Scalar Propagators and One-Loop φ⁴ Theory

The previous page developed the standard one-loop tools: Wick rotation, Feynman parameters, Schwinger parameters, and the logarithmic scalar bubble. We now put those tools to work in the simplest relativistic field theory with an interacting coupling: real scalar $\phi^4$ theory in four spacetime dimensions.

There are two lessons. First, the scalar propagator is a precise way of packaging the oscillator physics of each momentum mode. The $i0$ prescription selects the vacuum time ordering, while an occupied state adds extra on-shell pieces. Second, the one-loop correction to the four-point vertex is logarithmically ultraviolet sensitive. Its divergent part is independent of the external momenta, so it is absorbed by the same local operator $\phi^4$ that was already present in the Lagrangian.

The key calculation is

$${1\over2}\lambda_0^2 \int {d^4k\over(2\pi)^4} {1\over (k^2+m^2)[(q-k)^2+m^2]} \sim {\lambda_0^2\over2}\,{1\over16\pi^2}\log{\Lambda^2\over Q^2},$$

for each of the three four-point channels. Here $Q$ denotes the external momentum or mass scale that cuts off the infrared end of the logarithm. This is the first appearance of the coefficient that the renormalization group will organize and resum.

Required background. Euclidean Loop Integrals and Feynman Parameters supplies the bubble integral and its logarithmic coefficient. Scalar propagators, ordered correlators, and sources supplies the free-vacuum distribution and pole prescription used below.

Helpful background. Källén–Lehmann representation explains how the free pole generalizes to stable-particle poles and multiparticle continua in an interacting vacuum. Two propagators must be distinguished below: the vacuum Feynman propagator packages virtual off-shell fluctuations, whereas an occupied-state propagator also contains on-shell terms from real quanta in the state. Those medium terms can alter infrared physics, but the vacuum UV counterterm is fixed by the large off-shell-momentum region.

Scalar modes and the Feynman propagator

A free real scalar field is an infinite collection of harmonic oscillators, one for each spatial momentum. On this page

$$D_F(x-y)=\langle0|T\{\phi(x)\phi(y)\}|0\rangle,$$

so its momentum-space representation includes the numerator $i$. For a single oscillator with frequency $\omega$, the time-ordered two-point function in the vacuum is

$$D_F(t) =\langle0|Tq(t)q(0)|0\rangle ={1\over2\omega}e^{-i\omega |t|}.$$

For a field mode with momentum $\mathbf k$, the frequency is

$$\omega_{\mathbf k}=\sqrt{\mathbf k^2+m^2}.$$

The fixed-$\mathbf k$ propagator is therefore

$$D_F(\mathbf k;t) =\int {dk^0\over2\pi}\,{i e^{-ik^0t}\over (k^0)^2-\omega_{\mathbf k}^2+i0} ={1\over2\omega_{\mathbf k}}e^{-i\omega_{\mathbf k}|t|}.$$

The $i0$ prescription specifies where the poles sit:

$$k^0=+\omega_{\mathbf k}-i0, \qquad k^0=-\omega_{\mathbf k}+i0.$$

For $t>0$, the factor $e^{-ik^0t}$ damps the large semicircle in the lower half-plane, so the contour encloses the positive-energy pole. For $t<0$, the contour closes above and encloses the negative-energy pole. This is the frequency-space origin of the absolute value $|t|$.

The Feynman prescription places the positive-energy pole below the real $k^0$ axis and the negative-energy pole above it. Closing the contour below for $t>0$ and above for $t<0$ gives $D_F(\mathbf k;t)=e^{-i\omega_{\mathbf k}|t|}/(2\omega_{\mathbf k})$.

Fourier transforming the spatial momentum gives the vacuum scalar propagator in mixed form:

$$D_F(t,\mathbf x) =\int {d^3\mathbf k\over(2\pi)^3} {1\over2\omega_{\mathbf k}} \exp\left(i\mathbf k\cdot\mathbf x-i\omega_{\mathbf k}|t|\right).$$

This formula is often more physically transparent than the fully covariant one: each momentum mode propagates like an oscillator with frequency $\omega_{\mathbf k}$, and the vacuum time ordering is encoded by $|t|$.

Propagators in occupied states

For a free mode, the vacuum is the state with occupation number $N=0$ in the particle basis selected by positive frequency. If instead a single oscillator is in a number eigenstate $|N\rangle$, then

$$q(t)={1\over\sqrt{2\omega}}\left(ae^{-i\omega t}+a^\dagger e^{i\omega t}\right), \qquad [a,a^\dagger]=1,$$

and

$$\langle N|a a^\dagger|N\rangle=N+1, \qquad \langle N|a^\dagger a|N\rangle=N.$$

The time-ordered propagator becomes

$$D_N(t) ={1\over2\omega}\left[(N+1)e^{-i\omega |t|}+N e^{+i\omega |t|}\right].$$

The first term is the vacuum-like propagation of a quantum inserted into the state. The second term is possible because the state already contains quanta that can be removed and later replaced. In frequency space,

$$\boxed{ D_N(k^0) =(N+1){i\over(k^0)^2-\omega^2+i0} -N{i\over(k^0)^2-\omega^2-i0}. }$$

Equivalently,

$$D_N(k^0)=D_0(k^0)+2\pi N\,\delta((k^0)^2-\omega^2).$$

The difference between the occupied-state propagator and the vacuum propagator is purely on shell. This fact reappears in finite-temperature and finite-density field theory: a medium modifies propagators by adding terms that know about real particles present in the state. Vacuum ultraviolet divergences, by contrast, come from virtual momenta far off shell and are local.

For a scalar field, one replaces $\omega$ by $\omega_{\mathbf k}$ mode by mode. A Lorentz-invariant vacuum has $N=0$ for every $\mathbf k$. A thermal bath or many-body state has nonzero occupation numbers and therefore chooses a preferred rest frame.

The distinction also explains why renormalization constants are usually fixed in the vacuum theory. The extra term

$$2\pi N\delta((k^0)^2-\omega_{\mathbf k}^2)$$

forces a line on shell. It changes phase space and thermal response, but it does not produce the same large Euclidean four-momentum behavior as the vacuum loop. Vacuum UV divergences are local and state independent.

Vacuum spectral structure

The single free-particle pole is the simplest possible vacuum spectrum. In an interacting Poincaré-invariant vacuum, positivity and completeness organize the exact two-point function as

$$\boxed{ D_F(p)=\int_0^\infty d\mu^2\,\rho(\mu^2) {i\over p^2-\mu^2+i0}, \qquad \rho(\mu^2)\ge0. }$$

A stable one-particle state contributes $Z\delta(\mu^2-m_{\mathrm{phys}}^2)$, while multiparticle states contribute a continuum beginning at the relevant threshold. The same Feynman boundary prescription accompanies every spectral mass.

This vacuum spectral density should not be confused with the extra on-shell term in $D_N$. The former describes which invariant-mass states the field can create from the vacuum; the latter describes real quanta already occupying a chosen state and generally selects a preferred frame.

Short-distance behavior of the free scalar field

At equal time and at distances much shorter than the Compton wavelength $m^{-1}$, the mass is negligible. The vacuum two-point function becomes

$$D_F(0,\mathbf r) \simeq \int {d^3\mathbf k\over(2\pi)^3} {e^{i\mathbf k\cdot\mathbf r}\over2|\mathbf k|}.$$

Using spherical coordinates around $\mathbf r$, one has

$$\int d\Omega\,e^{i\mathbf k\cdot\mathbf r} =4\pi{\sin kr\over kr}, \qquad r=|\mathbf r|,$$

so

$$D_F(0,\mathbf r) \simeq {1\over(2\pi)^3}{4\pi\over2r} \int_0^\infty dk\,\sin(kr).$$

With the usual convergence factor, $\int_0^\infty dk\,\sin(kr)=1/r$. Hence

$$\boxed{ D_F(0,\mathbf r)\simeq {1\over4\pi^2 r^2} \qquad (mr\ll1). }$$

The same result is obtained from the Euclidean propagator,

$$G_E(x)=\int {d^4k\over(2\pi)^4}{e^{ik\cdot x}\over k^2+m^2} ={m\over4\pi^2 |x|}K_1(m|x|),$$

because $K_1(z)\sim1/z$ for $z\to0$:

$$G_E(x)\sim {1\over4\pi^2x^2}.$$

This $1/x^2$ singularity is the coordinate-space statement that a scalar field in four dimensions has engineering dimension one:

$$[\phi]=1.$$

It also explains why $\phi^4$ is marginal by power counting in four dimensions. The product of two propagators behaves as

$$G_E(x)^2\sim {1\over16\pi^4x^4},$$

and the four-dimensional volume element is $d^4x\sim r^3dr$. Thus a short-distance loop integral contains

$$\int {r^3dr\over r^4}=\int {dr\over r},$$

a logarithm. The momentum-space calculation below is the same physics written in the language of Feynman diagrams.

φ⁴ theory and the one-loop vertex

Consider Euclidean real scalar theory in four dimensions,

$$S_E[\phi]=\int d^4x\left[ {1\over2}(\partial_\mu\phi)^2+{1\over2}m^2\phi^2+{\lambda_0\over4!}\phi^4 \right],$$

with ultraviolet cutoff $\Lambda$. The Euclidean propagator is

$$G_E(k)={1\over k^2+m^2}.$$

To fix the sign once and for all, define $\Gamma_4$ as the coefficient of $\varphi^4/4!$ in the Euclidean 1PI effective action $\Gamma[\varphi]$. Thus $\Gamma_4=\lambda_0$ at tree level. We first isolate the positive integral multiplying one bubble channel; its symmetry factor is $1/2$:

$$\mathcal B(q) ={\lambda_0^2\over2} \int {d^4k\over(2\pi)^4} {1\over (k^2+m^2)[(q-k)^2+m^2]}.$$

There are three ways to pair four external legs into a bubble channel: $s$, $t$, and $u$. Therefore the total one-loop logarithmic weight is three times the channel answer.

The sign follows directly from the one-loop effective action. With $G_E^{-1}=-\partial^2+m^2$,

$$\begin{aligned} \Gamma_1[\varphi] &={1\over2}\operatorname{Tr}\log \left(G_E^{-1}+{\lambda_0\over2}\varphi^2\right) \\ &={1\over2}\operatorname{Tr}\log G_E^{-1} +{\lambda_0\over4}\operatorname{Tr}(G_E\varphi^2) -{\lambda_0^2\over16} \operatorname{Tr}(G_E\varphi^2G_E\varphi^2)+\cdots. \end{aligned}$$

Four derivatives of the last term produce the three pairings and give $-\lambda_0^2/2$ times the bubble integral in each channel. The factor $1/2$ and the minus sign therefore have different origins: the former is the bubble symmetry factor, while the latter is the quadratic term in the logarithm.

The positive bubble weight is the sum of the $s$, $t$, and $u$ channels, each with symmetry factor $1/2$. For $\Gamma_4$ defined as the Euclidean effective-action coefficient, this total weight is subtracted from the tree coupling.

For the ultraviolet logarithm, assume all external momenta and the mass are of order $Q$, with

$$Q\ll |k|\ll\Lambda.$$

Then

$$(k^2+m^2)[(q-k)^2+m^2] =k^4\left[1+O(Q/k)\right],$$

and the leading logarithmic region gives

$$\mathcal B(q)_{\log} ={\lambda_0^2\over2} \int_{Q<|k|<\Lambda}{d^4k\over(2\pi)^4}{1\over k^4} ={\lambda_0^2\over2}\ell(Q,\Lambda).$$

Using

$$d^4k=2\pi^2 k^3dk,$$

we get

$$\ell(Q,\Lambda) ={1\over(2\pi)^4}2\pi^2\int_Q^\Lambda {k^3dk\over k^4} ={1\over8\pi^2}\log{\Lambda\over Q} ={1\over16\pi^2}\log{\Lambda^2\over Q^2}.$$

Thus the logarithmic contribution from all three channels is

$${3\over2}\lambda_0^2\ell(Q,\Lambda) ={3\lambda_0^2\over32\pi^2}\log{\Lambda^2\over Q^2}.$$

In the 1PI effective-action convention just defined, the low-energy four-point coupling has the form

$$\boxed{ \Gamma_4(Q) =\lambda_0 -{3\over2}\lambda_0^2\ell(Q,\Lambda) +\text{subleading terms}. }$$

If one instead calls the Euclidean diagrammatic vertex factor itself the vertex, it is $-\Gamma_4$ and every displayed overall sign reverses. The definition above matches the running coupling on the following lessons; within that definition the minus sign is not optional. The next page turns this coefficient into a differential RG equation.

Why the divergence is local

The leading ultraviolet term does not depend on the external momenta. That is the decisive fact. In the hard region,

$${1\over (k^2+m^2)[(q-k)^2+m^2]} ={1\over k^4}+\text{terms suppressed by }Q/k.$$

After angular integration, the terms odd in $q\cdot k$ vanish. The remaining correction is at worst $O((m^2+q^2)/k^6)$ and is ultraviolet convergent:

$$\int^\Lambda d^4k\,{q^2\over k^6} \sim q^2\int^\Lambda {dk\over k^3},$$

which is ultraviolet finite as $\Lambda\to\infty$. The only logarithmic part is therefore independent of external momentum. In position space this says that the short-distance singularity occurs when the two interaction points collide, leaving a local four-field operator at the collision point.

When the loop momentum is much larger than all external scales, angular averaging leaves $1/k^4$ plus UV-convergent corrections. The logarithmic part has the same local four-leg structure as the original $\phi^4$ vertex and is absorbed into a local counterterm.

This locality is the practical reason renormalization works. The ultraviolet divergence is not an arbitrary nonlocal function of the external momenta. It is proportional to an operator already present in the low-energy effective action:

$$\Delta S_E=\int d^4x\,{\delta\lambda\over4!}\phi^4.$$

Renormalization means choosing $\delta\lambda$ so that physical low-energy quantities are insensitive to the artificial cutoff $\Lambda$.

A useful normalization check is obtained by differentiating only the leading logarithm. With

$$\Gamma_4(Q)=\lambda_0-{3\lambda_0^2\over16\pi^2}\log{\Lambda\over Q}+\cdots,$$

we find

$$Q{d\Gamma_4\over dQ}={3\lambda_0^2\over16\pi^2}+\cdots.$$

Replacing $\lambda_0$ by the running coupling inside the right-hand side gives the RG equation of page 06. This quick check is often the safest way to remember the sign: for positive $\lambda$, the coupling decreases toward the infrared and increases toward the ultraviolet.

Leading-log bookkeeping

The bubble integral is not just divergent; it is the first term in a structured series. At $n$ loops, the leading logarithms in the four-point coupling have the schematic form

$$\lambda_0(\lambda_0\ell)^n, \qquad \ell={1\over16\pi^2}\log{\Lambda^2\over Q^2}.$$

If $\lambda_0\ll1$ but $\lambda_0\ell$ is not small, then all powers of $\lambda_0\ell$ must be kept. Terms with fewer logarithms, such as

$$\lambda_0^{n+1}\ell^{n-1},$$

are subleading in this approximation.

This is why constants inside logarithms do not matter at leading-log accuracy. For example,

$$\log{\Lambda^2\over4Q^2} =\log{\Lambda^2\over Q^2}-\log4.$$

The first term can be large; the second is an ordinary constant. In a one-loop correction, the large part scales as $\lambda_0^2\ell$, while the constant part scales as $\lambda_0^2$. If $\lambda_0\ell\sim1$, the logarithmic correction is of order $\lambda_0$, but the constant correction is still of order $\lambda_0^2$ and is dropped at leading-log accuracy.

The next page will explain why the leading logarithms are not independent miracles. They are generated recursively by nested momentum regions and are summed by the renormalization-group equation.

Summary

The scalar Feynman propagator is the oscillator propagator of each momentum mode:

$$D_F(\mathbf k;t)={1\over2\omega_{\mathbf k}}e^{-i\omega_{\mathbf k}|t|}.$$

The $i0$ prescription chooses the vacuum time ordering by placing the positive-energy pole below and the negative-energy pole above the real $k^0$ axis.

An occupied state modifies the propagator by an on-shell term:

$$D_N(k^0)=D_0(k^0)+2\pi N\delta((k^0)^2-\omega^2).$$

An interacting vacuum instead replaces the free spectral delta function by a positive spectral density containing stable-particle poles and multiparticle continua.

At short distances, a four-dimensional scalar propagator behaves as

$$G_E(x)\sim {1\over4\pi^2x^2}.$$

Therefore a one-loop bubble in $\phi^4$ theory has a logarithmic short-distance singularity. In momentum space,

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

The one-loop four-point correction contains three bubble channels and a symmetry factor $1/2$ per channel. The universal leading logarithmic coefficient is therefore

$${3\over2}\lambda_0^2\ell(Q,\Lambda).$$

For the coefficient $\Gamma_4$ of $\varphi^4/4!$ in the Euclidean 1PI effective action,

$$\Gamma_4(Q)=\lambda_0-{3\over2}\lambda_0^2\ell(Q,\Lambda)+\cdots.$$

Because the logarithmic hard-loop contribution is independent of external momenta, it is local and is absorbed into the $\phi^4$ coupling.

Common pitfalls

Do not confuse the Feynman propagator with a causal response function. The Feynman propagator is time ordered and is generally nonzero at spacelike separation. Causality is expressed by commutators or retarded functions, not by $D_F$ itself.

The $N$ in the occupied-state propagator is an occupation number, not a cutoff or a loop order. Its contribution is on shell and state dependent.

The factor $3/2$ in the one-loop $\phi^4$ vertex is two ingredients multiplied together: three channels and a symmetry factor $1/2$ for the two identical internal lines in each bubble.

Do not change signs midway through a calculation. On this page $\Gamma_4$ is the coefficient of $\varphi^4/4!$ in the Euclidean 1PI effective action, so its bubble correction is negative. The Euclidean diagrammatic vertex factor is $-\Gamma_4$ and therefore has the opposite displayed sign.

A logarithm such as $\log(\Lambda^2/4Q^2)$ has the same leading-log content as $\log(\Lambda^2/Q^2)$. The difference is a finite constant and belongs to subleading accuracy.

A hard cutoff can obscure momentum-shift invariance in power-divergent terms. The logarithmic coefficient of the $\phi^4$ bubble is universal, but finite constants and power divergences are regulator dependent.

Exercises

Exercise 1: Vacuum contour integral

Evaluate

$$D_F(\mathbf k;t)=\int {dk^0\over2\pi}\,{i e^{-ik^0t}\over(k^0)^2-\omega_{\mathbf k}^2+i0}$$

by contour integration and show that

$$D_F(\mathbf k;t)={1\over2\omega_{\mathbf k}}e^{-i\omega_{\mathbf k}|t|}.$$

Solution

The poles are

$$k^0=\omega_{\mathbf k}-i0, \qquad k^0=-\omega_{\mathbf k}+i0.$$

For $t>0$, close the contour in the lower half-plane. The orientation is clockwise, so

$$D_F(\mathbf k;t) =-2\pi i\,{1\over2\pi}\operatorname{Res}_{k^0=\omega_{\mathbf k}} \left[{i e^{-ik^0t}\over(k^0)^2-\omega_{\mathbf k}^2+i0}\right].$$

The residue is

$${i e^{-i\omega_{\mathbf k}t}\over2\omega_{\mathbf k}},$$

so

$$D_F(\mathbf k;t)={e^{-i\omega_{\mathbf k}t}\over2\omega_{\mathbf k}} \qquad (t>0).$$

For $t<0$, close in the upper half-plane. The residue at $k^0=-\omega_{\mathbf k}$ is

$${i e^{+i\omega_{\mathbf k}t}\over-2\omega_{\mathbf k}},$$

and the contour is counterclockwise, giving

$$D_F(\mathbf k;t)={e^{+i\omega_{\mathbf k}t}\over2\omega_{\mathbf k}} ={e^{-i\omega_{\mathbf k}|t|}\over2\omega_{\mathbf k}}.$$

Combining the two time orderings gives the stated result.

Exercise 2: Occupied oscillator

For a harmonic oscillator in the number state $|N\rangle$, show that

$$D_N(t)={1\over2\omega}\left[(N+1)e^{-i\omega |t|}+N e^{+i\omega |t|}\right].$$

Then show that its frequency-space form can be written as

$$D_N(k^0)=D_0(k^0)+2\pi N\delta((k^0)^2-\omega^2).$$

Solution

Write

$$q(t)={1\over\sqrt{2\omega}}\left(ae^{-i\omega t}+a^\dagger e^{i\omega t}\right).$$

For $t>0$,

$$\langle N|q(t)q(0)|N\rangle ={1\over2\omega}\left[ \langle N|aa^\dagger|N\rangle e^{-i\omega t} +\langle N|a^\dagger a|N\rangle e^{i\omega t} \right].$$

Since

$$\langle N|aa^\dagger|N\rangle=N+1, \qquad \langle N|a^\dagger a|N\rangle=N,$$

we get

$$D_N(t)={1\over2\omega}\left[(N+1)e^{-i\omega t}+N e^{i\omega t}\right] \qquad (t>0).$$

For $t<0$, time ordering exchanges the two operators and gives the same expression with $t$ replaced by $|t|$ in the first exponential and by $-|t|$ in the second. Thus

$$D_N(t)={1\over2\omega}\left[(N+1)e^{-i\omega |t|}+N e^{+i\omega |t|}\right].$$

Using

$${1\over x+i0}-{1\over x-i0}=-2\pi i\delta(x),$$

and

$$D_N(k^0)=(N+1){i\over(k^0)^2-\omega^2+i0} -N{i\over(k^0)^2-\omega^2-i0},$$

we find

$$D_N(k^0)-D_0(k^0) =N i\left[{1\over x+i0}-{1\over x-i0}\right] =2\pi N\delta(x),$$

where $x=(k^0)^2-\omega^2$. Therefore

$$D_N(k^0)=D_0(k^0)+2\pi N\delta((k^0)^2-\omega^2).$$

Exercise 3: Short-distance scalar correlator

Show that the massless equal-time scalar propagator in four spacetime dimensions is

$$D_F(0,\mathbf r)=\int {d^3\mathbf k\over(2\pi)^3}{e^{i\mathbf k\cdot\mathbf r}\over2|\mathbf k|} ={1\over4\pi^2r^2},$$

with $r=|\mathbf r|$, understood as a regulated distribution.

Solution

Use spherical coordinates with the polar axis along $\mathbf r$:

$$\int d\Omega\,e^{i\mathbf k\cdot\mathbf r} =4\pi{\sin kr\over kr}.$$

Then

$$D_F(0,\mathbf r) ={1\over(2\pi)^3}\int_0^\infty dk\,k^2{1\over2k} 4\pi{\sin kr\over kr}.$$

This simplifies to

$$D_F(0,\mathbf r) ={1\over4\pi^2r}\int_0^\infty dk\,\sin kr.$$

With a convergence factor $e^{-\epsilon k}$,

$$\int_0^\infty dk\,e^{-\epsilon k}\sin kr ={r\over r^2+\epsilon^2} \longrightarrow {1\over r}$$

as $\epsilon\to0^+$. Hence

$$D_F(0,\mathbf r)={1\over4\pi^2r^2}.$$

Exercise 4: Three-channel logarithm

Compute the leading logarithmic part of

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

Then use it to find the coefficient of the one-loop logarithm in the four-point vertex of real $\phi^4$ theory.

Solution

In four Euclidean dimensions,

$$d^4k=2\pi^2k^3dk.$$

Therefore

$$I(Q,\Lambda) ={2\pi^2\over(2\pi)^4}\int_Q^\Lambda {k^3dk\over k^4} ={1\over8\pi^2}\log{\Lambda\over Q} ={1\over16\pi^2}\log{\Lambda^2\over Q^2}.$$

One bubble channel has symmetry factor $1/2$, so its logarithmic weight is

$${\lambda_0^2\over2}I(Q,\Lambda).$$

There are three channels. Thus the total one-loop logarithmic weight is

$$3\cdot {\lambda_0^2\over2}I(Q,\Lambda) ={3\lambda_0^2\over2}{1\over16\pi^2}\log{\Lambda^2\over Q^2}.$$

Equivalently,

$${3\lambda_0^2\over32\pi^2}\log{\Lambda^2\over Q^2}.$$

In the effective-action convention used on this page, the signed result is

$$\Gamma_4(Q)=\lambda_0 -{3\lambda_0^2\over32\pi^2}\log{\Lambda^2\over Q^2} +\cdots.$$

Equivalently, the standard Euclidean diagrammatic vertex factor $-\Gamma_4$ receives the opposite sign.

Exercise 5: Leading-log accuracy

Suppose a one-loop calculation gives

$$\lambda_0^2\log{\Lambda^2\over4Q^2}$$

instead of

$$\lambda_0^2\log{\Lambda^2\over Q^2}.$$

Explain why the difference is subleading in the leading-log approximation when $\lambda_0\ll1$ but $\lambda_0\log(\Lambda^2/Q^2)$ is order one.

Solution

The two logarithms differ by a constant:

$$\log{\Lambda^2\over4Q^2} =\log{\Lambda^2\over Q^2}-\log4.$$

Thus the difference in the one-loop correction is

$$-\lambda_0^2\log4.$$

In the leading-log regime,

$$\lambda_0 L\sim1, \qquad L=\log{\Lambda^2\over Q^2},$$

so

$$\lambda_0^2L\sim\lambda_0.$$

The logarithmic one-loop correction is therefore parametrically of order $\lambda_0$, while the constant correction is of order $\lambda_0^2$. It is smaller by one power of $\lambda_0$ and is not part of leading-log accuracy.

Exercise 6: One-loop beta function

Starting from the one-loop leading-log result

$$\Gamma_4(Q)=\lambda_0-{3\lambda_0^2\over16\pi^2}\log{\Lambda\over Q}+O(\lambda_0^3),$$

derive the one-loop beta-function coefficient in the convention

$$\beta(\lambda)=\mu{d\lambda\over d\mu}.$$

Solution

Introduce a renormalized coupling at scale $\mu$ by

$$\lambda_R=\Gamma_4(\mu).$$

The one-loop relation is

$$\lambda_R=\lambda_0-{3\lambda_0^2\over16\pi^2}\log{\Lambda\over\mu}+O(\lambda_0^3).$$

Invert to the same order:

$$\lambda_0=\lambda_R+{3\lambda_R^2\over16\pi^2}\log{\Lambda\over\mu}+O(\lambda_R^3).$$

Holding the bare coupling and cutoff fixed, differentiate with respect to $\log\mu$:

$$0=\beta(\lambda_R)-{3\lambda_R^2\over16\pi^2}+O(\lambda_R^3).$$

Therefore

$$\boxed{\beta(\lambda_R)={3\over16\pi^2}\lambda_R^2+O(\lambda_R^3).}$$

The positive sign means the coupling grows toward higher renormalization scales.

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. See the lectures on Wick diagrams, divergences, and coupling renormalization.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. See Chapters 6, 7, 14, 19, and 23.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007. See Sections 3, 8–10, 16, 27, and 28.
• Weinberg, Steven. The Quantum Theory of Fields. Vol. I, Foundations, Cambridge University Press, 1995, Chapters 6 and 12; Vol. II, Modern Applications, Cambridge University Press, 1996, Section 18.2.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 5th ed., Oxford University Press, 2021. See Chapters 2 and 8–13.