3.6 All-orders cancellation of corrections
3.6.2 Factorization of tree sums
By the decomposition lemma (3.3) the integrand of any fixed graph Γ is a product of trees. For the total correction to the 2-point function, one has to sum over all such Γ.
This can be done by fixing all except for one tree-component of Γ and summing over all possible rearrangements and permutations of the remaining tree. Since all graphs have to be produced, this is a sum over all trees with a given number of external onshell edges and (possibly) one external offshell edge. It is not trivial to see if this procedure really reproduces all graphs Γ with their correct symmetry factor.
But by lemma A.3 it actually does: Connecting trees in all possible ways produces just the right symmetry factors up to mul(Γ, C), i.e. the number of equivalent Cutkosky cuts. Indeed, this is what is needed here: Consider ΓC,4 from fig. 3.9which admits two equivalent Cutkosky cuts. Seemingly, in the sum over all connected trees, this graph will appear twice as often as it should. But the Feynman rules of the diffeomorphism vertex actually produce two terms: One in which k, k−l and l+p are uncancelled and one in which l, k−l and l+p are uncancelled, see eq. (3.31). These two terms directly correspond to the two different Cutkosky cuts ΓAand ΓB inexample 22. This phenomenon works even for bigger graphs: Admitting two equivalent Cutkosky cuts and admitting two different ways of (non-vanishing) cancellation of inner edges is the same thing.
Note that for bigger graphs contributing to the 2-point function there will be no decomposition into just two trees. But this is not necessary: By lemma 3.3 there is a decomposition into trees, so it is always safe to assume each of the external vertices belongs to a tree. Possibly there are more “intermediate” trees. Nonthelesslemma A.3 also holds for more than two components. One can even choose one of the external vertices, call it the left one, and use the corresponding tree as Γ1 and the whole rest of the graph as Γ2. Then, Γ2 generally is no tree, but still lemma A.3 can be applied.
Even if the graph does not decompose into two tree sums, it is still proportional to one tre esum.
Lemma 3.4 Factorization of interaction tree sums.
For a fixed loop number l, and if Γ(l) is the sum of all graphs contributing to the 2-point function, then its non-tadpole Feynman integrand I Γ(l)
is a sum of terms each of which is proportional to one of the tree sums b0n forn≤l+ 1.
Proof. By lemma 3.3 and the above discussion it is always possible to split off a tree on the left side of the graph and in the sum over all graphs this tree will resemble the sum over all trees, i.e. b0n. Using the Euler characteristic section 1.5.2, for a l-loop graph, there are never more than l+ 1 edges to be cut to decompose the graph into two connected components, henceb0l+1 is the biggest tree sum which can possibly appear.
Note that lemma 3.4 does not yet mean that the Feynman integral F Γ(l)
factorizes into some tree sum multiplied by an integral. Generically, since b0n involve kinematic variables, it is not possible to exclude them from the integration.
Example 11: Decomposition of a 3-loop graph
As an example forlemma 3.4, take Γ1 to be the trees fromb3 and Γ2 a graph with one internal loop.
Γ1 = +
Γ2 =
Let further Γ11 be the left and Γ21 the right connected component of Γ1. Then sym(Γ11) = 2
sym(Γ21) = 6 sym(Γ2) = 4,
the latter due to the exchange of the two leftmost external legs as well as the exchange of the two edges in the internal multiedge. The possible outcomes of connecting Γ1 and Γ2 are shown below.
ΓA=
k+p k
l−k
l k
m m−k
k
p −p
ΓB=
k+p l+p
l
l−k k
m m−k
k
p −p
ΓC =
k+p l
l−k k
m m−k
k
p −p
Since there are three edges to be connected, there are 3! = 6 possible ways of
assigning them to each other. With the definitions fromlemma A.3, F
Γ11,Γ2 = 2ΓA+ 4ΓB F
Γ21,Γ2 = 6ΓC. The correct resulting symmetry factors are
sym(ΓA) = 1
4 sym(ΓB) = 1
2 sym(ΓC) = 1 4. Using the procedure from lemma A.3indeed reproduces them:
1
sym(Γ11) sym(Γ2)F
Γ11,Γ2 = 1
2·4(2ΓA+ 4ΓB) (3.44) 1
sym(Γ21) sym(Γ2)F
Γ11,Γ2 = 1 6·46ΓC, so
1
sym(Γ11) sym(Γ2)F
Γ11,Γ2 + 1
sym(Γ21) sym(Γ2)F
Γ11,Γ2 (3.45)
= 1
4ΓA+1
2ΓB+1
4ΓC. (3.46)
The procedure however cannot be directly applied to sums of subgraphs in a naive way such as F
Γ11+ Γ21,Γ2 because each individual summand needs to be mul-tiplied by its corresponding inverse symmetry factor. The problem can be cured if one uses the graphs including their proper symmetry factors, i.e.
F
Γ11
sym(Γ11) + Γ21
sym(Γ21), Γ2 sym(Γ2)
will obviously reproduce the correct resulteq. (3.45). But this also is what actually happens if one usesb3instead of Γ1: As stated explicitly inexample 4,b3consists of Γ21 and three different versions of Γ11 where the unique external leg is assigned one of the three external momenta, respectively. But the operator F from lemma A.3 always sums over all possible ways of connecting edges, hence each of these three graphs in b3 must necessarily give the same result ifF acts on them. Explicitly,
F({b3,Γ2}) =F
3Γ11+ Γ21,Γ2
= 3F
Γ11,Γ2 +F
Γ21,Γ2 .
The relative factor of 3 between the summands just reproduces the prefactors 2·41 and 6·41 ineq. (3.44) up to an overall factor.
This phenomenon is no random coincidence forb3 but actually follows from the orbit-stabilizer-theorem: The operatorF sums over just all ways to connect edges, which is the symmetric group Sn forn edges withn! permutations. On the other hand the symmetry factor sym(Γ) is the number of permutations of the n edges
which do not alter a planar version of the graph, i.e. exchanges of equivalent edges in the usual sense. Then, summing overSnand dividing by sym(Γ) as inlemma A.3 is the same as multiplying with the number of permutations which exchange non-equivalent edges. But in bn, it is precisely these permutations of non-equivalent edges which the graphs are summed over. Hence using bn instead of the naive sum of trees (where each tree would be weighted equally) will always reproduce the symmetry factors up to an overall constant factor. The same is true for b0n instead of bn since the combinatorics there is just the same. Therefore lemma 3.4 is true even if the operation performed there is slightly different from the one in lemma A.3.
Lemma 3.5 Factorization of integration.
For a fixed loop numberl > 1, if the diffeomorphism parameters aj have been chosen such that all contributions Γ(s) to the 2-point function with loopnumber s < l vanish, then the remaining non-tadpole Feynman integrand of Γ(l) reduces to
I Γ(l)
= i3+l
(l+ 1)!x2p· bl+12
·M(l),
where M(l) is the Feynman Integrand of a l-loop multiedge with vertex amplitudes 1 and external momentump.
Proof. The statement holds for 1 and 2 loops by the examplessections 3.4and 3.5, see eqs. (3.20)and(3.42). Assume it holds forl−1 loops. This means,aj forj≤l−1 have been chosen such thatb0k= 0 for allk≤l. Bylemma 3.2this also impliesbk= 0∀k≤l.
Then by lemma 3.4, the only remaining part of the Feynman integrand of the l-loop graphs is proportional to b0l+1. Since all lower bk are already zero, by lemma 3.1 it is actually proportional to the pure diffeomorphism tree sum bl+1. But when the edge cancellation imposed by the diffeomorphism vertex Feynman rules is carried out, any pure diffeomorphism tree sumbk reduces to a (k+ 1)-valent vertex. Hence the graphs in Γ(l+1) which are proportional tobl+1 must, after cancellation, have a topology where there is a (l+ 2)-valent vertex at at least one side. The only l-loop topology without tadpoles where this can happen is the l-loop multiedge with integrand I(i). Hence all remaining Feynman integrals involve the same topology and the integration factors.
The l-loop multiedge contains two (l+ 2)-valent vertices, hence the prefactor must be (bl+1)2. The factor (l+1)!1 is the symmetry factor of thel-loop multiedge. M(l) contains l+1 edges giving (i)l+1if the factors ofiare not included into the integrandM(l). There is no factorifrom vertices since viewing bl+1 as a vertex, it actually has Feynman rule bl+1 and not ibl+1 (all factors i have already been included in the computation of bn, see e.g. example 4) Finally, the external propagators were included in bl+1, they must be cancelling by multiplication with (−ixp)2.
Example 12: 2-loop graphs
Consider the 2-loop graphs from section 3.5. There, a1 was chosen in such a way that 1-loop graphs Γ(1) are cancelled, therefore the conditions of lemma 3.5 are fulfilled. Indeed, the explicitly computed result eq. (3.42)is
I Γ(2)
= 1 6i
9λ4
x2p −36λ2a2+ 36a22x2p
M(2)
= i5 3!x2p
9λ4
x4p −36λ2a2
x2p + 36a22
M(2). Usingeq. (3.21)
a1 = λ
2xp ⇒ λ
xp = 2a1, the parenthesis becomes
9(2a1)4−36(2a1)2a2+ 36a22= 144a41−144a21a2+ 36a22
= 12a21−6a2
2
. According to example 4 this isb3, thus
I Γ(2)
= i5
3!x2p(b3)2M(2).
Example 13: Topology of 3-loop graphs
Consider again the above example example 11. On first sight it might seem ques-tionable that the three Feynman integrands ΓA,ΓB,ΓC indeed cancel, i.e. that they reduce to some prefactor multiplied by an integral over the same topology.
Assume for convenience that no edge in Γ2 is cancelled. But in Γ1, the internal edge in the tree consisting of two vertices will be cancelled: We know b3 does not contain any internal propagator, but is just b3 = 12a21 −6a2. Consequently, in terms of topology, b3 is a single vertex. After executing the cancellation, all three graphs have the same topology ΓC. This is not a 3-loop multiedge as promised by lemma 3.5 because the order is wrong: The multiedge is produced by connecting b4 with a rest (which, as discussed, has to be b4, too, to produce a 3-loop graph).
Conversely, in the conditions of lemma 3.5 the assumption that 2-loop integrals cancel already impliesb3= 0 as can be checked fromeqs. (3.21)and(3.43). There-fore, the topology ΓC does not appear in the final result of the 3-loop integrand and lemma 3.5is not violated.
3.6.3 Parameters of the adiabatic diffeomorphism