obviously acts to create a new vertex by acting on the [tex] \int J \Delta J [/tex][tex] \int J \Delta J [/tex] to label both ends of the propagator by x, i.e. attach both end of the propagator to this new type of 2 valent vertex, that we symbolise with an X. Since A, B are O(g^2) the first counterm is that shown on the right side of my picture.

But then why does this diagram also appear as the rightmost figure of Srednicki's 14.3 as an O(g^2) contributor, how can it possibly be both?

I have a couple of questions on this (mostly based on Srednicki ch14).

If I understand things correctly the difference between the exact propagator (always shown in bold in Srednicki) and the "usual propagator" from earlier chapter (i.e. the one corresponding to lines in the Feynman graphs of Chapter 9 say) is that propagator was for the free theory and this exact propagator is for the interacting theory.

I understand that by the definition [tex] \frac{1}{i} {\bf \Delta}(x_1-x_2) =\delta_1 \delta_2 iW(J) \mid_{J=0} [/tex], that this means sum over all connected diagrams which originally had two sources (2 external lines), with the sources removed(by the [tex] \delta_i [/tex]'s), since all other diagrams are either killed by the differentiation (i.e. those diagrams with less than two J's) or are killed when we put J=0 at the end, because they still contain J's. But J=2 to start with survives.

It is obvious that the simplest and lowest order in g diagram is just the barbell type diagram. Since this has two external lines to start with, and has no vertices (so it is of order ~1), after the [tex] \delta_i [/tex]'s act, we are left with just the free propagator.

To get the next order of g, which will be O(g^2), since to have a connected diagram with two external sources, this means at least 2 propagators hanging around with the end not connected to the source, connected to a vertex. Considering if we just had one vertex it would have the two source propagators connected but then just a loose line if we didn't have another vertex, so next contributing term must be of order O(g^2).

So far so good, however I don't understand now why my 2nd, on left, diagram isn't included by Srednicki as a O(g^2) correction? Is this something to do with the fact we don't need to include diagrams with tadpoles (diagrams that when a single line is cut falls into two diagrams one of which has no sources)?

Now we also have counterterms at this order, from chapter 9 we have:

[tex] Z(J)=exp[-\frac{i}{2} \int d^4x (\frac{1}{i}\frac{\delta}{\delta J(x)}) (-A\partial^{2}_x+Bm^2)(\frac{\delta}{\delta J(x)})] Z_Y(J) [/tex]

This obviously acts to create a new vertex by acting on the [tex] \int J \Delta J [/tex][tex] \int J \Delta J [/tex] to label both ends of the propagator by x, i.e. attach both end of the propagator to this new type of 2 valent vertex, that we symbolise with an X. Since A, B are O(g^2) the first counterm is that shown on the right side of my picture.

But then why does this diagram also appear as the rightmost figure of Srednicki's 14.3 as an O(g^2) contributor, how can it possibly be both?