From 63966407338ed0bdb061ddfd67f8940c2ccb51d2 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Wed, 1 Dec 2021 19:38:28 +0100 Subject: Expand knowledge base --- content/know/concept/self-energy/index.pdc | 9 ++++----- 1 file changed, 4 insertions(+), 5 deletions(-) (limited to 'content/know/concept/self-energy/index.pdc') diff --git a/content/know/concept/self-energy/index.pdc b/content/know/concept/self-energy/index.pdc index c86f8c5..7e67143 100644 --- a/content/know/concept/self-energy/index.pdc +++ b/content/know/concept/self-energy/index.pdc @@ -172,7 +172,7 @@ $$\begin{aligned} &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \bigg[ \sum_{m = 0}^{n} \frac{n!}{m! (n \!-\! m)!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]} - {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}} + {-\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}} \end{aligned}$$ Where the total order is the sum of the orders of all considered diagrams, @@ -186,8 +186,7 @@ $$\begin{aligned} &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}} \bigg[ \sum_{n = 0}^\infty \frac{1}{2^{n-m} (n \!-\! m)!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]} - {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} - \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}} + {-\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}} \end{aligned}$$ Since both $n$ and $m$ start at zero, @@ -195,7 +194,7 @@ and the sums include all possible diagrams, we see that the second sum in the numerator does not actually depend on $m$: $$\begin{aligned} - G_{ba} + -\hbar G_{ba} &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}} \bigg[ \sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}} \bigg]} {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}} @@ -245,7 +244,7 @@ you can convince youself that $G(b,a)$ obeys a [Dyson equation](/know/concept/dyson-equation/) involving $\Sigma(y, x)$: - + This makes sense: in the "normal" Dyson equation -- cgit v1.2.3