Appendix A: stability analysis of the semi-implicit scheme for Abelian gauge fields
In this section of the appendix we prove stability for the semi-implicit scheme for Abelian fields derived in Sect. 3.3. Using temporal gauge, \(A_{x,0}=0\), the equations of motion (86) and (87) read
$$\begin{aligned} -\partial _{0}^{2}A_{x,1}&= \frac{1}{2}\sum _{i}\partial _{i}^{B}\left( W_{x,1i}+M_{x,1i}\right) , \end{aligned}$$
(A.1)
$$\begin{aligned} -\partial _{0}^{2}A_{x,i}&= \sum _{j\ne i}\partial _{j}^{B}M_{x,ij}+\frac{1}{2}\partial _{1}^{B}\left( W_{x,i1}+F_{x,i1}\right) , \end{aligned}$$
(A.2)
where \(W_{x,1i}\) reduces to
$$\begin{aligned} W_{x,1i}=\partial _{1}^{F}A_{x,i}-\partial _{i}^{F}{\bar{A}}_{x,1}. \end{aligned}$$
(A.3)
The Gauss constraint (85) reads
$$\begin{aligned} \sum _{i=1}^{d}\partial _{i}^{B}\partial _{0}^{F}A_{x,i}+\left( \frac{a^{0}}{2}\right) ^{2}\sum _{i}\partial _{1}^{B}\partial _{i}^{B}\partial _{0}^{F}\left( \partial _{1}^{F}A_{x,i}-\partial _{i}^{F}A_{x,1}\right) =0.\nonumber \\ \end{aligned}$$
(A.4)
Splitting the equations of motion into Laplacian terms and mixed derivative terms we find
$$\begin{aligned} -\partial _{0}^{2}A_{x,1}&= -\sum _{i}\partial _{i}^{2}{\overline{A}}_{x,1} +\frac{1}{2}\sum _{i}\partial _{i}^{B}\partial _{1}^{F}\left( A_{x,i}+{\overline{A}}_{x,i}\right) , \end{aligned}$$
(A.5)
$$\begin{aligned} -\partial _{0}^{2}A_{x,i}&= -\sum _{j\ne i}\partial _{j}^{2}{\overline{A}}_{x,i}-\partial _{1}^{2}A_{x,i} \nonumber \\&\quad +\sum _{j\ne i}\partial _{j}^{B}\partial _{i}^{F}{\overline{A}}_{x,j}+\frac{1}{2}\partial _{1}^{B}\partial _{i}^{F} \left( {\overline{A}}_{x,1}+A_{x,1}\right) . \end{aligned}$$
(A.6)
Using a plane wave ansatz
$$\begin{aligned} A_{x,i} = A_i e^{i \left( \omega x^0 - \sum _i k^i x^i \right) }, \end{aligned}$$
(A.7)
we will use the Gauss constraint to first reduce the number of degrees of freedom and then compute the dispersion relation \(\omega (k)\). Inserting the ansatz into the Gauss constraint we find
$$\begin{aligned} \left( 1+\sum _{i}\chi _{i}^{2}\right) \chi _{1}^{B}A_{1}+\left( 1-\chi _{1}^{2}\right) \sum _{i}\chi _{i}^{B}A_{i}=0, \end{aligned}$$
(A.8)
where we use the dimensionless lattice momenta (63). The constraint equation can alternatively be written as
$$\begin{aligned} \chi _{1}^{B}A_{1}=-\beta \sum _{i}\chi _{i}^{B}A_{i}, \end{aligned}$$
(A.9)
where \(\beta \) is a momentum-dependent factor given by
$$\begin{aligned} \beta = \frac{1-\chi _{1}^{2}}{1+\sum _{i}\chi _{i}^{2}}. \end{aligned}$$
(A.10)
The temporal average \({\overline{A}}_{x,i}\) reduces to a multiplication with a frequency dependent factor:
$$\begin{aligned} {\overline{A}}_{x,i} = \cos \left( \omega a^0 \right) A_{x,i} = c A_{x,i}, \end{aligned}$$
(A.11)
where we used the shorthand \(c = \cos \left( \omega a^0 \right) \). Inserting the plane wave ansatz into the EOM yields
$$\begin{aligned} \chi _{0}^{2}A_{1}&= c \sum _{i} \chi _{i}^{2} A_{1} - \frac{1}{2} \left( 1+c \right) \sum _{i} \chi _{i}^{B} \chi _{1}^{F} A_{i}, \end{aligned}$$
(A.12)
$$\begin{aligned} \chi _{0}^{2}A_{i}&= c \sum _{j\ne i} \chi _{j}^{2} A_{i} + \chi _{1}^{2} A_{i} - c \sum _{j\ne i} \chi _{j}^{B} \chi _{i}^{F} A_{j} \nonumber \\&\quad - \frac{1}{2}\left( 1+c\right) \chi _{1}^{B}\chi _{i}^{F}A_{1}. \end{aligned}$$
(A.13)
Note that both \(\chi _0\) and c depend on \(\omega \). After making use of the Gauss constraint the longitudinal EOM reads
$$\begin{aligned} \chi _{0}^{2}A_{1}=c\left( \chi _{2}^{2}+\chi _{3}^{2}\right) A_{1}+\frac{1}{2}\left( 1+c\right) \beta ^{-1}\chi _{1}^{2}A_{1}, \end{aligned}$$
(A.14)
and the two transverse equations read
$$\begin{aligned} \chi _{0}^{2}A_{2}&= c\chi _{3}^{2}A_{2} + \chi _{1}^{2}A_{2} - c\chi _{3}^{B}\chi _{2}^{F}A_{3} - \frac{1}{2}\left( 1+c\right) \chi _{1}^{B}\chi _{2}^{F}A_{1} \nonumber \\&= c \chi _{3}^{2}A_{2}+\chi _{1}^{2}A_{2}+\frac{1}{2}\left( 1+c\right) \beta \chi _{2}^{2}A_{2} \nonumber \\&\quad + \left( \frac{1}{2}\left( 1+c\right) \beta -c\right) \chi _{2}^{F}\chi _{3}^{B}A_{3}, \end{aligned}$$
(A.15)
$$\begin{aligned} \chi _{0}^{2}A_{3}&= c\chi _{2}^{2}A_{3} + \chi _{1}^{2}A_{3} - c\chi _{2}^{B}\chi _{3}^{F}A_{2}-\frac{1}{2}\left( 1+c\right) \chi _{1}^{B}\chi _{3}^{F}A_{1} \nonumber \\&= c\chi _{2}^{2}A_{3}+\chi _{1}^{2}A_{3}+\frac{1}{2}\left( 1+c\right) \beta \chi _{3}^{2}A_{3} \nonumber \\&\quad +\left( \frac{1}{2}\left( 1+c\right) \beta -c\right) \chi _{2}^{B}\chi _{3}^{F}A_{2}. \end{aligned}$$
(A.16)
This system of equations can be written in matrix notation as an eigenvalue problem
$$\begin{aligned} M \mathbf {A} = \chi ^2_0 \mathbf {A}, \end{aligned}$$
(A.17)
where the vector \(\mathbf {A}\) is simply
$$\begin{aligned} \mathbf {A} = \left( \begin{array}{c} A_{1}\\ A_{2}\\ A_{3} \end{array} \right) . \end{aligned}$$
(A.18)
The eigenvectors of M are
$$\begin{aligned} \left\{ \mathbf {A}_{L},\mathbf {A}_{T,1},\mathbf {A}_{T,2}\right\} = \left\{ \left( \begin{array}{c} 1\\ 0\\ 0 \end{array} \right) , \left( \begin{array}{c} 0\\ -\chi _{3}^{B}\\ \chi _{2}^{B} \end{array} \right) , \left( \begin{array}{c} 0\\ \chi _{2}^{F}\\ \chi _{3}^{F} \end{array} \right) \right\} ,\nonumber \\ \end{aligned}$$
(A.19)
where we find two transverse, momentum dependent eigenvectors \(\mathbf {A}_{T,1}\), \(\mathbf {A}_{T,2}\) and the longitudinal unit vector \(\mathbf {A}_L\). The two transverse vectors \(\mathbf {A}_{T,1}\) and \(\mathbf {A}_{T,2}\) can be interpreted as transverse polarization modes and are orthogonal in the sense of \(\left( \mathbf {A}_{T,1}\right) \cdot \left( \mathbf {A}_{T,2}\right) ^{\dagger }=0\). The three eigenvectors yield three different equations for the eigenvalue problem, namely
$$\begin{aligned} M \mathbf {A}_L&= \lambda _L(c, \chi _i) \mathbf {A}_L = \chi ^2_0 \mathbf {A}_L, \end{aligned}$$
(A.20)
$$\begin{aligned} M \mathbf {A}_{T,l}&= \lambda _{T,l}(c, \chi _i) \mathbf {A}_{T,l} = \chi ^2_0 \mathbf {A}_{T,l}, \quad l \in {1, 2}, \end{aligned}$$
(A.21)
where \(\lambda _L(c, \chi _i)\) and \(\lambda _{T,k}(c, \chi _i)\) are expressions which depend on the momenta \(\chi _i\) and the frequency \(\omega \) via \(c=\cos {\left( \omega a^0 \right) }\). Solving the first equation \(\lambda _L(c, \chi _i) = \chi ^2_0\) for \(\omega \) yields the dispersion relation of the longitudinal component:
$$\begin{aligned} \omega _{L}a^{0}=\arccos \left( \frac{1-\chi _{1}^{2}\left( 2+\chi _{2}^{2}+\chi _{3}^{2}\right) }{1+\chi _{2}^{2}\left( 2-\chi _{1}^{2}\right) +\chi _{3}^{2}\left( 2-\chi _{1}^{2}\right) }\right) .\nonumber \\ \end{aligned}$$
(A.22)
Solving the two transverse equations \(\lambda _{T,l}(c, \chi _i) = \chi ^2_0\) yields
$$\begin{aligned} \omega _{T,1}a^{0}=\arccos \left( \frac{1-2\chi _{1}^{2}}{1+2\chi _{2}^{2}+2\chi _{3}^{2}}\right) , \end{aligned}$$
(A.23)
and
$$\begin{aligned} \omega _{T,2}a^{0}&= \arccos \left( \frac{1-\chi _{1}^{2}\left( 2+\chi _{2}^{2}+\chi _{3}^{2}\right) }{1+\chi _{2}^{2}\left( 2-\chi _{1}^{2}\right) +\chi _{3}^{2}\left( 2-\chi _{1}^{2}\right) }\right) \nonumber \\&= \omega _{L}a^{0}. \end{aligned}$$
(A.24)
It turns out that the expressions for \(\omega \) associated with the different eigenvectors are not the same, although \(\mathbf {A}_L\) and \(\mathbf {A}_{T,2}\) share the same dispersion relation. We interpret this as numerical (or artificial) birefringence. Given a momentum k, the amplitude of an arbitrary wave has to be split into two components: a part which is projected into the plane spanned by \(\mathbf {A}_L\) and \(\mathbf {A}_{T,2}\) which oscillates with \(\omega _L=\omega _{T,2}\) and a part parallel to \(\mathbf {A}_{T,1}\) which oscillates with frequency \(\omega _{T,1}\).
We require the propagation of a wave to be stable, i.e. we require the frequencies \(\omega \) to be real-valued. This is guaranteed if the arguments of the \(\arccos \) expressions in Eqs. (A.22) and (A.23) are restricted to \([-1,1]\). Both dispersion relations remain stable if the CFL condition
$$\begin{aligned} \chi ^2_1 \le 1, \end{aligned}$$
(A.25)
holds. Using
$$\begin{aligned} \chi ^2_1 = \left( \frac{a^0}{a^1} \right) \sin ^2 \left( \frac{k^1 a^1}{2} \right) , \end{aligned}$$
(A.26)
and requiring stability for all values of \(k^1\) yields the constraint
$$\begin{aligned} a^0 \le a^1. \end{aligned}$$
(A.27)
This concludes the proof that the semi-implicit scheme, even though exhibiting peculiar wave propagation phenomena, is stable.
Appendix B: variation of gauge links
We introduce the infinitesimal variation of a gauge link variable
$$\begin{aligned} \delta U_{x,\mu } = i \delta A_{x,\mu } U_{x,\mu }, \end{aligned}$$
(B.28)
where the variation of the gauge field \(\delta A_{x,\mu }\) is traceless and hermitian. In the continuum limit \(\delta A_{x,\mu }\) becomes the infinitesimal variation of the gauge field \(A_{\mu }(x)\). The infinitesimal variation \(\delta U_{x,\mu }\) preserves the unitary of gauge links. Let \(U'_{x,\mu } = U_{x,\mu } + \delta U_{x,\mu }\) then we find
$$\begin{aligned} U'_{x,\mu } U'^\dagger _{x,\mu }&= \left( U_{x,\mu } + \delta U_{x,\mu } \right) \left( U^\dagger _{x,\mu } + \delta U^\dagger _{x,\mu } \right) \nonumber \\&\simeq \mathbb {1}+ \delta U_{x,\mu } U^\dagger _{x,\mu } + U_{x,\mu } \delta U^\dagger _{x,\mu } +{\mathscr {O}} \left( \delta A^2 \right) \nonumber \\&\simeq \mathbb {1}+ i \delta A_{x,\mu } - i \delta A_{x\,\mu } +{\mathscr {O}} \left( \delta A^2 \right) \nonumber \\&\simeq \mathbb {1}+ {\mathscr {O}} \left( \delta A^2 \right) . \end{aligned}$$
(B.29)
The determinant is also unaffected for infinitesimal variations.
$$\begin{aligned} \det U'_{x,\mu }&= \det \left( U_{x,\mu } + \delta U_{x,\mu } \right) \nonumber \\&\simeq 1 + {{\mathrm{tr}}}\left( {{\mathrm{adj}}}\left( U_{x,\mu } \right) \delta U_{x,\mu } \right) + {\mathscr {O}} \left( \delta A^2 \right) \nonumber \\&\simeq 1 + {{\mathrm{tr}}}\left( U^\dagger _{x,\mu } \delta U_{x,\mu } \right) + {\mathscr {O}} \left( \delta A^2 \right) \nonumber \\&\simeq 1 + {\mathscr {O}} \left( \delta A^2 \right) . \end{aligned}$$
(B.30)
The variation \(\delta U_{x,\mu }\) therefore preserves the constraints and allows us to vary the action without Lagrange multipliers, which dramatically simplifies the derivation of equations of motion.
Appendix C: variation of the leapfrog action
In this section of the appendix we give a derivation of the discrete equations of motion obtained from the standard Wilson action (109) using the constraint preserving variation of link variables of the previous section. To make the calculation more organized, we first split the action into two parts: a part containing temporal plaquettes \(S_E[U]\) (“E” for electric) and a part containing spatial plaquettes \(S_B[U]\) (“B” for magnetic). We write
$$\begin{aligned} S[U] = S_E[U] - S_B[U], \end{aligned}$$
(C.31)
where
$$\begin{aligned} S_E[U]&= \frac{V}{g^2} \sum _{x,i} \frac{1}{\left( a^0 a^i \right) ^2} {{\mathrm{tr}}}\left( C_{x,0i} C^\dagger _{x,0i} \right) , \end{aligned}$$
(C.32)
$$\begin{aligned} S_B[U]&= \frac{V}{g^2} \sum _{x,i,j} \frac{1}{2} \frac{1}{\left( a^i a^j \right) ^2} {{\mathrm{tr}}}\left( C_{x,ij} C^\dagger _{x,ij} \right) . \end{aligned}$$
(C.33)
Appendix C.1: Gauss constraint
For the Gauss constraint in the leapfrog scheme we only have to consider the variation \(\delta _t S_E[U]\) as \(S_B[U]\) does not contain any temporal links. In the following sections we make use of the “\(\sim \)” symbol, denoting equality under the sum over lattice sites x and under the trace. We then have
$$\begin{aligned}&\delta _t \left( C_{x,0i} C^\dagger _{x,0i} \right) = \left( \delta U_{x,0} U_{x+0,i} - U_{x,i} \delta U_{x+i,0} \right) C^\dagger _{x,0i} + \text {h.c.}\nonumber \\&\quad \sim \delta U_{x,0} \left( U_{x+0,i} C^\dagger _{x,0i} - C^\dagger _{x-i,0i} U_{x-i,i}\right) + \text {h.c.}\nonumber \\&\quad = i \delta A_{x,0} \left( U_{x,0} \left( U_{x+0,i} C^\dagger _{x,0i} - C^\dagger _{x-i,0i} U_{x-i,i} \right) \right) + \text {h.c.}\end{aligned}$$
(C.34)
To go from the first to the second line, we applied a shift \(x \rightarrow x-i\) in the right term of the first line and made use of the cyclicity of the trace. In the third line we simply used the definition of the variation of gauge links. The variation of the action therefore reads
$$\begin{aligned} \delta _t S_E[U]&= -\frac{V}{g^{2}} \sum _{x,i,a} \frac{1}{\left( a^{0}a^{i}\right) ^{2}} P^{a} \bigg ( U_{x,0} \big ( U_{x+0,i} C_{x,0i}^{\dagger } \nonumber \\&\quad -C_{x-i,0i}^{\dagger }U_{x-i,i}\big )\bigg ) \delta A_{x,0}^{a}, \end{aligned}$$
(C.35)
where we used
$$\begin{aligned} P^a\left( U \right) \equiv 2 \text {Im}{{\mathrm{tr}}}\left( t^a U \right) = -i {{\mathrm{tr}}}\left( t^a \left( U - U^\dagger \right) \right) . \end{aligned}$$
(C.36)
Replacing the \(C_{x,ij}\) terms with link variables we find
$$\begin{aligned} U_{x,0}\left( U_{x+0,i}C_{x,0i}^{\dagger }-C_{x-i,0i}^{\dagger }U_{x-i,i}\right) = 2-U_{x,0i}-U_{x,0-i},\nonumber \\ \end{aligned}$$
(C.37)
and subsequently
$$\begin{aligned} \delta _t S_E[U] = \frac{V}{g^{2}}\sum _{x,i,a} \frac{1}{\left( a^{0}a^{i}\right) ^{2}}P^{a}\left( U_{x,0i}+U_{x,0-i}\right) \delta A_{x,0}^{a}.\nonumber \\ \end{aligned}$$
(C.38)
We require that the variation vanishes, i.e. \(\delta S[U] = 0\). Since all gauge links can be varied independently we find
$$\begin{aligned} \sum _{i}\frac{1}{\left( a^{0}a^{i}\right) ^{2}}P^{a}\left( U_{x,0i}+U_{x,0-i}\right) =0. \end{aligned}$$
(C.39)
Appendix C.2: equations of motion
First we consider the variation of \(S_E[U]\) w.r.t spatial links. We find a result that is similar to the expression for the Gauss constraint
$$\begin{aligned} \delta _{s}\left( C_{x,0i}C_{x,0i}^{\dagger }\right)&\sim i\delta A_{x,i}\left( U_{x,i}\left( C_{x-0,0i}^{\dagger } U_{x-0,0}-U_{x+i,0} C_{x,0i}^{\dagger }\right) \right) \nonumber \\&\quad \,\, + \text {h.c.}\end{aligned}$$
(C.40)
and
$$\begin{aligned}&\delta _s S_E[U] \nonumber \\&\quad = -\frac{V}{g^{2}}\sum _{x,i,a}\frac{1}{\left( a^{0}a^{i}\right) ^{2}} \delta A_{x,i}^{a} P^{a} \bigg ( U_{x,i}\big (C_{x-0,0i}^{\dagger }U_{x-0,0} -U_{x+i,0}C_{x,0i}^{\dagger }\big )\bigg )\nonumber \\&\quad = \frac{V}{g^{2}}\sum _{x,i,a}\frac{1}{\left( a^{0}a^{i}\right) ^{2}} \delta A_{x,i}^{a} P^{a}\left( U_{x,i0}+U_{x,i-0}\right) . \end{aligned}$$
(C.41)
For the variation of \(S_B[U]\) we use
$$\begin{aligned}&\sum _{i,j} \delta _s \left( C_{x,ij} C^\dagger _{x,ij} \right) \nonumber \\&\quad \sim \sum _{i, j} \bigg (\delta U_{x,i}U_{x+i,j}+U_{x,i}\delta U_{x+i,j} \nonumber \\&\qquad - \delta U_{x,j} U_{x+j,i} - U_{x,j} \delta U_{x+j,i}\bigg )C_{x,ij}^{\dagger }+\text {h.c.}\nonumber \\&\quad \sim i \sum _{i, j} 2 \delta A_{x,i}U_{x,i}\bigg (U_{x+i,j}C_{x,ij}^{\dagger }+C_{x-j,ji}^{\dagger }U_{x-j,j}\bigg ) +\text {h.c.}\end{aligned}$$
(C.42)
The variation then reads
$$\begin{aligned} \delta _s S_B[U]= & {} -\frac{V}{g^{2}}\sum _{x,i,j,a}\frac{1}{\left( a^{i}a^{j}\right) ^{2}} \delta A_{x,i}^{a}\nonumber \\&\times P^{a}\bigg (U_{x,i}\big (U_{x+i,j}C_{x,ij}^{\dagger }+C_{x-j,ji}^{\dagger }U_{x-j,j}\big )\bigg ).\nonumber \\ \end{aligned}$$
(C.43)
We set \(\delta S = 0\) and after canceling some constants we find the discrete EOM
$$\begin{aligned}&\frac{1}{\left( a^{0}a^{i}\right) ^{2}}P^{a}\left( U_{x,i0}+U_{x,i-0}\right) \nonumber \\&\quad = -\sum _{j}\frac{1}{\left( a^{i}a^{j}\right) ^{2}}P^{a}\left( U_{x,i}\left( U_{x+i,j}C_{x,ij}^{\dagger }+C_{x-j,ji}^{\dagger }U_{x-j,j}\right) \right) , \end{aligned}$$
(C.44)
which can also be written as
$$\begin{aligned} \frac{1}{\left( a^0 a^i \right) ^2} P^a \left( U_{x,i0} + U_{x,i-0}\right) = \sum _j \frac{1}{\left( a^i a^j \right) ^2} P^a \left( U_{x,ij} + U_{x,i-j}\right) . \end{aligned}$$
(C.45)
Appendix D: conservation of the Gauss constraint in the leapfrog scheme
We now explicitly show that the leapfrog EOM preserve the associated Gauss constraint. We use the identity for the fundamental representation of SU(N)
$$\begin{aligned} \sum _a t^{a}P^{a}\left( X\right) = \frac{1}{2i}\left( X-X^{\dagger }\right) -\frac{1}{2i N}{{\mathrm{tr}}}\left( X-X^{\dagger }\right) \mathbb {1},\nonumber \\ \end{aligned}$$
(D.46)
which can be shown using the Fierz identity for the generators of \(\mathfrak {su}(\mathrm {N})\)
$$\begin{aligned} \sum _a t_{ij}^{a}t_{kl}^{a}=\frac{1}{2}\left( \delta _{il}\delta _{jk}-\frac{1}{N}\delta _{ij}\delta _{kl}\right) , \end{aligned}$$
(D.47)
where i, j, k, l are fundamental representation matrix indices. Using a shorthand we can write
$$\begin{aligned} \sum _a t^{a}P^{a}\left( X\right) = \left[ X \right] _\text {ah}, \end{aligned}$$
(D.48)
where “ah” denotes the anti-hermitian traceless part of X. The constraint and the equations of motion then read (including external charges)
$$\begin{aligned}&\sum _{i}\frac{1}{\left( a^{0}a^{i}\right) ^{2}}\left[ U_{x,0i}+U_{x,0-i} \right] _\text {ah} = \frac{g}{a^0} \rho _x, \end{aligned}$$
(D.49)
$$\begin{aligned}&\frac{1}{\left( a^0 a^i \right) ^2} \left[ U_{x,i0} + U_{x,i-0} \right] _\text {ah} \nonumber \\&\quad = \sum _j \frac{1}{\left( a^i a^j \right) ^2} \left[ U_{x,ij} + U_{x,i-j} \right] _\text {ah} - \frac{g}{a^i} j_{x,i}. \end{aligned}$$
(D.50)
We take (D.50) and sum over i. Due to \(\left[ U_{x,ij} \right] _\text {ah}\) being antisymmetric in the index pair i, j we find
$$\begin{aligned}&\sum _i \frac{1}{\left( a^0 a^i \right) ^2} \left[ U_{x,i0} + U_{x,i-0} \right] _\text {ah}\nonumber \\&\quad = \sum _{i,j} \frac{1}{\left( a^i a^j \right) ^2} \left[ U_{x,i-j} \right] _\text {ah} - \sum _i \frac{g}{a^i} j_{x,i}. \end{aligned}$$
(D.51)
Doing the same at \(x-i\) and parallel transporting from \(x-i\) to x yields
$$\begin{aligned}&\sum _i \frac{1}{\left( a^0 a^i \right) ^2} \left[ U_{x,0-i} + U_{x,-0-i} \right] _\text {ah} \nonumber \\&\quad =\sum _{i,j} \frac{1}{\left( a^i a^j \right) ^2} \left[ U_{x,j-i} \right] _\text {ah} - \sum _i \frac{g}{a^i} U^\dagger _{x-i,i} j_{x-i,i} U_{x-i,i}. \end{aligned}$$
(D.52)
Subtracting the above two equations gives
$$\begin{aligned}&\sum _i \frac{1}{\left( a^0 a^i \right) ^2} \left[ U_{x,i0} + U_{x,i-0} - U_{x,0-i} - U_{x,-0-i} \right] _\text {ah} \nonumber \\&\quad =- \sum _i \frac{g}{a^i} \left( j_{x,i} - U^\dagger _{x-i,i} j_{x-i,i} U_{x-i,i} \right) . \end{aligned}$$
(D.53)
Using antisymmetry we have
$$\begin{aligned} \left[ U_{x,i0} \right] _\text {ah} = - \left[ U_{x,0i} \right] _\text {ah} \end{aligned}$$
(D.54)
and
$$\begin{aligned} \left[ U_{x,-0-i} \right] _\text {ah} = - \left[ U_{x,-i-0} \right] _\text {ah}. \end{aligned}$$
(D.55)
Moreover, in temporal gauge we have \(U_{x,i-0} = U_{x-0,0i}\) and \(U_{x,-i-0}=U_{x-0,0-i}\), which leads to
$$\begin{aligned}&\sum _i \frac{1}{\left( a^0 a^i \right) ^2} \big [U_{x,i0} + U_{x,i-0} - U_{x,0-i} - U_{x,-0-i} \big ]_{\text {ah}} \nonumber \\&\quad = \sum _i \frac{1}{\left( a^0 a^i \right) ^2} \big [\left( U_{x-0,0i} + U_{x-0,0-i} \right) -\left( U_{x,0i} + U_{x,0-i} \right) \big ]_{\text {ah}} \nonumber \\&\quad = -\frac{g}{a^0} \left( \rho _x -\rho _{x-0} \right) , \end{aligned}$$
(D.56)
where we used the Gauss constraint in the last line to replace the temporal plaquette terms with charge densities. This yields the gauge-covariant continuity equation
$$\begin{aligned} \frac{1}{a^0} \left( \rho _x -\rho _{x-0} \right) = \sum _i \frac{1}{a^i} \left( j_{x,i} - U^\dagger _{x-i,i} j_{x-i,i} U_{x-i,i} \right) .\nonumber \\ \end{aligned}$$
(D.57)
If there are no external charges, we simply have the conservation of the Gauss constraint: assume that the constraint in the previous time slice holds, i.e.
$$\begin{aligned} \sum _i \frac{1}{\left( a^0 a^i \right) ^2} \left[ U_{x-0,0i} + U_{x-0,0-i} \right] _\text {ah} = 0, \end{aligned}$$
(D.58)
then the EOM guarantee that it will also hold in the next one, i.e.
$$\begin{aligned} \sum _i \frac{1}{\left( a^0 a^i \right) ^2} \left[ U_{x,0i} + U_{x,0-i} \right] _\text {ah} = 0. \end{aligned}$$
(D.59)
Appendix E: variation of the implicit action
We now consider the action
$$\begin{aligned} S[U] = S_E[U] - S_B[U], \end{aligned}$$
(E.60)
where
$$\begin{aligned} S_E[U]&= \frac{V}{g^2} \sum _{x,i} \frac{1}{\left( a^0 a^i \right) ^2} {{\mathrm{tr}}}\left( C_{x,0i} C^\dagger _{x,0i} \right) , \end{aligned}$$
(E.61)
$$\begin{aligned} S_B[U]&= \frac{V}{g^2} \sum _{x,i,{|}j{|}} \frac{1}{4} \frac{1}{\left( a^i a^j \right) ^2} {{\mathrm{tr}}}\left( C_{x,ij} M^\dagger _{x,ij} \right) . \end{aligned}$$
(E.62)
The electric part is the same as in the leapfrog scheme. However, the magnetic part \(S_B[U]\) now contains temporal gauge links and gives a new contribution to the Gauss constraint.
Before we vary this action, we must verify that \(S_B[U]\) is indeed real-valued. While it is easy to see that the original leapfrog action is real-valued because of the obvious hermicity of \(C_{x,ij} C^\dagger _{x,ij}\), the term \(C_{x,ij} M^\dagger _{x,ij}\) is not hermitian in general. Still, we can show that the action is real: we have
$$\begin{aligned} C_{x,ij} M^\dagger _{x,ij} = \frac{1}{2} C_{x,ij} \left( C^{(+0)}_{x,ij} + C^{(-0)}_{x,ij} \right) ^\dagger , \end{aligned}$$
(E.63)
where we can rewrite
$$\begin{aligned} C_{x,ij} C^{(+0)\dagger }_{x,ij}&= C_{x,ij} \left( U_{x,0} C_{x+0,ij} U^\dagger _{x+i+j,0}\right) ^\dagger \nonumber \\&= C_{x,ij} U_{x+i+j,0} C^\dagger _{x+0,ij} U^\dagger _{x,0} \nonumber \\&\sim U^\dagger _{x,0} C_{x,ij} U_{x+i+j,0} C^\dagger _{x+0,ij} \nonumber \\&= C^{(-0)}_{x+0,ij} C^\dagger _{x+0,ij}. \end{aligned}$$
(E.64)
Here we used the cyclicity of the trace. Then using a shift \(x \rightarrow x-0\) we have
$$\begin{aligned} C_{x,ij} C^{(+0) \dagger }_{x,ij} \sim C^{(-0)}_{x,ij} C^\dagger _{x,ij}. \end{aligned}$$
(E.65)
Likewise we have
$$\begin{aligned} C_{x,ij} C^{(-0) \dagger }_{x,ij} \sim C^{(+0)}_{x,ij} C^\dagger _{x,ij}, \end{aligned}$$
(E.66)
which leads to
$$\begin{aligned} C_{x,ij} M^\dagger _{x,ij} \sim M_{x,ij} C^\dagger _{x,ij}. \end{aligned}$$
(E.67)
Incidentally, the RHS term is exactly the hermitian conjugate of the LHS term. In other words
$$\begin{aligned} \left( C_{x,ij} M^\dagger _{x,ij} \right) ^\dagger = M_{x,ij} C^\dagger _{x,ij} \sim C_{x,ij} M^\dagger _{x,ij}. \end{aligned}$$
(E.68)
This means that under the sum over x and the trace, the expression \(C_{x,ij} M^\dagger _{x,ij}\) is indeed real-valued and by extension \(S_B[U]\) is real-valued as well. We have also shown that the time-average in \(M_{x,ij}\) can be “shifted” to the other term \(C_{x,ij}\) under the sum and trace, i.e.
$$\begin{aligned} C_{x,ij} {\overline{C}}^\dagger _{x,ij} \sim {\overline{C}}_{x,ij} C^\dagger _{x,ij}. \end{aligned}$$
(E.69)
This is a useful property that we will need in the following derivation.
Appendix E.1: Gauss constraint
We perform the variation of \(S_E[U]\) and \(S_B[U]\) w.r.t. temporal links to derive the Gauss constraint in the implicit scheme. Since \(S_E[U]\) is the same for all schemes, we do not have to repeat it. On the other hand, the variation of the magnetic part involves terms like \(\delta _t \left( C_{x,ij} M^\dagger _{x,ij} \right) \), which we now discuss explicitly. First, we make use of
$$\begin{aligned}&\delta _t \left( C_{x,ij} M^\dagger _{x,ij} \right) \sim \delta _t M_{x,ij} C^\dagger _{x,ij}\nonumber \\&\quad = \frac{1}{2} \left( \delta _t C^{(+0)}_{x,ij} + \delta _t C^{(-0)}_{x,ij} \right) C^\dagger _{x,ij}. \end{aligned}$$
(E.70)
Then, after some algebra we find
$$\begin{aligned} \delta _t C^{(+0)}_{x,ij} C^\dagger _{x,ij} \sim i \delta A_{x,0} \left( C^{(+0)}_{x,ij} C^{\dagger }_{x,ij} - C_{x,-i-j} \left( C^{(+0)}_{x,-i-j} \right) ^\dagger \right) \nonumber \\ \end{aligned}$$
(E.71)
and
$$\begin{aligned}&\delta _t C^{(-0)}_{x,ij} C^\dagger _{x,ij} \sim i \delta A_{x,0} \left( - C_{x,ij} \left( C^{(+0)}_{x,ij} \right) ^\dagger + C^{(+0)}_{x,-i-j} C^\dagger _{x,-i-j} \right) \nonumber \\&\quad = \left( \delta _t C^{(+0)}_{x,ij} C^\dagger _{x,ij} \right) ^\dagger , \end{aligned}$$
(E.72)
which yields
$$\begin{aligned}&\delta _t M_{x,ij} C^\dagger _{x,ij} \sim \frac{i}{2} \delta A_{x,0} \bigg ( \left( C^{(+0)}_{x,ij} C^{\dagger }_{x,ij} - \text {h.c.}\right) \nonumber \\&\quad +\left( C^{(+0)}_{x,-i-j} C^\dagger _{x,-i-j} -\text {h.c.}\right) \bigg ). \end{aligned}$$
(E.73)
The variation of the magnetic part therefore reads
$$\begin{aligned} \delta _t S_B[U]= & {} - \frac{V}{g^2} \sum _{x,a,i,{|}j{|}} \frac{1}{8} \frac{1}{\left( a^i a^j \right) ^2} \delta A^a_{x,0}\nonumber \\&\times P^a \bigg (C^{(+0)}_{x,ij} C^{\dagger }_{x,ij} + C^{(+0)}_{x,-i-j} C^\dagger _{x,-i-j}\bigg )\nonumber \\= & {} - \frac{V}{g^2} \sum _{x,a,{|}i{|},{|}j{|}} \frac{1}{8} \frac{1}{\left( a^i a^j \right) ^2} \delta A^a_{x,0} P^a\left( C^{(+0)}_{x,ij} C^{\dagger }_{x,ij} \right) .\nonumber \\ \end{aligned}$$
(E.74)
In the last line we consolidated terms with index i and \(-i\) into a single term using the sum \(\sum _{{|}i{|}}\). Taking the result for \(\delta _t S_E[U]\) from the leapfrog scheme, we find the Gauss constraint
$$\begin{aligned}&\sum _{i}\frac{1}{\left( a^{0}a^{i} \right) ^{2}}P^{a}\left( U_{x,0i}+U_{x,0-i}\right) \nonumber \\&\quad = - \sum _{{|}i{|},{|}j{|}} \frac{1}{8} \frac{1}{\left( a^i a^j \right) ^2} P^a\left( C^{(+0)}_{x,ij} C^{\dagger }_{x,ij} \right) . \end{aligned}$$
(E.75)
Appendix E.2: equations of motion
For the EOM we vary S[U] w.r.t spatial links. Again, we already have the result for \(\delta _s S_E[U]\) from the leapfrog scheme and only need to calculate \(\delta _s S_B[U]\). In particular we consider the term
$$\begin{aligned} \delta _s \left( C_{x,ij} M^\dagger _{x,ij} \right) = \delta _s C_{x,ij} M^\dagger _{x,ij} + C_{x,ij} \delta _s M^\dagger _{x,ij}. \end{aligned}$$
(E.76)
Since the variation only acts on spatial links we can shift the time-average of the right term from \(M^\dagger _{x,ij}\) to \(C_{x,ij}\). This gives
$$\begin{aligned} \delta _s \left( C_{x,ij} M^\dagger _{x,ij} \right)&\sim \delta C_{x,ij} M^\dagger _{x,ij} + M_{x,ij} \delta C^\dagger _{x,ij} \nonumber \\&= \delta C_{x,ij} M^\dagger _{x,ij} + \text {h.c.}\end{aligned}$$
(E.77)
The variation then proceeds analogously to the derivation of the leapfrog scheme. We find
$$\begin{aligned} \delta _s S_B[U]&= -\frac{V}{g^2} \sum _{x,i,{|}j{|}} \frac{1}{2} \frac{1}{\left( a^{i}a^{j}\right) ^{2}} \delta A_{x,i}^{a} \nonumber \\&\quad \times P^{a}\left( U_{x,i}\left( U_{x+i,j} M_{x,ij}^{\dagger }+M_{x-j,ji}^{\dagger } U_{x-j,j}\right) \right) . \end{aligned}$$
(E.78)
With the result for \(\delta _s S_E[U]\) we obtain the discrete EOM
$$\begin{aligned}&\frac{1}{\left( a^0 a^i \right) ^2} P^a \left( U_{x,i0} + U_{x,i-0}\right) \nonumber \\&\quad =- \frac{1}{2} \sum _{{|}j{|}} \frac{1}{\left( a^i a^j \right) ^2} P^a \left( U_{x,i} \left( U_{x+i,j} M^\dagger _{x,ij} + M^\dagger _{x-j,ji} U_{x-j,j}\right) \right) . \end{aligned}$$
(E.79)
Introducing the shorthand
$$\begin{aligned} K_{x,ij}[U, M] = - \frac{1}{2} \frac{1}{\left( a^i a^j \right) ^2} \left( U_{x+i,j} M^\dagger _{x,ij} - M^\dagger _{x-j,ij} U_{x-j,j}\right) , \end{aligned}$$
(E.80)
allows us to write the EOM rather compactly as
$$\begin{aligned} \frac{1}{\left( a^0 a^i \right) ^2} P^a \left( U_{x,i0} + U_{x,i-0}\right) = \sum _{{|}j{|}} P^a \left( U_{x,i} K_{x,ij}[U,M] \right) .\nonumber \\ \end{aligned}$$
(E.81)
Appendix F: variation of the semi-implicit action
In the semi-implicit scheme the action reads
$$\begin{aligned} S[U] = S_E[U] - S_B[U], \end{aligned}$$
(F.82)
where \(S_E[U]\) is the same as before. The magnetic part comprises of \(S_B[U] = S_{B,M}[U] + S_{B,W}[U]\), where \(S_{B,M}[U]\) is the same as \(S_B[U]\) from the implicit scheme except that \(\sum _{i {|}j{|}}\) only runs through transverse components:
$$\begin{aligned} S_{B,M}[U] = \frac{V}{g^2} \sum _{x,i,{|}j{|}} \frac{1}{4} \frac{1}{\left( a^i a^j \right) ^2} {{\mathrm{tr}}}\left( C_{x,ij} M^\dagger _{x,ij} \right) . \end{aligned}$$
(F.83)
Therefore we can take the results from the previous section for \(\delta S_{B,M}[U]\), Eqs. (E.74) and (E.78). The new part is given by
$$\begin{aligned} S_{B,W}[U] = \frac{V}{g^2} \sum _{x, {|}j{|}} \frac{1}{4} \frac{1}{\left( a^1 a^j \right) ^2} {{\mathrm{tr}}}\left( C_{x,1j} W^\dagger _{x,1j} + \text {h.c.}\right) .\nonumber \\ \end{aligned}$$
(F.84)
Appendix F.1: Gauss constraint
We already know \(\delta _t S_E[U]\) and \(\delta _t S_{B,M}[U]\) from previous sections, so we only have to compute \(\delta _t S_{B,W}[U]\). The relevant terms are
$$\begin{aligned}&\sum _{{|}j{|}} \frac{1}{\left( a^j \right) ^2}\delta _t W_{x,1j} C^\dagger _{x,1j} \nonumber \\&\quad =\sum _{{|}j{|}} \frac{1}{\left( a^j \right) ^2} \left( \delta _t {\overline{U}}_{x,1} U_{x+1,j} - U_{x,j} \delta _t {\overline{U}}_{x+j,1} \right) C^\dagger _{x,1j} \nonumber \\&\quad \sim \delta _t {\overline{U}}_{x,1} \sum _{{|}j{|}} \frac{1}{\left( a^j \right) ^2} \left( U_{x+1,j} C^\dagger _{x,1j} - C^\dagger _{x-j,1j} U_{x-j,j} \right) \nonumber \\&\quad = \delta _t {\overline{U}}_{x,1} T^{\dagger }_{x,1}, \end{aligned}$$
(F.85)
where we defined
$$\begin{aligned} T^{\dagger }_{x,1} = \sum _{{|}j{|}} \frac{1}{\left( a^j \right) ^2} \left( U_{x+1,j} C^\dagger _{x,1j} - C^\dagger _{x-j,1j} U_{x-j,j} \right) . \end{aligned}$$
(F.86)
Using the same techniques as before we find
$$\begin{aligned}&\delta _t {\overline{U}}_{x,1} T^{\dagger }_{x,1} \sim \frac{i}{2} \delta A_{x,0} \bigg (U^{(+0)}_{x,1} T^\dagger _{x,1} - U_{x,1} T^{(+0) \dagger }_{x,1} \nonumber \\&\quad - T^\dagger _{x-1,1} U^{(+0)}_{x-1,1} + T^{(+0) \dagger }_{x-1,1} U_{x-1,1} \bigg ), \end{aligned}$$
(F.87)
and the variation of \(S_{B,W}[U]\) reads
$$\begin{aligned}&\delta _t S_{B,W}[U] = - \frac{V}{g^2} \sum _{x,a} \frac{1}{8\left( a^1 \right) ^2} \delta A^a_{x,0} \times \nonumber \\&\quad \times P^a \bigg (U^{(+0)}_{x,1} T^\dagger _{x,1} + T^{(+0)}_{x,1} U^\dagger _{x,1} + U^{(+0)\dagger }_{x-1,1} T_{x-1,1} + T^{(+0) \dagger }_{x-1,1} U_{x-1,1}, \bigg ) \end{aligned}$$
(F.88)
which (with the previous results taken into account) yields the Gauss constraint in the semi-implicit scheme, see Eq. (154).
Appendix F.2: equations of motion
For the variation w.r.t. spatial links we have to distinguish two cases: the longitudinal and the transverse links. Starting with the variation of longitudinal links we find
$$\begin{aligned}&\delta _1 \left( C_{x,1j} W^\dagger _{x,1j} \right) +\text {h.c.}\nonumber \\&\quad =\delta _1 C_{x,1j} W^\dagger _{x,1j} + C_{x,1j} \delta _1 W^\dagger _{x,1j} + \text {h.c.}\nonumber \\&\quad \sim i \delta A_{x,1} U_{x,1} \bigg ( \left( U_{x+1,j} W^\dagger _{x,1j} + W^\dagger _{x-j,j1} U_{x-j,j} \right) \nonumber \\&\qquad + {\overline{\left( U_{x+1,j} C^\dagger _{x,1j} + C^\dagger _{x-j,j1} U_{x-j,j} \right) }}\bigg ) + \text {h.c.}\end{aligned}$$
(F.89)
Again we used the fact that the time-average can be shifted to other terms by exploiting the sum over x and the cyclicity of the trace. The variation of \(S_{B,W}[U]\) then reads
$$\begin{aligned} \delta _1 S_{B,W}[U]&= -\frac{V}{g^2} \sum _{x,a,j} \frac{1}{4} \frac{1}{\left( a^1 a^j \right) ^2} \delta A^a_{x,1} \nonumber \\&\quad \times P^a \bigg ( U_{x,1} \bigg ( \left( U_{x+1,j} W^\dagger _{x,1j} + W^\dagger _{x-j,j1} U_{x-j,j} \right) \nonumber \\&\quad +{\overline{\left( U_{x+1,j} C^\dagger _{x,1j} + C^\dagger _{x-j,j1} U_{x-j,j} \right) }}\bigg ) \bigg ). \end{aligned}$$
(F.90)
Combining the above with \(\delta _1 S_E[U]\) gives the longitudinal EOM Eq. (156). No contributions from \(S_{B,M}[U]\) are necessary because it does not include any longitudinal links.
For the transverse components of the EOM we vary w.r.t. \(U_{x,j}\), where j is a transverse index. The relevant terms for \(j>0\) are
$$\begin{aligned} \delta _j W_{x,1j} C^\dagger _{x,1j}\sim & {} i \delta A_{x,j} U_{x,j}\nonumber \\&\quad \times \left( {\overline{U}}_{x+j,1} C^\dagger _{x,j1} - C^\dagger _{x-1,j1} {\overline{U}}_{x-1,1} \right) ,\nonumber \\ \end{aligned}$$
(F.91)
and
$$\begin{aligned} \delta _j C_{x,1j} W^\dagger _{x,1j}\sim & {} i \delta A_{x,j} U_{x,j}\nonumber \\&\times \left( U_{x+j,1} W^\dagger _{x,j1} - W^\dagger _{x-1,j1} U_{x-1,1} \right) . \end{aligned}$$
(F.92)
For terms with negative component indices \(j<0\) we can show that they are identical (under the sum over x and the trace) to the last two terms except for the substitution \(1 \rightarrow -1\).
$$\begin{aligned}&\delta _j \sum _{-j} \left( W_{x,1j} C^\dagger _{x,1j} + \text {h.c.}\right) = \delta _j \sum _{j} \left( W_{x,1-j} C^\dagger _{x,1-j} + \text {h.c.}\right) \nonumber \\&\quad \sim \sum _{j} \left( \delta _j W_{x,-1j} C^\dagger _{x,-1j} + \delta _j C_{x,-1j} W^\dagger _{x,-1j} + \text {h.c.}\right) . \end{aligned}$$
(F.93)
This allows us to write
$$\begin{aligned} \sum _{{|}j{|}} \delta _j W_{x,1j} C^\dagger _{x,1j}\sim & {} \sum _{{|}1{|}} i \delta A_{x,j} U_{x,j}\nonumber \\&\times \bigg ( {\overline{U}}_{x+j,1} C^\dagger _{x,j1} - C^\dagger _{x-1,j1} {\overline{U}}_{x-1,1} \bigg ), \nonumber \\ \end{aligned}$$
(F.94)
$$\begin{aligned} \sum _{{|}j{|}} \delta _j C_{x,1j} W^\dagger _{x,1j}\sim & {} \sum _{{|}1{|}} i \delta A_{x,j} U_{x,j}\nonumber \\&\times \bigg ( U_{x+j,1} W^\dagger _{x,j1} - W^\dagger _{x-1,j1} U_{x-1,1} \bigg ),\nonumber \\ \end{aligned}$$
(F.95)
where \(\sum _{{|}1{|}}\) stands for summing over terms with component indices 1 and \(-1\). The variation of \(S_{B,W}[U]\) then reads
$$\begin{aligned}&\delta _j S_{B,W}[U] = \frac{V}{g^2} \sum _{x, {|}j{|}} \frac{1}{4} \frac{1}{\left( a^1 a^j \right) ^2} \delta _j {{\mathrm{tr}}}\left( C_{x,1j} W^\dagger _{x,1j} + \text {h.c.}\right) \nonumber \\&\quad = - \frac{V}{g^2} \sum _{x,j,a} \frac{1}{4} \frac{1}{\left( a^1 a^j \right) ^2} \delta A^a_{x,j} \nonumber \\&\qquad \times P^a \bigg ( U_{x,j} \sum _{{|}1{|}}\bigg ( \left( {\overline{U}}_{x+j,1} C^\dagger _{x,j1} - C^\dagger _{x-1,j1} {\overline{U}}_{x-1,1} \right) \nonumber \\&\qquad + \left( U_{x+j,1} W^\dagger _{x,j1} - W^\dagger _{x-1,j1} U_{x-1,1} \right) \bigg ) \bigg ). \end{aligned}$$
(F.96)
With the expressions for \(\delta _j S_E[U]\) and \(\delta _j S_{W,M}[U]\) we find the transverse components of the EOM Eq. (157).