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+---
+title: "Self-energy"
+date: 2021-11-21
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+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 $\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$:
+
+$$\begin{aligned}
+ G_{ba}
+ &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \int\cdots\int_0^{\hbar \beta}
+ \Expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
+ {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \int\cdots\int_0^{\hbar \beta}
+ \Expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ G_{ba}
+ &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n+1}
+ \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{num} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+ {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n}
+ \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{den} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ G_{ba}
+ &= -\frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+ \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n + 1} G^0_{(p,m)} \Big) \dd{1}' \dd{1} \cdots \dd{n'} \dd{n}}
+ {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+ \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n} G^0_{(p,m)} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+\end{aligned}$$
+
+These integrals over products of interactions and Green's functions
+are the perfect place to apply [Feynman diagrams](/know/concept/feynman-diagram/).
+Conveniently, it 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 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**.
+
+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:
+
+$$\begin{aligned}
+ G_{ba}
+ &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!}
+ \bigg[ \sum_{m = 0}^{n} \frac{n!}{m! (n \!-\! m)!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+ \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+ {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+\end{aligned}$$
+
+Where the total order is the sum of the orders of all considered diagrams,
+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:
+
+$$\begin{aligned}
+ G_{ba}
+ &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+ \bigg[ \sum_{n = 0}^\infty \frac{1}{2^{n-m} (n \!-\! m)!}
+ \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+ {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+\end{aligned}$$
+
+Since both $n$ and $m$ start at zero,
+and the sums include all possible diagrams,
+we see that the second sum in the numerator does not actually depend on $m$:
+
+$$\begin{aligned}
+ \hbar G_{ba}
+ &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+ \bigg[ \sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}} \bigg]}
+ {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+ \\
+ &= \sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+\end{aligned}$$
+
+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 $b$. 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 **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 $b$:
+
+
+
+
+
+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 $\Sigma(y, x)$:
+
+
+
+
+
+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.
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