Ok, so I am just looking at a regular, scalar \(\varphi ^4 \) theory here. The generating functional is: \(Z(J) =\int D\varphi e^{i\int d^{4}x\{\frac{1}{2}[(\partial \varphi )^{2}-m^{2}\varphi ^{2}]-(\lambda /4!)\varphi ^{4}+J\varphi \}} \) When we expand the J term in a power series we can rearrange it to this: \(Z(J) =\sum_{s=0}^{\infty }\frac{1}{s!}\int \ldots \int d^{4}x_{1} \ldots d^{4}x_{s}J(x_{1}) \ldots J(x_{s})\int D\varphi e^{i\int d^{4}x\{\frac{1}{2} [(\partial \varphi )^{2}-m^{2}\varphi ^{2}]-(\lambda /4!)\varphi ^{4}\}}\varphi (x_{1}) \ldots \varphi (x_{s}) \) From which we can define the n-point Green's functions: \(Z(J) =Z(J=0, \lambda = 0)\sum_{s=0}^{\infty }\frac{1}{s!}\int \ldots \int d^{4}x_{1}\ldots d^{4}x_{s}J(x_{1})\ldots J(x_{s})G^{(s)}(x_{1},\ldots ,x_{s}) \) With the 4-point function being this (for example): \(G(x_{1},x_{2},x_{3},x_{4}) \equiv \frac{1}{Z(J=0 , \lambda = 0)}\int D\varphi e^{i\int d^{4}x\{\frac{1}{2}[(\partial \varphi )^{2}-m^{2}\varphi ^{2}]-(\lambda /4!)\varphi ^{4}\}}\varphi (x_{1})\varphi (x_{2})\varphi (x_{3})\varphi (x_{4}) \) If we want to examine a 4-particle scattering process (say), then we need to calculate the 4-point Green's function. My question is this: if we know there are only 4 particles involved in the process, i.e. there are 4 sources/sinks for particles to be created/destroyed at, how is it that we know the 4-point Green's function is all we need? This implies to me that all the orders of the expansion of Z(J) except s=4 are zero in such a scenario. I can't really see how this can be the case. The (perhaps slightly naive) thing I am trying to do is define the source term as 4 localised functions, i.e. delta functions: \( J(x) = \delta (s_1 - x) + \delta (s_2 - x) + \delta (s_3 - x) + \delta (s_4 - x) \) where the s's are the locations of the sources. I then expected that substituting this into \(Z(J)\) should somehow kill all terms except the 4th order term. Perhaps this won't help, because I guess there is no reason two particles can't come from the same source. Anyway, I hope that illustrates my general concern. I feel like perhaps there is something simple I am overlooking. Any ideas? edit: I just noticed I left out a couple of \(i^s\)'s, but they aren't important I hope.
I just finished chapter 9 of Peskin & Schroeder, where they go over the path integral formalism. I don't remember the textbook discussing your method of expanding the generating functional, but the same method was in my course notes and in the homework I just submitted yesterday. I'll take a look at my notes and see what I can find.
Is it possible to write \(G(x_1,x_2,x_3,x_4)\) in explicit form (without functional integration)? If not, try writing \(G(x_1,x_2)\) and you may see some explanation. Why are you expanding like this? What is the definition of "4 particles involved in the process"?
Yes, this is possible (perturbatively anyway). Expanding \(G(x_1,x_2,x_3,x_4)\) in powers of \(\lambda\) lets you write out a bunch of terms which can then be evaluated by wick contraction, getting rid of the nasty path integral. Thus we have the double perturbation expansion of Z which can be expressed in terms of Feynman diagrams. The \(G(x_1,x_2,x_3,x_4)\) terms (4th order in J) are all the diagrams with 4 external lines, the expansion in \(\lambda[tex] then gives diagrams with 1,2,3...etc interaction vertices. Having said all that I'm not super sure why that is true, leading back to the original question. I have discovered that if you expand [tex]G(x_1,x_2,x_3,x_4)\) as you suggested (I did it only to first order) and substitute in my composite delta function source term then you get a bunch of terms describing particles getting created/destroyed at the various sources and interacting at a single vertex, in various permutations, so that is ok. I have not yet found why this won't work for e.g. the 6 point Greens function. I think I know what to try now though, I just haven't done it yet because it is kind of tedious. Ahh yes, I just meant that when you finish everything there should be 4 external propagators. As I understand it, each order of the expansion in J gives a different number of external propagators, but I have yet to get my head exactly around how this goes through or what is going on. I only just figured out how to do the wick contractions though, so now that I know that I may have a better chance. I may also be asking the wrong question here, Z describes ALL processes in some sense so maybe I can't just contrive whatever source term I like to kill off the diagrams I'm not interested in. It may not be the right way to think about this whole thing. I guess what I am mainly interested in is identifying terms in the expansion of Z with specific Feynman diagrams and then understanding physically why we need only worry about those particular diagrams for a given physical situation, i.e. we know we start with two particles and end with two particles, or some such.
I thought I had an answer for you, but then I realized my understanding of n-point correlation functions wasn't as good as I thought. When computing Feynman diagrams for particle scattering and all the particles are coming in with well-defined momenta, it's easy to interpret the results. But when you're calculating quantities like \(<\Omega|\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)| \Omega >\), I'm not even sure what that's supposed to correspond to. For a free field theory I know you'll get a sum of products of propagators, and in interacting theories you can do perturbation expansions, but what exactly do the n-point functions correspond to? Are they just amplitudes for a collection of particles to be detected/measured at various points in time, whether they're initial state or final state particles?
Yeah I had wondered about that too. I'm not really sure. I think you are more or less right though. The 2-point function is just the propagator, so it's the amplitude for a particle to be detected at x_2 given it started from x_1, or some time permutation of that, so I guess the n-point function is some generalisation of that.
Well in any case, in a \(\phi^4\) theory that just means you have 4 scalar particles entering into each intermediate vertex for self-interactions. I'm pretty sure you could describe as many particles as you want scattering and propagating, doesn't just have to be 4, as long as each vertex in the process has 4 legs.
Hi. I've followed a course on quantum field theory that followed the path integral approach about a year ago. Unfortunately I've since been busy with other things and this was a while ago, so not only have I not yet had a chance to think about the subject beyond what I "needed for the exam", but it's all getting a little hazy. That said I think I can still answer a couple of questions and no-one else has really answered, but just don't ask me for any marvellous insights: The source should be zero. It's the analogue of what classical electric charges/currents are to the electromagnetic field, and aren't any sort of "initial condition". As far as I've understood, the only real reason for the inclusion of \(J\) in the partition functional is because: \( -i \frac{\delta}{\delta J(x_{1})} \, e^{i \int \text{d}^{4}x J \phi} \,=\, \phi(x_{1}) \, e^{i \int \text{d}^{4}x J \phi} \)where \(\frac{\delta}{\delta J(x_{1})}\) is the functional derivative with respect to \(J(x_{1})\). It lets you perform a few slick mathematical tricks, such as: \( \int \mathcal{D}\phi \, \phi(x_{1}) \, e^{i \int \text{d}^{4}x \mathcal{L}} \,=\, \left. -i \frac{\delta}{\delta J(x_{1})} \, \int \mathcal{D}\phi \, e^{i \int \text{d}^{4}x \bigl\{ \mathcal{L} + J \phi \bigr\}} \right|_{J=0} \)ie. you can symbolically replace terms like \(\phi(x_{n})\) with \(-i\frac{\delta}{\delta J(x_{n})}\), which you can pull outside of your path integrals. This allows you to write your n-point correlation functions as functional derivatives of \(Z\) evaluated at \(J=0\): \( G^{(n)}(x_{1}, \,\ldots\, ,\, x_{n}) \,=\, \left. \Bigl( -i \frac{\delta}{\delta J(x_{1})} \Bigr) \,\cdots\, \Bigl( -i \frac{\delta}{\delta J(x_{n})} \Bigr) \, Z \, \right|_{J=0} \quad (1) \)(which is why they call it the generating functional). If you can split your Lagrangian into a "free" part and an "interaction" part that only depends on \(\phi\) (and not its derivatives): \( \mathcal{L}(\phi,\, \partial_{\mu}\phi) \,=\, \mathcal{L}_{0}(\phi,\, \partial_{\mu}\phi) \,+\, \mathcal{L}_{\text{int}}(\phi) \)then you can use the same trick to extract the interaction part from the partition functional, and you can write \(Z[J]\) in terms of the functional derivatives of the free partition functional \(Z_{0}[J]\): \( Z[J_{0}] \,=\, \left. e^{i \int \text{d}^{4}x \mathcal{L}_{\text_{int}}\Bigl( -i \frac{\delta}{\delta J(x)} \Bigr)} \, Z_{0} \, \right|_{J=J_{0}} \quad(2) \)The point is that \(Z_{0}[J]\) is exactly soluble, so you can develop the exponential in (2) and evaluate \(Z[J]\) to any desired order. If I remember correctly, you end up with an expression containing sums of terms like, for example: \( Z[J] \,=\, \cdots \,+\, \int \text{d}^{4}x_{1} \cdots \text{d}^{4}x_{4} \, \Bigl\{ \text{something} \Bigr\} \, J(x_{1}) \cdots J(x_{4}) \,+\, \cdots \,+\, \int \text{d}^{4}x_{1} \cdots \text{d}^{4}x_{8} \, \Bigl\{ \text{something else} \Bigr\} \, J(x_{1}) \cdots J(x_{8}) \,+\, \cdots \)and then (1) lets you extract your correlation functions. I think what you and kurros want to look up is the "LSZ reduction formula" (in Peskin it's presented in section 7.2). It gives the general expression of S-matrix elements in terms of the n-point correlation functions, which seems to be the missing piece of your puzzle. By the way, for anyone grappling with the path integral formulation of QFT, I remember Bailin and Love's "Introduction to Gauge Field Theory" giving quite a good intuitive introduction to the manipulation of path integrals in its first few chapters.
