Expand knowledge base

1 files changed, 29 insertions, 26 deletions

diff --git a/content/know/concept/self-energy/index.pdc b/content/know/concept/self-energy/index.pdc
index d2908eb..c86f8c5 100644
--- a/content/know/concept/self-energy/index.pdc
+++ b/content/know/concept/self-energy/index.pdc
@@ -50,8 +50,6 @@ and $\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$:
 
 $$\begin{aligned}
     G_{ba}
-    %&= - \frac{\expval{\mathcal{T}\Big\{ \hat{K}(\hbar \beta, 0) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}}}{\expval{\hat{K}(\hbar \beta, 0)}}
-    %\\
     &= - \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
     \expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
     {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
@@ -141,12 +139,15 @@ $$\begin{aligned}
 
 These integrals over products of interactions and Green's functions
 are the perfect place to apply [Feynman diagrams](/know/concept/feynman-diagram/).
-Conveniently, it even turns out that the factor $(-1)^p$
-is exactly equivalent to the rule that each diagram is multiplied by $(-1)^F$,
+Conveniently, it turns out that the factor $(-1)^p$
+is equivalent to the rule that each diagram must be multiplied by $(-1)^F$,
 with $F$ the number of fermion loops.
+Keep in mind that fermion lines absorb a factor $-\hbar$ each (see above),
+and interactions $-1/\hbar$.
 
-The denominator thus turns into a sum of all possible diagrams for each total order $n$
-(the order of a diagram is the number of interaction lines it contains).
+The denominator turns into a sum of all possible diagrams
+(including equivalent ones) for each total order $n$
+(the order is the number of interaction lines).
 The endpoints $a$ and $b$ do not appear here,
 so we conclude that all those diagrams only have internal vertices;
 we will therefore refer to them as **internal diagrams**.
@@ -168,24 +169,25 @@ We thus find:
 
 $$\begin{aligned}
     G_{ba}
-    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \bigg[ \sum_\mathrm{all\;ext}^{n} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m \!\le\! n}
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)} \bigg]}
-    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!}
+    \bigg[ \sum_{m = 0}^{n} \frac{n!}{m! (n \!-\! m)!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
 \end{aligned}$$
 
-Where the total order refers to the sum of the orders of all disconnected diagrams.
-Note that the external diagram does not directly depend on $n$.
-We can therefore reorganize:
+Where the total order is the sum of the orders of all considered diagrams,
+and the new factor is needed for all the possible choices
+of vertices to put in the external part.
+Note that the external diagram does not directly depend on $n$,
+so we reorganize:
 
 $$\begin{aligned}
     G_{ba}
-    &= \frac{\displaystyle\sum_\mathrm{all\;ext}^{\infty} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
-    \bigg[ \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)} \bigg]}
-    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+    &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{2^{n-m} (n \!-\! m)!}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
 \end{aligned}$$
 
 Since both $n$ and $m$ start at zero,
@@ -194,18 +196,19 @@ we see that the second sum in the numerator does not actually depend on $m$:
 
 $$\begin{aligned}
     G_{ba}
-    &= \frac{\displaystyle\sum_\mathrm{all\;ext} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
-    \bigg[ \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n} \bigg]}
-    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
-    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+    &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
     \\
-    &= \sum_\mathrm{all\;ext} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
+    &= \sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
 \end{aligned}$$
 
 In other words, all the disconnected diagrams simply cancel out,
 and we are left with a sum over all possible fully connected diagrams
-that contain $a$ and $b$. Let $G(b,a) = G_{ba}$:
+that contain $a$ and $b$. Furthermore, it can be shown using combinatorics
+that exactly $2^m m!$ diagrams at each order are topologically equivalent,
+so we are left with non-equivalent diagrams only.
+Let $G(b,a) = G_{ba}$:
 
 <a href="fullgf.png">
 <img src="fullgf.png" style="width:90%;display:block;margin:auto;">