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-rw-r--r--content/know/concept/self-energy/index.pdc9
1 files changed, 4 insertions, 5 deletions
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 $\Sigma(y, x)$:
<a href="dyson.png">
-<img src="dyson.png" style="width:90%;display:block;margin:auto;">
+<img src="dyson.png" style="width:95%;display:block;margin:auto;">
</a>
This makes sense: in the "normal" Dyson equation