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Often one needs to sum over all intermediate states. Starting with equation 4.56 and multipling it by $f_a^{(\pm)}(x)$ and suming over all states, we have

i\int d^3x^\prime \sum_a f_a^{(\pm)}(x) {f_a^{(\pm)}}^*(x^\p...
...e) = \pm\sum_a
f_a^{(\pm)}(x)\delta_{ab} = \pm f_b^{(\pm)}(x).
\end{displaymath} (4.176)

Comparison with equation 4.170 shows

\sum_a f_a^{(\pm)}(x) {f_a^{(\pm)}}^*(x^\prime) = \mp i\triangle_{\pm}
(x-x^\prime) .
\end{displaymath} (4.177)

Douglas M. Gingrich (gingrich@