2026-10-01 · chapter Yang–Mills and the mass gapPart 1 Mathematical preparationProbability and measure

Probability and measure — the mean settles down, and the fluctuations become Gaussian

Moving quantum field theory to imaginary time (Euclideanisation, 4-02) attaches a weight $e^{-S}$ ($S$ is the action) to each configuration of the field, and after normalisation it can be read as a probability. The problem statement too lists 'non-Gaussian, Euclidean-invariant measures' among the by-products of constructing field theories Known. In lattice Yang–Mills a group element $U_\ell$ sits on each edge $\ell$ (5-01), and the probability has the following form.

$dU_\ell$ is the uniform weight on the group (§03), and $Z$ the constant that makes the total 1 (2-08). Example: if there is one edge, the group is the circle $U(1)$ and $S=-\beta\cos\theta$ ($\beta$ is the inverse of the lattice coupling; the same letter as the inverse temperature of statistical mechanics), then (1) has density $e^{\beta\cos\theta}/Z$, and the larger $\beta$ is, the more it gathers near $\theta=0$. This chapter assembles the language for this.

A probability is a rule that distributes weights from 0 to 1 over subsets (events) of the whole $\Omega$ of what can happen. The family $\mathcal F$ of events that receive weights is assumed to contain $\Omega$ and to be closed under complements and countable unions (a σ-algebra). The triple $(\Omega,\mathcal F,P)$ is a probability space, and on $P$ we impose the following Known.

A non-negative set function with $\mu(\emptyset)=0$ satisfying the second half of (2) (countable additivity) is a measure, and a measure with total 1 is a probability (Durrett §1.1; the axiomatisation is said to begin with Kolmogorov's book of 1933). $\mathcal F$ is not taken to be all subsets because a construction using the axiom of choice produces subsets to which no length invariant under rotation and satisfying (2) can be assigned (Vitali 1905) Known. Example: a die is $P(A)=|A|/6$. The uniform distribution on $[0,1]$ is $P([a,b])=b-a$, that is, length (Lebesgue measure). The probability of each point is 0, yet some point always comes out; but (2) cannot be used for uncountable unions, so there is no contradiction. Something that holds with probability 1 is said to hold "almost surely".

A random variable is a real-valued function $X$ on $\Omega$ such that every $\{X\le a\}$ is an event (measurable). Its distribution is the measure obtained by pushing the weight of $\Omega$ onto the number line by $X$.

$F_X$ is the distribution function, and $\rho$ the density. Example: the number of heads in $n$ tosses of a coin with probability of heads $p$ follows the binomial distribution $\binom nk p^k(1-p)^{n-k}$, and keeping $np=\lambda$ while $n\to\infty$ gives the Poisson distribution $e^{-\lambda}\lambda^k/k!$.

A group of matrices that is closed and bounded, such as $SU(2)$, is called a compact group. A compact group $G$ has exactly one probability measure unchanged by multiplication from either left or right, that is, satisfying $\mu(gA)=\mu(Ag)=\mu(A)$ ($gA=\{ga : a\in A\}$) for every $g\in G$ and subset $A$; it is called the Haar measure (for the circle, $d\theta/2\pi$) Known. The $dU_\ell$ of (1) is this. Example: elements of $SU(2)$ are $U=a_0+i(a_1\sigma_1+a_2\sigma_2+a_3\sigma_3)$ ($\sigma_i$ are the Pauli matrices, $|a|=1$), and $SU(2)$ is the sphere $S^3$ (1-02). The Haar measure is the uniform distribution on $S^3$, and $\operatorname{tr}U=2a_0$. The cross-section at height $a_0$ is a sphere of radius $r=\sqrt{1-a_0^2}$ (area $\propto r^2$), and the width of the band is $da_0/r$, so the density of $a_0$ is proportional to $r$, and $\operatorname{tr}U$ has the semicircle density Known.

Making a histogram (20 bins) of $\operatorname{tr}U$ from $10^6$ uniform points on $S^3$, the difference between the fraction falling in each bin and the probability of each bin given by (4) (0.019–0.064) was within $6.4\times10^{-4}$ Computed. Figure 1 shows how it goes as the points increase. The same semicircle also appears as the distribution of eigenvalues of a large symmetric matrix with independent Gaussian entries, in the limit where it is suitably normalised and the size of the matrix goes to infinity (7-04).

ComputedThe histogram of Figure 1 embeds the counts per bin of the first $n$ points of a sample made on this machine (the same $10^6$ points as in the text above). The curve and the bin probabilities are computed from formula (4). If JavaScript is disabled, it can be read from the numbers in the text above.

The expectation is the total of "value × weight", and for continuous cases an integral. The Lebesgue integral slices the values instead, and adds up $c$ times "the weight of the set where the value is near $c$". As a metaphor, it is counting a pile of coins by denomination; to be exact, it is the limit of the sums $\sum_ic_iP(A_i)$ for functions taking only finitely many values.

Taking the expectation of $\mathbf 1_{\{|Y|\ge t\}}\le Y^2/t^2$ ($\mathbf 1_A$ is 1 on $A$ and 0 outside) as $Y=X-E[X]$ gives Chebyshev's inequality Known.

The characteristic function $\varphi_X(t)=E[e^{itX}]$ determines the distribution uniquely (Durrett §3.3). Example: for a die, $E=7/2$, $\operatorname{Var}=35/12$. The $E[(-1)^N]=e^{-2\lambda}$ of the Poisson distribution is used in the counting of centre vortices (7-02). For (4), $E[\operatorname{tr}U]=0$, $E[(\operatorname{tr}U)^2]=1$.

$X$ and $Y$ being independent means that the joint probability splits into a product.

Then the distribution of $(X,Y)$ is the measure that gives the weight $P(X\in A)P(Y\in B)$ to the rectangle $A\times B$ in the plane (the product measure), and $E[XY]=E[X]E[Y]$, $\operatorname{Var}(X+Y)=\operatorname{Var}X+\operatorname{Var}Y$, $\varphi_{X+Y}=\varphi_X\varphi_Y$ Known. If the covariance $E[XY]-E[X]E[Y]$ is 0 they are uncorrelated; independent implies uncorrelated, but the converse is false. Example: if $X$ takes $\{-1,0,1\}$ with equal probability and $Y=X^2$, the covariance is 0 but $P(X=0,Y=0)=1/3\ne1/9$.

In field theory we would like the fields at two distant points to be "almost independent". The problem statement gives clustering as a consequence of the mass gap $\Delta$ (4-11) Known. It is the property that the size of the correlation of local quantities $O(\vec x)$, $O(\vec y)$ with expectation 0 placed at points $\vec x,\vec y$ of space $\mathbb R^3$ (since the expectations are 0, the covariance above is just the expectation of the product) is, if $0\lt C\lt\Delta$, at most $e^{-C|\vec x-\vec y|}$ where $|\vec x-\vec y|$ is large enough. To be exact, $O$ are operators of quantum theory, and the correlation is the expectation in the ground state (the vacuum) (an example is in §07).

A Gaussian distribution with mean 0 in $n$ dimensions is fixed by the covariance matrix $C_{ij}=E[x_ix_j]$ alone (a positive definite symmetric matrix, 1-01).

There are three properties Known. (i) The image under a linear map $y=Mx$ is also Gaussian, with covariance $MCM^{\mathsf T}$. (ii) If jointly Gaussian, uncorrelated implies independent (if $C$ is diagonal the density splits into a product). (iii) Higher moments are the sum over all ways of pairing up the variables of products of $C$ (Isserlis 1918; in physics a formula of the same form is used as Wick's theorem (1950)).

For random $C$ of $4\times4$, against the right-hand side $-0.0769$ of (9), $4\times10^6$ samples gave $-0.0774\pm0.0005$ Computed. In one variable it is $E[x^{2k}]=(2k-1)!!\,\sigma^{2k}$. $(2k-1)!!=1\cdot3\cdot5\cdots(2k-1)$ (the double factorial, multiplying only the odd numbers) is the number of ways of pairing up $2k$ points (the partner of point 1 can be chosen in $2k-1$ ways, and the same is repeated for the rest). On the other hand, the moments of the semicircle (4) are the number of ways of joining $2k$ points on a circle by chords that do not cross, the Catalan numbers $\mathrm{Cat}_k=\binom{2k}{k}/(k+1)$ Paper (the detour computes both). Counting the non-crossing pairings and numerically integrating (4) both agreed with $\mathrm{Cat}_k$ for $k\le6$ (the integral to 6 digits) Computed. Choosing a row of Table 1 lines up those ways of pairing for $k$ in Figure 2.

ComputedThe pairings of Figure 2 are listed in the browser, and their numbers have been checked to agree with the counts of Table 1 (computed on this machine). If JavaScript is disabled, they can be read from Table 1.

(9) is the prototype of Feynman diagrams (4-03), and non-crossing pairings lead to the planar diagrams of large $N$ (7-04).

Detour: counting the Catalan numbers in two ways

To $N$ variables $x_0,\dots,x_{N-1}$ ($x_N=x_0$) arranged on a ring, attach a weight under which neighbours prefer close values and the whole is pulled back towards 0.

$(\Delta_{\mathrm{lat}}x)_j=x_{j+1}-2x_j+x_{j-1}$ is the lattice Laplacian. It is different from the mass gap $\Delta$ of §05, so we distinguish it with the subscript lat. Comparing with (8), (10) is the Gaussian distribution with covariance $C=(-\Delta_{\mathrm{lat}}+m^2)^{-1}$. The problem statement calls the Gaussian measure on the continuum $\mathbb R^d$ with a covariance of the same form (with the lattice Laplacian replaced by the continuum one) 'the free, mass-m0 field'. (10) is its analogue on a one-dimensional lattice (a ring), and on a $d$-dimensional lattice too the covariance has the same form $(-\Delta_{\mathrm{lat}}+m^2)^{-1}$ Known. If the ring is infinitely long, $C(0,x)=e^{-|x|/\xi}/(2\sinh(1/\xi))$ exactly Known.

$\xi$ is the correlation length, and the mass $m$ is roughly its inverse (both measured in units of the lattice spacing). This rate of decrease corresponds to the "gap" of quantum theory (2-06). (10) is the discretised imaginary-time path integral of an oscillator, and $m$ corresponds to the $a\omega$ of 2-07 (there $\omega=1$, so it is the step $a$ itself). The general theory of correlation length is in 2-08.

ComputedThe numbers of Figure 3 and Table 2 were computed on this machine (the figure is drawn from the exact correlations and fixed random numbers; the slope from $x=4$–12). If JavaScript is disabled, they can be read from Table 2.

For $m=0.05$ the $\xi$ read from the slope is larger than 20 because the ring is only about 3 times the correlation length, and the contribution $e^{-(N-x)/\xi}$ from going the other way round matters (a finite-box effect, 5-09). (10) is Gaussian, so by (9) all correlations are fixed by $C$, and there is no interaction (a free field).

Let $S_n$ be the sum of independent $X_1,X_2,\dots$ following the same distribution (mean $\mu$, variance $\sigma^2$). From §05, $\operatorname{Var}(S_n/n)=\sigma^2/n$, and putting this into (6) gives the weak law of large numbers Known.

Moreover, with $E|X_1|\lt\infty$ alone, $S_n/n\to\mu$ holds almost surely (the strong law, Durrett Theorem 2.4.1) Known. In "As is" in Figure 4, you can see the distribution of $S_n/n$ shrinking to $p$ and the left-hand side of (12) ($\varepsilon=0.1$).

ComputedThe numbers of Figure 4 and Table 3 were computed on this machine (the figure computes on the spot from the binomial distribution and an approximation formula for $\Phi$, agreeing with the Python values to 4 digits). If JavaScript is disabled, they can be read from the next paragraph and Table 3.

For $p=1/2$, $\varepsilon=0.1$, the left-hand side of (12) is 0.754, 0.0569, $2.7\times10^{-10}$ at $n=10,100,1000$, and the right-hand side $25/n$ is 2.5, 0.25, 0.025 Computed. The right-hand side is a crude upper bound, but enough for the limit. The practical reading is that the error of a sample mean is about $\sigma/\sqrt N$, so ten times the precision needs a hundred times the samples. Estimating $E[(\operatorname{tr}U)^2]=1$ by sample means gave $0.916\pm0.092$ for $N=10^2$ and $1.0015\pm0.0010$ for $10^6$ (a different sample from Table 1; the error bar is the sample standard deviation $/\sqrt N$) Computed. In lattice Monte Carlo, successively generated configurations are not independent, so the effective $N$ decreases (5-07).

The deviation $S_n/n-\mu$ is of size $1/\sqrt n$. Magnifying it by $\sqrt n$ makes the Gaussian distribution appear, whatever the original distribution (the central limit theorem, Durrett Theorem 3.4.1) Known.

The reason is visible through characteristic functions. If $\mu=0$, then $\varphi(s)=1-\sigma^2s^2/2+o(s^2)$, and by independence $\varphi(t/\sigma\sqrt n)^n=(1-t^2/2n+o(1/n))^n\to e^{-t^2/2}$, which is the characteristic function of the one-dimensional standard Gaussian of (8). The continuity theorem, which derives convergence of distributions from convergence of characteristic functions (Durrett Theorem 3.3.17), gives (13) (the classical proof under general conditions is Lindeberg 1922). The rate of approach is given by the Berry–Esseen inequality ($\rho=E|X-\mu|^3$).

Durrett proves it with $C_0=3$, and Shevtsova (2011) has shown $C_0\lt0.4748$ Known. Both sides can be compared in the magnified view of Figure 4 and in Table 3.

For $p=1/2$, $D_n\sqrt n$ settles at $1/\sqrt{2\pi}\approx0.399$. If $n$ is even, the distribution function has a step of height about $\sqrt{2/\pi n}$ at $a=0$, and the smooth $\Phi$ passes through its middle, so half the step always remains (Durrett §3.4.4). If $n$ is odd there is no step in the centre, and between the two central steps ($a=\pm1/\sqrt n$) the distribution function stays flat at $1/2$, leaving a deviation of the same size at its ends ($D_n\sqrt n=0.393$ at $n=11$, 0.399 at $n=199$) Computed. Either way it is not faster than $1/\sqrt n$. For $p=0.02$, $n=100$ we have $np=2$: the distribution is close to an asymmetric Poisson distribution, and still far from Gaussian. Limit theorems do not say "when it is close", so an estimate of the rate is needed.

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