- #1
maverick280857
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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.
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.
