2026-10-01 · chapter Yang–Mills and the mass gapPart 2 Physical preparationthe harmonic oscillator and Fock space

The harmonic oscillator and creation and annihilation — the levels are equally spaced, and the quanta of a field are counted as particles

Quantum field theory begins as a collection of harmonic oscillators. Near an equilibrium point $x_0$ any potential energy can be approximated as $V(x_0)+\frac12V''(x_0)(x-x_0)^2$, and the string too was a collection of independent simple oscillations (2-02). The position $x$ and the momentum $p=-i\hbar\,d/dx$ are self-adjoint operators on $L^2(\mathbb R)$, with $[x,p]=i\hbar$ (2-05).

Measuring length in $\ell=\sqrt{\hbar/m\omega}$ and energy in $\hbar\omega$, and putting $\xi=x/\ell$, $\eta=p\ell/\hbar$, we get $H=\frac{\hbar\omega}{2}(\xi^2+\eta^2)$, $[\xi,\eta]=i$. Example. ComputedFor a spring with 1 g and 1 Hz, $\ell\approx1.3\times10^{-16}$ m and $\hbar\omega\approx6.6\times10^{-34}$ J, a value 13 orders of magnitude smaller than $k_BT\approx4.1\times10^{-21}$ J at room temperature.

The eigenvalues of (1) can be found by algebra alone, without solving a differential equation ($a^\dagger$ is the adjoint of $a$).

Since $[a,a^\dagger]=1$, $a^\dagger a=\frac12(\xi^2+\eta^2-1)$,

From $[N,a]=-a$, $[N,a^\dagger]=a^\dagger$: if $N\psi=\nu\psi$, then $N(a\psi)=(\nu-1)a\psi$, $N(a^\dagger\psi)=(\nu+1)a^\dagger\psi$; $a$ lowers by 1 and $a^\dagger$ raises by 1. Since $\lVert a\psi\rVert^2=\langle\psi,N\psi\rangle=\nu\lVert\psi\rVert^2\ge0$, if $\nu$ is not an integer we keep going down and reach a contradiction; if it is an integer we stop at $a\psi_0=0$ with $\nu=0$. Going up never stops, by $\lVert a^\dagger\psi\rVert^2=(\nu+1)\lVert\psi\rVert^2\gt0$.

KnownThe levels are equally spaced, and the gap is $\hbar\omega$. Normalising, $a^\dagger\psi_n=\sqrt{n+1}\,\psi_{n+1}$, $a\psi_n=\sqrt n\,\psi_{n-1}$. This is the same argument as the ladder of 1-04, but that was finite-dimensional, stopping at both ends. No finite matrices satisfy $[a,a^\dagger]=1$ (the trace, 0 against the dimension, does not match), so this algebra is realised only in infinite dimensions.

In coordinates $a\psi_0=0$ is the first-order equation $(\xi+\frac{d}{d\xi})\psi_0=0$, and its solution is unique up to a constant multiple (no degeneracy).

The ground state is a Gaussian, and $\langle x^2\rangle=\hbar/2m\omega$, $\Delta x\,\Delta p=\hbar/2$ is exactly the lower limit of the uncertainty relation. We call $E_0=\hbar\omega/2$ the zero-point energy; it cannot be made 0 because position and momentum cannot both be brought towards 0 at the same time. Precisely, $\langle H\rangle\ge\omega\sqrt{\langle p^2\rangle\langle x^2\rangle}\ge\omega\,\Delta x\,\Delta p\ge\hbar\omega/2$. $\psi_n$ is the product of a Hermite polynomial and $e^{-\xi^2/2}$.

ComputedThe curves are computed on the spot from the recurrence $\psi_{n+1}=\sqrt{\frac{2}{n+1}}\,\xi\psi_n-\sqrt{\frac{n}{n+1}}\,\psi_{n-1}$ (which follows from (2) and the ladder relations) and the classical density $1/(\pi\sqrt{2n+1-\xi^2})$.

ComputedThe probability of being in the forbidden region, found by numerical integration, is 0.1573, 0.1116, 0.0951, 0.0740, 0.0602, 0.0425 for $n=0,1,2,5,10,30$ ($n=0$ is $\operatorname{erfc}(1)=\frac{2}{\sqrt\pi}\int_1^\infty e^{-t^2}\,dt$). For large $n$, $|\psi_n|^2$ with its wiggles averaged out approaches the classical density.

An eigenstate only has its phase turning, and its density does not move. For $\psi=c_0\psi_0+c_1\psi_1$, with $\Delta=E_1-E_0$ (the diagonal terms are 0 by parity),

The gap is the speed at which observable quantities oscillate, and for the oscillator it is the classical $\omega$. The weights of a coherent state $\psi_\alpha=e^{-|\alpha|^2/2}\sum_n\frac{\alpha^n}{\sqrt{n!}}\psi_n$ ($a\psi_\alpha=\alpha\psi_\alpha$) follow a Poisson distribution with mean $|\alpha|^2$, and if $\alpha$ is real the density keeps its shape and goes back and forth along the classical orbit.

ComputedComputed on the spot from formula (7) and the Poisson distribution.

In imaginary time, the gap is the rate at which correlations decrease

We compute (8) of 2-05 with the $A=x$ of the oscillator. From (2), $x=\ell\,(a+a^\dagger)/\sqrt2$, so $x\psi_0=\frac{\ell}{\sqrt2}\psi_1$, and there are no other components.

The rate of decrease is the gap $(E_1-E_0)/\hbar=\omega$, and (8) of 2-05 is an equality for the oscillator. For a general $V$ it becomes $G(\tau)=\sum_{n\ge1}|\langle\psi_n,x\psi_0\rangle|^2e^{-(E_n-E_0)\tau/\hbar}$, and the term $n=1$ remains at $\tau\to\infty$ (when $\langle\psi_1,x\psi_0\rangle\ne0$). Correlations in imaginary time decrease at the rate of the gap. Measuring glueball masses on the lattice (5-08) and restating the mass gap as the decay of correlations (4-11) both have this form (for the meaning of $t=-i\tau$ see 2-04; for the path integral, 2-07).

We add a quartic term ($\hbar=m=\omega=1$).

$x^4=\frac14(a+a^\dagger)^4$ moves $n$ by up to 4, so the ladder no longer closes. At first order of perturbation (replacing by the expectation value), $E_n\approx n+\frac12+\frac{3\lambda}{4}(2n^2+2n+1)$, and the gap is $1+3\lambda$.

ComputedTable 1 The levels of formula (9) found by finite differences (the change on halving the step is within $3\times10^{-6}$).

The estimate fits at $\lambda=0.01$ and fails at $\lambda=1$. The spacing widens further up, and the gap stays positive. KnownIn one dimension, if $V$ is continuous and $V\to\infty$ as $|x|\to\infty$, the spectrum consists only of eigenvalues (discrete spectrum, 1-10), and no eigenvalue is degenerate. Why there is no degeneracy can be said briefly. For eigenfunctions $\psi,\chi$ with the same $E$, the Wronskian $W=\psi\chi'-\psi'\chi$ is constant by the equation, $W'=0$, and is 0 far away, so it is identically 0; that is, $\psi$ and $\chi$ are proportional. The proof of discreteness is in the textbook in the table of sources (for the ground state see section 04 of 2-05).

If there are $K$ independent oscillators, $[a_k,a_l^\dagger]=\delta_{kl}$, $[a_k,a_l]=0$, and

The eigenstates are labelled by the tuple of occupation numbers $(n_1,\dots,n_K)$, and $n_k$ is "the number of quanta in mode $k$". All zeros is the vacuum $|0\rangle$, and $a_k^\dagger|0\rangle$ is a one-quantum state. Since $a_k^\dagger a_l^\dagger|0\rangle=a_l^\dagger a_k^\dagger|0\rangle$, quanta carry no name tags; only "how many in which mode" has meaning. Quantising the chain of 2-02 gives this form.

The form that does not depend on the number of oscillators is Fock space. Let the one-quantum space be $\mathfrak h$ (here $\mathbb C^K$; for a field, a space of functions). For $\mathfrak h=\mathbb C^K$, $\mathrm{Sym}^r\mathfrak h$ may be read as the space whose orthonormal basis is the tuples of occupation numbers above summing to $r$, $(n_1,\dots,n_K)$ (precisely, it is the symmetrisation of the tensor product of $r$ copies). If there are infinitely many modes, we take as basis the tuples with finitely many non-zero $n_k$ summing to $r$.

The direct sum is the set of all sequences whose sums of squared norms are finite, and the total number of quanta $\hat N=\sum_ka_k^\dagger a_k$ has value $r$ on $\mathrm{Sym}^r\mathfrak h$. Example. The number of ways to distribute $r$ quanta over $K$ modes is $\binom{r+K-1}{r}$, and $\dim\mathrm{Sym}^2\mathbb C^3=6$. KnownThe method of treating light with creation and annihilation is said to go back to Dirac (1927), and this space to Fock (1932).

On a ring of length $L$ (periodic boundary), we quantise a field with a mass term $m$ (2-02 §06) ($\hbar=c=1$). Placing an oscillator for each mode $e^{ikx}$ ($k=2\pi n/L$, $\omega_k=\sqrt{k^2+m^2}$; the dispersion relation $\omega^2=c^2k^2+\Omega^2$ of 2-02 with $c=1$, $\Omega=m$),

The total momentum is $P=\sum_kk\,a_k^\dagger a_k$. One quantum of mode $k$ carries energy $\sqrt{k^2+m^2}$ and momentum $k$, which is exactly the relation $E^2=p^2+m^2$ of a particle of mass $m$ (2-04). A particle is a quantum of a normal mode of the field. In the $(P,E)$ plane (Figure 4) a single particle lies on the hyperbola $E=\sqrt{P^2+m^2}$, and two or more particles fill the side $E\ge\sqrt{P^2+4m^2}$. Next after the vacuum is the single particle with $k=0$, and even when the box is widened the gap is $m$. KnownIn infinite volume $k$ ranges over all real numbers, the single-particle energies $\sqrt{k^2+m^2}$ cover $[m,\infty)$, and multi-particle states lie in $[2m,\infty)$, so the spectrum of the Hamiltonian is $\{0\}\cup[m,\infty)$. For $m=0$ the gap $2\pi/L$ shrinks to 0 (excluding $k=0$, which is not an oscillator).

ComputedThe dots are tuples of occupation numbers enumerated on the spot (the same procedure as Table 3).

ComputedTable 3 The number of states with $E\le3$ and the gap, for the free field in a box of period $L$.

The vacuum energy $\sum_k\omega_k/2$ of (15) diverges. Cutting off at $|k|\le\Lambda$ and putting $\Lambda\gg m$,

It is proportional to the volume and grows like $\Lambda^2$. Since what can be measured is differences, we discard the constant and put $H=\sum_k\omega_ka_k^\dagger a_k$ (normal ordering). The gap is a difference, so it does not change. PhysicsThe difference of vacuum energies when the box is changed can remain finite, and the Lüscher term is held to be an example (7-03).

Providing operators satisfying $[a_k,a_l^\dagger]=\delta_{kl}$ on a Hilbert space is called a representation of the commutation relations. KnownFor finitely many oscillators, when the commutation relations are written in exponential form (the Weyl relations; for one oscillator $e^{isx}e^{itp}=e^{-i\hbar st}e^{itp}e^{isx}$), the unitary irreducible representations (those whose only invariant closed subspaces are trivial) are unique up to unitary equivalence (being carried into each other by a unitary map) (Stone–von Neumann). For infinitely many, representations not unitarily equivalent to the Fock representation can be built. For example, the state with one quantum in every mode has an infinite sum of occupation numbers and is not in Fock space, and another representation with it as the vacuum can be considered. Fields with interactions are treated together with Haag's theorem in 4-09.

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