--- 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 **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.