diff options
Diffstat (limited to 'content/know/concept/self-energy')
| -rw-r--r-- | content/know/concept/self-energy/index.pdc | 55 |
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;"> |