Leading Logarithms and Nested Subgraphs

The previous page found the first logarithm in four-dimensional $\phi^4$ theory. A one-loop bubble gives

$$\int_{Q<|k|<\Lambda}{d^4k\over(2\pi)^4}{1\over k^4} ={1\over16\pi^2}\log{\Lambda^2\over Q^2},$$

so the four-point vertex contains a correction of order $\lambda_0^2\log(\Lambda/Q)$. A single logarithm is already enough to warn us that perturbation theory is not organized only by powers of $\lambda_0$. If the ratio of scales is very large, then

$$\lambda_0\log{\Lambda\over Q}$$

may be order one even when $\lambda_0\ll1$.

This page explains where the higher powers of logarithms come from. The answer is not “every complicated diagram produces as many logarithms as loops.” The leading powers come from strongly ordered momentum regions and from nested logarithmically divergent subgraphs. In such a region, a hard subgraph shrinks to a local vertex, then a softer subgraph uses that vertex, then an even softer subgraph uses the result, and so on. This is the diagrammatic seed of the renormalization group.

Required background. Scalar Propagators and One-Loop φ⁴ supplies the bubble symmetry factor, the three crossing channels, and the sign convention for the one-loop four-point vertex. A leading logarithm requires more than a large overall momentum: a loop momentum must become hard relative to a subgraph so that the subgraph contracts to a local operator, and this hierarchy must repeat over strongly ordered scales. Without that nested hierarchy, a multiloop graph need not produce one large logarithm per loop.

What counts as a leading logarithm

Logarithmic variables and one-loop coefficient

We write

$$L\equiv \log{\Lambda\over Q}, \qquad \ell(Q,\Lambda) \equiv {1\over16\pi^2}\log{\Lambda^2\over Q^2} ={1\over8\pi^2}L.$$

For real scalar $\phi^4$ theory in four Euclidean dimensions, the one-loop logarithmic correction to the four-point coupling is written as

$$\Gamma_4(Q)=\lambda_0-a\lambda_0^2L+O(\lambda_0^2L^0), \qquad a={3\over16\pi^2}.$$

The coefficient $a$ includes the three $s,t,u$ channels and the bubble symmetry factor $1/2$. The sign convention is the same as on the previous page: $\Gamma_4(Q)$ is the low-energy coupling at scale $Q$ with the bare coupling fixed at $\Lambda$.

At $n$ loops, a typical contribution to a marginal coupling has the schematic form

$$\lambda_0^{n+1}\left(c_{n,n}L^n+c_{n,n-1}L^{n-1}+\cdots+c_{n,0}\right).$$

The leading-logarithmic approximation keeps only the highest power at each order:

$$\lambda_0^{n+1}L^n.$$

It is designed for the scaling regime

$$\lambda_0\ll1, \qquad \lambda_0 L\sim1.$$

In this regime,

$$\lambda_0^{n+1}L^n \sim \lambda_0,$$

so infinitely many loop orders can contribute at the same parametric size. Terms such as $\lambda_0^{n+1}L^{n-1}$ are smaller by one explicit power of $\lambda_0$ and belong to next-to-leading-log accuracy.

This is why constants inside logarithms are unimportant at leading-log order. For example,

$$\log{\Lambda\over 2Q} =\log{\Lambda\over Q}-\log 2.$$

The first term may be large; the second is an ordinary finite number. At one loop the difference between $\lambda_0^2\log(\Lambda/Q)$ and $\lambda_0^2\log(\Lambda/2Q)$ is only $O(\lambda_0^2)$, while the leading-log term is treated as $O(\lambda_0)$ when $\lambda_0L\sim1$.

The leading-log approximation is therefore not a statement about computing diagrams sloppily. It is a controlled asymptotic expansion in which logarithmic scale dependence is kept and finite matching constants are postponed.

Ordered logarithmic integrals

The simplest source of a logarithm is the scale-invariant integral

$$I_1(\Lambda,Q)=\int_Q^\Lambda {dk\over k}=L.$$

If two independent variables each ranged freely from $Q$ to $\Lambda$, the integral would give $L^2$. But loop momenta in a leading region are often not merely independent; they are ordered. For two ordered scales,

$$Q<k_2<k_1<\Lambda,$$

we find

$$I_2(\Lambda,Q) =\int_Q^\Lambda {dk_1\over k_1}\int_Q^{k_1}{dk_2\over k_2} ={1\over2}L^2.$$

The factor $1/2$ is the area of a triangle in logarithmic variables. Set

$$u_i=\log{k_i\over Q}, \qquad 0<u_i<L.$$

Then $k_1>k_2$ means $u_1>u_2$, and the ordered region is half of the square $0<u_1,u_2<L$.

Two logarithmic integrations become an area in the plane of logarithmic momenta. The strongly ordered region $Q<k_2<k_1<\Lambda$ occupies a triangle of area $L^2/2$, where $L=\log(\Lambda/Q)$.

The triangle is one leading region, not the whole two-loop integration domain. If the graph also permits the opposite hierarchy $k_2\gg k_1$, that is a second region; if the loop labels are related by a graph symmetry, its multiplicity must be included only once. Channel sums and insertion sites supply further combinatorial factors. The simplex volume $L^2/2$ therefore does not by itself determine the coefficient of a Feynman graph. It determines the logarithmic volume of one specified ordering.

For $n$ strongly ordered scales,

$$Q<k_n<k_{n-1}<\cdots<k_1<\Lambda,$$

one obtains the volume of an $n$-simplex:

$$\boxed{ \int_Q^\Lambda {dk_1\over k_1} \int_Q^{k_1}{dk_2\over k_2}\cdots \int_Q^{k_{n-1}}{dk_n\over k_n} ={L^n\over n!}. }$$

This factorial is not an accident. It is the same combinatorial structure that appears when a differential RG equation is integrated repeatedly. The RG equation will be the efficient way of summing these ordered regions.

Toy integrals and the loss of constants

It is useful to see how a logarithmic approximation forgets irrelevant constants. Consider

$$J_1(\Lambda)=\int_0^\Lambda {dx\over\sqrt{x^2+1}} =\operatorname{arsinh}\Lambda.$$

For large $\Lambda$,

$$J_1(\Lambda)=\log(\Lambda+\sqrt{\Lambda^2+1}) =\log(2\Lambda)+O(\Lambda^{-2}).$$

At leading-log accuracy this is simply $\log\Lambda$. The additive constant $\log2$ is finite and cannot be distinguished from other finite short-distance details.

Now consider the ordered two-variable version

$$J_2(\Lambda) =\int_0^\Lambda {dx_1\over\sqrt{x_1^2+1}} \int_0^{x_1}{dx_2\over\sqrt{x_2^2+1}}.$$

Since the inner integral is $\operatorname{arsinh}x_1$, we have

$$J_2(\Lambda) ={1\over2}\left(\operatorname{arsinh}\Lambda\right)^2 ={1\over2}\log^2\Lambda+\text{subleading logs and constants}.$$

The lesson is not tied to this particular integral. Whenever a loop region reduces to a scale-invariant measure $dk/k$, an ordered chain of such regions produces powers of logarithms. Leading logs remember only the large logarithmic volume; finite endpoints and smooth details are subleading.

Momentum shells in four dimensions

In four Euclidean dimensions,

$${d^4k\over(2\pi)^4}{1\over k^4} ={2\pi^2k^3dk\over(2\pi)^4}{1\over k^4} ={1\over8\pi^2}{dk\over k}$$

after angular integration. Thus a one-loop logarithmic subgraph behaves like a shell integral:

$$\int_{Q<|k|<\Lambda}{d^4k\over(2\pi)^4}{1\over k^4} ={1\over8\pi^2}\int_Q^\Lambda {dk\over k}.$$

For the one-loop four-point vertex in real $\phi^4$ theory, there are three channels and a symmetry factor $1/2$, so the logarithmic shell coefficient becomes

$$a={3\over16\pi^2}$$

when we use $L=\log(\Lambda/Q)$. This coefficient is the elementary building block for the leading logarithms of the four-point coupling.

At two loops and beyond, a large logarithm appears when at least one loop momentum can move through a wide range of scales while the rest of the integrand is approximately scale invariant. A double logarithm appears when two such scale variables can range over a two-dimensional logarithmic region. The crucial distinction is whether this two-dimensional region is broad in both directions or squeezed into a diagonal band.

Nested bubbles in φ⁴ theory

The most transparent two-loop leading-log region in $\phi^4$ theory is a bubble correction inserted into another bubble. Let $K$ be the hard loop momentum inside a subgraph and $k$ the softer loop momentum of the graph that contains it. The leading region is

$$\Lambda\gg K\gg k\gg Q.$$

The hard subgraph cannot resolve the external momenta of the softer graph. To the softer loop, it looks like a local correction to the four-point vertex. Thus the two-loop integral factorizes into two one-loop logarithmic integrations:

$$\int_Q^\Lambda {dk\over k} \int_k^\Lambda {dK\over K} ={1\over2}L^2.$$

In the region $K\gg k\gg Q$, the hard loop with momentum $K$ shrinks to a local four-point vertex for the softer loop with momentum $k$. This staged collapse is the diagrammatic origin of leading-log factorization.

To connect this with the coefficient $a=3/(16\pi^2)$, write the one-loop result schematically as

$$\Gamma_4(Q)=\lambda_0-a\lambda_0^2L+\cdots.$$

If a loop at scale $k$ uses a vertex already corrected by harder momenta $K>k$, then to leading-log accuracy the effective vertex entering that softer loop is

$$\Gamma_4(k)=\lambda_0-a\lambda_0^2\log{\Lambda\over k}+\cdots.$$

The one-loop shell correction from the softer scale is proportional to $-a\Gamma_4(k)^2\,d\log k$ in the natural logarithmic measure. Keeping terms through order $\lambda_0^3$ gives

$$-a\Gamma_4(k)^2 =-a\lambda_0^2+2a^2\lambda_0^3\log{\Lambda\over k}+\cdots.$$

Integrating from $Q$ to $\Lambda$ gives

$$\Gamma_4(Q) =\lambda_0-a\lambda_0^2L +2a^2\lambda_0^3\int_Q^\Lambda {dk\over k}\log{\Lambda\over k} +\cdots.$$

Since

$$\int_Q^\Lambda {dk\over k}\log{\Lambda\over k} ={1\over2}L^2,$$

we get

$$\Gamma_4(Q) =\lambda_0-a\lambda_0^2L+a^2\lambda_0^3L^2+\text{subleading terms}.$$

The two-loop leading logarithm is therefore determined by the one-loop logarithm. This is the first visible hint that the entire leading-log series is not new information at every order.

Locality at each stage

The factorization above relies on locality. A practical test is this: after the hard subgraph is Taylor-expanded in its external momenta, the leading term must have the form of an operator already present in the effective action, or of another allowed local operator. That local replacement is what lets a softer loop treat the hard subgraph as a corrected vertex.

When $K\gg k$, the hard bubble has characteristic size $1/K$, while the softer graph varies over distances of order $1/k$. Since

$${1\over K}\ll {1\over k},$$

the hard subgraph is effectively pointlike from the viewpoint of the softer graph.

In momentum space this means that the hard subgraph can be Taylor-expanded in the small external momenta flowing through it:

$$\mathcal A_\gamma(k_i;K) =C_0(K)+C_2(K){k_i^2\over K^2}+\cdots.$$

For leading logarithms of a marginal operator, only the local leading term $C_0(K)$ matters. Terms with extra powers of $k_i/K$ correspond to higher-derivative operators and do not contribute to the same leading logarithm of the original marginal coupling.

This is why the nested graph reduces to an iteration of the one-loop four-point correction: the hard bubble produces the same local operator $\phi^4$, and the softer loop then treats it as an ordinary vertex.

Subgraphs, quotient graphs, and forests

A subgraph $\gamma$ is a collection of vertices and internal lines of a Feynman graph. A subgraph is important for renormalization when it is superficially divergent and has the external-leg structure of an operator allowed in the effective action. In $\phi^4$ theory in four dimensions, the central example is the logarithmically divergent four-point subgraph.

If $\gamma$ is much harder than the rest of the graph, it can be collapsed to a point. The resulting graph is called the quotient graph and is denoted $G/\gamma$. Symbolically,

$$G\quad\longrightarrow\quad (\text{local coefficient from }\gamma)\times (G/\gamma).$$

A collection of divergent subgraphs that are mutually disjoint or nested is called a forest. Leading logarithms are associated with forests of logarithmic subgraphs. For example, a chain

$$\gamma_1\subset\gamma_2\subset\cdots\subset G$$

corresponds to ordered scales

$$\Lambda\gg k_1\gg k_2\gg\cdots\gg Q.$$

A logarithmic subgraph nested inside an overall logarithmic graph supplies two ordered scale variables and can generate $L^2$. A primitive logarithmic graph, with no logarithmic subdivergence, has only one overall scaling variable and therefore contributes a single logarithm.

The word “forest” is more than terminology. It encodes the reason leading logs are recursively computable. Each hard subgraph in a forest is replaced by a local operator before the next softer integration is performed. Nonlocal dependence on external momenta is suppressed by powers of scale ratios and belongs to subleading terms.

Subtracting a nested subdivergence

The same locality that permits factorization also dictates the counterterm subtraction. Let $G$ contain a proper logarithmically divergent four-point subgraph $\gamma$. The operator $T_\gamma$ extracts the momentum-independent local part of $\gamma$—the coefficient with the same external-leg structure as $\phi^4$. Define its counterterm recursively by

$$C(\gamma)=-T_\gamma\,\overline R\,\gamma.$$

Here $\overline R\,\gamma$ means that any subdivergences inside $\gamma$ have already been removed. For a graph with one proper nested subdivergence, the incomplete subtraction is

$$\overline R\,G =G+C(\gamma){G\over\gamma},$$

where $G/\gamma$ is the quotient graph obtained by contracting $\gamma$ to a local vertex. Only after this inner subtraction do we remove the overall local divergence:

$$R G=(1-T_G)\,\overline R\,G, \qquad C(G)=-T_G\,\overline R\,G.$$

The order matters. Subtracting the overall graph first would leave the ultraviolet divergence of $\gamma$ hidden inside it. For a longer chain, one proceeds from the innermost hard subgraph outward. Mutually nested or disjoint subgraphs may occur together in a forest; genuinely overlapping subgraphs cannot occur together in the same forest.

There are two equivalent ways to use this structure at leading-log accuracy. In a renormalized graph, the local counterterm cancels the cutoff-dependent part of the hard subgraph, leaving a logarithm of physical or subtraction scales. In a Wilsonian calculation, the hard shell is integrated out and its local contribution is absorbed into the running vertex $\Gamma_4(k)$. The shell equation on the next page uses the second language. In both descriptions, the hard contribution is inserted once in $G/\gamma$; this is what prevents double counting of the nested region.

Why primitive graphs give fewer logs

A primitive divergent graph is divergent only when all its loop momenta become large together. It has no divergent proper subgraph. Such a graph can produce a logarithm, but it cannot produce the highest possible power of logarithms.

The reason is simple in logarithmic variables. Suppose a two-loop primitive graph is logarithmically divergent overall. Introduce an overall scale $\rho$ and dimensionless ratios $r$:

$$k=\rho\,\hat k(r), \qquad p=\rho\,\hat p(r).$$

The logarithmic divergence comes from

$$\int^\Lambda {d\rho\over\rho}.$$

If the ratio integral over $r$ is finite, then there is only one large logarithm. A second logarithm would require the ratio integral itself to become logarithmically singular, for example when $k/p\to0$ or $p/k\to0$. But precisely such a singular ratio limit is the signal that a proper subgraph has become logarithmically divergent. In other words:

$$\text{extra logarithm} \quad\Longleftrightarrow\quad \text{subdivergent scale hierarchy}.$$

A useful toy model makes the distinction sharp. Let

$$I_{\text{diag}}(L)=\int_0^L du\int_0^L dv\,{1\over\cosh^2(u-v)}.$$

The kernel is broad neither in $u$ nor in $v$ independently; it is concentrated near the diagonal $u=v$. For large $L$,

$$I_{\text{diag}}(L) =2L+O(1),$$

not $L^2$. By contrast, the ordered region $0<v<u<L$ has area $L^2/2$. The first integral represents a graph with one overall logarithmic scale; the second represents two separated scales.

This explains the practical rule: complicated-looking diagrams often do not contribute to leading logs. If their loop momenta are tied together by the denominator structure and there is no chain of logarithmic subgraphs, they give at most lower powers of $L$.

This is one of the most useful calculation rules: leading logs are usually easier than full loop integrals because they ask only for scale-separated local limits. The full graph knows about constants, thresholds, tensor numerators, and regulator details; the leading log asks for the part that survives when each hard subgraph is shrunk to a point.

Overlapping subgraphs and leading logs

Not all divergent subgraphs are neatly nested. Some overlap: they share internal lines but neither contains the other. Overlapping divergences are important for full renormalization, but they are not the basic source of a simple ordered leading-log chain.

At leading-log accuracy, the dominant regions can still be described by choosing scale hierarchies. A particular hierarchy may select one subgraph as hard and local, while another hierarchy selects a different one. Each allowed region corresponds to a forest of compatible subgraphs, and the overlap subtractions organize their sum without counting a shared region twice. This is the intuition behind the forest formula in perturbative renormalization.

For the present course, the essential point is less formal:

$$\text{leading logarithms come from repeated local renormalization of subgraphs.}$$

Once a hard subgraph has been replaced by its local counterterm or local effective vertex, the remaining softer graph has the same form as a lower-loop problem. This recursive structure is why the renormalization-group equation will be first order in the scale.

Analytic continuation and cuts

There is another way to recognize the coefficient of a logarithm. In Euclidean calculations the answer contains terms such as

$$\log{\Lambda^2\over Q^2}.$$

For a Minkowski scattering invariant $s$, the corresponding analytic continuation often produces

$$\log{\Lambda^2\over -s-i0}.$$

For $s>0$,

$$\log(-s-i0)=\log s-i\pi,$$

so

$$\log{\Lambda^2\over -s-i0} =\log{\Lambda^2\over s}+i\pi.$$

With the discontinuity convention

$$\operatorname{Disc}_s F \equiv F(s+i0)-F(s-i0),$$

the logarithm obeys

$$\operatorname{Disc}_s\log(-s)=-2\pi i.$$

After analytic continuation, a Euclidean logarithm becomes a logarithm with a branch cut in the physical scattering variable. The imaginary part across the cut is fixed by on-shell intermediate states.

The imaginary part is not arbitrary. By unitarity, the discontinuity across a physical cut is determined by products of lower-order amplitudes. Schematically,

$$2\operatorname{Im}\mathcal A =\sum_X \int d\Pi_X\,\mathcal A(i\to X)\mathcal A^*(f\to X).$$

Thus the coefficient of the logarithm is related to a lower-order on-shell process. This gives a physical interpretation of leading-log recursion: the logarithmic scale dependence is tied to the repeated opening of lower-order processes across scale-separated momentum regions.

This unitarity viewpoint is especially useful in scattering problems, but the Wilsonian viewpoint is more general. Whether one reads the coefficient from a hard Euclidean subgraph or from a cut in Minkowski space, the result is the same local data that enters the RG equation.

Example: the first three terms of the leading-log series

Let

$$a={3\over16\pi^2}, \qquad L=\log{\Lambda\over Q}.$$

The one-loop result is

$$\Gamma_4(Q)=\lambda_0-a\lambda_0^2L+\text{subleading terms}.$$

The leading two-loop term is generated by inserting the one-loop corrected vertex into the one-loop shell calculation, giving

$$+a^2\lambda_0^3L^2.$$

Repeating once more gives

$$-a^3\lambda_0^4L^3.$$

Thus the first leading-log terms organize as

$$\Gamma_4(Q) =\lambda_0 -a\lambda_0^2L +a^2\lambda_0^3L^2 -a^3\lambda_0^4L^3+\cdots,$$

which is the expansion of

$$\boxed{ \Gamma_4(Q) ={\lambda_0\over 1+a\lambda_0 L} }$$

at leading-log accuracy. The next page derives this result directly as a differential equation in the scale. For now, the important lesson is diagrammatic: the powers of $L$ are produced by nested local subgraphs, not by mysterious new ultraviolet structures at every loop order.

Summary

A leading logarithm at $n$ loops has the form

$$\lambda_0^{n+1}L^n, \qquad L=\log{\Lambda\over Q}.$$

Such terms must be resummed when $\lambda_0\ll1$ but $\lambda_0L\sim1$.

A single logarithm comes from a scale-invariant shell integral $dk/k$. Multiple logarithms come from multiple scale variables. If the variables are strongly ordered,

$$Q<k_n<\cdots<k_1<\Lambda,$$

the logarithmic volume is

$${L^n\over n!}.$$

In Feynman diagrams, the ordered regions correspond to nested logarithmically divergent subgraphs. A hard subgraph shrinks to a local vertex, and the softer graph uses that local vertex. This is the physical origin of leading-log factorization.

Primitive graphs without logarithmic subdivergences have only one overall scale integration and therefore produce at most one logarithm. Extra logarithms arise only from singular ratio limits, which are precisely subdivergent scale hierarchies.

For $\phi^4$ theory, the one-loop coefficient

$$a={3\over16\pi^2}$$

recursively generates the leading-log series

$$\Gamma_4(Q)=\lambda_0-a\lambda_0^2L+a^2\lambda_0^3L^2-\cdots.$$

This recursion is the diagrammatic form of the RG flow derived on the next page.

Common pitfalls

Loop order is not logarithmic power. A two-loop graph can give $L^2$, $L$, or no large logarithm, depending on its divergent subgraphs and momentum regions.

Comparable momenta do not give a nested double logarithm. The region $K\sim k\gg Q$ gives one logarithm from the common overall scale. A second logarithm appears only when a ratio such as $K/k$ can itself range logarithmically.

Do not read a diagram coefficient from $L^n/n!$ alone. The simplex is one ordered region. Include every allowed ordering, insertion site, channel, and graph symmetry exactly once.

Finite constants are not leading logarithms. Constants inside logarithms are matching data and enter at lower logarithmic accuracy.

Local contraction requires scale separation. Replace a hard subgraph by a local vertex only when its external momenta are much smaller than its internal momenta.

Subdivergences are subtracted from the inside out. Remove proper nested subdivergences before the overall divergence. Overlapping subgraphs require separate compatible forests and cannot be subtracted simultaneously as though they were nested.

Exercises

Exercise 1 — The logarithmic simplex

Show that

$$I_n(L)=\int_0^L du_1\int_0^{u_1}du_2\cdots\int_0^{u_{n-1}}du_n$$

is equal to $L^n/n!$.

Solution

The integral is the volume of the simplex

$$0<u_n<u_{n-1}<\cdots<u_1<L$$

inside the $n$-dimensional cube $0<u_i<L$. The cube can be partitioned into $n!$ equal regions according to the ordering of the $u_i$. Each region has the same volume by symmetry. Since the cube volume is $L^n$, the ordered region has volume

$$I_n(L)={L^n\over n!}.$$

Equivalently, one can prove it recursively. Since

$$I_n(L)=\int_0^L du_1\,I_{n-1}(u_1),$$

and $I_{n-1}(u_1)=u_1^{n-1}/(n-1)!$, one gets

$$I_n(L)=\int_0^L du_1\,{u_1^{n-1}\over(n-1)!} ={L^n\over n!}.$$

Exercise 2 — A nested two-scale integral

Let

$$L=\log{\Lambda\over Q}.$$

Evaluate the leading logarithmic integral

$$I=\int_Q^\Lambda {dk\over k}\int_k^\Lambda {dK\over K}.$$

Interpret the result as the logarithmic volume of the region $\Lambda>K>k>Q$.

Solution

The inner integral is

$$\int_k^\Lambda {dK\over K}=\log{\Lambda\over k}.$$

Therefore

$$I=\int_Q^\Lambda {dk\over k}\log{\Lambda\over k}.$$

Set

$$u=\log{\Lambda\over k}, \qquad {dk\over k}=-du.$$

When $k=Q$, $u=L$; when $k=\Lambda$, $u=0$. Thus

$$I=\int_0^L du\,u={1\over2}L^2.$$

In variables $x=\log(K/Q)$ and $y=\log(k/Q)$, the region is

$$0<y<x<L,$$

which is a triangle of area $L^2/2$.

Exercise 3 — Why a diagonal band gives one logarithm

Consider the “diagonal-band” toy integral

$$I_{\mathrm{diag}}(L)=\int_0^Ldu\int_0^Ldv\,{1\over\cosh^2(u-v)}.$$

Show that for $L\gg1$,

$$I_{\mathrm{diag}}(L)=2L+O(1).$$

Why does this model produce one logarithm rather than two?

Solution

Use variables

$$x=u-v.$$

For a fixed $x\in[-L,L]$, the length of the diagonal segment inside the square is $L-|x|$. Therefore

$$I_{\mathrm{diag}}(L) =\int_{-L}^{L}dx\,{L-|x|\over\cosh^2x}.$$

For $L\gg1$, the kernel $1/\cosh^2x$ is localized near $x=0$ with width of order one. Thus

$$I_{\mathrm{diag}}(L) =L\int_{-\infty}^{\infty}{dx\over\cosh^2x}+O(1).$$

Since

$$\int_{-\infty}^{\infty}{dx\over\cosh^2x} =\tanh x\bigg|_{-\infty}^{\infty}=2,$$

we get

$$I_{\mathrm{diag}}(L)=2L+O(1).$$

The integral does not scale like $L^2$ because the kernel forces $u$ and $v$ to remain within $O(1)$ of each other. The allowed region is a band of width $O(1)$ and length $L$, not a two-dimensional region of area $L^2$. In diagram language, this corresponds to a primitive graph with one overall scale but no logarithmically wide hierarchy between two loop momenta.

Exercise 4 — Iterating the corrected local vertex

Starting from

$$\Gamma_4(k)=\lambda_0-a\lambda_0^2\log{\Lambda\over k}+O(\lambda_0^3),$$

show that inserting this running local vertex into the one-loop shell correction

$$\Gamma_4(Q)=\lambda_0-a\int_Q^\Lambda {dk\over k}\,\Gamma_4(k)^2+\text{subleading terms}$$

gives

$$\Gamma_4(Q)=\lambda_0-a\lambda_0^2L+a^2\lambda_0^3L^2+\text{subleading terms}.$$

Solution

Square the effective vertex through the required order:

$$\Gamma_4(k)^2 =\lambda_0^2-2a\lambda_0^3\log{\Lambda\over k}+O(\lambda_0^4).$$

Substitute into the shell integral:

$$-a\int_Q^\Lambda {dk\over k}\Gamma_4(k)^2 =-a\lambda_0^2\int_Q^\Lambda {dk\over k} +2a^2\lambda_0^3\int_Q^\Lambda {dk\over k}\log{\Lambda\over k} +\cdots.$$

The first integral is

$$\int_Q^\Lambda {dk\over k}=L,$$

and the second is

$$\int_Q^\Lambda {dk\over k}\log{\Lambda\over k}={1\over2}L^2.$$

Therefore

$$\Gamma_4(Q)=\lambda_0-a\lambda_0^2L+a^2\lambda_0^3L^2+\text{subleading terms}.$$

The factor $2$ from expanding $\Gamma_4(k)^2$ cancels the ordered-integration factor $1/2$.

Exercise 5 — The physical branch of the logarithm

For $s>0$, prove that

$$\log(-s-i0)=\log s-i\pi.$$

Then find the imaginary part of

$$\mathcal A(s)=\lambda_0+c\lambda_0^2\log{\Lambda^2\over -s-i0}$$

for real $c$.

Solution

The point $-s-i0$ lies just below the negative real axis. Its polar form is

$$-s-i0=s e^{-i\pi}.$$

On the principal branch,

$$\log(-s-i0)=\log s-i\pi.$$

Thus

$$\log{\Lambda^2\over -s-i0} =\log\Lambda^2-\log(-s-i0) =\log{\Lambda^2\over s}+i\pi.$$

Therefore

$$\operatorname{Im}\mathcal A(s)=\pi c\lambda_0^2.$$

The imaginary part is fixed by the coefficient of the logarithm. In a unitary scattering theory, the same imaginary part is computed from on-shell intermediate states, which is why cuts can determine logarithmic coefficients.

Exercise 6 — All-order leading-log coefficients

Suppose the leading-log running coupling has the form

$$\Gamma(Q)={\lambda_0\over1+a\lambda_0L}, \qquad L=\log{\Lambda\over Q}.$$

Show that the coefficient of $\lambda_0^{n+1}L^n$ is $(-a)^n$. Explain why no finite one-loop constant can change this coefficient.

Solution

Expand the denominator as a geometric series:

$${1\over1+a\lambda_0L} =\sum_{n=0}^\infty (-a\lambda_0L)^n.$$

Multiplying by $\lambda_0$ gives

$$\Gamma(Q)=\sum_{n=0}^\infty (-a)^n\lambda_0^{n+1}L^n.$$

Thus the coefficient of $\lambda_0^{n+1}L^n$ is $(-a)^n$.

Now suppose the one-loop logarithm were replaced by

$$L+c,$$

where $c$ is finite. At order $n$, powers of $(L+c)^n$ include

$$L^n+n c L^{n-1}+\cdots.$$

The coefficient of the highest power $L^n$ is unchanged. Finite constants affect next-to-leading logs and matching terms, not the leading-log coefficient.

Further reading

• Coleman, Sidney. Lectures of Sidney Coleman on Quantum Field Theory. Edited by Bryan Gin-ge Chen, David Derbes, David Griffiths, Brian Hill, Richard Sohn, and Yuan-Sen Ting. World Scientific, 2019.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007.
• Weinberg, Steven. The Quantum Theory of Fields, Volume II: Modern Applications. Cambridge University Press, 1996.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 4th ed. Oxford University Press, 2002.