2026-10-01 · chapter Yang–Mills and the mass gapPart 4 QuantisationΛ, exp(−1/g²)

Dimensional transmutation and Λ — quantum theory brings one scale into a classical theory without scales

With $\hbar=c=1$, length is the inverse of mass, and every quantity can be measured in powers of mass. The dimension of the coupling in the action $S=\frac{1}{2g^2}\int\mathrm{tr}\,F_{\mu\nu}F_{\mu\nu}\,d^dx$ is

(2-04, 1-12). In three dimensions $g^2$ is itself a mass, and one can write $m=c\,g^2$ (6-06). In four dimensions $g$ is dimensionless, so no mass can be made from the classical ingredients. Classical waves have the dispersion $\omega=|\mathbf k|$, and the problem statement also writes 'massless waves that travel at the speed of light' (§1) Known. There are no static finite-energy solutions (lumps) either (3-03; Deser 1976, Coleman 1977) Known.

Example. The action $8\pi^2/g^2$ of the BPST instanton does not depend on its size $\rho$ (3-05). Classical solutions have no preferred size. If $\Delta\gt0$ in the quantum theory, that scale must come from outside the classical action. This is what 'the mass gap is not classically visible' in §6.6 of the problem statement is about (§05).

The $\beta(g)=\mu\,dg/d\mu$ of 4-06 is a first-order ordinary differential equation. Separating the variables and integrating gives

The freedom in the lower limit on the right-hand side, taken to the left-hand side as a constant, is $\Lambda$; being an integration constant, it has the same value whichever point of the flow it is computed from. At one loop ($\beta=-b_0g^3$), choosing the constant to be 0 gives

and differentiating $\ln\Lambda$ with respect to $\ln\mu$ gives $1+\beta/(b_0g^3)=0$. Rewritten for SU(2) on the lattice ($\beta=4/g^2$) it is $1/(a\Lambda)\propto e^{3\pi^2\beta/11}$, the same formula as the one-loop part of the bookkeeping length $\xi_{\rm RG}$ of Part 10 (10-01 The line of construction and Assumption H) Known. From two loops on, a convention for the lower limit is needed. The one widely used is formula (2.7) of Capitani et al. 1999.

The bracket is finite as $x\to0$, so 0 can be used as the lower limit. The PDG review explains $\Lambda$ as 'a constant of integration, which corresponds to the scale where the perturbatively-defined coupling would diverge' Known.

Example (Table 2). Treating the two-loop β as exact, we integrated the flow numerically for SU(3) from $g^2(\mu_0)=2$. Formula (4) is constant to 9 digits. Formula (3) drifts by a factor of 2.7 while $\mu$ grows by a factor of $10^{16}$, and the formula with only the prefactor power attached merely shrinks the difference from 4.6% to 0.43%. The numbers are given with their defining formulas.

Computedpython3 (RK4). The integral in formula (4) agrees between the closed form and numerical integration to 9 digits.

Turning formula (3) around, the whole flow is fixed by $\Lambda$ alone.

$\Lambda$ enters only in the form $\mu/\Lambda$. Doubling $\Lambda$ is the same as halving the unit of mass, and the one-parameter family "choose the coupling at $\mu_0$" is, up to units, a single theory. The dimensionless $g$ has been replaced by the dimensionful $\Lambda$. This is called dimensional transmutation Physics.

Tong writes that this idea was known from the 1973 work of Coleman and E. Weinberg (Tong §2.9). E. Weinberg's thesis of the same year writes, for the electrodynamics of a massless scalar field, that changing the vacuum expectation value $\langle\phi\rangle$ generated by quantum effects (the value of the scalar field in the vacuum, which serves as the scale of mass) is 'completely equivalent to changing the scale in which masses are measured' Known. Pure Yang–Mills has no dimensionless coupling left, and the only remaining inputs are the group ($N$) and the θ angle (3-05).

ComputedThe numbers in this figure were computed on this machine (the figure evaluates formulas (3)(5) on the spot). $\Lambda/\mu_0$ of the thin lines ranges from $3.44\times10^{-4}$ to $0.106$, but $g^2$ at $\mu=10\Lambda$, $10^3\Lambda$ and $10^6\Lambda$ is always 3.117, 1.039 and 0.520.

The scale comes from the cutoff. On the lattice $\Lambda$ is made from the pair $(a,g_0)$ (formula (8) of 4-06).

Computing a physical quantity with the dimension of mass, dimensional analysis gives the form $m=\mu\,F(g(\mu))$. $m$ does not depend on the choice of $\mu$, so

and the solution is $F(g)=C\exp\!\big(-\int^g dg'/\beta(g')\big)$. Comparing with formula (2),

where $C$ is a pure number. Formula (6) is the renormalisation group equation (4-04) for a physical quantity without anomalous dimension. For the string tension $\sigma=C'\Lambda^2$, and the ratio $m/\sqrt\sigma$ becomes a pure number fixed by $N$ and θ alone, with no value left to adjust Physics. This is what "one scale" means (7-01, 7-12).

The solutions of formula (6) include $C=0$. What can be said is only "if there is a mass, it is a constant multiple of $\Lambda$", and $C\gt0$ cannot be said. The mass gap problem can be read as the problem of constructing the theory and showing $C\gt0$ for the state next to the vacuum (0-03, 4-11).

Example (Table 3). Lattice numbers come out as dimensionless ratios ($r_0$ is the length of 4-06). $m(0^{++})$ is the mass of the lightest glueball with spin 0, positive parity and positive charge conjugation (5-08).

ComputedThe last row was divided with python3, propagating the errors as independent. PhysicsAll are extrapolations from the lattice, to be read on the assumption that the continuum theory exists (7-01).

Rewriting formula (7) at one loop gives

$f(x)=e^{-1/x}$ ($x\gt0$), $f(x)=0$ ($x\le0$) can be differentiated any number of times at $x=0$, and all its derivatives are 0. This is because $f^{(n)}(x)=P_n(1/x)\,e^{-1/x}$ ($P_1(u)=u^2$, $P_2(u)=u^4-2u^3$), and polynomials cannot beat $e^{-u}$ Known. Expanding (8) in powers of $g^2$, all the coefficients are 0.

Perturbation theory (4-03) computes a power series in $g^2$, so a quantity of the form (8) looks like 0 at every order. §6.6 of the problem statement writes that, since the gap is not classically visible, showing it may need a non-classical change of variables ('change of variables or “duality transformation.”') Known. The abstract of Gross–Neveu 1974 also says that a mass generated dynamically in an asymptotically free theory 'must depend on the coupling constants in a nonanalytic fashion, vanishing exponentially when these vanish' (at the level of the renormalisation group). The tunnelling splitting of the double well is of the same type (2-05).

Example (Table 4). As a rough guide with coefficients set to 1, the order $n^*=\ln(\Lambda/\mu)/\ln\alpha$ at which the $n$-th power of $\alpha=g^2/4\pi$ drops to $\Lambda/\mu$ is 2.8 at $g^2=1$, and grows without bound as $g\to0$. The actual coefficients grow with the order, and the series is thought to diverge (4-12).

Computedpython3. The rightmost column (§08) agrees with $(\Lambda/\mu)^{11}$ in all digits. $P_n$ was checked up to $n=6$.

An exact example exists in two-dimensional quantum mechanics. In units with $\hbar^2/2m=1$, consider the delta-function well in $d$ dimensions

$\delta^d$ has the dimension of length to the power $-d$, and $\nabla^2$ to the power $-2$, so $g$ has the dimension of length to the power $d-2$, and is dimensionless at $d=2$. This is the same situation as four-dimensional Yang–Mills. Taking the Fourier transform as $\hat\psi(\mathbf k)=\int\psi(\mathbf x)\,e^{-i\mathbf k\cdot\mathbf x}\,d^dx$ (1-11), the transform of $\delta^d$ is 1. A bound state $E=-\kappa^2$ satisfies $(k^2+\kappa^2)\,\hat\psi(\mathbf k)=g\,\psi(0)$, so dividing by $k^2+\kappa^2$ and integrating gives

($K$ is the momentum cutoff). For $d=2$ the right-hand side is $(g/4\pi)\ln(1+K^2/\kappa^2)$, which diverges logarithmically, as does one-loop four-dimensional Yang–Mills (4-04). Solving,

the same form as formula (8). Renormalisation is the operation of fixing $E_B$ and taking $K\to\infty$, which sends the coupling towards 0 as $g(K)=4\pi/\ln(1+K^2/E_B)$. For $K^2\gg E_B$,

the sign is negative, that is, asymptotically free. The only parameter left in the limit is $E_B$. Up to here this is an elementary calculation with a momentum cutoff Computed.

On the mathematical side, it is known that in the limit where the cutoff is removed the Hamiltonian converges (in the sense of the resolvent $(H-z)^{-1}$) to a self-adjoint operator with exactly one bound state (energy $-E_B$) (Albeverio et al., Part I, Chapter 5; bibliographic data only) Known. The limit there is taken by shrinking a short-range potential while renormalising the coupling logarithmically, a different route from the momentum cutoff above. As teaching material there is Gosdzinsky–Tarrach 1991 (bibliographic data only).

ComputedThe numbers in this figure were computed on this machine. The closed form of formula (10) agrees with numerical integration to 11 digits, and $K\,dg/dK$ agrees with $-g^2/2\pi$ for $K\ge10^4$ (ratio 1.000000). The factor by which $E_B$ changes when $g_2$ is moved by 1% is 1.05 at $K=10$ and 1.73 at $10^{12}$.

For $d=1$, $E_B=g^2/4$ is a power of $g$, and $g(K)$ settles at 2. For $d=3$, $g$ has to be brought to the critical value $2\pi^2/K$ as a power. These are the cases where the mass dimension of the coupling is positive, zero and negative; four-dimensional Yang–Mills is on the $d=2$ side, and three-dimensional Yang–Mills on the $d=1$ side (an analogy).

Detour: Cooper pairs in a metal

The value of $\Lambda$ changes with the definition of the coupling (the scheme). If $g$ of scheme A and $\tilde g$ of scheme B are related by $\tilde g=g+c\,g^3+O(g^5)$, then $1/\tilde g^2=1/g^2-2c+O(g^2)$, so from formula (3)

Since $b_0,b_1$ do not depend on the scheme (4-06), this ratio holds exactly for the $\Lambda$ of formula (4). The abstract of Celmaster–Gonsalves 1979 writes that the ratio of $\Lambda$ in two schemes is 'a pure number which we show is exactly deducible from our one-loop results' Known.

Example (Table 6). MS (minimal subtraction: the scheme that subtracts only the poles appearing in dimensional regularisation; 4-04) and $\overline{\mathrm{MS}}$, the PDG's standard (the scheme that also subtracts the constant $\ln4\pi-\gamma_E$ along with the poles; Bardeen et al. 1978), differ only by the rescaling $\mu^2\to\mu^2e^{\gamma_E}/4\pi$, and the ratio is $\sqrt{4\pi e^{-\gamma_E}}=2.656$. The ratio 28.81 to the lattice Wilson action (SU(3), 4-06) also comes from a shift in $1/g^2$ of only 0.468. The one-loop difference enters the exponent multiplied by $1/b_0$, so $\Lambda$ changes by an order of magnitude or more.

Computedpython3. Building the β of scheme B from $\tilde g=g(1+cg^2)$ and integrating formula (4) numerically, the ratio is constant to 6 digits for $\mu/\mu_0=1$ to $10^{16}$. The ratio for the formula with the prefactor power only is slow: 6.76→23.08 (SU(3)).

Ratios of physical quantities such as $m/\sqrt\sigma$ do not depend on the scheme. $r_0\Lambda$ is a single number only once the scheme is named. To attach MeV, another quantity is borrowed (7-01).

The classical scale symmetry of §01 leaves traces in two places in the quantum theory. One is the size of instantons: $\rho$, free in the classical theory, is distinguished by the running of the coupling (Tong §2.3.1–§2.3.2). Using the coupling at the scale $1/\rho$, from formula (3) and $16\pi^2b_0=11N/3$, at one loop

holds identically (checked against Table 4 Computed). Small instantons are suppressed by a power of $\rho\Lambda$, and the suppression stops working at $\rho\sim1/\Lambda$. The one-loop calculation is 't Hooft 1976 Physics.

The other is the trace of the energy–momentum tensor. The divergence of the conserved current of scale transformations (2-01) is the trace $T^\mu{}_\mu$, which is 0 classically in four dimensions (3-03). In the quantum theory a term in $F^a_{\mu\nu}F^{a\,\mu\nu}$ proportional to $\beta(g)$ remains (Collins–Duncan–Joglekar 1977; for the formula, 8-04) Physics. §4 of the problem statement also includes 'the existence of a stress tensor' in the short-distance requirements Known.

On the lattice, a test from formula (7) is whether ratios of correlation lengths $\xi_{\mathrm{lat}}=1/(am)$ approach constant values (5-05, 5-09). We list the degree of certainty example by example. The top four are two-dimensional models with a dimensionless coupling, and what has become a theorem is either a model with finitely many degrees of freedom or a form that keeps the cutoff.

Row by row, we separate what is a theorem from what is physics. For the delta well, the binding condition (10) with a cutoff is an elementary calculation, and the operator in the limit with the cutoff removed is in the mathematical literature. For the two-dimensional Gross–Neveu model (a model of interacting fermions), keeping the ultraviolet cutoff and taking the number of components $N$ large (but finite), it is a theorem that the two-point function decays exponentially, that is, that a mass is generated. For the two-dimensional O(N) σ model (a model of fields valued in $N$-component vectors of length 1), mass generation in a regularised version is also a theorem, but there is still no construction of the theory with the cutoff removed. The $8/e$ of O(3) is a physical value obtained by matching the result of the Bethe ansatz (a method that uses the exact solvability of scattering in two-dimensional integrable models) with perturbation theory. For pure SU(3) in four dimensions, the ratio $m/\Lambda$ comes only from lattice numbers, and neither the construction of the theory nor $\Delta\gt0$ has been shown.

The problem statement does not mention $\Lambda$; it asks for $\Delta\gt0$ and for agreement with the predictions of asymptotic freedom at short distances (0-03). If the theory exists and agrees with those predictions, in physics one considers that $\Lambda$ can be read off from them Physics. The mass gap is then the claim that $C$, which shows up neither in perturbation theory nor in the classical theory, is positive. How this single scale is used is treated in 7-01 and 7-12.

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