Running Couplings and Critical Free Energy

The previous page turned the one-loop logarithm in four-dimensional $\phi^4$ theory into a renormalization-group equation. The four-point vertex became a running coupling,

$$\lambda(q)={\lambda_0\over 1+a\lambda_0\log(\Lambda/q)}, \qquad a={3\over16\pi^2},$$

and the quadratic operator insertion acquired a running normalization,

$$\tau(q)=\left(1+{3\lambda_0\over16\pi^2}\log{\Lambda\over q}\right)^{-1/3}.$$

This page explains why those two formulas are not merely scattering-amplitude technology. They also control critical thermodynamics. In four Euclidean dimensions, the scalar $\phi^4$ coupling is marginal by engineering dimension and marginally irrelevant in the infrared. That is exactly the upper-critical-dimension situation: the leading critical exponents keep their mean-field values, but observables acquire calculable powers of logarithms.

The main thermodynamic result derived below is the logarithmic singularity of the specific heat in the one-component theory,

$$\boxed{ C_{\rm sing}(T)\propto \left[\log {1\over |T-T_c|}\right]^{1/3} }$$

up to nonuniversal constants inside the logarithm and in the overall normalization. Equivalently, the singular part of the free-energy density behaves as

$$\boxed{ f_{\rm sing}(T) \propto (T-T_c)^2 \left[\log {1\over |T-T_c|}\right]^{1/3} }$$

at leading-log accuracy. The power $1/3$ is not a new mean-field exponent. It is the ratio of two one-loop RG coefficients. The same RG applies whether $\phi$ is interpreted as a relativistic field or a statistical order parameter. Near criticality the inverse correlation length $m=\xi^{-1}$ supplies the infrared cutoff, so the thermodynamic singularity follows by running to $q\sim m$ and then expressing $m$ in terms of the reduced temperature.

Sliding the cutoff

Quartic and thermal running

This page repeatedly uses the previous page’s leading-log equations in the normalization

$$S_E\supset\int d^4x\,{\lambda_0\over4!}\phi^4.$$

For one real scalar field,

$$a={3\over16\pi^2}, \qquad b={1\over16\pi^2},$$

where $a$ appears in the four-point coupling and $b$ appears in the $\phi^2$ insertion:

$${d\lambda\over d\log q}=a\lambda^2, \qquad q{d\tau\over dq}=b\lambda(q)\tau(q).$$

It is often cleaner to use

$$L=\log{\Lambda\over q},$$

so that

$$\lambda(L)={\lambda_0\over1+a\lambda_0L}, \qquad \tau(L)=(1+a\lambda_0L)^{-b/a}=(1+a\lambda_0L)^{-1/3}.$$

The running coupling can be read in two equivalent ways. First, it is the effective four-point vertex observed at momentum scale $q$ with a fixed microscopic cutoff $\Lambda$. Second, it tells us how to change the bare coupling if we change the cutoff but demand the same low-energy physics.

Indeed,

$${1\over\lambda(k)}={1\over\lambda_0}+a\log{\Lambda\over k}.$$

Suppose we lower the cutoff from $\Lambda$ to $\Lambda'$, with $k<\Lambda'<\Lambda$. To leave the low-energy vertex at $k$ unchanged, the new bare coupling $\lambda_0'$ must satisfy

$${1\over\lambda(k)}={1\over\lambda_0'}+a\log{\Lambda'\over k}.$$

Therefore

$$\boxed{ {1\over\lambda_0'}={1\over\lambda_0}+a\log{\Lambda\over\Lambda'}. }$$

The integrated-out modes between $\Lambda'$ and $\Lambda$ have been hidden inside the new bare coupling. This is the Wilsonian idea in its most economical form: changing the resolution is compensated by changing the coordinates on the space of local actions.

The same low-energy vertex $\lambda(k)$ can be computed with cutoff $\Lambda$ and bare coupling $\lambda_0$, or with cutoff $\Lambda'$ and bare coupling $\lambda_0'$. RG flow is the rule for keeping physics fixed while changing this split.

There is a sign convention hiding here, and it is worth making it explicit. If $q$ is the probe momentum, then

$$q{d\lambda(q)\over dq}=a\lambda(q)^2.$$

For positive $\lambda$, raising the probe scale increases the coupling, while lowering the probe scale decreases it. If instead we use the logarithmic coarse-graining variable

$$L=\log{\Lambda\over q},$$

then

$${d\lambda\over dL}=-a\lambda^2.$$

These are the same statement. The sign changes because increasing $L$ means moving toward the infrared.

Landau–Ginzburg theory as Euclidean field theory

To apply this to critical phenomena, consider a one-component order parameter $M(x)$, such as the coarse-grained magnetization of an Ising-like ferromagnet. Near a continuous transition, the Landau–Ginzburg functional has the local form

$$\mathcal F[M]=\int d^d x\left[ {1\over2}(\nabla M)^2+{1\over2}r_0M^2+{\lambda_0\over4!}M^4+\cdots \right].$$

The classical statistical partition function is

$$Z=\int \mathcal D M\,\exp\left[-{\mathcal F[M]\over k_BT}\right].$$

In practice one absorbs the factor $k_BT$ into the parameters of the functional. The mathematical object is then a Euclidean scalar field theory. The mass parameter is the thermal tuning parameter:

$$r_0-r_{0,c}\propto T-T_c.$$

The subtraction by $r_{0,c}$ is essential. Ultraviolet fluctuations shift the critical temperature. The physical transition is not located by the naive condition $r_0=0$, but by the condition that the inverse correlation length vanish.

The two-point function of the order parameter diagnoses this correlation length. In a transfer-matrix language, if one spatial direction is treated as Euclidean time, then

$$\langle M(0)M(x)\rangle =\sum_n |\langle0|M|n\rangle|^2 e^{-(E_n-E_0)|x|}.$$

A finite gap $E_1-E_0$ gives exponential decay. A continuous transition occurs when the gap closes and the correlation length

$$\xi={1\over m}$$

becomes large. Thus the same field-theoretic mass $m$ that regulates infrared loop integrals is the inverse correlation length of the statistical system.

The upper critical dimension of $\phi^4$ theory is $d=4$. In $d=4$, the quartic coupling has zero engineering dimension. It is not simply irrelevant by power counting, but it becomes irrelevant logarithmically:

$$\lambda(k)\sim {1\over a\log(\Lambda/k)} \qquad (k\ll\Lambda).$$

This slow drift to zero is why mean-field powers survive but logarithms do not disappear.

Reduced temperature and the flow-stopping scale

Let

$$t={T-T_c\over T_c}$$

be the reduced temperature. At mean-field level the renormalized mass parameter is proportional to $t$,

$$m^2\propto |t|,$$

so the RG flow stops at

$$q\sim m\sim |t|^{1/2}.$$

In four dimensions, the relation between $m$ and $t$ itself receives logarithmic corrections because the thermal scaling field runs. For the leading powers of large logarithms derived in this page, this distinction only changes subleading $\log\log$ terms inside

$$L_m=\log{\Lambda\over m}.$$

Indeed, replacing $m\sim |t|^{1/2}$ gives

$$L_m={1\over2}\log{1\over |t|}+O(\log\log(1/|t|)),$$

and the leading power $L_m^{1/3}$ becomes the same leading power of $\log(1/|t|)$. This is why the page can safely write the final answer in terms of $|T-T_c|$ while computing the logarithm as an RG flow down to $m$.

The thermal operator

The singular specific heat is obtained by differentiating the free energy twice with respect to temperature. Since the temperature enters the Landau–Ginzburg functional through $r_0$, the operator conjugate to the reduced temperature is the quadratic operator

$$\mathcal E(x)={\partial\mathcal F\over\partial r_0(x)}={1\over2}M(x)^2.$$

This is often called the thermal operator or energy-density operator. Up to nonuniversal constants,

$$C_{\rm sing}\propto \int d^d x\,\langle \mathcal E(x)\mathcal E(0)\rangle_c,$$

where the connected correlator is understood. In momentum space, define

$$C(q)=\int d^d x\,e^{iq\cdot x}\langle \mathcal E(x)\mathcal E(0)\rangle_c.$$

The physical heat capacity is obtained by evaluating the critical correlator down to the infrared scale set by the mass. At criticality, the external momentum $q$ is the infrared cutoff. Away from criticality, the mass $m=\xi^{-1}$ stops the RG flow, so one should set $q\sim m$ in the leading logarithm.

The thermal operator is precisely the $\phi^2$ insertion from the previous page. Its dimensionless two-leg insertion factor is $\tau(k)$. The RG equation

$${d\tau\over dL}=-b\lambda(L)\tau, \qquad b={1\over16\pi^2},$$

has the solution

$$\tau(L)=(1+a\lambda_0L)^{-1/3}.$$

The quadratic operator insertion $\mathcal E=\phi^2/2$ is dressed by a logarithmic shell containing one quartic vertex. The one-loop ratio $b/a=1/3$ gives $\tau(L)=(1+a\lambda_0L)^{-1/3}$.

The factor $\tau$ should not be confused with a new coupling. It is a normalization factor for a local operator insertion. Depending on convention, one may say that the source $r$ runs, or that the operator $\mathcal E$ runs, or that the vertex associated with the insertion runs. Correlation functions only depend on the combined normalization.

Specific heat from two thermal insertions

The leading contribution to $C(q)$ is the bubble with two $\mathcal E$ insertions. In the free massless theory in four dimensions,

$$C_0(q)\propto \int^\Lambda {d^4p\over(2\pi)^4}\,{1\over p^2(p+q)^2} \sim \log{\Lambda\over q}.$$

The interacting leading-log result is obtained by slicing this logarithmic integral into shells. Each shell at scale $k$ sees two dressed thermal insertions, so

$${dC\over dL}=A\,\tau(L)^2,$$

where $A$ is a nonuniversal positive constant depending on the normalization of $\mathcal E$. The logarithmic dependence is universal; the overall coefficient is not.

This formula is intentionally differential in $L$. A single shell contributes the free-theory logarithmic measure, while the accumulated effect of harder shells is already contained in the two factors of $\tau(L)$. This avoids double-counting the same logarithms.

The singular specific heat is the integrated connected two-point function of the thermal operator $\mathcal E=\phi^2/2$. At leading-log accuracy, a logarithmic shell contributes an amount proportional to $\tau(L)^2dL$.

Using

$$\tau(L)^2=(1+a\lambda_0L)^{-2/3},$$

we find

$$C(L)-C(0)=A\int_0^L dL'\,(1+a\lambda_0L')^{-2/3}.$$

The integral is elementary:

$$\int_0^L dL'\,(1+a\lambda_0L')^{-2/3} ={3\over a\lambda_0}\left[(1+a\lambda_0L)^{1/3}-1\right].$$

Thus, for large $L$,

$$\boxed{ C(q)\propto \left(\log{\Lambda\over q}\right)^{1/3} }$$

at leading-log accuracy. The exponent is positive even though each individual insertion factor decreases. The reason is simple: the heat capacity integrates over all logarithmic shells. The integrand decreases as $L^{-2/3}$, but the accumulated integral still grows as $L^{1/3}$.

Away from the critical point, the RG stops when $q$ reaches the physical mass

$$m=\xi^{-1}.$$

Therefore

$$C_{\rm sing}(T)\propto \left(\log{\Lambda\over m}\right)^{1/3}.$$

To leading mean-field accuracy,

$$m^2\propto |T-T_c|,$$

so constants and factors of $2$ inside the logarithm are unimportant, and

$$\boxed{ C_{\rm sing}(T)\propto \left[\log {1\over |T-T_c|}\right]^{1/3}. }$$

At nonzero reduced temperature, the inverse correlation length $m=\xi^{-1}$ stops the critical RG flow. The singular heat capacity accumulates logarithmic shell contributions only down to $k\sim m$.

The singular free-energy density follows by integrating twice with respect to the thermal scaling variable. Since the mean-field value of the specific-heat exponent is $\alpha=0$, the leading power is $f_{\rm sing}\sim r^2$, with $r\propto T-T_c$. The logarithm is inherited from $C$:

$$\boxed{ f_{\rm sing}(r) \propto r^2\left[\log{1\over |r|}\right]^{1/3} }$$

up to subleading powers of $1/\log(1/|r|)$ and nonuniversal constants. This is the precise sense in which the Gaussian fixed point controls the critical powers while the marginally irrelevant coupling controls logarithmic corrections.

Why four dimensions is special

The logarithm in the heat capacity is already visible in the free bubble integral

$$I_d(q)=\int^\Lambda {d^d p\over(2\pi)^d}\,{1\over p^2(p+q)^2}.$$

By dimensional analysis, the nonlocal part scales as

$$I_d(q)_{\rm nonlocal}\sim q^{d-4}$$

when $d\ne4$, while at $d=4$ this power becomes a logarithm. Equivalently, in the region $q\ll p\ll\Lambda$,

$$I_4(q)\sim\int_q^\Lambda {p^3dp\over p^4} =\int_q^\Lambda {dp\over p}.$$

Thus $d=4$ is the dimension in which every decade of momenta contributes comparably. That is exactly when the RG is needed to sum a long chain of logarithmic shells.

For $d>4$, the quartic coupling has negative engineering dimension:

$$[\lambda]=4-d<0.$$

It is irrelevant already by power counting. Loop corrections shift local parameters, including the critical temperature, but they do not generate the same universal logarithmic scaling. In five dimensions, for example,

$$\int^\Lambda {d^5p\over p^2(p+q)^2}$$

is dominated by cutoff-dependent local terms. These terms must be absorbed into the location of the critical point and other nonuniversal coefficients.

For $d<4$, the quartic coupling is relevant at the Gaussian fixed point. The long-distance theory is not governed by the Gaussian fixed point with logarithmic corrections, but by an interacting Wilson–Fisher fixed point. That case produces non-mean-field power laws rather than the four-dimensional logarithms derived here.

O(N) aside

For the $O(N)$ theory, the one-loop coefficients become

$$a_N={N+8\over48\pi^2}, \qquad b_N={N+2\over48\pi^2},$$

where $b_N$ is the coefficient for the thermal operator $\phi_i\phi_i/2$. Hence

$$\tau_N(L)=(1+a_N\lambda_0L)^{-(N+2)/(N+8)}.$$

The specific heat receives two thermal insertions, so

$$C_N(L)\propto\int^L dL'\,(L')^{-2(N+2)/(N+8)}.$$

For $N<4$, this gives the divergent logarithmic singularity

$$\boxed{ C_N\propto L^{(4-N)/(N+8)}. }$$

The Ising case $N=1$ gives $1/3$. The $N=4$ case is marginal inside the logarithmic correction itself and gives a further logarithm, $C_4\propto\log L$, at this level of approximation. For $N>4$, the shell integral converges as $L\to\infty$: the heat capacity approaches a nonuniversal constant, with a leading nonanalytic correction whose magnitude scales as $L^{(4-N)/(N+8)}$. Thus a negative power here describes a finite cusp correction, not a heat capacity that literally vanishes.

Summary

Changing the cutoff from $\Lambda$ to $\Lambda'$ is compensated by changing the bare coupling so that

$${1\over\lambda_0'}={1\over\lambda_0}+{3\over16\pi^2}\log{\Lambda\over\Lambda'}.$$

The physical low-energy vertex is unchanged. This is the simplest form of Wilsonian RG invariance.

The Landau–Ginzburg theory of a scalar order parameter is a Euclidean field theory. Its quadratic coefficient is the thermal tuning parameter,

$$r_0-r_{0,c}\propto T-T_c,$$

and the thermal operator is

$$\mathcal E={1\over2}M^2.$$

At the four-dimensional upper critical dimension, the quartic coupling is marginally irrelevant:

$$\lambda(k)={\lambda_0\over1+{3\lambda_0\over16\pi^2}\log(\Lambda/k)}.$$

The thermal operator insertion has the leading-log normalization

$$\tau(k)=\left(1+{3\lambda_0\over16\pi^2}\log{\Lambda\over k}\right)^{-1/3}.$$

The singular specific heat is the integrated connected correlator of two thermal operators. Since each logarithmic shell contributes $\tau^2dL$,

$$C_{\rm sing}\propto\int^{\log(\Lambda/m)} dL\,\tau(L)^2 \propto\left(\log{\Lambda\over m}\right)^{1/3}.$$

Using $m^2\propto |T-T_c|$ at leading mean-field order gives

$$C_{\rm sing}(T)\propto \left[\log {1\over |T-T_c|}\right]^{1/3},$$

up to subleading $\log\log$ changes inside the large logarithm. Integrating twice with respect to temperature gives

$$f_{\rm sing}(T)\propto (T-T_c)^2 \left[\log {1\over |T-T_c|}\right]^{1/3}.$$

Common pitfalls

Do not write $\log|T-T_c|$ as if it were a positive large quantity. Near the critical point the useful large logarithm is $\log(1/|T-T_c|)$, or equivalently $\log(\Lambda/m)$.

Do not set the critical point by the bare condition $r_0=0$. Fluctuations shift the critical temperature. The correct tuning is $r_0=r_{0,c}(\Lambda,\lambda_0)$, where the physical mass vanishes.

Do not conclude that because $\lambda(k)\to0$ in the infrared, the result is exactly Gaussian. The approach to the Gaussian fixed point is logarithmically slow, and operator insertions accumulate anomalous logarithmic factors.

Do not confuse the probe scale $q$ with the mass $m$. At criticality, $q$ can serve as the infrared cutoff. Away from criticality, the correlation length is finite and the flow stops at $q\sim m=\xi^{-1}$.

Do not treat cutoff-dependent constants in the free energy as universal. The location of $T_c$, additive constants in $f$, and analytic background terms depend on microscopic physics. The logarithmic singularity is the universal object.

Exercises

Exercise 1 — Cutoff redefinition invariance

Show explicitly that the running coupling

$$\lambda(k)={1\over {1/\lambda_0}+a\log(\Lambda/k)}$$

is invariant under the replacement $\Lambda\to\Lambda'$ if $\lambda_0$ is replaced by

$${1\over\lambda_0'}={1\over\lambda_0}+a\log{\Lambda\over\Lambda'}.$$

Solution

With the new cutoff and new bare coupling, the low-energy coupling is

$$\lambda'(k)={1\over {1/\lambda_0'}+a\log(\Lambda'/k)}.$$

Substitute the proposed relation:

$${1\over\lambda_0'}+a\log{\Lambda'\over k} = {1\over\lambda_0}+a\log{\Lambda\over\Lambda'}+a\log{\Lambda'\over k}.$$

The logarithms combine:

$$\log{\Lambda\over\Lambda'}+ \log{\Lambda'\over k} = \log{\Lambda\over k}.$$

Therefore

$${1\over\lambda_0'}+a\log{\Lambda'\over k} ={1\over\lambda_0}+a\log{\Lambda\over k},$$

so $\lambda'(k)=\lambda(k)$.

Exercise 2 — Running of the thermal insertion

Let

$${d\tau\over dL}=-b\lambda(L)\tau, \qquad \lambda(L)={\lambda_0\over1+a\lambda_0L}, \qquad \tau(0)=1.$$

Solve for $\tau(L)$, and evaluate the exponent for

$$a={3\over16\pi^2}, \qquad b={1\over16\pi^2}.$$

Solution

Divide by $\tau$:

$${d\log\tau\over dL}=-b{\lambda_0\over1+a\lambda_0L}.$$

Integrating from $0$ to $L$ gives

$$\log\tau(L)=-b\int_0^L {\lambda_0\,dL'\over1+a\lambda_0L'}.$$

The integral is

$$\int_0^L {\lambda_0\,dL'\over1+a\lambda_0L'} ={1\over a}\log(1+a\lambda_0L).$$

Therefore

$$\tau(L)=(1+a\lambda_0L)^{-b/a}.$$

For the one-component theory,

$${b\over a}={1/(16\pi^2)\over3/(16\pi^2)}={1\over3},$$

so

$$\tau(L)=\left(1+{3\lambda_0\over16\pi^2}L\right)^{-1/3}.$$

Exercise 3 — The specific-heat logarithm

Assume the singular part of the heat-capacity correlator obeys

$${dC\over dL}=A(1+a\lambda_0L)^{-2/3}, \qquad C(0)=C_0.$$

Show that $C(L)\propto L^{1/3}$ for large $L$.

Solution

Integrate:

$$C(L)-C_0=A\int_0^L dL'\,(1+a\lambda_0L')^{-2/3}.$$

Set

$$\nu=1+a\lambda_0L', \qquad d\nu=a\lambda_0dL'.$$

Then

$$C(L)-C_0={A\over a\lambda_0}\int_1^{1+a\lambda_0L}d\nu\,\nu^{-2/3}.$$

Since

$$\int d\nu\,\nu^{-2/3}=3\nu^{1/3},$$

we get

$$C(L)-C_0={3A\over a\lambda_0} \left[(1+a\lambda_0L)^{1/3}-1\right].$$

For large $L$,

$$C(L)\sim L^{1/3},$$

up to a nonuniversal multiplicative constant.

Exercise 4 — The four-dimensional massless bubble

Use Feynman parameters to show that the massless four-dimensional bubble

$$I(q)=\int^{\Lambda}{d^4p\over(2\pi)^4}{1\over p^2(p+q)^2}$$

has a logarithmic dependence on $q$ for $q\ll\Lambda$.

Solution

Use

$${1\over AB}=\int_0^1 dx\,{1\over[xA+(1-x)B]^2}.$$

With $A=p^2$ and $B=(p+q)^2$, shift

$$\ell=p+(1-x)q.$$

Then

$$xp^2+(1-x)(p+q)^2=\ell^2+x(1-x)q^2.$$

Thus

$$I(q)=\int_0^1 dx\int^{\Lambda}{d^4\ell\over(2\pi)^4} {1\over[\ell^2+x(1-x)q^2]^2}$$

up to cutoff-shape effects that only change nonlogarithmic terms. The standard logarithmic integral gives

$$\int^{\Lambda}{d^4\ell\over(2\pi)^4}{1\over(\ell^2+\Delta)^2} ={1\over16\pi^2}\log{\Lambda^2\over\Delta}+\text{nonlogarithmic terms}.$$

Therefore

$$I(q)={1\over16\pi^2}\int_0^1 dx\, \log{\Lambda^2\over x(1-x)q^2}+\cdots.$$

The $x$-dependent part contributes only a constant, so

$$I(q)={1\over16\pi^2}\log{\Lambda^2\over q^2}+\cdots ={1\over8\pi^2}\log{\Lambda\over q}+\cdots.$$

The precise additive constant depends on the regulator, but the logarithmic dependence does not.

Exercise 5 — The O(N) logarithmic exponent

For the $O(N)$ model, take

$$a_N={N+8\over48\pi^2}, \qquad b_N={N+2\over48\pi^2}.$$

Assuming

$$\tau_N(L)\sim L^{-b_N/a_N}$$

at large $L$, determine the leading heat-capacity behavior for $N<4$, $N=4$, and $N>4$. For $N<4$, show that the divergent part is

$$C_N(L)\sim L^{(4-N)/(N+8)}.$$

Solution

The thermal correlator has two thermal insertions, so

$${dC_N\over dL}\propto \tau_N(L)^2.$$

At large $L$,

$$\tau_N(L)^2\sim L^{-2b_N/a_N}.$$

The ratio of coefficients is

$${b_N\over a_N}={N+2\over N+8}.$$

Therefore

$${dC_N\over dL}\sim L^{-2(N+2)/(N+8)}.$$

For $N\ne4$, an antiderivative has the power

$$C_N(L)\sim L^{1-2(N+2)/(N+8)}.$$

The exponent is

$$1-{2(N+2)\over N+8} ={N+8-2N-4\over N+8} ={4-N\over N+8}.$$

Thus, for $N<4$,

$$C_N(L)\sim L^{(4-N)/(N+8)}.$$

When $N=4$, the exponent of the integrand is $-1$, so the integral gives a logarithm of a logarithm:

$$C_4(L)\sim\log L.$$

For $N>4$, the large-$L$ integrand is integrable. The full heat capacity approaches a cutoff-dependent constant $C_\infty$, while the leading critical correction obeys

$$C_N(L)-C_\infty\sim -L^{(4-N)/(N+8)}.$$

The negative exponent makes this correction vanish at criticality; it describes a finite cusp rather than a divergence.

Further reading

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014, chapter 23.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007, sections 28–29.
• Weinberg, Steven. The Quantum Theory of Fields, Volume II: Modern Applications. Cambridge University Press, 1996, sections 18.2, 18.5, and 18.8.
• Wilson, Kenneth G., and Michael E. Fisher. “Critical Exponents in 3.99 Dimensions.” Physical Review Letters 28 (1972): 240–243.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 4th ed., Oxford University Press, 2002, chapters 8–19.