A gap made by the box

Prerequisites: the relativistic dispersion relation and the finite-volume lesson.

After this lab: you can explain why every finite box may have a positive lowest mode while no positive lower bound survives as its size grows, and identify the different behavior when a physical mass is already present.

A standing wave must fit between the ends of a string. A field constrained to vanish at both ends of an interval has the same restriction on its spatial wavelengths. That restriction can create a positive lowest frequency even when the input mass is zero. Enlarging the interval tests whether this positivity comes from the boundary or from the mass.

THE BOX AND THE MASS

A positive level can vanish in a limit

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Free relativistic modes on the interval [0,L], with the field zero at both endpoints. Units: ℏ = c = 1; mass and energy are inverse reference lengths.

Mass m

0.00

Inverse reference length; m = 0 is allowed

Interval length L

8.0

Reference lengths; Dirichlet endpoints stay fixed

Lowest mode E₁(L)0.392699Inverse reference length

Limit as L grows without bound0.000000The specified input mass m

Lowest mode against interval lengthSolid curve: E₁(L). Dashed line: its limiting value m. Marker: your selected interval.

Three modes at several interval sizes (same selected mass)

L	E₁	E₂	E₃
2	1.570796	3.141593	4.712389
4	0.785398	1.570796	2.356194
8	0.392699	0.785398	1.178097
16	0.196350	0.392699	0.589049
32	0.098175	0.196350	0.294524

The finite box creates this positive energy.

At m = 0, E₁ = π/L is positive for each finite L but tends to zero. There is no positive bound uniform in L.

Download mode table (CSV)

Exact mode model, not Yang–Mills data. Eₙ(L) = √[m² + (nπ/L)²], n = 1,2,… . Dirichlet boundaries exclude a constant spatial mode by definition; no periodic zero mode has been discarded. This spectrum does not construct an interacting field theory or certify its infinite-volume limit.

Derive the mode energies

Use a free scalar mode on $0<x<L$ with Dirichlet boundary conditions, $f(0)=f(L)=0$. Set $\hbar=c=1$. The spatial eigenfunctions and momenta are

$$f_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L},\qquad k_n=\frac{n\pi}{L},\qquad n=1,2,\ldots.$$

Applying $-d^2/dx^2$ to $f_n$ gives $k_n^2 f_n$. The positive relativistic one-particle operator $\sqrt{m^2-d^2/dx^2}$ therefore has eigenvalues

$$E_n(L)=\sqrt{m^2+\left(\frac{n\pi}{L}\right)^2}.$$

Here $L$ is measured in reference length units; $m$, $k_n$ and $E_n$ are in their inverse. The graph shows this exact formula, not simulated configurations or measured glueball energies. For the corresponding free finite-interval oscillator system, these are excitation energies above the vacuum, with the vacuum energy subtracted.

Worked examples: the quantifiers change the claim

First set $m=0$. At $L=\pi$, the lowest energy is $E_1=1$; at $L=2\pi$, it is $1/2$. For every finite positive $L$, $E_1(L)>0$. But for any proposed fixed lower bound $\Delta>0$, choosing $L>\pi/\Delta$ makes $E_1(L)<\Delta$. Thus “positive for each box” does not imply “bounded below by one positive number for all sufficiently large boxes.”

Now set $m=1$. At $L=\pi$, $E_1=\sqrt2$, and at $L=2\pi$, $E_1=\sqrt{5/4}$. Increasing the interval lowers the boundary contribution while retaining the mass:

$$\lim_{L\to\infty}E_1(L)=m,\qquad E_1(L)-m=\frac{(\pi/L)^2}{\sqrt{m^2+(\pi/L)^2}+m}.$$

For fixed positive $m$ and large $L$, the excess is approximately $\pi^2/(2mL^2)$. This expansion assumes $\pi/L\ll m$; it is inappropriate at $m=0$, where the exact falloff is $1/L$ instead.

Boundary conditions are part of the model

There is no constant Dirichlet mode: a constant function vanishing at both ends is zero. Periodic boundaries instead permit momentum $k=0$, so simply taking their first nonzero momentum and calling it the full gap would be wrong. The massless periodic zero mode also raises separate quantization questions. This lab keeps Dirichlet boundaries throughout rather than hiding that issue.

Exercises with solutions

Exercise 1. With $m=0$, what happens to the first three energies if you double $L$? Does adding more plotted modes prevent the lowest one from approaching zero?

Solution. Every $E_n=n\pi/L$ is halved. Adding modes only fills in more of the spectrum; it does not change $E_1$ or create a positive bound independent of $L$.

Exercise 2. At fixed $m>0$, why is $E_1$ decreasing in $L$, although it never drops below $m$?

Solution. The derivative is $-\pi^2/(E_1 L^3)<0$. Meanwhile $E_1^2=m^2+\pi^2/L^2>m^2$, so the mass is a lower bound for every interval. Its origin is the specified free dispersion relation, not evidence about an interacting gauge theory.

What this model does not establish

The mode calculation demonstrates a quantifier problem. It does not construct an infinite-volume massless scalar field, control Yang–Mills boundary effects, or prove a physical Yang–Mills gap. An interacting construction needs the relevant uniform estimates and compatible limits for its actual observables.

The free-field dispersion relation is explained in David Tong’s Quantum Field Theory, 2006/2007 notes, §2.2, equations (2.18)–(2.23). The interval boundary-value calculation is supplied explicitly above. For the target requiring a theory on unbounded four-dimensional space, see Jaffe–Witten, §4. Return to uniform estimates or compare with the independent cutoff limit.