4  Scalar QED on the Lattice

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4.1 Scalar QED in the Continuum

The action of scalar QED is \[ S = \int d^4x\left\{ \frac{1}{4e^2}F_{\mu\nu}^2 + |D_\mu\varphi|^2 + m^2|\varphi|^2 + \frac{\lambda}{4}|\varphi|^4 \right\}, \] where \(\varphi:\mathbb{R}^4\to\mathbb{C}\) is a complex scalar field, \(e\) is the gauge coupling, and

\[ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu, \qquad D_\mu = \partial_\mu + iA_\mu. \]

The action is invariant under gauge transformations parametrised by an arbitrary function \(\lambda(x)\): \[ A_\mu'(x) = A_\mu(x) - \partial_\mu\lambda(x), \qquad \varphi'(x) = e^{i\lambda(x)}\varphi(x). \]

One verifies that \(F_{\mu\nu}\) and the covariant derivative are gauge covariant: \[ F_{\mu\nu}' = \partial_\mu(A_\nu-\partial_\nu\lambda) - \partial_\nu(A_\mu-\partial_\mu\lambda) = F_{\mu\nu}, \] \[ D_\mu'\varphi' = (\partial_\mu + iA_\mu - i\partial_\mu\lambda)\,e^{i\lambda}\varphi = e^{i\lambda}D_\mu\varphi. \]

Goal: discretize the theory on a 4d lattice while preserving exact gauge invariance.

4.2 Two Failed Attempts

The difficulty is that the forward difference operator \(\partial_\mu^f\) does not satisfy the usual Leibniz rule with a spatially varying phase. Specifically, \[ \partial_\mu^f[e^{i\lambda}\varphi](x) = \frac{e^{i\lambda(x+ae_\mu)}\varphi(x+ae_\mu) - e^{i\lambda(x)}\varphi(x)}{a} \neq ie^{i\lambda(x)}\partial_\mu^f\lambda(x)\,\varphi(x) + e^{i\lambda(x)}\partial_\mu^f\varphi(x), \] because \(e^{i\lambda}\) is evaluated at \(x+ae_\mu\) in the first term of the numerator, not at \(x\).

First attempt. Try \(A_\mu' = A_\mu - \partial_\mu^f\lambda\), \(\varphi' = e^{i\lambda}\varphi\), and \(D_\mu\varphi = (\partial_\mu^f + iA_\mu)\varphi\). Then \[ D_\mu'\varphi'(x) = \partial_\mu^f[e^{i\lambda}\varphi](x) - ie^{i\lambda(x)}\partial_\mu^f\lambda(x)\,\varphi(x) + ie^{i\lambda(x)}A_\mu(x)\varphi(x) = \frac{e^{i\lambda(x+ae_\mu)}\varphi(x+ae_\mu) - e^{i\lambda(x)}\varphi(x)}{a} + ie^{i\lambda(x)}A_\mu(x)\varphi(x) - ie^{i\lambda(x)}\partial_\mu^f\lambda(x)\,\varphi(x). \] This is not equal to \(e^{i\lambda(x)}D_\mu\varphi(x)\). Consequently \(|D_\mu'\varphi'|^2 \neq |D_\mu\varphi|^2\), so this discretization is not gauge invariant. Failed.

Second attempt. Keep \(\varphi' = e^{i\lambda}\varphi\) and \(D_\mu\varphi = (\partial_\mu^f + iA_\mu)\varphi\), but try to determine \(A_\mu'\) by imposing \(D_\mu'\varphi' = e^{i\lambda}D_\mu\varphi\). Solving for \(A_\mu'(x)\) yields \[ A_\mu'(x) = A_\mu(x) + i\,\frac{e^{i[\lambda(x+ae_\mu)-\lambda(x)]}-1}{a}\,\frac{\varphi(x+ae_\mu)}{\varphi(x)}. \] The transformation of \(A_\mu\) depends on \(\varphi\), so \(A_\mu\) and \(\varphi\) cannot be treated as independent degrees of freedom. Failed.

4.3 Geometric Meaning of the Covariant Derivative

The two failures have a common root: at two neighboring sites \(x\) and \(x+ae_\mu\), the field transforms with different phase factors, \[ \varphi(x) \;\to\; e^{i\lambda(x)}\varphi(x), \qquad \varphi(x+ae_\mu) \;\to\; e^{i\lambda(x+ae_\mu)}\varphi(x+ae_\mu). \] From the point of view of gauge transformations, it does not make sense to subtract \(\varphi(x)\) from \(\varphi(x+ae_\mu)\) directly. One must first transport \(\varphi(x+ae_\mu)\) to \(x\), i.e. construct from \(\varphi(x+ae_\mu)\) a quantity that transforms with the same phase \(e^{i\lambda(x)}\) as \(\varphi(x)\).

The key is to understand the covariant derivative geometrically. The ordinary derivative is \[ \partial_\mu\varphi(x) = \frac{d}{da}\varphi(x+ae_\mu)\Big|_{a=0} = \lim_{a\to 0}\frac{\varphi(x+ae_\mu)-\varphi(x)}{a}. \] The covariant derivative is defined analogously but with a connection factor \(W(x\to x+ae_\mu)\) that “parallel transports” the field from \(x+ae_\mu\) back to \(x\): \[ D_\mu\varphi(x) = \frac{d}{da}\bigl[W(x\to x+ae_\mu)\,\varphi(x+ae_\mu)\bigr]\Big|_{a=0} = \lim_{a\to 0}\frac{W(x\to x+ae_\mu)\,\varphi(x+ae_\mu) - \varphi(x)}{a}. \] Dropping the \(a\to 0\) limit gives a natural lattice discretization.

4.4 The Wilson Line (Parallel Transporter)

We construct \(W(x\to x+av)\) — the parallel transporter (or Wilson line) along the straight line from \(x\) to \(x+av\) in the direction of a generic vector \(v\) — and require:

1. \(W(x\to x+av)\,\varphi(x+av)\) transforms as \(e^{i\lambda(x)}\varphi(x)\) under a gauge transformation.
2. \(W\) is a function of \(A_\mu\) alone, independent of \(\varphi\).

The unique solution is \[ W(x\to x+av) \stackrel{\mathrm{def}}{=} \exp\!\left\{i\int_0^a ds\,\sum_\mu v_\mu A_\mu(x+sv)\right\} \in U(1). \]

Proof of covariance. Under \(A_\mu\to A_\mu-\partial_\mu\lambda\) and \(\varphi\to e^{i\lambda}\varphi\): \[ W(x\to x+av)\,\varphi(x+av) \;\to\; e^{i\int_0^a ds\,\sum_\mu v_\mu[A_\mu-\partial_\mu\lambda](x+sv)}\, e^{i\lambda(x+av)}\varphi(x+av). \] Using \(\sum_\mu v_\mu\partial_\mu\lambda(x+sv) = \frac{d}{ds}\lambda(x+sv)\), the extra phase in the exponent telescopes: \[ \int_0^a ds\,\sum_\mu v_\mu\partial_\mu\lambda(x+sv) = \int_0^a ds\,\frac{d}{ds}\lambda(x+sv) = \lambda(x+av) - \lambda(x). \] Therefore \[ W(x\to x+av)\,\varphi(x+av) \;\to\; e^{i\lambda(x)}\,W(x\to x+av)\,\varphi(x+av), \] which transforms like \(\varphi(x)\), as required. In particular: \[ W'(x\to x+av) = e^{i\lambda(x)}\,W(x\to x+av)\,e^{-i\lambda(x+av)}. \]

Adjoint (reversed) Wilson line. Reversing the path and changing variables \(s' = a-s\): \[ W(x+av\to x) = \exp\!\left\{-i\int_0^a ds'\,\sum_\mu v_\mu A_\mu(x+s'v)\right\} = W(x\to x+av)^*. \]

4.5 Lattice Covariant Derivative

Dropping the limit \(a\to 0\) in the geometric definition of the covariant derivative gives the forward discrete covariant derivative: \[ D_\mu^f\varphi(x) \stackrel{\mathrm{def}}{=} \frac{W(x\to x+ae_\mu)\,\varphi(x+ae_\mu) - \varphi(x)}{a}. \]

Gauge covariance. Using the transformation laws of \(W\) and \(\varphi\): \[ D_\mu^{f\,\prime}\varphi'(x) = \frac{e^{i\lambda(x)}W(x\to x+ae_\mu)\,e^{-i\lambda(x+ae_\mu)}\cdot e^{i\lambda(x+ae_\mu)}\varphi(x+ae_\mu) - e^{i\lambda(x)}\varphi(x)}{a} = e^{i\lambda(x)}\,D_\mu^f\varphi(x). \] Therefore \(|D_\mu^f\varphi|^2\) is gauge invariant and can enter the lattice action.

4.6 The Plaquette

To discretize the kinetic term \(\frac{1}{4e^2}F_{\mu\nu}^2\) we need a gauge-invariant object built from the gauge field. Since \(F_{\mu\nu} = -i[D_\mu, D_\nu]\), it is natural to look for such an object in terms of Wilson lines.

Consider the plaquette: the Wilson line around the elementary square in the \(\mu\nu\)-plane (\(\mu\neq\nu\)) with corner at \(x\), \[ W_{\mu\nu}(x) \stackrel{\mathrm{def}}{=} W(x\to x+ae_\mu)\, W(x+ae_\mu\to x+ae_\mu+ae_\nu)\, W(x+ae_\mu+ae_\nu\to x+ae_\nu)\, W(x+ae_\nu\to x). \]

Property 1: Gauge invariance. Under a gauge transformation, consecutive phase factors cancel: \[ W_{\mu\nu}(x) \;\to\; e^{i\lambda(x)}\,W_{\mu\nu}(x)\,e^{-i\lambda(x)} = W_{\mu\nu}(x). \]

Property 2: Plaquette as an exponential of \(F_{\mu\nu}\). \[ W_{\mu\nu}(x) = e^{\,i\int_0^a ds\,dt\;F_{\mu\nu}(x+se_\mu+te_\nu)} = 1 + ia^2 F_{\mu\nu}(x) + O(a^3), \] from which \(F_{\mu\nu}(x) = \lim_{a\to 0}\dfrac{W_{\mu\nu}(x)-1}{ia^2}\).

Property 3: Real part of the plaquette. \[ \mathrm{Re}\,W_{\mu\nu}(x) = 1 - \frac{a^4}{2}F_{\mu\nu}^2(x) + O(a^5). \]

Properties 2 and 3 are proved below.

4.7 Proof of Property 2

We compute the surface integral of \(F_{\mu\nu}\) over the plaquette. Using \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\), together with \(\partial_\mu A_\nu(x+se_\mu+te_\nu) = \frac{d}{ds}A_\nu(x+se_\mu+te_\nu)\) and \(\partial_\nu A_\mu = \frac{d}{dt}A_\mu\): \[ \int_0^a\!ds\,dt\;F_{\mu\nu}(x+se_\mu+te_\nu) = \int_0^a dt\,\bigl[A_\nu(x+ae_\mu+te_\nu) - A_\nu(x+te_\nu)\bigr] - \int_0^a ds\,\bigl[A_\mu(x+se_\mu+ae_\nu) - A_\mu(x+se_\mu)\bigr]. \] Exponentiating: \[ e^{\,i\int_0^a ds\,dt\;F_{\mu\nu}} = \underbrace{e^{\,i\int_0^a ds\;A_\mu(x+se_\mu)}}_{W(x\to x+ae_\mu)} \underbrace{e^{\,i\int_0^a dt\;A_\nu(x+ae_\mu+te_\nu)}}_{W(x+ae_\mu\to x+ae_\mu+ae_\nu)} \underbrace{e^{-i\int_0^a ds\;A_\mu(x+se_\mu+ae_\nu)}}_{W(x+ae_\mu+ae_\nu\to x+ae_\nu)} \underbrace{e^{-i\int_0^a dt\;A_\nu(x+te_\nu)}}_{W(x+ae_\nu\to x)} \] where the last two factors use the property \(W(x+av\to x) = W(x\to x+av)^*\). The right-hand side is exactly \(W_{\mu\nu}(x)\), proving Property 2.

Expanding to leading order in \(a\), the integral \(\int_0^a ds\,dt\;F_{\mu\nu}(x+se_\mu+te_\nu) = a^2F_{\mu\nu}(x) + O(a^3)\), so \[ W_{\mu\nu}(x) = e^{ia^2F_{\mu\nu}(x)+O(a^3)} = 1 + ia^2F_{\mu\nu}(x) + O(a^3). \]

4.8 Proof of Property 3

Expanding the exponential to second order: \[ W_{\mu\nu}(x) = 1 + \left(i\int_0^a\!ds\,dt\;F_{\mu\nu}\right) + \frac{1}{2}\!\left(i\int_0^a\!ds\,dt\;F_{\mu\nu}\right)^{\!2} + O(a^6). \] The first-order term is purely imaginary (it does not contribute to the real part), while the square of the integral gives \(-a^4F_{\mu\nu}^2(x) + O(a^5)\). Therefore: \[ \mathrm{Re}\,W_{\mu\nu}(x) = 1 - \frac{1}{2}\!\left[\int_0^a\!ds\,dt\;F_{\mu\nu}(x+se_\mu+te_\nu)\right]^{\!2} + O(a^6) = 1 - \frac{a^4}{2}F_{\mu\nu}^2(x) + O(a^5). \]

This means that summing \((1 - \mathrm{Re}\,W_{\mu\nu}(x))\) over all plaquettes and dividing by \(a^4\) reproduces \(\frac{1}{2}F_{\mu\nu}^2(x)\) in the continuum limit, providing a natural gauge-invariant lattice discretization of the field-strength kinetic term.