Expand knowledge base

7 files changed, 499 insertions, 7 deletions

diff --git a/content/know/concept/feynman-diagram/index.pdc b/content/know/concept/feynman-diagram/index.pdc
index eee6bf3..dfb63c1 100644
--- a/content/know/concept/feynman-diagram/index.pdc
+++ b/content/know/concept/feynman-diagram/index.pdc
@@ -1,7 +1,7 @@
 ---
 title: "Feynman diagram"
 firstLetter: "F"
-publishDate: 2021-11-15
+publishDate: 2021-11-18
 categories:
 - Physics
 - Quantum mechanics
@@ -105,7 +105,7 @@ are instead represented by a special vertex:
 <img src="perturbation.png" style="width:35%;display:block;margin:auto;">
 </a>
 $$\begin{aligned}
-    = \frac{1}{i \hbar} V_I(\vb{r}, t, \sigma)
+    = \frac{1}{i \hbar} V_s(\vb{r}, t)
 \end{aligned}$$
 
 Other graphical components exist representing
diff --git a/content/know/concept/imaginary-time/index.pdc b/content/know/concept/imaginary-time/index.pdc
index 55f163a..b68afce 100644
--- a/content/know/concept/imaginary-time/index.pdc
+++ b/content/know/concept/imaginary-time/index.pdc
@@ -145,20 +145,30 @@ $$\begin{aligned}
     \hat{A}_I(\tau) \hat{K}_I(\tau, \tau') \hat{B}_I(\tau') \hat{K}_I(\tau', 0) \!\Big)
 \end{aligned}$$
 
-Assuming $\tau > \tau'$,
-we introduce a time-ordering $\mathcal{T}$,
-allowing us to reorder the operators inside,
+We now introduce a time-ordering $\mathcal{T}$,
+letting us reorder the (bosonic) $\hat{K}_I$-operators inside,
 and thereby reduce the expression considerably:
 
 $$\begin{aligned}
-    \expval*{\hat{A}_H \hat{B}_H}
+    \expval{\mathcal{T}\Big\{\hat{A}_H \hat{B}_H\Big\}}
     &= \frac{1}{Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{K}_I(\hbar \beta, \tau) \hat{K}_I(\tau, \tau') \hat{K}_I(\tau', 0)
     \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp\!(-\beta \hat{H}_{0,S}) \Big)
     \\
-    &= \frac{1}{Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp\!(-\beta \hat{H}_{0,S}) \Big)
+    &= \frac{1}{Z} \Tr\!\Big( \mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp\!(-\beta \hat{H}_{0,S}) \Big)
 \end{aligned}$$
 
 Where $Z = \Tr\!\big(\exp\!(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp\!(-\beta \hat{H}_{0,S})\big)$.
+If we now define $\expval{}_0$ as the expectation value with respect
+to the unperturbed equilibrium involving only $\hat{H}_{0,S}$,
+we arrive at the following way of writing this time-ordered expectation:
+
+$$\begin{aligned}
+    \boxed{
+        \expval{\mathcal{T}\Big\{\hat{A}_H \hat{B}_H\Big\}}
+        = \frac{\expval{\mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\}}_0}{\expval{\hat{K}_I(\hbar \beta, 0)}_0}
+    }
+\end{aligned}$$
+
 For another application of imaginary time,
 see e.g. the [Matsubara Green's function](/know/concept/matsubara-greens-function/).
 
diff --git a/content/know/concept/self-energy/dyson.png b/content/know/concept/self-energy/dyson.png
new file mode 100644
index 0000000..efa7a63
--- /dev/null
+++ b/content/know/concept/self-energy/dyson.png
diff --git a/content/know/concept/self-energy/fullgf.png b/content/know/concept/self-energy/fullgf.png
new file mode 100644
index 0000000..3c88c6a
--- /dev/null
+++ b/content/know/concept/self-energy/fullgf.png
diff --git a/content/know/concept/self-energy/index.pdc b/content/know/concept/self-energy/index.pdc
new file mode 100644
index 0000000..d2908eb
--- /dev/null
+++ b/content/know/concept/self-energy/index.pdc
@@ -0,0 +1,311 @@
+---
+title: "Self-energy"
+firstLetter: "S"
+publishDate: 2021-11-21
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-11-15T21:02:02+01:00
+draft: false
+markup: pandoc
+---
+
+# Self-energy
+
+Suppose we have a time-independent Hamiltonian $\hat{H} = \hat{H}_0 + \hat{W}$,
+consisting of a simple $\hat{H}_0$ and a difficult interaction $\hat{W}$,
+for example describing Coulomb repulsion between electrons.
+
+The concept of [imaginary time](/know/concept/imaginary-time/)
+exists to handle such difficult time-independent Hamiltonians
+at nonzero temperatures. Therefore, we know that the
+[Matsubara Green's function](/know/concept/matsubara-greens-function/)
+$G$ can be written as follows, where $\mathcal{T}$ is the
+[time-ordered product](/know/concept/time-ordered-product/),
+and $\beta = 1 / (k_B T)$:
+
+$$\begin{aligned}
+    G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)
+    = - \frac{\expval{\mathcal{T}\Big\{ \hat{K}(\hbar \beta, 0) \hat{\Psi}_{s_b}(\vb{r}_b, \tau_b) \hat{\Psi}_{s_a}^\dagger(\vb{r}_a, \tau_a) \Big\}}}
+    {\hbar \expval{\hat{K}(\hbar \beta, 0)}}
+\end{aligned}$$
+
+Where we know that the time evolution operator $\hat{K}$
+is as follows in the [interaction picture](/know/concept/interaction-picture/):
+
+$$\begin{aligned}
+    \hat{K}(\tau_2, \tau_1)
+    &= \mathcal{T}\bigg\{ \exp\!\bigg( \!-\!\frac{1}{\hbar} \int_{\tau_1}^{\tau_2} \hat{W}(\tau) \dd{\tau} \bigg) \bigg\}
+    \\
+    &= \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n
+    \mathcal{T}\bigg\{ \bigg( \int_{\tau_1}^{\tau_2} \hat{W}(\tau) \dd{\tau} \bigg)^n \bigg\}
+\end{aligned}$$
+
+Where $\hat{W}$ is the two-body operator in the interaction picture.
+We insert this into the full Green's function above,
+and abbreviate
+$G_{ba} \equiv G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)$
+and $\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$:
+
+$$\begin{aligned}
+    G_{ba}
+    %&= - \frac{\expval{\mathcal{T}\Big\{ \hat{K}(\hbar \beta, 0) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}}}{\expval{\hat{K}(\hbar \beta, 0)}}
+    %\\
+    &= - \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
+    \expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
+    \expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
+\end{aligned}$$
+
+Next, we write out the interaction operator $\hat{W}$
+in the [second quantization](/know/concept/second-quantization/),
+assuming there is no spin-flipping,
+and that $W(\vb{r}_1, \vb{r}_2) = W(\vb{r}_2, \vb{r}_1)$
+(hence $1/2$ to avoid double-counting):
+
+$$\begin{aligned}
+    \hat{W}(\tau_1)
+    &= \frac{1}{2} \sum_{s_1 s_2} \iint_{-\infty}^\infty \hat{\Psi}_{s_1}^\dagger(\vb{r}_1, \tau_1) \hat{\Psi}_{s_2}^\dagger(\vb{r}_2, \tau_1)
+    W(\vb{r}_1, \vb{r}_2) \hat{\Psi}_{s_2}(\vb{r}_2, \tau_1) \hat{\Psi}_{s_1}(\vb{r}_1, \tau_1) \dd{\vb{r}_1} \dd{\vb{r}_2}
+\end{aligned}$$
+
+We integrate this over $\tau_1$ and over a dummy $\tau_2$.
+Defining $W_{j'j} \equiv W(\vb{r}_j', \vb{r}_j) \: \delta(\tau_1 \!-\! \tau_2)$ we get:
+
+$$\begin{aligned}
+    \int_0^{\hbar \beta} \hat{W}(\tau_1) \dd{\tau_1}
+    &= \frac{1}{2} \iint \hat{\Psi}_{s_1}^\dagger(\vb{r}_1, \tau_1) \hat{\Psi}_{s_2}^\dagger(\vb{r}_2, \tau_2)
+    \: W_{1,2} \: \hat{\Psi}_{s_2}(\vb{r}_2, \tau_2) \hat{\Psi}_{s_1}(\vb{r}_1, \tau_1) \dd{\tau_2} \dd{\vb{r}_1} \dd{\vb{r}_2}
+    \\
+    &= \frac{1}{2} \iint \hat{\Psi}_1^\dagger \hat{\Psi}_2^\dagger  W_{1,2} \hat{\Psi}_2 \hat{\Psi}_1 \dd{1} \dd{2}
+\end{aligned}$$
+
+Where we have further abbreviated $\int \dd{j} \equiv \sum_{s_j} \int \dd{\vb{r}_j} \int \dd{\tau_j}$.
+The full $G_{ba}$ thus becomes:
+
+$$\begin{aligned}
+    G_{ba}
+    &= - \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n+1}
+    \idotsint W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{num} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n}
+    \idotsint W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{den} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+\end{aligned}$$
+
+Where we have realized that both the numerator and denominator
+contain many-particle non-interacting Green's functions, defined as:
+
+$$\begin{aligned}
+    G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)
+    &= \Big( \!-\!\frac{1}{\hbar} \Big)^{2 n + 1}
+    \expval{\mathcal{T}\Big\{ \hat{\Psi}_{1'}^\dagger \hat{\Psi}_{1}^\dagger \hat{\Psi}_{1} \hat{\Psi}_{1'} \cdots
+    \hat{\Psi}_{n'}^\dagger \hat{\Psi}_{n}^\dagger \hat{\Psi}_{n} \hat{\Psi}_{n'} \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}}
+    \\
+    G^0_\mathrm{den}(1'1 \cdots n'n; 1'1 \cdots n'n)
+    &= \Big( \!-\!\frac{1}{\hbar} \Big)^{2 n}
+    \expval{\mathcal{T}\Big\{ \hat{\Psi}_{1'}^\dagger \hat{\Psi}_{1}^\dagger \hat{\Psi}_{1} \hat{\Psi}_{1'} \cdots
+    \hat{\Psi}_{n'}^\dagger \hat{\Psi}_{n}^\dagger \hat{\Psi}_{n} \hat{\Psi}_{n'} \Big\}}
+\end{aligned}$$
+
+By applying [Wick's theorem](/know/concept/wicks-theorem/),
+we can rewrite these as a sum of products of single-particle Green's functions,
+so for instance $G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)$ becomes:
+
+$$\begin{aligned}
+    G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)
+    = \mathrm{det} \begin{bmatrix}
+        G^0_{ba} & G^0_{b1'} & G^0_{b1} & G^0_{b2'} & \cdots & G^0_{bn'} & G^0_{bn} \\
+        G^0_{1'a} & G^0_{1'1'} & G^0_{1'1} & G^0_{1'2'} & \cdots & G^0_{1'n'} & G^0_{1'n} \\
+        \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+        G^0_{n'a} & G^0_{n'1'} & G^0_{n'1} & G^0_{n'2'} & \cdots & G^0_{n'n'} & G^0_{n'n} \\
+        G^0_{na} & G^0_{n1'} & G^0_{n1} & G^0_{n2'} & \cdots & G^0_{nn'} & G^0_{nn}
+    \end{bmatrix}
+\end{aligned}$$
+
+And analogously for $G^0_\mathrm{den}$.
+If we are studying bosons instead of fermions,
+the above determinant would need to be replaced by a *permanent*.
+We assume fermions from now on.
+
+We thus have sums over all permutations $p$
+of products of single-particle Green's function,
+times $(-1)^p$ to account for swaps of fermionic operators:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \idotsint W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n + 1} G^0_{(p,m)} \Big) \dd{1}' \dd{1} \cdots \dd{n'} \dd{n}}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \idotsint W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n} G^0_{(p,m)} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+\end{aligned}$$
+
+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 $(-1)^F$,
+with $F$ the number of fermion loops.
+
+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 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**.
+
+And in the numerator, we sum over all diagrams of total order $n$
+containing the external vertices $a$ and $b$.
+Some of them are **connected**,
+so all vertices (including $a$ and $b$) are in the same graph,
+but most are **disconnected**.
+Because disconnected diagrams have no shared lines or vertices to integrate over,
+they can simply be factored into separate diagrams.
+
+If it contains $a$ and $b$, we call it an **external diagram**,
+and then clearly all disconnected parts must be internal diagrams
+($a$ and $b$ are always connected,
+since they are the only vertices with just one fermion line;
+all internal vertices must have two).
+We thus find:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \bigg[ \sum_\mathrm{all\;ext}^{n} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m \!\le\! n}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_\mathrm{all\;ext}^{\infty} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+\end{aligned}$$
+
+Since both $n$ and $m$ start at zero,
+and the sums include all possible diagrams,
+we see that the second sum in the numerator does not actually depend on $m$:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_\mathrm{all\;ext} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}}
+    \\
+    &= \sum_\mathrm{all\;ext} \binom{1 \: \mathrm{external}}{\mathrm{order} \; m}
+\end{aligned}$$
+
+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 $G(b,a) = G_{ba}$:
+
+<a href="fullgf.png">
+<img src="fullgf.png" style="width:90%;display:block;margin:auto;">
+</a>
+
+A **reducible diagram** is a Feynman diagram
+that can be cut in two valid diagrams
+by removing just one fermion line,
+while an **irreducible diagram** cannot be split like that.
+
+At last, we define the **self-energy** $\Sigma(y,x)$
+as the sum of all irreducible terms in $G(b,a)$,
+after removing the two external lines from/to $a$ and $b$:
+
+<a href="selfenergy.png">
+<img src="selfenergy.png" style="width:90%;display:block;margin:auto;">
+</a>
+
+Despite its appearance, the self-energy has the semantics of a line,
+so it has two endpoints over which to integrate if necessary.
+
+By construction, by reattaching $G^0(x,a)$ and $G^0(b,y)$ to the self-energy,
+we get all irreducible diagrams,
+and by connecting multiple irreducible diagrams with single fermion lines,
+we get all fully connected diagrams containing the endpoints $a$ and $b$.
+
+In other words, the full $G(b,a)$ is constructed
+by taking the unperturbed $G^0(b,a)$
+and inserting one or more irreducible diagrams between $a$ and $b$.
+We can equally well insert a single irreducible diagram
+as a sequence of connected irreducible diagrams.
+Thanks to this recursive structure,
+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;">
+</a>
+
+This makes sense: in the "normal" Dyson equation
+we have a one-body perturbation instead of $\Sigma$,
+while $\Sigma$ represents a two-body effect
+as an infinite sum of one-body diagrams.
+Interpreting this diagrammatic Dyson equation yields:
+
+$$\begin{aligned}
+    \boxed{
+        G(b, a)
+        = G^0(b, a) + \iint G^0(b, y) \: \Sigma(y, x) \: G(x, a) \dd{x} \dd{y}
+    }
+\end{aligned}$$
+
+Keep in mind that $\int \dd{x} \equiv \sum_{s_x} \int \dd{\vb{r}_x} \int \dd{\tau_x}$.
+In the special case of a system with continuous translational symmetry
+and no spin dependence, this simplifies to:
+
+$$\begin{aligned}
+    \boxed{
+        G_{s}(\tilde{\vb{k}})
+        = G_{s}^0(\tilde{\vb{k}}) + G_{s}^0(\tilde{\vb{k}}) \: \Sigma_{s}(\tilde{\vb{k}}) \: G_{s}(\tilde{\vb{k}})
+    }
+\end{aligned}$$
+
+Where $\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n)$,
+with $\omega_n$ being a fermionic Matsubara frequency.
+Note that conservation of spin, $\vb{k}$ and $\omega_n$,
+together with the linear structure of the Dyson equation,
+makes $\Sigma$ diagonal in all of those quantities.
+Isolating for $G$:
+
+$$\begin{aligned}
+    G_{s}(\tilde{\vb{k}})
+    = \frac{G_{s}^0(\tilde{\vb{k}})}{1 - G_{s}^0(\tilde{\vb{k}}) \: \Sigma_{s}(\tilde{\vb{k}})}
+    = \frac{1}{1 / G_{s}^0(\tilde{\vb{k}}) - \Sigma_{s}(\tilde{\vb{k}})}
+\end{aligned}$$
+
+From [equation-of-motion theory](/know/concept/equation-of-motion-theory/),
+we already know an expression for $G$ in diagonal $\vb{k}$-space:
+
+$$\begin{aligned}
+    G_s^0(\vb{k}, i \omega_n)
+    = \frac{1}{i \hbar \omega_n - \varepsilon_\vb{k}}
+    \quad \implies \quad
+    G_{s}(\vb{k}, i \omega_n)
+    = \frac{1}{i \hbar \omega_n - \varepsilon_\vb{k} - \Sigma_{s}(\vb{k}, i \omega_n)}
+\end{aligned}$$
+
+The self-energy thus corrects the non-interacting energies for interactions.
+It can therefore be regarded as the energy
+a particle has due to changes it has caused in its environment.
+
+Unfortunately, in practice, $\Sigma$ is rarely as simple as
+in the translationally-invariant example above;
+in fact, it does not even need to be Hermitian,
+i.e. $\Sigma(y,x) \neq \Sigma^*(x,y)$,
+in which case it resists the standard techniques for analysis.
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/content/know/concept/self-energy/selfenergy.png b/content/know/concept/self-energy/selfenergy.png
new file mode 100644
index 0000000..8eaffff
--- /dev/null
+++ b/content/know/concept/self-energy/selfenergy.png
diff --git a/sources/know/concept/self-energy/main.tex b/sources/know/concept/self-energy/main.tex
new file mode 100644
index 0000000..114b433
--- /dev/null
+++ b/sources/know/concept/self-energy/main.tex
@@ -0,0 +1,171 @@
+\documentclass[11pt]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{physics}
+
+\usepackage{feynmp}
+\DeclareGraphicsRule{*}{mps}{*}{}
+
+\begin{document}
+
+\begin{center}
+	{\Huge \sc Feynman diagrams}
+\end{center}
+
+\noindent
+Full Green's function
+\begin{align}
+	G(b,a)
+&= \quad
+\begin{fmffile}{0order1}
+	\begin{fmfgraph*}(30,20)
+		\fmfipair{i,o}
+		\fmfiequ{i}{(0w,0.3h)}
+		\fmfiequ{o}{(1w,0.3h)}
+		\fmfi{fermion}{i -- o}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$a$,label.angle=-90}{i}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$b$,label.angle=-90}{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\quad + \quad
+\begin{fmffile}{1order1}
+	\begin{fmfgraph*}(50,20)
+		\fmfipair{i,m,t,t',o}
+		\fmfiequ{i}{(0w,0.3h)}
+		\fmfiequ{m}{(0.5w,0.3h)}
+		\fmfiequ{t}{(0.5w,0.9h)}
+		\fmfiequ{t'}{(0.5w,1.3h)}
+		\fmfiequ{o}{(1w,0.3h)}
+		\fmfi{fermion}{i -- m}
+		\fmfi{boson}{m -- t}
+		\fmfi{plain}{t{left} .. {right}t'}
+		\fmfi{plain}{t'{right} .. {left}t}
+		\fmfi{fermion}{m -- o}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$a$,label.angle=-90}{i}
+		\fmfiv{decor.shape=circle,decor.size=4}{m}
+		\fmfiv{decor.shape=circle,decor.size=4}{t}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$b$,label.angle=-90}{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\quad + \quad
+\begin{fmffile}{1order2}
+	\begin{fmfgraph*}(70,20)
+		\fmfipair{i,a,b,o}
+		\fmfiequ{i}{(0w,0.3h)}
+		\fmfiequ{a}{(0.33w,0.3h)}
+		\fmfiequ{b}{(0.67w,0.3h)}
+		\fmfiequ{o}{(1w,0.3h)}
+		\fmfi{fermion}{i -- a}
+		\fmfi{boson}{a -- b}
+		\fmfi{fermion}{a{up} .. {down}b}
+		\fmfi{fermion}{b -- o}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$a$,label.angle=-90}{i}
+		\fmfiv{decor.shape=circle,decor.size=4}{a}
+		\fmfiv{decor.shape=circle,decor.size=4}{b}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$b$,label.angle=-90}{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\quad + \:\: \cdots
+\end{align}
+Self-energy definition
+\begin{align}
+\Sigma(y,x)
+=\!\!\!\!
+\begin{fmffile}{selfenergy}
+	\begin{fmfgraph*}(20,20)
+		\fmfright{v}
+		\fmfv{label=$x\!\!\to\!\!y$,label.angle=-90,label.dist=10}{v}
+		\fmfblob{15}{v}
+	\end{fmfgraph*}
+\end{fmffile}
+&\quad\: \equiv \:\:\:\:
+\begin{fmffile}{hartree}
+	\begin{fmfgraph*}(25,20)
+		\fmfleft{i}
+		\fmfright{o}
+		\fmf{boson}{i,o}
+		\fmf{fermion}{o,o}
+		\fmfv{label=$\:\:x\!\!=\!\!y$,label.angle=-90}{i}
+		\fmfdot{i}
+		\fmfdot{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\qquad + \:\:\:
+\begin{fmffile}{fock}
+	\begin{fmfgraph*}(40,20)
+		\fmfipair{i,o}
+		\fmfiequ{i}{(0,0.3h)}
+		\fmfiequ{o}{(w,0.3h)}
+		\fmfi{boson}{i -- o}
+		\fmfi{fermion}{i{up} .. tension 1.5 .. {down}o}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$x$,label.angle=-90,label.dist=5}{i}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$y$,label.angle=-90,label.dist=5}{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\:\:\: + \:\:\:
+\begin{fmffile}{pairbubble}
+	\begin{fmfgraph*}(40,20)
+		\fmfipair{i,a,b,o}
+		\fmfiequ{i}{(0,0h)}
+		\fmfiequ{a}{(0,1h)}
+		\fmfiequ{b}{(1w,1h)}
+		\fmfiequ{o}{(1w,0h)}
+		\fmfi{fermion}{i -- o}
+		\fmfi{boson}{i -- a}
+		\fmfi{boson}{b -- o}
+		\fmfi{fermion}{a{up} .. tension 2.5 .. {down}b}
+		\fmfi{fermion}{b{down} .. tension 2.5 .. {up}a}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$x$,label.angle=-90,label.dist=5}{i}
+		\fmfiv{decor.shape=circle,decor.size=4,label.angle=-90,label.dist=3}{a}
+		\fmfiv{decor.shape=circle,decor.size=4,label.angle=-90,label.dist=3}{b}
+		\fmfiv{decor.shape=circle,decor.size=4,label=$y$,label.angle=-90,label.dist=5}{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\:\:\: + \: \cdots
+\end{align}
+Dyson equation
+\begin{align}
+G(b,a)
+= \quad
+\begin{fmffile}{fullgf}
+	\begin{fmfgraph*}(50,20)
+		\fmfleft{i}
+		\fmfright{o}
+		\fmf{heavy}{i,o}
+		\fmfv{label=$a$,label.angle=-90}{i}
+		\fmfv{label=$b$,label.angle=-90}{o}
+		\fmfdot{i}
+		\fmfdot{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\quad = \quad
+\begin{fmffile}{freegf}
+	\begin{fmfgraph*}(50,20)
+		\fmfleft{i}
+		\fmfright{o}
+		\fmf{fermion}{i,o}
+		\fmfv{label=$a$,label.angle=-90}{i}
+		\fmfv{label=$b$,label.angle=-90}{o}
+		\fmfdot{i}
+		\fmfdot{o}
+	\end{fmfgraph*}
+\end{fmffile}
+\quad + \quad
+\begin{fmffile}{dyson}
+	\begin{fmfgraph*}(110,20)
+		\fmfleft{i}
+		\fmfright{o}
+		\fmf{fermion}{i,v}
+		\fmf{heavy}{v,o}
+		\fmfv{label=$a$,label.angle=-90}{i}
+		\fmfv{label=$b$,label.angle=-90}{o}
+		\fmfdot{i}
+		\fmfdot{o}
+		\fmfv{label=$x\!\to\!y$,label.angle=-90,label.dist=10}{v}
+		\fmfblob{15}{v}
+	\end{fmfgraph*}
+\end{fmffile}
+\end{align}
+
+\end{document}