Four Point Correlation function from Generating Functional

[maverick280857]

Hi everyone,

I'm working through Section 9.2 (Functional Quantization of Scalar Fields) from Peskin and Schroeder. I have trouble understanding the absence of a term in equation 9.41 which I get but the authors do not.

Define \phi_i \equiv \phi(x_i), J_{x} \equiv J(x), D_{xi} \equiv D_{F}(x-x_i) (the Feynman propagator). Repeated subscripts are integrated over implicitly.

Equation 9.41 in the book reads

\langle 0|T\phi_1\phi_2\phi_3\phi_4|0\rangle = \frac{\delta}{\delta J_1}\frac{\delta}{\delta J_2}\frac{\delta}{\delta J_3}\frac{\delta}{\delta J_4}e^{-\frac{1}{2}J_x D_{xy} J_{y}}

= \frac{\delta}{\delta J_1}\frac{\delta}{\delta J_2}\frac{\delta}{\delta J_3}\left[-J_x D_{x4}\right]e^{-\frac{1}{2}J_x D_{xy} J_{y}}
= \frac{\delta}{\delta J_1}\frac{\delta}{\delta J_2}\left[-D_{34}+J_{x}D_{x4}J_{y}D_{y3}\right]e^{-\frac{1}{2}J_x D_{xy} J_{y}}
= \frac{\delta}{\delta J_1}\left[D_{34}J_{x}D_{x2}+D_{24}J_{y}D_{y3} +J_{x}D_{x4}D_{23}\right]e^{-\frac{1}{2}J_x D_{xy} J_{y}}
= D_{34}D_{12} + D_{24}D_{13} + D_{14}D_{23}

where J is set equal to zero after all the 4 functional derivatives have been evaluated.

When I do this by hand, I get (in the second last step), an extra term:

= \frac{\delta}{\delta J_1}\left[D_{34}J_{x}D_{x2}+D_{24}J_{y}D_{y3} +J_{x}D_{x4}D_{23}-J_x D_{x4} J_{y}D_{y3} J_z D_{z2}\right]e^{-\frac{1}{2}J_x D_{xy} J_{y}}

which isn't given in the book. I just follow the prescription of differentiating with proper order and bringing down a -J*D type of term every time the exponent is differentiated. What happened to this term? Please help..

Thanks in advance.

 

[maverick280857]

Wait...I think I got it...the terms that are O(J^2) survive the derivatives but go to zero when J is set equal to zero, whereas the O(J) terms are robbed of their J dependence by the last functional derivative. The authors just don't show the steps.

I should've read what I was writing