2026-10-01 · chapter Yang–Mills and the mass gapPart 6 Constructive field theorysix stages

What constructive field theory is — splitting the "making" of quantum field theory into six stages

Physics textbooks compute assuming the theory exists. The mathematical question comes before that: whether there is an object satisfying the axioms (4-09, 4-10). The problem statement writes that not a single non-trivial relativistic field theory satisfying the axioms is known in four dimensions (JW §6, 2000).

The problem statement divides the ways of showing existence into two (JW §6.1). (i) Find an exact solution in closed form. (ii) Solve a sequence of approximate problems and show that the solutions converge to a limit. Constructive field theory (constructive quantum field theory) refers to (ii), and at its centre are inequalities independent of the approximation: uniform a priori estimates. Known

In the Euclidean form (4-02), the object to be made is a probability measure $\mu$ on the space of distributions $\mathcal S'(\mathbb R^d)$. If its moments

satisfy the OS axioms, the reconstruction theorem gives a Hilbert space and a Hamiltonian ($\phi(f)$ is the field smeared with a test function $f$). In physics the same measure is written

This is the target form, but none of the symbols on the right is a mathematical object as it stands.

Example: the free field of mass $m$ is the Gaussian measure with mean 0 and covariance $C=(-\Delta+m^2)^{-1}$ ($\Delta$ is the Laplacian, different from the $\Delta$ of the mass gap). It satisfies the OS axioms, but has no interaction and is trivial (JW §6.2). The work is to make a non-Gaussian measure. According to the problem statement, the first constructions of the form satisfying the axioms (as non-Gaussian measures) came in the 1970s, for fields in two dimensions with small polynomial interactions (JW §6.2).

A figure lining up, in eleven rows, what switches with the dimension is 1-12 §10 Figure 4 (opens at d = 4; the bottom row is the YM row of Table 2).

Expression (2) is broken in three places. (i) $\mathcal D\phi$: on an infinite-dimensional space, the only measure that is unchanged by translation and gives every ball a finite weight is 0 (1-09 §02). (ii) Values at points: in two or more dimensions samples of the free field are distributions, and $\phi(x)^4$ has no meaning (1-11). (iii) Volume: on the whole of $\mathbb R^d$ the action is infinite. The action of a scalar field is

and for Yang–Mills the field is a connection $A$ and the action is the integral of the square of the curvature $F$ (3-03).

(ii) can be seen in numbers. The variance at a point of the free field on the lattice $a\mathbb Z^d$ is

where the denominator is the lattice version of $m^2+|k|^2$, and the momenta are cut off at $|k_j|\le\pi/a$.

ComputedTable 1 $G_a(0)$ ($m=1$). Numerical integration of the heat-kernel representation $G_a(0)=a^{2-d}\int_0^\infty e^{-m^2a^2 s}\,[e^{-2s}I_0(2s)]^d\,\mathrm ds$ ($I_0$ is a modified Bessel function). For $d=1$ it agrees to 5 digits with the closed form $1/(m\sqrt{m^2a^2+4})$.

In two or more dimensions the variance goes to infinity, and values at points are not random variables. $\phi^4$ has to be redefined with a subtraction depending on $a$, and this is the entrance to the renormalisation of stage 3.

Replace the broken places with finite things (regularisation and finite volume), and remove them at the end (the two limits). Once the limits are taken, check the axioms and the spectral gap. The problem statement too writes, for the lattice, that one should establish the limits of expectation values as the lattice spacing goes to 0 and the volume to infinity (JW §6.5).

Stages 1–4 make the double limit

and stages 5 and 6 are properties of $\mu$. There is also the road, as in strong-coupling results on the lattice, of holding $a$ fixed and taking $L\to\infty$ first, in which case stage 3 is left for later. However, the strong-coupling region (where the lattice coupling $g$ is large and $\beta=2N/g^2$ small) is far from the continuum limit ($\beta\to\infty$), so this road does not as it stands connect to stage 3. Known

KnownTable 2 How far each theory has got (Closed = there is a theorem / Partial / Not yet / — = the question is hard to pose / Not taken = that road does not pass through the stage). From the problem statement and primary sources. Years are the publication years of representative papers (the timeline in 9-01 counts by first public appearance, so Sengupta is 1992–93 and Lévy 2001). Selecting a row changes Figure 1.

The "—" in Table 2 means that the question is hard to pose in its plain form. Two-dimensional Yang–Mills has no propagating degrees of freedom (waves that travel far, the counterpart of light in electromagnetism), so the question of a spectral gap cannot be posed in the same form as in four dimensions (6-02). Known "Not taken" means that the road does not pass through that stage. At strong coupling on the lattice $a$ stays fixed, so stage 3 is not passed. Stage 2 for three- and four-dimensional Yang–Mills is closed on the lattice by formula (6) (§04). Making the finite-volume measure while staying in the continuum has not yet been done, and that part is written as "Partial" in stage 3 and in the summary column.

Regularisation can be done with a lattice (putting the field on lattice points or edges) or a momentum cutoff (restricting Fourier components to $|k|\le\kappa$), and both break something.

On the lattice the finite-volume measure has meaning at once. If the gauge group $G$ is compact,

where $\ell$ are the edges of the lattice, $U_\ell\in G$, $\mathrm dU_\ell$ is the Haar measure (the invariant measure of total weight 1), and $S_W$ is the Wilson action

The sum is over the smallest squares of the lattice (plaquettes) $P$, and $U_P$ is the product of the $U_\ell$ on the four edges of $P$ taken in order around it (5-01). The integrand is bounded and continuous and the domain compact, so $Z_{a,L}$ is finite and positive (JW §6.5). Known

Example: a four-dimensional periodic lattice with 16 points on a side has $4\times16^4=262144$ edges, and SU(3) is 8-dimensional over the reals, so formula (6) is an integral in 2,097,152 dimensions. Computed

Making it directly in the continuum is hard, and even three-dimensional Yang–Mills has not yet been constructed on the torus $T^3$ (Chevyrev 2022). On the lattice, stage 2 is closed by formula (6) ("Closed (lattice)" in Table 2), and instead the recovery of rotational symmetry is passed on to stage 5.

As $a\to0$, unless the coefficients of the action (the bare couplings) are moved according to $a$, the limit diverges or collapses to something trivial. This moving is renormalisation (4-04). According to the problem statement, for $\phi^4$, if the cutoff-dependent constants are chosen in the form indicated by perturbation theory, expectation values converge even though the constants diverge. In two dimensions a few constants suffice; in three dimensions one needs all the coefficients plus wave-function renormalisation, that is, renormalising the state vectors (vectors of the Fock space of 2-06) by a constant that diverges with the cutoff (JW §6.2). Known

What is needed can be guessed by counting dimensions. Known Since the action has no units, in units of mass

($g$ is the Yang–Mills coupling). For $d\lt4$ the coupling has positive mass dimension (super-renormalisable), and there are finitely many types of superficially divergent diagrams (pictures of the terms of perturbation theory; 4-03). "Superficially" means that divergence is judged only by counting the powers of momentum in the integral attached to the diagram (4-04; as a textbook, Rivasseau 1991). $d=4$ is the boundary where the coupling has no units. Table 1 can be read by the same counting. Since $[\phi^2]=d-2$, $G_a(0)$ grows as $a^{2-d}$, and for $d=2$ logarithmically (1-12).

Example: in two dimensions it suffices to redefine the product as ${:}\phi^4{:}_a=\phi^4-6G_a(0)\phi^2+3G_a(0)^2$ (the Wick-ordered product), together with a constant for the vacuum energy (the $\alpha,\gamma$ of JW §6.2).

What one wants to show is that expectation values are bounded by constants independent of $a$,

With a uniform estimate (8), one can take a subsequence of the measures that converges in a weak sense (the measure version of taking a convergent subsequence from a bounded sequence of numbers). The limit, however, need not be unique, and the axioms and the gap are separate questions. The problem statement does not accept weak existence by compactness alone as a solution unless the properties of the limit are shown separately (JW footnote 2). Four dimensions is treated in 6-07.

$L\to\infty$ needs estimates independent of $L$, and the key is in particular that quantities far apart become nearly independent (clustering). The problem statement lists as ways to set up estimates (a) correlation inequalities, (b) symmetries, and (c) convergent expansions (cluster expansions), and says that (a) tends to be limited to scalar fields and abelian gauge theories, and (c) became the central tool (JW §6.1). The form one wants to show is

($B_x$ is $B$ shifted by $x$). Given a decay rate $m$, the influence of the edge of the box reaches only about $e^{-mL}$.

Example: for the one-dimensional chain, the difference between the variance at a point of a ring of length $L$ and the value for the infinite chain is, for large $L$, almost $e^{-mL}/m$. For $m=1$ it is $1.1\times10^{-7}$ at $L=16$; for $m=0.1$ it is still 2.5. The rate of approach is set by $m$. Computed

In four-dimensional Yang–Mills the coupling is thought to become strong at long distances. Physics This view comes from the renormalisation group of perturbation theory (4-04). MRS 1993 §I writes, as the reason for not removing the infrared cutoff, 'this would lead to large values of the coupling constant', and further 'there is no easy solvable model of these phenomenons around which to expand'.

As a metaphor, the scaffolding for the expansion is gone. Precisely, there is neither a solvable model to serve as the starting point of an expansion nor a small parameter to expand in. The problem statement too says that present methods do not suggest how to obtain $T^4\to\mathbb R^4$ (JW §6).

Axioms that hold with the regularisation tend to pass to the limit, and what was broken has to be recovered. The key is reflection positivity: for time reversal $\Theta:t\mapsto-t$, for $F$ depending only on the field at $t\gt0$,

holds. This is the condition for extracting a Hilbert space and a Hamiltonian (5-03).

Formula (10) is an inequality, so it passes to the limit. Known There are three conditions. The $F$ for which the limit is taken can be approximated by functions depending only on the lattice field on the positive-time side. Their expectation values converge. The reflection plane is a lattice plane (a plane through lattice points, or a plane between neighbouring lattice points). Then it is a limit of non-negative values in each $\mu_{a,L}$, so it is non-negative in the limit too. For lattice gauge theory, reflection positivity with respect to lattice planes has been shown (Osterwalder–Seiler 1978; MRS 1993 §I also cites it).

Conversely, ways to recover lost reflection positivity are scarce, and the problem statement writes that setting up the Hilbert space is itself part of the problem (JW §6.5). Rotational symmetry broken by the lattice has to be shown to recover separately. Known

Example: for a Gaussian measure, reflection positivity is the same as positivity of $\Theta C$ restricted to the positive-time side (JW §6.2). In one dimension $C(t)=e^{-m|t|}/(2m)$, so for $t_i,t_j\gt0$

becomes a square. On a lattice chain with 40 points, trying 20000 random $f$ gives a minimum of $1.4\times10^{-8}$, never negative. Being a square, this is to be expected; it is a check of the formula. As a control, with the sign-reversed covariance $-C(t_i+t_j)$ the smallest eigenvalue of the matrix is $-2.49$, and reflection positivity fails. Computed

Stochastic quantisation (6-09) makes the measure as the invariant measure of an equation, so reflection positivity is not part of the output, and has to be shown as a separate task.

Next one asks about the gap in the spectrum of the Hamiltonian (4-11). On the Euclidean side it is the rate of decay of correlations. On the lattice the decay rate in the time direction is the gap in lattice units $m_{\rm lat}$, and the gap in physical units is

One wants to show that this is positive and finite, and moreover uniform in the volume.

On a lattice in a finite box the gap is there automatically. Known The transfer matrix $T$ is the integral operator that carries weight from the slice at time $t$ (the set $u$ of values of the lattice field at that time) to the slice $v$ at time $t+a$

The kernel $K(u,v)$ is made as $e^{-S_{t,t+a}(u,v)}$ from the part $S_{t,t+a}(u,v)$ of the action connecting the two times. Reflection positivity makes $T$ a positive operator, and the Hamiltonian is fixed by $T=e^{-aH}$. If the range of field values is bounded and closed (lattice gauge theory with a compact gauge group is such a case) and the kernel is continuous and positive, the largest eigenvalue $\lambda_0$ is not degenerate. This is the infinite-dimensional version (a generalisation of Jentzsch's theorem) of the Perron–Frobenius theorem for matrices with all entries positive Known (the generalisation is Schaefer 1974; sources in 5-03). So the ratio to the next eigenvalue $\lambda_1$ is $\lambda_1/\lambda_0\lt1$, and there is a gap $-\log(\lambda_1/\lambda_0)/a\gt0$ (5-03; the positive self-adjoint transfer matrix of lattice gauge theory is Osterwalder–Seiler 1978). What is asked is uniformity under $a\to0$ and $L\to\infty$, and the problem statement writes that a gap uniform in the volume needs new ideas (JW §6.5).

Example: for the chain with $m=1$, $m_{\rm lat}/a$ approaches 1 as 0.9624 ($a=1$), 0.9974 ($a=0.25$), 0.99990 ($a=0.05$), while $m_{\rm lat}$ in lattice units itself goes towards 0, about 0.05 at $a=0.05$. In the continuum limit the gap in lattice units closing is the correct picture, and what should be looked at is the ratio (12) (5-09). The ratio of eigenvalues of the transfer matrix, $-\log(\lambda_1/\lambda_0)/a$, agrees with the decay rate of correlations, the difference being at most $5\times10^{-14}$. Computed

Each row of Table 2 shows the stages being closed in order from stage 1. The chapters of Part 6 follow, stage by stage, which tools gave each "Closed", and where the "Partial" and "Not yet" stop. Reading two-dimensional Yang–Mills and scalar fields as examples where the stages went through, the cluster expansion and the renormalisation group as tools for the stages where they work, and the chapters on three and four dimensions as records of the stages that are stuck, shows where each chapter stands. One chapter may touch several stages, so the correspondence below gives the main stages treated.

• Stages 1–4: 6-02 two-dimensional Yang–Mills
• Stages 1–6: 6-03 $\phi^4_2$, $\phi^4_3$ (examples where all six stages went through)
• Stages 4 and 6: 6-04 the cluster expansion
• Stage 3: 6-05 the renormalisation group, 6-06 three dimensions, 6-07 four dimensions, 6-08 triviality, 6-12 Detour: 1/0
• Stages 2 and 3 by another road: 6-09 stochastic quantisation
• All: 6-10 where four dimensions stops, 6-11 obstacles other than divergence, 9-01 the timeline

What this chapter can say

Sources and reproduction

Next chapter to read: 6-02 Solvability in two dimensions and the Yang–Mills measure (an example in which stages 1–4 close).

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