gaussian integral

[Jim Kata]

Sorry in advance for the length of this question. Can someone show me the solution to the integral

[math]

\int\limits_{ - \infty }^\infty {\prod\limits_r {d\xi _r } } \exp ( - 1/2\sum\limits_{r,s} {K_{rs} } \xi _r \xi _s + V(\xi ))

[/math]

Where [math]

K_{rs}

[/math] is positive definite, non-singular, and symmetric and

[math]

V(\xi )

[/math] is analytic? The reason I ask is because in quantum field theory you encounter integrals like these in the Feynman path integral formalism. I want to show through a path integral integration for the photon propagator you get [math]

- i\Delta '_{\mu \tau } (x,y) = \left\langle {T\{ A_\mu (x)A_\tau (y)\} } \right\rangle _0

[/math] Where [math]

\Delta ' = \Delta [1 - \Pi ^* \Delta ]^{ - 1}

[/math] with [math]

\Delta _{\mu x,\tau y} = (2\pi )^{ - 4} \int {d^4 } q\frac{{\eta _{\mu \tau } }}{{q^2 - i\varepsilon }}e^{iq \cdot (x - y)}

[/math] in the Feynman gauge and [math]

i(2\pi )^4 \Pi ^{*\rho \sigma } (q)

[/math] equal to the sum of the one particle irreducible graphs (with two external photon lines). I would like to see two things. One that [math]

\int\limits_{ - \infty }^\infty {\prod\limits_r {d\xi _r } } \exp ( - 1/2\sum\limits_{r,s} {K_{rs} } \xi _r \xi _s + V(\xi ))= \exp (Z)

[/math]. Where Z is the sum disconnected diagrams. That is I would like to see the exponentiation of disconnected diagrams. Also, if someone could show me perturbatively why the tadpole diagrams in QED disapear, through a path integral. I know this integral can be solved since [math]

\int {\prod\limits_r {d\xi _r } } \xi _{r_1 } \xi _{r_2 } \ldots \xi _{2n} \exp ( - 1/2\sum\limits_{r,s} {K_{rs} } \xi _r \xi _s ) = [Det(\frac{K}{{2\pi }})]^{ - \frac{1}{2}} \sum\limits_{\scriptstyle pairings \hfill \atop

\scriptstyle r_1 \ldots r_{2n} \hfill} {\prod\limits_{pairs} {K^{ - 1} } }

[/math] with the sum being over all ways of pairing indices [math]

r_1 \ldots r_{2n}

[/math], with two pairings being considered the same if they differ only in order of pairs, or by the order of indices within the pair. So for example, [MATH]

\int {\prod\limits_r {d\xi _r } } \xi _{s_{_1 } } \xi _{s_2 } \exp ( - 1/2\sum\limits_{rs} {K_{rs} \xi _r } \xi _s ) = [Det(\frac{K}{{2\pi }})]^{ - \frac{1}{2}} [(K^{ - 1} )_{s_1 s_2 } ]

[/MATH]. I do not understand in quantum field theory when say [math]

\Delta '(q) = \Delta (q) + \Delta (q)\Pi ^* (q)\Delta (q) + \Delta (q)\Pi ^* (q)\Delta (q)\Pi ^* (q)\Delta (q) + \cdots

[/math]. Can this be shown directly from the path integral equation [math]

\frac{{\int {\prod\limits_r {d\xi _r \xi _\mu \xi _\tau \exp [ - 1/2\sum\limits_{rs} {K_{rs} \xi _r \xi _s + V(\xi )]} } } }}{{\int {\prod\limits_r {d\xi _r \exp [ - 1/2\sum\limits_{rs} {K_{rs} \xi _r \xi _s + V(\xi )]} } } }}

[/math] and the perturbative properties of QED, not Furry's theorem. Here's the sketch I see, let [math]

I_{\mu \tau } = \frac{{\int {\prod\limits_r {d\xi _r \xi _\mu \xi _\tau \exp [ - 1/2\sum\limits_{rs} {K_{rs} \xi _r \xi _s + V(\xi )]} } } }}{{\int {\prod\limits_r {d\xi _r \exp [ - 1/2\sum\limits_{rs} {K_{rs} \xi _r \xi _s + V(\xi )]} } } }}

[/math] so [math]

I_{\mu \tau } = \frac{{(\Delta _{\mu \tau } + \Delta _{\mu \rho } \Pi ^{\rho \sigma } \Delta _{\sigma \tau } + \Delta _{\mu \rho } \Pi ^{\rho \sigma } \Delta _{\sigma \lambda } \Pi ^{\lambda \delta } \Delta _{\delta \tau } + \cdots )\exp (Z)}}{{\exp (Z)}}

[/math] while some where along the line showing that the sum of the diagrams with one external photon line disappear. I think that can be shown from the trace properties of the Dirac matrices.

[Norman Albers]

No I cannot, and am just barely approaching what you so beautifully express here. Thank you for showing this.

[solidspin]

I concur, Norm -

 

beautiful -

 

solidspin

[BenTheMan]

Jim---do you have a QFT book? Any good book on quantum field theory will help you with this question.

 

A free one is available by Mark Srednicki:

http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

 

Also check out Peskin and Schroeder, which is pretty much the standard now. Weinberg's book is pretty uncomprehensible unless you are Weinberg. Other good books are by Ryder, and Itzykson and Zuber. The IZ textbook I saw in Barnes and Noble for $40 the other day, which is a steal. For a more pedagogical approach, try Tony Zee's ``QFT in a Nutshell''.