Bold Feynman Diagrams and the Luttinger–Ward Formalism via Gibbs Measures: Perturbative Approach

Abstract

Many-body perturbation theory (MBPT) is widely used in quantum physics, chemistry, and materials science. At the heart of MBPT is the Feynman diagrammatic expansion, which is, simply speaking, an elegant way of organizing the combinatorially growing number of terms of a certain Taylor expansion. In particular, the construction of the ‘bold Feynman diagrammatic expansion’ involves the partial resummation to infinite order of possibly divergent series of diagrams. This procedure demands investigation from both the combinatorial (perturbative) and the analytical (non-perturbative) viewpoints. In this paper, we illustrate how the combinatorial properties of Feynman diagrams in MBPT can be studied in the simplified setting of Gibbs measures (known as the Euclidean lattice field theory in the physics literature) and provide a self-contained explanation of Feynman diagrams in this setting. We prove the combinatorial validity of the bold diagrammatic expansion, with methods generalizable to several types of field theories and interactions. Our treatment simplifies the presentation and numerical study of known techniques in MBPT such as the self-consistent Hartree–Fock approximation (HF), the second-order Green’s function approximation (GF2), and the GW approximation. The bold diagrams are closely related to the Luttinger-Ward (LW) formalism, which was proposed in 1960 but whose analytic properties have not been rigorously established. The analytical study of the LW formalism in the setting of Gibbs measures will be the topic of an accompanying paper.

Appendices

Appendix A. Proof of Theorem 2.2 (Isserlis-Wick)

Proof

For any $b\in {\mathbb {R}}^{N}$, define

$$\begin{aligned} Z_{0}(b) = \int _{{\mathbb {R}}^{N}} e^{-\frac{1}{2} x^{T} Ax + b^{T}x} \,\mathrm {d}x. \end{aligned}$$

(A.1)

Clearly $Z_{0}=Z_{0}(0)$. Performing the change of variable $\widetilde{x}=x+A^{-1}b$, we find

$$\begin{aligned} Z_{0}(b) = e^{\frac{1}{2} b^{T} A^{-1} b} Z_{0}. \end{aligned}$$

(A.2)

Observe that, for integers $1\le \alpha _{1},\ldots ,\alpha _{m}\le N$,

$$\begin{aligned} \frac{\partial ^{m} Z_{0}(b)}{\partial b_{\alpha _{1}}\cdots \partial b_{\alpha _{m}}}\Big \vert _{b=0} = \int _{{\mathbb {R}}^{N}} x_{\alpha _{1}}\cdots x_{\alpha _{m}} e^{-\frac{1}{2} x^{T} Ax} \,\mathrm {d}x. \end{aligned}$$

(A.3)

Combining with Eq. (A.2), we have that

$$\begin{aligned} \left\langle x_{\alpha _{1}}\cdots x_{\alpha _{m}}\right\rangle _{0} = \frac{\partial ^{m} }{\partial b_{\alpha _{1}}\cdots \partial b_{\alpha _{m}}} e^{\frac{1}{2} b^{T} A^{-1} b} \Big \vert _{b=0}. \end{aligned}$$

(A.4)

Now write the Taylor expansion

$$\begin{aligned} e^{\frac{1}{2} b^{T} A^{-1} b} = \sum _{n=0}^{\infty } \frac{1}{n! \, 2^{n}} \left( b^{T}A^{-1} b \right) ^{n}. \end{aligned}$$

(A.5)

Since the expansion contains no odd powers of the components of b, the right-hand side of Eq. (A.4) vanishes whenever m is odd. If m is an even number, only the term $n=m/2$ in the Taylor expansion will contribute to the right-hand side of Eq. (A.4). This gives

$$\begin{aligned} \frac{\partial ^{m} }{\partial b_{\alpha _{1}}\cdots \partial b_{\alpha _{m}}} e^{\frac{1}{2} b^{T} A b} \Big \vert _{b=0} = \frac{1}{2^{m/2} (m/2)!} \frac{\partial ^{m} }{\partial b_{\alpha _{1}}\cdots \partial b_{\alpha _{m}}} \left( b^{T}A^{-1} b \right) ^{m/2}. \end{aligned}$$

(A.6)

Now one can write

$$\begin{aligned} \frac{\partial ^{m} }{\partial b_{\alpha _{1}}\cdots \partial b_{\alpha _{m}}} = \frac{\partial ^{m} }{\partial ^{m_1} b_{\beta _{1}}\cdots \partial ^{m_p} b_{\beta _{p}}}, \end{aligned}$$

where the indices $\beta _1,\ldots ,\beta _p$ are distinct and $\sum _{j=1}^p m_j = m$. Then to further simplify the right-hand side of Eq. (A.6), we are interested in computing the coefficient of the $b_{\beta _1}^{m_1} \cdots b_{\beta _p}^{m_p}$ term of the polynomial $(b^T A^{-1} b)^{m/2}$. Expand $(b^T A^{-1} b)^{m/2}$ into a sum of monomials as

$$\begin{aligned} (b^T A^{-1} b)^{m/2} = \sum _{j_1, k_1, \ldots , j_{m/2}, k_{m/2} } \prod _{i=1}^{m/2} (A^{-1})_{j_i, k_i} b_{j_i} b_{k_i}. \end{aligned}$$

Thus each distinct permutation $(j_1, k_1, \ldots , j_{m/2}, k_{m/2})$ of the multiset $\{\alpha _1, \ldots ,\alpha _m \}$⁵ contributes $\prod _{i=1}^{m/2} (A^{-1})_{j_i, k_i}$ to our desired coefficient.

Since

$$\begin{aligned} \frac{\partial ^{m} }{\partial ^{m_1} b_{\beta _{1}}\cdots \partial ^{m_p} b_{\beta _{p}}} b_{\beta _1}^{m_1} \cdots b_{\beta _p}^{m_p} = m_1! \cdots m_p!, \end{aligned}$$

we obtain, from Eq. (A.4) and Eq. (A.6), that

$$\begin{aligned} \left\langle x_{\alpha _{1}}\cdots x_{\alpha _{m}}\right\rangle _{0} = \frac{m_1! \cdots m_p!}{2^{m/2} (m/2)!} \sum _{(j_1, k_1, \ldots , j_{m/2}, k_{m/2})} \prod _{i=1}^{m/2} (A^{-1})_{j_i, k_i}, \end{aligned}$$

(A.7)

where the sum is understood to be taken over multiset permutations of $\{\alpha _1, \ldots ,\alpha _m \}$.

Now every permutation of $\mathcal {I}_m$ can be thought of as inducing a permutation of the multiset $\{\alpha _1, \ldots ,\alpha _m \}$ via the map $(l_1, \ldots , l_m) \mapsto (\alpha _{l_1}, \ldots , \alpha _{l_m})$. By this map each multiset permutation of $\{\alpha _1, \ldots ,\alpha _m \}$ is associated with $m_1! \cdots m_p!$ different permutations of $\mathcal {I}_m$.

Moreover, every permutation of $\mathcal {I}_m$ can be thought of as inducing a pairing on $\mathcal {I}_m$ by pairing the first two indices in the permutation, then the next two, etc. For example, if $m=4$, then the permutation (1, 4, 3, 2) induces the pairing (1, 4)(3, 2). By this map, each pairing on $\mathcal {I}_m$ is associated with $2^{m/2} (m/2)!$ permutations of $\mathcal {I}_m$ since a pairing does not distinguish the ordering of the pairs, nor the order of the indices within each pair.

Finally, observe that if we take any two permutations of $\mathcal {I}_m$ associated with a pairing on $\mathcal {I}_m$ and then consider the (possibly distinct) multiset permutations $(j_1, k_1, \ldots , j_{m/2}, k_{m/2})$ and $(j_1', k_1', \ldots , j_{m/2}', k_{m/2}')$ of $\{\alpha _1, \ldots ,\alpha _m \}$ associated to these permutations, then in fact $\prod _{i=1}^{m/2} (A^{-1})_{j_i, k_i} = \prod _{i=1}^{m/2} (A^{-1})_{j_i', k_i'}$.

Thus we can replace the sum over multisets in Eq. (A.7) with a sum over pairings on $\mathcal {I}_m$. Each term must be counted $\frac{2^{m/2} (m/2)!}{m_1! \cdots m_p!}$ times because each pairing on $\mathcal {I}_m$ induces $2^{m/2} (m/2)!$ permutations of $\mathcal {I}_m$, each of which is redundant by a factor of $m_1! \cdots m_p!$. This factor cancels with the factor in Eq. (A.7) to yield

$$\begin{aligned} \left\langle x_{\alpha _{1}}\cdots x_{\alpha _{m}}\right\rangle _{0} = \sum _{\sigma \in \Pi (\mathcal {I}_{m})} \prod _{i \in \mathcal {I}_{m}/\sigma } (A^{-1})_{\alpha _{i},\alpha _{\sigma (i)} }. \end{aligned}$$

Recalling that $G^{0}=A^{-1}$, this completes the proof of Theorem 2.2. $\quad \square$

Remark A.1

In field theories, the auxiliary variable b is often interpreted as a coupling external field.

Appendix B. Proof of Proposition 4.4 (Skeleton decomposition)

Proof

As suggested in the statement of the proposition, let $\Gamma ^{(k)}$ be the maximal insertions admitted by $\Gamma$, where we do not count separately any insertions that differ only by their external labeling (see Remark 4.5). Let $h_1^{(k)}$ and $h_2^{(k)}$ be half edges such that $\Gamma$ admits the insertion $\Gamma ^{(k)}$ at $(h_1^{(k)},h_2^{(k)})$. Let $e_1^{(k)},e_2^{(k)}$ be the external half-edges of the truncated Green’s function diagram $\Gamma ^{(k)}$ paired with $h_1^{(k)},h_2^{(k)}$, respectively, in the overall diagram $\Gamma$.

First we aim to show that the $\Gamma ^{(k)}$ are disjoint, that is, share no half-edges. In this case we say that the $\Gamma ^{(k)}$ do not overlap. In fact we will show additionally that the external half-edges of the $\Gamma ^{(k)}$ do not touch one another (that is, are not paired in $\Gamma$), and accordingly we say that the $\Gamma ^{(k)}$ do not touch.

To this end, let $j \ne k$. The idea is to consider the diagrammatic structure formed by collecting all of the half-edges appearing in $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ and then arguing by maximality that $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ cannot overlap or touch, let this structure admit $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ as insertions.

We now formalize this notion. Let $H^{(j)}$ and $H^{(k)}$ be the half-edge sets of $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$, respectively, and consider the union $H^{(j,k)} := H^{(j)} \cup H^{(k)}$, together with a partial pairing $\Pi ^{(j,k)}$ on $H^{(j,k)}$, that is, a collection of disjoint pairs in $H^{(j,k)}$, defined by the rule that $\{h_1,h_2\} \in \Pi ^{(j,k)}$ if and only if $\{h_1,h_2\} \in \Pi _{\Gamma }$ and $h_1,h_2 \in H^{(j,k)}$. In other words, the structure $(H^{(j,k)},\Pi ^{(j,k)})$ is formed by taking all of the half-edges appearing in $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ and then pairing the ones that are paired in the overall diagram $\Gamma$.

Let $E^{(j,k)}$ be the subset of unpaired half-edges in $H^{(j,k)}$, that is, those half-edges that do not appear in any pair in $\Pi ^{(j,k)}$. Since all half-edges in $\Gamma ^{(k)}$ besides $e_1^{(k)},e_2^{(k)}$ are paired in the diagram $\Gamma ^{(k)}$, we must have $E^{(j,k)} \subset \{ e_1^{(j)},e_2^{(j)}, e_1^{(k)},e_2^{(k)} \}$. $\quad \square$

Lemma B.1

$\vert E^{(j,k)} \vert = 4$.

Proof

We claim that $\vert E^{(j,k)} \vert$ is even. To see this, we first establish that $\vert H^{(j,k)} \vert$ is even. Notice that a truncated Green’s function diagram (in particular, $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$) contains an even number of half-edges (more specifically 4n, where n is the order of the diagram). Thus $\vert H^{(j)} \vert , \vert H^{(k)}\vert$ are even. Moreover, $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ share a half-edge if and only if they share the vertex (or interaction line) associated with this half-edge, in which case $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ share all four half-edges emanating from this vertex. Thus the number $\vert H^{(j)} \cap H^{(k)} \vert$ of half-edges common to $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ is four times the number of common interaction lines, in particular an even number. So $\vert H^{(j,k)} \vert = \vert H^{(j)} \vert + \vert H^{(k)} \vert - \vert H^{(j)} \cap H^{(k)} \vert$ is even, as desired. Now any partial pairing on $H^{(j,k)}$ includes an even number of distinct elements, so the number of leftover elements, that is, $\vert E^{(j,k)} \vert$, must be even as well, as claimed.

Next we rule out the cases $\vert E^{(j,k)} \vert \in \{ 0, 2 \}$.

Suppose that $\vert E^{(j,k)} \vert = 0$. Then the structure $(H^{(j,k)},\Pi ^{(j,k)})$ defines a closed sub-diagram of $\Gamma$, disconnected from the rest of $\Gamma$. Since $\Gamma$ is not closed, the sub-diagram cannot be all of $\Gamma$. This conclusion contradicts the connectedness of $\Gamma$.⁶

Next suppose that $\vert E^{(j,k)} \vert = 2$. Then the structure $(H^{(j,k)},\Pi ^{(j,k)})$ has two unpaired half-edges, hence defines a truncated Green’s function diagram $\Gamma ^{(j,k)}$ that contains both $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ and admits both as insertions. But since $\Gamma ^{(j)} \ne \Gamma ^{(k)}$, $\Gamma ^{(j,k)}$ is neither $\Gamma ^{(j)}$ nor $\Gamma ^{(k)}$ (that is, the containment is proper). This conclusion contradicts the maximality of $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$.

From Lemma B.1 it follows that $E^{(j,k)} = \{ e_1^{(j)},e_2^{(j)}, e_1^{(k)},e_2^{(k)} \}$. (And moreover $e_1^{(j)},e_2^{(j)}, e_1^{(k)},e_2^{(k)}$ are distinct, which was not assumed!) This establishes one of our original claims, that is, that the external half-edges of $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ do not touch (that is, are not paired.)

We need to establish the other part of our original claim, that is, that the half-edge sets $H^{(j)}$ and $H^{(k)}$ are disjoint. To see this, suppose otherwise, so $\Gamma ^{(j)}$ and $\Gamma ^{(k)}$ share some half-edge h. Since $\Gamma ^{(j)} \ne \Gamma ^{(k)}$, one of $H^{(j)}$ and $H^{(k)}$ does not contain the other, so assume without loss of generality that $H^{(j)}$ does not contain $H^{(k)}$, so there is some half-edge $h' \in H^{(k)} \backslash H^{(j)}$. Since $\Gamma ^{(k)}$ is connected there must be some path in $\Gamma ^{(k)}$ connecting h with $h'$.⁷ Now $\Gamma ^{(j)}$ can be disconnected from the rest of $\Gamma$ by deleting the links $\{e_1^{(j)},h_1^{(j)}\}$ and $\{e_2^{(j)},h_2^{(j)}\}$ from the pairing $\Pi _{\Gamma }$, so evidently our path in $\Gamma ^{(k)}$ connecting h and $h'$ must contain either $(e_1^{(j)},h_1^{(j)})$ or $(e_2^{(j)},h_2^{(j)})$ as a ‘subpath’. But then either $e_1^{(j)}$ or $e_2^{(j)}$ is paired in $\Gamma ^{(k)}$, hence also by $\Pi ^{(j,k)}$, contradicting its inclusion in $E^{(j,k)}$.

In summary we have shown that the $\Gamma ^{(k)}$ are disjoint, that is, share no half-edges, and that the external half-edges of the $\Gamma ^{(k)}$ do not touch one another, that is, are not paired in $\Gamma$. We can define a truncated Green’s function diagram $\Gamma _{\mathrm {s}}$ by considering $\Gamma$, then omitting all half-edges and vertices appearing in the $\Gamma ^{(k)}$. Since the $\Gamma ^{(k)}$ are disjoint and do not touch, this leaves behind the half-edges $h_1^{(k)},h_2^{(k)}$ for all k, which are now left unpaired. Then we complete the construction of $\Gamma _{\mathrm {s}}$ by adding the pairings $\{h_1^{(k)},h_2^{(k)}\}$. In short, we form $\Gamma _{\mathrm {s}}$ from $\Gamma$ by replacing each insertion $\Gamma ^{(k)}$ with the edge $\{h_1^{(k)},h_2^{(k)}\}$. Eq. (4.2) then holds by construction, for a suitable external labeling of the $\Gamma ^{(k)}$.

Moreover, we find that $\Gamma _{\mathrm {s}}$ is 2PI. It is not hard to check first that $\Gamma _{\mathrm {s}}$ is 1PI. Indeed, the unlinking of any half-edge pair in $\Gamma _{\mathrm {s}}$ that is not one of the $\{h_1^{(k)},h_2^{(k)}\}$ can be lifted to the unlinking of the same half-edge pair in the original diagram $\Gamma$. Since $\Gamma$ is 1PI, this unlinking does not disconnect $\Gamma$. Re-collapsing the maximal insertions once again does not affect the connectedness of the result, so $\Gamma _{\mathrm {s}}$ does not become disconnected by the unlinking. On the other hand, the unlinking of a half-edge pair $\{h_1^{(k)},h_2^{(k)}\}$ were to disconnect $\Gamma _{\mathrm {s}}$, then necessarily the unlinking of either $\{e_1^{(k)},h_1^{(k)}\}$ or $\{e_2^{(k)},h_2^{(k)}\}$ would disconnect $\Gamma$, which contradicts the premise that $\Gamma$ is 1PI.

Thus $\Gamma _{\mathrm {s}}$ is 1PI, and two-particle irreducibility is equivalent to the absence of any insertions. But if $\Gamma _{\mathrm {s}}$ admits an insertion containing either $h_1^{(k)}$ or $h_2^{(k)}$, then this contradicts the maximality of the insertion $\Gamma ^{(k)}$ in $\Gamma$. Moreover, if $\Gamma _{\mathrm {s}}$ admits an insertion containing none of the $h_1^{(k)},h_2^{(k)}$, then this insertion lifts to an insertion in the original diagram $\Gamma$, hence this insertion (that is, all of its interaction lines and half-edges) must have been omitted from $\Gamma _{\mathrm {s}}$ (contradiction). We conclude that $\Gamma _{\mathrm {s}}$ admits no insertions, hence is 2PI.

Finally it remains to prove the uniqueness of the decomposition of Eq. (4.2). To this end, let $\Gamma _{\mathrm {s}} \in \mathfrak {F}_2^{\mathrm {2PI}}$ and $\Gamma ^{(k)} \in \mathfrak {F}_2^{\mathrm {c,t}}$ for $k=1,\ldots ,K$, and let $\left\{ h_1^{(k)},h_2^{(k)}\right\}$ be distinct half-edge pairs in $\Gamma _{\mathrm {s}}$ for $k=1,\ldots ,K$. Then define $\Gamma$ via Eq. (4.2). The uniqueness claim then follows if we can show that the $\Gamma ^{(k)}$ are the maximal insertions in $\Gamma$.

Suppose that $\Gamma ^{(k)}$ is not maximal for some k. Then by definition the diagram $\Gamma '$ formed from $\Gamma$ by collapsing the insertion $\Gamma ^{(k)}$ admits an insertion containing $h_1^{(k)}$ or $h_2^{(k)}$ (assume $h_1^{(k)}$ without loss of generality). In fact let $\Gamma ^{(k)}_{\mathrm {m}}$ be a maximal insertion containing $h_1^{(k)}$. Note also that $\Gamma '$ still admits the $\Gamma ^{(j)}$ for $j\ne k$ as insertions.

Then for $j\ne k$ form the structure $(H^{(j,k)},\Pi ^{(j,k)})$ by ‘merging’ $\Gamma ^{(j)}$ and $\Gamma ^{(k)}_{\mathrm {m}}$ via the same construction as above (now within the overall diagram $\Gamma '$). By the same reasoning as in Lemma B.1, the set of unpaired half-edges $E^{(j,k)}$ must be of even cardinality, and we can rule out the case $\vert E^{(j,k)} \vert =0$.

In the case $\vert E^{(j,k)} \vert = 2$, the structure $(H^{(j,k)},\Pi ^{(j,k)})$ once again defines a truncated Green’s function diagram $\Gamma ^{(j,k)}$ admitting both $\Gamma ^{(j)}$ and $\Gamma ^{(k)}_{\mathrm {m}}$ as insertions. But since $\Gamma ^{(k)}_{\mathrm {m}}$ is maximal, $\Gamma ^{(j,k)} = \Gamma ^{(k)}_{\mathrm {m}}$, and $\Gamma ^{(j)}$ is contained in $\Gamma ^{(k)}_{\mathrm {m}}$.

In the case $\vert E^{(j,k)} \vert = 4$, since $\Gamma ^{(j)}$ does not contain $\Gamma ^{(k)}_{\mathrm {m}}$ (that is, the half-edge set of the former does not contain that of the latter), the same reasoning as above guarantees that $\Gamma ^{(j)}$ and $\Gamma ^{(k)}_{\mathrm {m}}$ do not overlap or touch.

Then consider the diagram $\Gamma ''$ formed from $\Gamma '$ by collapsing the insertion $\Gamma ^{(j)}$. In both of our cases (namely, that $\Gamma ^{(j)}$ is contained in $\Gamma ^{(k)}_{\mathrm {m}}$ and that $\Gamma ^{(j)}$ and $\Gamma ^{(k)}_{\mathrm {m}}$ do not overlap or touch), the insertion $\Gamma ^{(k)}_{\mathrm {m}}$ descends to an insertion in $\Gamma ''$ containing $h_1^{(k)}$.

Iteratively repeating this procedure for all $j\ne k$ (that is, collapsing all of the insertions $\Gamma ^{(j)}$ to obtain the original skeleton diagram $\Gamma _{\mathrm {s}}$), we find that $\Gamma ^{(k)}_{\mathrm {m}}$ descends to an insertion in $\Gamma _{\mathrm {s}}$ containing $h_1^{(k)}$, contradicting the fact that $\Gamma _{\mathrm {s}}$ is 2PI.

$\quad \square$