The energy cost of stretching an electric-flux string

Hold the matter still

Separate a positive and a negative unit source by $d$ links. Keep matter in the neutral reference pattern and set both boundary fluxes to zero. Gauss’s law requires unit flux between the sources and zero outside. Add the identical electric costs of those $d$ links.

The straight line has a simple origin: a one-dimensional field cannot spread sideways. A three-dimensional point charge can. The formation of a flux tube in a three-dimensional gauge theory requires its own dynamics; this one-dimensional argument does not supply that derivation.

V_{\mathrm{string}}(d)=\frac{g^2}{2}\,d

This is the electric energy for unit sources, zero boundary background, neutral fixed matter and lattice spacing one. The shared reference mass energy has been subtracted.

Field energy

V/g^2

Source separation

d

Analytic model reference with fixed neutral matter and unit sources. Every added link stores the same field energy. These points are calculated values.

Reading: David Tong · Sandbox Physics FR04 finite Hamiltonian

What can one particle and one hole save?

For positive mass, a negative hole starts on an occupied odd site and a positive particle occupies an even site. Place the hole near the positive source and the particle near the negative source. Their charges can cancel the original field on $\ell$ links, saving $g^2\ell/2$ in field energy. The mass cost is $2m$ .

Add the two changes to compare the definite configurations. Once hopping is restored, a ground state generally mixes several configurations. The crossing of their separate diagonal energies becomes a mixed finite-system problem.

\Delta E=E_{\mathrm{screened}}-E_{\mathrm{string}}=2m-\frac{g^2}{2}\ell

A negative

\Delta E

favors this screened configuration energetically. The charge positions and Gauss’s law determine how many links it actually clears.

Two pictures solve the four-site example

Take the four-site chain with endpoint sources. The initial occupations $(0,1,0,1)$ give flux $(1,1,1)$ . Moving the middle particle right gives $(0,0,1,1)$ and flux $(1,0,1)$ . Only the middle link clears, so $\ell=1$ .

Choose $g^2=1$ . Initial electric energy is $3/2$ and screened electric energy is $1$ . At mass $m=1/4$ , the pair costs $1/2$ , exactly the field energy saved. Both configurations have the same total energy. Increasing mass raises the pair cost; reducing mass makes this screening configuration cheaper.

The remaining field on the two edge links follows from the staggered charge positions and the source placement. Screening does not require every link in the picture to vanish. It requires checking where the allowed matter charges can reside.

Four-site energies after subtracting the common reference mass energy.
Configuration	Added mass energy	Electric energy
Initial string	$0$	$3g^2/2$
One link screened	$2m$	$g^2$

An energy comparison becomes a dynamical question

At zero hopping, even an energetically expensive initial string remains unchanged. No matrix element can move its occupations. Restoring hopping lets configurations exchange probability amplitude. The transition rate then depends on couplings, detunings and the other configurations along the way.

Slowly changing parameters asks another question: can the state follow the changing ground state? A sudden string release instead evolves one chosen initial state with a fixed Hamiltonian. The next chapter solves that latter problem and follows both matter and electric field in time.

Try the prediction

Predict the field change before releasing the string

Select “Frozen matter” and explain why an energized string cannot change at zero hopping.
In advanced settings choose four Chain sites, Electric cutoff one and fixed external charges. Set the right source to site three, mass to 0.25, electric coupling to one and hopping to one. Predict which link can screen.
Increase mass with the other parameters fixed, then compare the field change and matter-pair count.

Exercises and explanations

An allowed pair clears three unit-flux links. For $m=1$ and $g^2=2$ , what is its energy relative to the initial string?

Read the explanation

The mass cost is $2m=2$ and the field-energy saving is $g^2\ell/2=3$ . The relative energy is $-1$ , one reference energy unit below the initial string. This comparison still gives no transition time.

If the central electric field decreases in an isolated open chain, has the system reached its screened ground state?

Read the explanation

Not necessarily. A quenched state usually contains several energy eigenstates. Their phases can reduce a local field and later restore it. Identifying a ground state requires energy, state overlap or a specified preparation protocol; one local snapshot is insufficient.

Sources and calculation scope

Sandbox Physics FR04 finite Hamiltonian, conventions and calculation scope

Defines the course’s zero-based staggered charge, open boundaries, integer-flux hard cutoff, source geometry and deliberate gauge-breaking error. The accompanying paper-reproduction and numerical-validation notes specify which finite-model calculations were checked independently. This is our model definition and calculation record.

David Tong, Gauge Theory (2018 lecture notes)

Sections 2.5.1 and 2.7 discuss static probes and dynamical matter; sections 4.2.1 and 4.3.1 introduce lattice links and two-dimensional staggered fermions; section 7.1 gives the electrostatic energy in one spatial dimension. Our finite Hamiltonian, indexing and hard cutoff are defined explicitly in this course.