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Feynman Path Integral Formulation

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8.5 Wilson Lines and Static Potential 281<br />

8.5 Wilson Lines and Static Potential<br />

In a gauge theory such as QED the static potential can be computed from the manifestly<br />

gauge invariant Wilson loop. To this end one considers the process where<br />

a particle-antiparticle pair are created at time zero, separated by a fixed distance<br />

R, and re-annihilated at a later time T . In QED the amplitude for such a process<br />

associated with the closed loop Γ is given by the Wilson loop<br />

{ ∮<br />

W(Γ )=< exp ie A μ (x)dx μ} >, (8.35)<br />

Γ<br />

which is a manifestly gauge invariant quantity. Performing the required Gaussian<br />

average using the (Euclidean) free photon propagator one obtains<br />

{ ∮<br />

< exp ie A μ dx μ} ∮ ∮<br />

}<br />

> = exp<br />

{− 1 2 e2 dx μ dy ν Δ μν (x − y) . (8.36)<br />

Γ<br />

Γ Γ<br />

For a rectangular loop of sides R and T one has after a short calculation<br />

{ ∮<br />

< exp ie A μ dx μ} > ≃ exp<br />

{− e2 e2 T<br />

(T + R)+<br />

Γ<br />

4πε 4π R + e2<br />

)+···}<br />

2π 2 log(T ε<br />

[ ]<br />

(8.37)<br />

∼ exp −V (R) T ) , (8.38)<br />

T ≫R<br />

where ε → 0 is an ultraviolet cutoff. In the last line use has been made of the fact<br />

that for large imaginary times the exponent in the amplitude involves the energy for<br />

the process multiplied by the time T . Thus for V (R) itself one obtains<br />

1<br />

{ ∮<br />

V (R)=− lim<br />

T→∞ T log < exp ie A μ dx μ} > ∼ cst. − e2<br />

Γ<br />

4πR , (8.39)<br />

which is the correct Coulomb potential for two oppositely charged particles.<br />

To obtain the static potential it is not necessary to consider closed loops. Alternatively,<br />

in a periodic box of length T one can introduce two long oppositely oriented<br />

parallel lines in the time direction, separated by a distance R and closed by the periodicity<br />

of the lattice, and associated with oppositely charged particles,<br />

{ ∫<br />

< exp ie A μ dx μ} { ∫<br />

exp ie A νdy ν} > (8.40)<br />

Γ<br />

Γ ′<br />

In the large time limit one can then show that the result for the potential V (R) is the<br />

same.<br />

In the gravitational case there is no notion of “oppositely charged particles”, so<br />

one cannot use the closed Wilson loop to extract the potential (Modanese, 1995).<br />

One is therefore forced to consider a process in which one introduces two separate<br />

world-lines for the two particles. It is well known that the equation for free fall can<br />

be obtained by extremizing the space-time distance travelled. Thus the quantity

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