Dimensional Transmutation and Mass Gaps

The previous page ended with the one-loop asymptotically free flow

$$\beta(g)=-{b_0\over16\pi^2}g^3+O(g^5), \qquad b_0>0,$$

and therefore

$${1\over g^2(\mu)}={1\over g^2(\mu_0)}+{b_0\over8\pi^2}\log{\mu\over\mu_0}.$$

This formula looks modest, but it contains one of the deepest lessons of quantum field theory. A classically scale-invariant theory with a dimensionless coupling can produce a dimensionful scale. The process is called dimensional transmutation: instead of specifying a dimensionless coupling at some arbitrary subtraction scale, one may specify a physical mass scale such as $\Lambda_{\mathrm{YM}}$ or $\Lambda_{\mathrm{QCD}}$.

The same idea also explains why perturbation theory can be excellent at short distances while giving no direct perturbative expansion for the masses of bound states. If the generated scale is

$$\Lambda\sim \mu\exp\left[-{8\pi^2\over b_0g^2(\mu)}\right],$$

then it is non-analytic at $g=0$. No finite Taylor series in $g$ can see it. This is the first warning that mass gaps in asymptotically free theories are nonperturbative, even though their scale is already visible in perturbative RG.

Required background. QED and Yang–Mills Beta Functions fixes the beta-function normalization, derives asymptotic freedom, and supplies the one-loop running used throughout this page.

The RG-invariant scale

Scale variables. Momentum scales such as $\mu$ and $p$ denote positive Euclidean magnitudes. On this page

$$\beta(g)=\mu {dg\over d\mu}, \qquad \beta(g)=-{b_0\over16\pi^2}g^3+O(g^5)$$

for an asymptotically free coupling. The UV cutoff is written as $\Lambda_0=a^{-1}$, where $a$ may be a lattice spacing or any short-distance regulator length. When the precise one-loop coefficient is unimportant, we write positive constants as $c$.

Start from the one-loop equation

$${d\over d\log\mu}{1\over g^2(\mu)}={b_0\over8\pi^2}.$$

Integrating gives

$${1\over g^2(\mu)}={b_0\over8\pi^2}\log{\mu\over\Lambda},$$

where the integration constant has been written as a scale $\Lambda$. Solving for $g^2(\mu)$ gives

$$\boxed{ g^2(\mu)={8\pi^2\over b_0\log(\mu/\Lambda)} } \qquad (\mu\gg\Lambda).$$

Equivalently,

$$\boxed{ \Lambda=\mu\exp\left[-{8\pi^2\over b_0g^2(\mu)}\right]. }$$

This is the RG-invariant scale. At one loop it is exactly independent of $\mu$. Beyond one loop its definition is convention-dependent by a multiplicative constant, but the statement that the theory generates a scale is not convention-dependent. Changing the renormalization scheme changes the numerical value assigned to $\Lambda$; it does not change dimensionless physical predictions such as mass ratios.

The all-orders construction makes this precise. In a specified renormalization scheme $\mathcal S$, define formally

$$\boxed{ \Lambda_{\mathcal S} =\mu\exp\left[-\int^{g(\mu)}{dg'\over\beta_{\mathcal S}(g')}\right] }$$

with the finite part of the integration constant included in the definition of the scheme. Differentiating shows $d\Lambda_{\mathcal S}/d\log\mu=0$. If

$$\beta(g)=-{b_0g^3\over16\pi^2} -{b_1g^5\over(16\pi^2)^2}+\cdots,$$

then at weak coupling

$$\begin{aligned} \Lambda_{\mathcal S}={}& \mu\exp\left[-{8\pi^2\over b_0g^2(\mu)}\right]\\ &\times \left({b_0g^2(\mu)\over16\pi^2}\right)^{-b_1/(2b_0^2)} \left[1+O(g^2)\right]. \end{aligned}$$

A redefinition $g^2\mapsto g^2+c_1g^4+\cdots$ changes the numerical value called $\Lambda_{\mathcal S}$ by a finite multiplicative factor. It cannot change dimensionless physical predictions such as mass ratios $M_i/M_j$, nor can it remove the essential singularity $e^{-\mathrm{const}/g^2}$.

The one-loop asymptotically free flow trades a dimensionless coupling for a dimensionful scale. In inverse-coupling variables, the RG trajectory is a straight line that reaches strong coupling near $\mu\sim\Lambda$. With a UV cutoff $a^{-1}$, the physical scale behaves as $a\Lambda\sim e^{-\mathrm{const}/g_0^2}$.

The name “dimensional transmutation” is literal. Suppose the bare Lagrangian of pure Yang–Mills theory contains only

$$S={1\over4g_0^2}\int d^4x\,F^a_{\mu\nu}F^a_{\mu\nu}$$

at the cutoff scale $\Lambda_0=a^{-1}$. Classically, $g_0$ is dimensionless and there is no mass parameter. Quantum mechanically, the pair $(a,g_0)$ determines

$$\boxed{ \Lambda_{\mathrm{YM}}\sim {1\over a}\exp\left[-{8\pi^2\over b_0g_0^2}\right] }$$

up to scheme-dependent multiplicative factors and higher-loop corrections. To keep $\Lambda_{\mathrm{YM}}$ finite while taking $a\to0$, one must tune

$$g_0^2(a)\simeq {8\pi^2\over b_0\log(1/a\Lambda_{\mathrm{YM}})}.$$

Thus the continuum limit is not obtained by keeping the bare coupling fixed. It is obtained by approaching the UV fixed point $g_0=0$ along a trajectory that keeps the physical scale fixed.

The cutoff viewpoint

The same result can be written in the language of a running coupling measured at momentum $p$ below the cutoff. At one loop,

$${1\over g^2(p)}={1\over g_0^2}+{b_0\over8\pi^2}\log{p\over\Lambda_0}, \qquad \Lambda_0={1\over a}.$$

Since $p<\Lambda_0$, the logarithm is negative. As $p$ is lowered, $1/g^2(p)$ decreases and the coupling grows. The formal scale at which the one-loop denominator vanishes is

$$\Lambda=\Lambda_0\exp\left[-{8\pi^2\over b_0g_0^2}\right].$$

It is tempting to say “the coupling becomes infinite at $p=\Lambda$.” That sentence is too literal. One-loop perturbation theory has already failed before the denominator vanishes. The reliable statement is instead:

$$\text{the theory becomes strongly coupled at momenta of order }\Lambda.$$

For $p\gg\Lambda$,

$$\boxed{ g^2(p)={8\pi^2\over b_0\log(p/\Lambda)} }$$

is small and perturbation theory is meaningful. For $p\sim\Lambda$, the coupling is order one and one needs nonperturbative physics.

This is why Yang–Mills theory and QCD can be weakly coupled at short distance and strongly coupled at long distance. Short-distance gauge-invariant observables admit a weakly coupled partonic description. Confinement, bound states, and a mass gap concern the infrared and require nonperturbative input.

QED and the opposite flow

It is useful to contrast this with QED. For massless charged matter, the perturbative beta function has the opposite sign,

$$\beta(e)=+{b\over16\pi^2}e^3+O(e^5), \qquad b>0.$$

In momentum variables, the charge grows toward the ultraviolet. In coordinate-space variables, with $R$ a distance scale and $\mu\sim1/R$,

$${d e^2(R)\over d\log R}=-{b\over8\pi^2}e^4(R)+O(e^6).$$

Therefore

$$e^2(R)={e^2(a)\over 1+{b\over8\pi^2}e^2(a)\log(R/a)}$$

at large $R$ in the massless theory. The potential between static charges has the schematic form

$$V(R)\sim {e^2(R)\over R}.$$

If the beta function has no nonzero fixed point, then

$$e^2(R)\sim {1\over \log(R/a)} \qquad (R\to\infty),$$

so the long-distance charge is screened. In the opposite direction, the one-loop formula develops a UV Landau pole. In the absence of a separate non-Gaussian ultraviolet fixed point, perturbative QED therefore does not furnish the asymptotically free continuum limit available in Yang–Mills theory.

This comparison is conceptually important. Both theories have logarithmic running. But in QED, the massless photon remains in the spectrum and the perturbative flow does not by itself produce a confinement scale. In pure Yang–Mills theory, the coupling grows in the infrared, and the natural expectation is a spectrum of gauge-invariant massive states with masses of order $\Lambda_{\mathrm{YM}}$.

What a mass gap means

A relativistic quantum theory has a mass gap if the vacuum is separated from the rest of the physical spectrum by a positive energy. After choosing a vacuum sector and subtracting its energy, the infinite-volume definition is

$$\boxed{ M_{\mathrm{gap}} =\inf\bigl(\operatorname{spec}H\setminus\{0\}\bigr)>0. }$$

This formulation does not assume that the first excitation is a discrete finite-volume level. For a Lorentz-invariant theory, $M_{\mathrm{gap}}$ is the smallest invariant mass in the physical Hilbert space. In Euclidean correlation functions, the same statement appears as exponential decay. If $\mathcal O$ is a local gauge-invariant operator with overlap with the lightest state in its channel, then at large Euclidean separation $r=|x|$,

$$\langle \mathcal O(x)\mathcal O(0)\rangle_c \sim {e^{-M_{\mathcal O}|x|}\over |x|^\alpha}$$

for some power $\alpha$. The smallest such $M_{\mathcal O}$ over all nontrivial gauge-invariant channels is the mass gap.

Equivalently, when the asymptotic coefficient is nonzero,

$$M_{\mathcal O} =-\lim_{r\to\infty}{1\over r} \log\left|\langle\mathcal O(r)\mathcal O(0)\rangle_c\right|.$$

A mass gap is a positive separation between the vacuum and the lightest physical excitation. In pure Yang–Mills theory the expected particles are gauge-invariant glueballs, with masses $M_n=c_n\Lambda_{\mathrm{YM}}$.

For pure Yang–Mills theory, the expected physical particles are glueballs: gauge-invariant excitations created, for example, by local operators such as $\operatorname{tr}F_{\mu\nu}F_{\mu\nu}$. Their masses should have the form

$$M_n=c_n\Lambda_{\mathrm{YM}},$$

where the $c_n$ are dimensionless numbers. Perturbative RG determines the existence and scaling of $\Lambda_{\mathrm{YM}}$, but it does not determine the constants $c_n$. Those constants are genuinely nonperturbative.

There is a useful distinction here:

$$\text{RG perturbation theory: }\quad \text{sets the scale }\Lambda,$$

but

$$\text{strong dynamics: }\quad \text{decides the spectrum in units of }\Lambda.$$

This distinction keeps the logic honest. Asymptotic freedom strongly suggests where the mass gap should live, but the existence of a mass gap in four-dimensional pure Yang–Mills theory is a nonperturbative statement. The RG argument supplies the only possible scale; it does not by itself prove which gauge-invariant states exist or that the lightest one has strictly positive mass.

In practice one diagnoses a gap by looking for exponential decay in Euclidean correlators of gauge-invariant operators, not by looking for a pole in a gauge-dependent gluon propagator. The latter can be useful in a fixed gauge, but it is not itself the definition of a physical particle. A mass gap and confinement are also logically distinct properties: pure Yang–Mills theory is expected to have both, but the definition of either one does not imply the other in every quantum field theory.

Scaling forms after transmutation

Consider a dimensionless physical amplitude $\mathcal A$ depending on a characteristic Euclidean momentum $p$ and a coupling $g(\mu)$ defined at a subtraction scale $\mu$. RG invariance says schematically

$$\left(\mu{\partial\over\partial\mu}+\beta(g){\partial\over\partial g}\right)\mathcal A\left({p\over\mu},g(\mu)\right)=0.$$

For a single-scale observable, the solution may be written in terms of the running coupling at the physical momentum,

$$\mathcal A=F(g(p)).$$

For $p\gg\Lambda$, this becomes a perturbative expansion in

$$g^2(p)={8\pi^2\over b_0\log(p/\Lambda)}.$$

But the same RG equation also says that, after the arbitrary subtraction scale is eliminated, the only dimensionless argument left is

$${p\over\Lambda}.$$

Thus

$$\boxed{ \mathcal A(p)=\mathcal F\left({p\over\Lambda}\right) }$$

for a dimensionless observable in a massless one-coupling theory. More generally, if an observable $\mathcal O$ has mass dimension $d_{\mathcal O}$, then

$$\boxed{ \mathcal O(p_i)=\Lambda^{d_{\mathcal O}}\, \mathcal F\left({p_i\over\Lambda}\right). }$$

This is the practical content of dimensional transmutation. At finite cutoff, the same dimensionless amplitude can be parametrized as

$$\mathcal A=\mathcal A(g_0^2,ap),$$

but after taking the continuum limit along a trajectory of fixed $\Lambda$,

$$\mathcal A\longrightarrow \mathcal F\left({p\over\Lambda}\right).$$

Examples fix the dimensions. A glueball mass has $d_{\mathcal O}=1$, so $M=c\Lambda$. A string tension has $d_{\mathcal O}=2$, so $\sigma=c_\sigma\Lambda^2$. A dimensionless short-distance scattering amplitude depends on $p/\Lambda$ and becomes expandable in $1/\log(p/\Lambda)$ at large momentum.

For QCD with quark masses, additional dimensionless ratios remain:

$$\mathcal A=\mathcal F\left({p\over\Lambda_{\mathrm{QCD}}},{m_1\over\Lambda_{\mathrm{QCD}}},{m_2\over\Lambda_{\mathrm{QCD}}},\ldots\right).$$

Pure Yang–Mills has no such mass ratios. Once the overall scale is fixed, all mass ratios are predictions of the theory.

Continuum limits as critical limits

The continuum limit of an asymptotically free gauge theory is structurally similar to the continuum limit of a statistical system at a second-order critical point. In both cases a microscopic length $a$ is sent to zero while a physical correlation length is held fixed.

In a critical spin system,

$$\xi_{\mathrm{lat}}\sim |T-T_c|^{-\nu},$$

where $\xi_{\mathrm{lat}}$ is measured in lattice units. The physical correlation length is

$$\xi_{\mathrm{phys}}=a\xi_{\mathrm{lat}},$$

so the physical mass scale is

$$M={1\over \xi_{\mathrm{phys}}} \sim {1\over a}|T-T_c|^\nu.$$

To keep $M$ finite as $a\to0$, one tunes $T\to T_c$.

In asymptotically free Yang–Mills theory, the weak-coupling UV fixed point plays the role of the critical point. The physical scale is

$$M\sim \Lambda_{\mathrm{YM}} \sim {1\over a}\exp\left[-{8\pi^2\over b_0g_0^2}\right].$$

To keep $M$ finite as $a\to0$, one tunes $g_0^2\to0$.

The continuum limits of asymptotically free Yang–Mills theory and a critical spin system have the same architecture: send the microscopic spacing $a$ to zero while tuning a bare parameter so that a physical mass $M$ remains finite.

The analogy is not perfect. The Ising temperature perturbation is relevant at the critical point and produces power-law scaling. The Yang–Mills coupling is marginally relevant in the infrared, so the scaling is exponential in $1/g_0^2$. But the logic is the same: a continuum QFT is obtained by making the correlation length large compared with the cutoff.

Essential singularities and invisible perturbative masses

The exponential dependence

$$M a\sim \exp\left[-{8\pi^2\over b_0g_0^2}\right]$$

has an essential singularity at $g_0=0$. For every nonnegative integer $n$,

$$\lim_{g_0\to0} {\exp[-\mathrm{const}/g_0^2]\over g_0^n}=0.$$

Therefore a mass gap of this type is invisible in ordinary perturbation theory. Any finite perturbative calculation around $g_0=0$ gives a power series in $g_0$ and logarithms; it cannot produce $e^{-1/g_0^2}$.

This fact explains a common paradox. The short-distance theory looks almost free, and perturbation theory becomes better as $a\to0$ at fixed physical momentum far above $M$. Yet the long-distance spectrum is massive and strongly coupled. There is no contradiction because the two claims concern different scales. Perturbation theory controls

$$p\gg M,$$

while the mass gap concerns

$$p\sim M.$$

The RG flow connects the two regimes, but it does not make the infrared perturbative.

A Fermi-surface preview

A similar exponential scale appears near a Fermi surface. The details belong to later pages, but the analogy is too useful to ignore. Suppose an attractive interaction in a Cooper channel has a dimensionless coupling $\lambda<0$. Let $\ell=\log(\Lambda_F/E)$ increase toward the infrared. The one-loop RG equation has the schematic form

$${d\lambda\over d\ell}=-N(0)\lambda^2,$$

where $N(0)$ is the density of states at the Fermi surface and $\ell$ is the logarithm of the scale ratio. Solving gives a strong-coupling scale

$$\Delta\sim \Lambda_F\exp\left[-{1\over N(0)|\lambda_0|}\right].$$

The scale $\Delta$ is the superconducting or pairing gap. Again, a marginal interaction generates a dimensionful scale by running logarithmically until perturbation theory fails.

Near a Fermi surface, an attractive marginal interaction runs to strong coupling at an exponentially small scale. In the paired phase the Bogoliubov spectrum $E_k=\sqrt{\xi_k^2+\Delta^2}$ has minimum excitation energy $\Delta$. The mathematical pattern resembles dimensional transmutation.

The analogy has limits. A Fermi surface is not Lorentz invariant, and a gapped superconductor has different symmetry structure from a confining gauge theory. The shared lesson is narrower and sharper: logarithmic RG flow can turn a small dimensionless coupling into a nonperturbative energy scale.

What is and is not proved by the one-loop flow

The one-loop beta function is a UV statement. It proves that the coupling becomes weak at short distances when $b_0>0$. It also identifies the scale at which the perturbative description stops being reliable. It does not by itself prove confinement, a discrete glueball spectrum, or a positive mass gap.

For pure Yang–Mills theory, the physical expectation is

$$M_{\mathrm{gap}}\sim \Lambda_{\mathrm{YM}},$$

and lattice gauge theory strongly supports this picture. But from the viewpoint of continuum perturbation theory, the statement remains nonperturbative. This is exactly why dimensional transmutation is so important: it shows how the scale can exist without pretending that weak-coupling diagrams calculate the infrared spectrum.

A good slogan is:

$$\text{Asymptotic freedom explains why the scale is generated;}$$ $$\text{strong dynamics decides what happens at that scale.}$$

Summary

An asymptotically free theory with

$$\beta(g)=-{b_0\over16\pi^2}g^3, \qquad b_0>0,$$

has the one-loop running

$${1\over g^2(\mu)}={b_0\over8\pi^2}\log{\mu\over\Lambda}.$$

The integration constant may be written as the RG-invariant scale

$$\Lambda=\mu\exp\left[-{8\pi^2\over b_0g^2(\mu)}\right].$$

This is dimensional transmutation: the dimensionless coupling is traded for a dimensionful scale.

With a cutoff $a$, the continuum limit of pure Yang–Mills theory is obtained by taking

$$a\to0, \qquad g_0^2\to0, \qquad {1\over a}\exp\left[-{8\pi^2\over b_0g_0^2}\right] =\text{fixed}.$$

This resembles the critical continuum limit of a statistical system, where $T\to T_c$ while $a\xi_{\mathrm{lat}}$ is held fixed.

A mass gap means a positive separation between the vacuum and the lightest physical state. In pure Yang–Mills theory, the expected spectrum consists of gauge-invariant massive states with

$$M_n=c_n\Lambda_{\mathrm{YM}}.$$

The scale $\Lambda_{\mathrm{YM}}$ is visible from perturbative RG, but the constants $c_n$ and the existence of the gap are nonperturbative.

Common pitfalls

Claiming that perturbation theory computes the gap. Perturbation theory determines the high-momentum running and identifies the RG-invariant scale. The gap itself is infrared and nonperturbative.

Treating a one-loop pole as an exact singularity. The pole is a warning that the chosen perturbative variables have failed, not a controlled statement about the exact coupling.

Confusing scheme dependence with physical arbitrariness. A coupling redefinition changes the numerical value assigned to $\Lambda_{\mathcal S}$, but physical masses and mass ratios do not depend on that convention. One fixes a scheme for $\Lambda$ and quotes nonperturbative quantities in that scheme or as dimensionless ratios.

Holding the bare coupling fixed while removing the cutoff. Fixed $g_0$ leaves the correlation length finite in cutoff units. The continuum limit requires $g_0\to0$ so that $a\Lambda\to0$ while $\Lambda$ in physical units remains fixed.

Equating a mass gap with confinement. They are distinct infrared properties. Pure Yang–Mills theory is expected to possess both, but one-loop asymptotic freedom proves neither.

Assuming every marginal interaction generates a gap. Some marginal interactions are exactly marginal or marginally irrelevant. The sign of the beta function and the interaction channel matter.

Exercises

Exercise 1: Verify one-loop RG invariance of the transmutation scale

Let

$$\beta(g)=-{b_0\over16\pi^2}g^3, \qquad b_0>0.$$

Show that

$$\Lambda=\mu\exp\left[-{8\pi^2\over b_0g^2(\mu)}\right]$$

is independent of $\mu$ at one loop.

Solution

First compute

$${d\over d\log\mu}{1\over g^2} =-{2\over g^3}{dg\over d\log\mu} =-{2\over g^3}\beta(g).$$

Using the beta function,

$${d\over d\log\mu}{1\over g^2} =-{2\over g^3}\left[-{b_0\over16\pi^2}g^3\right] ={b_0\over8\pi^2}.$$

Now differentiate

$$\log\Lambda=\log\mu-{8\pi^2\over b_0g^2(\mu)}.$$

This gives

$${d\log\Lambda\over d\log\mu} =1-{8\pi^2\over b_0}{d\over d\log\mu}{1\over g^2} =1-{8\pi^2\over b_0}{b_0\over8\pi^2}=0.$$

Thus $\Lambda$ is RG invariant at one loop.

Exercise 2: Tune the bare coupling toward the continuum limit

With cutoff $\Lambda_0=a^{-1}$, suppose

$$\Lambda=\Lambda_0\exp\left[-{8\pi^2\over b_0g_0^2}\right].$$

Solve for $g_0^2$ in terms of $a$ and fixed $\Lambda$. What happens as $a\to0$?

Solution

Since $\Lambda_0=a^{-1}$,

$$\Lambda={1\over a}\exp\left[-{8\pi^2\over b_0g_0^2}\right].$$

Multiply by $a$ and take the logarithm:

$$\log(a\Lambda)=-{8\pi^2\over b_0g_0^2}.$$

Equivalently,

$$\log{1\over a\Lambda}={8\pi^2\over b_0g_0^2}.$$

Therefore

$$\boxed{ g_0^2(a)={8\pi^2\over b_0\log(1/a\Lambda)}. }$$

As $a\to0$ at fixed $\Lambda$, the logarithm diverges and hence

$$g_0^2(a)\to0.$$

The continuum limit approaches the weak-coupling UV fixed point.

Exercise 3: Derive the scaling form after dimensional transmutation

Assume a dimensionless physical observable satisfies the RG equation

$$\left(\mu{\partial\over\partial\mu}+\beta(g){\partial\over\partial g}\right) \mathcal A\left({p\over\mu},g(\mu)\right)=0.$$

Explain why, in a massless one-coupling asymptotically free theory, it can be written as

$$\mathcal A=\mathcal F\left({p\over\Lambda}\right).$$

Solution

The observable is dimensionless, so before solving the RG equation it can only depend on the dimensionless ratio $p/\mu$ and on the running coupling $g(\mu)$.

The RG equation says that changing $\mu$ while moving $g(\mu)$ along the RG trajectory leaves $\mathcal A$ unchanged. Therefore the only invariant data on the RG trajectory can appear. For a one-coupling asymptotically free theory, the invariant data are encoded in

$$\Lambda=\mu\exp\left[-{8\pi^2\over b_0g^2(\mu)}\right]$$

at one loop, with higher-loop refinements changing only the precise scheme definition of $\Lambda$.

Since $p$ is the only external scale and $\Lambda$ is the only intrinsic scale, dimensional analysis then gives

$$\boxed{ \mathcal A=\mathcal F\left({p\over\Lambda}\right). }$$

At $p\gg\Lambda$, the same statement can be rewritten as a perturbative expansion in $g(p)$.

Exercise 4: Solve the coordinate-space QED flow

In coordinate-space QED, suppose

$${d e^2(R)\over d\log R}=-B e^4(R), \qquad B>0.$$

Solve for $e^2(R)$ with boundary condition $e(a)=e_0$. What is the large-$R$ behavior?

Solution

Write

$${d\over d\log R}{1\over e^2(R)} =-{1\over e^4(R)}{d e^2(R)\over d\log R} =B.$$

Integrating from $a$ to $R$ gives

$${1\over e^2(R)}={1\over e_0^2}+B\log{R\over a}.$$

Thus

$$\boxed{ e^2(R)={e_0^2\over1+Be_0^2\log(R/a)}. }$$

For $R\gg a$,

$$e^2(R)\sim {1\over B\log(R/a)}.$$

The charge is screened at long distance. This running alone does not generate a mass gap for the photon.

Exercise 5: Tune a statistical system to its critical limit

Near an Ising critical point,

$$\xi_{\mathrm{lat}}\sim |T-T_c|^{-\nu}.$$

The physical correlation length is $\xi_{\mathrm{phys}}=a\xi_{\mathrm{lat}}$ and the physical mass is $M=1/\xi_{\mathrm{phys}}$. Determine how $T-T_c$ must scale with $a$ if $M$ is held fixed while $a\to0$.

Solution

The physical mass is

$$M={1\over a\xi_{\mathrm{lat}}} \sim {1\over a}|T-T_c|^\nu.$$

Holding $M$ fixed gives

$$|T-T_c|^\nu\sim aM.$$

Therefore

$$\boxed{ |T-T_c|\sim (aM)^{1/\nu}. }$$

As $a\to0$, the temperature must be tuned to $T_c$. This is the statistical-mechanics analog of tuning $g_0\to0$ in an asymptotically free gauge theory while keeping the physical mass scale fixed.

Exercise 6: Derive the exponentially small BCS scale

A marginal attractive BCS coupling obeys

$${d\lambda\over d\ell}=-N(0)\lambda^2, \qquad \lambda_0<0.$$

Find the scale $\ell_*$ at which perturbation theory breaks down and express the corresponding gap $\Delta$ in terms of the UV scale $\Lambda_F$.

Solution

Integrate

$${d\lambda\over\lambda^2}=-N(0)d\ell.$$

This gives

$$-{1\over\lambda(\ell)}=-{1\over\lambda_0}-N(0)\ell,$$

or

$${1\over\lambda(\ell)}={1\over\lambda_0}+N(0)\ell.$$

Since $\lambda_0<0$, the denominator reaches zero at

$$\ell_*={1\over N(0)|\lambda_0|}.$$

If the running energy scale is $E=\Lambda_F e^{-\ell}$, then the strong-coupling scale is

$$\boxed{ \Delta\sim \Lambda_F e^{-\ell_*} =\Lambda_F\exp\left[-{1\over N(0)|\lambda_0|}\right]. }$$

This is the same essential-singularity pattern as dimensional transmutation, though the physical setting is a Fermi surface rather than a relativistic vacuum.

Further reading

• Coleman, Sidney, and Erick Weinberg. “Radiative Corrections as the Origin of Spontaneous Symmetry Breaking.” Physical Review D 7, no. 6 (1973): 1888–1910.
• Gell-Mann, Murray, and Francis E. Low. “Quantum Electrodynamics at Small Distances.” Physical Review 95, no. 5 (1954): 1300–1312.
• Gross, David J., and Frank Wilczek. “Ultraviolet Behavior of Non-Abelian Gauge Theories.” Physical Review Letters 30, no. 26 (1973): 1343–1346.
• Politzer, H. David. “Reliable Perturbative Results for Strong Interactions?” Physical Review Letters 30, no. 26 (1973): 1346–1349.
• Polyakov, Alexander M. Gauge Fields and Strings. Chur: Harwood Academic Publishers, 1987, Chapter 2.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge: Cambridge University Press, 2014, Chapters 23 and 26.
• Srednicki, Mark. Quantum Field Theory. Cambridge: Cambridge University Press, 2007, Sections 28, 73, and 82.
• Weinberg, Steven. The Quantum Theory of Fields. Volume II: Modern Applications. Cambridge: Cambridge University Press, 1996, Chapter 18.
• Wilson, Kenneth G., and John Kogut. “The Renormalization Group and the $\epsilon$ Expansion.” Physics Reports 12, no. 2 (1974): 75–199.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 4th ed. Oxford: Clarendon Press, 2002.