Categories: Physics, Quantum mechanics.


Suppose we have a time-independent Hamiltonian H^=H^0+W^\hat{H} = \hat{H}_0 + \hat{W}, consisting of a simple H^0\hat{H}_0 and a difficult interaction W^\hat{W}, for example describing Coulomb repulsion between electrons.

The concept of imaginary time exists to handle such difficult time-independent Hamiltonians at nonzero temperatures. Therefore, we know that the Matsubara Green’s function GG can be written as follows, where T\mathcal{T} is the time-ordered product, and β=1/(kBT)\beta = 1 / (k_B T):

Gsbsa(rb,τb;ra,τa)=T{K^(β,0)Ψ^sb(rb,τb)Ψ^sa(ra,τa)}K^(β,0)\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 K^\hat{K} is as follows in the interaction picture:

K^(τ2,τ1)=T{exp ⁣( ⁣ ⁣1τ1τ2W^(τ)dτ)}=n=01n!( ⁣ ⁣1)nT{(τ1τ2W^(τ)dτ)n}\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 W^\hat{W} is the two-body operator in the interaction picture. We insert this into the full Green’s function above, and abbreviate GbaGsbsa(rb,τb;ra,τa)G_{ba} \equiv G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a) and Ψ^aΨ^sa(ra,τa)\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a):

Gba=n=01n!( ⁣ ⁣1)n0βT{W^(τ1)W^(τn)Ψ^bΨ^a}dτ1dτnn=01n!( ⁣ ⁣1)n0βT{W^(τ1)W^(τn)}dτ1dτn\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 W^\hat{W} in the second quantization, assuming there is no spin-flipping, and that W(r1,r2)=W(r2,r1)W(\vb{r}_1, \vb{r}_2) = W(\vb{r}_2, \vb{r}_1) (hence 1/21/2 to avoid double-counting):

W^(τ1)=12s1s2Ψ^s1(r1,τ1)Ψ^s2(r2,τ1)W(r1,r2)Ψ^s2(r2,τ1)Ψ^s1(r1,τ1)dr1dr2\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 τ1\tau_1 and over a dummy τ2\tau_2. Defining WjjW(rj,rj)δ(τ1 ⁣ ⁣τ2)W_{j'j} \equiv W(\vb{r}_j', \vb{r}_j) \: \delta(\tau_1 \!-\! \tau_2) we get:

0βW^(τ1)dτ1=12Ψ^s1(r1,τ1)Ψ^s2(r2,τ2)W1,2Ψ^s2(r2,τ2)Ψ^s1(r1,τ1)dτ2dr1dr2=12Ψ^1Ψ^2W1,2Ψ^2Ψ^1d1d2\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 djsjdrjdτj\int \dd{j} \equiv \sum_{s_j} \int \dd{\vb{r}_j} \int \dd{\tau_j}. The full GbaG_{ba} thus becomes:

Gba=n=01n!( ⁣ ⁣12)n()2n+1W11Wnn(Gnum0)d1d1dndnn=01n!( ⁣ ⁣12)n()2nW11Wnn(Gden0)d1d1dndn\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:

Gnum0(b11nn;a11nn)=( ⁣ ⁣1)2n+1T{Ψ^1Ψ^1Ψ^1Ψ^1Ψ^nΨ^nΨ^nΨ^nΨ^bΨ^a}Gden0(11nn;11nn)=( ⁣ ⁣1)2nT{Ψ^1Ψ^1Ψ^1Ψ^1Ψ^nΨ^nΨ^nΨ^n}\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, we can rewrite these as a sum of products of single-particle Green’s functions, so for instance Gnum0(b11nn;a11nn)G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n) becomes:

Gnum0(b11nn;a11nn)=det[Gba0Gb10Gb10Gb20Gbn0Gbn0G1a0G110G110G120G1n0G1n0Gna0Gn10Gn10Gn20Gnn0Gnn0Gna0Gn10Gn10Gn20Gnn0Gnn0]\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 Gden0G^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 pp of products of single-particle Green’s function, times (1)p(-1)^p to account for swaps of fermionic operators:

Gba=n=01n!( ⁣ ⁣2)nW11Wnn(p(1)pm=12n+1G(p,m)0)d1d1dndnn=01n!( ⁣ ⁣2)nW11Wnn(p(1)pm=12nG(p,m)0)d1d1dndn\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. Conveniently, it turns out that the factor (1)p(-1)^p is equivalent to the rule that each diagram must be multiplied by (1)F(-1)^F, with FF the number of fermion loops. Keep in mind that fermion lines absorb a factor -\hbar each (see above), and interactions 1/-1/\hbar.

The denominator turns into a sum of all possible diagrams (including equivalent ones) for each total order nn (the order is the number of interaction lines). The endpoints aa and bb 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 nn containing the external vertices aa and bb. Some of them are connected, so all vertices (including aa and bb) 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 aa and bb, we call it an external diagram, and then clearly all disconnected parts must be internal diagrams (aa and bb are always connected, since they are the only vertices with just one fermion line; all internal vertices must have two). We thus find:

Gba=n=012nn![m=0nn!m!(n ⁣ ⁣m)!(1  externalorder  m) ⁣Σall(0  or  more  internaltotal  order  (n ⁣ ⁣m)) ⁣Σall]n=012nn!(0  or  more  internaltotal  order  n) ⁣Σall\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 nn, so we reorganize:

Gba=m=012mm!(1  externalorder  m) ⁣Σall[n=012nm(n ⁣ ⁣m)!(0  or  more  internaltotal  order  (n ⁣ ⁣m)) ⁣Σall]n=012nn!(0  or  more  internaltotal  order  n) ⁣Σall\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 nn and mm start at zero, and the sums include all possible diagrams, we see that the second sum in the numerator does not actually depend on mm:

Gba=m=012mm!(1  externalorder  m) ⁣Σall[n=012nn!(0  or  more  internaltotal  order  n) ⁣Σall]n=012nn!(0  or  more  internaltotal  order  n) ⁣Σall=m=012mm!(1  externalorder  m) ⁣Σall\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 aa and bb. Furthermore, it can be shown using combinatorics that exactly 2mm!2^m m! diagrams at each order are topologically equivalent, so we are left with non-equivalent diagrams only. Let G(b,a)=GbaG(b,a) = G_{ba}:

Full expansion of G in Feynman diagrams

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 Σ(y,x)\Sigma(y,x) as the sum of all irreducible terms in G(b,a)G(b,a), after removing the two external lines from/to aa and bb:

Definition of the self-energy

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 G0(x,a)G^0(x,a) and G0(b,y)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 aa and bb.

In other words, the full G(b,a)G(b,a) is constructed by taking the unperturbed G0(b,a)G^0(b,a) and inserting one or more irreducible diagrams between aa and bb. 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)G(b,a) obeys a Dyson equation involving Σ(y,x)\Sigma(y, x):

Dyson equation in Feynman diagrams

This makes sense: in the “normal” Dyson equation we have a one-body perturbation instead of Σ\Sigma, while Σ\Sigma represents a two-body effect as an infinite sum of one-body diagrams. Interpreting this diagrammatic Dyson equation yields:

G(b,a)=G0(b,a)+G0(b,y)Σ(y,x)G(x,a)dxdy\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 dxsxdrxdτx\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:

Gs(k~)=Gs0(k~)+Gs0(k~)Σs(k~)Gs(k~)\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 k~(k,iωn)\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n), with ωn\omega_n being a fermionic Matsubara frequency. Note that conservation of spin, k\vb{k} and ωn\omega_n, together with the linear structure of the Dyson equation, makes Σ\Sigma diagonal in all of those quantities. Isolating for GG:

Gs(k~)=Gs0(k~)1Gs0(k~)Σs(k~)=11/Gs0(k~)Σs(k~)\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, we already know an expression for GG in diagonal k\vb{k}-space:

Gs0(k,iωn)=1iωnεk    Gs(k,iωn)=1iωnεkΣs(k,iωn)\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, Σ\Sigma is rarely as simple as in the translationally-invariant example above; in fact, it does not even need to be Hermitian, i.e. Σ(y,x)Σ(x,y)\Sigma(y,x) \neq \Sigma^*(x,y), in which case it resists the standard techniques for analysis.


  1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.