---
title: "Self-energy"
sort_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.