# Gauge-invariant discretization
We do the discretization of the Usadel equation as follows. We first
discretize the action $`S`$. The discrete saddle-point equation or expressions
for currents are not derived analytically, but instead obtained from the computer
implementation of the action via automatic differentiation (AD).
This simplifies things somewhat, because proper handling of the gauge fields is
simpler to do in the action formulation.
The action is
```math
S[Q] = \frac{i\pi\nu}{8} \mathrm{Tr}[D(\hat{\nabla}Q)^2 + \eta F_{ij}Q\hat{\nabla}_iQ\hat{\nabla}_jQ + 4 i \Omega Q]
\,,
```
where $`\hat{\nabla}_j Q = \partial_j Q - i[\mathcal{A}_j, Q]`$
and $`\Omega = \epsilon\tau_3 + i\Delta`$, $`\Delta^\dagger = \Delta`$.
We'd like the discretized functional to be also gauge invariant.
We subdivide the space to cells, centered on a rectangular lattice
$`\vec{r}_j=(h j_x, h j_y, h j_z)`$ with $`j_{x,y,z}\in\mathbb{Z}`$
and *h* are the lattice spacings. See the figure below:
We choose $`Q(j) = Q(\vec{r}_j)`$ to be values of $`Q`$ at the lattice
sites. We define the Wilson link matrices $`U_{ij} =U_{ji}^{-1}
=U(\vec{r}_{i},\vec{r}_{j}) =
\mathrm{Pexp}[i\int_{L(\vec{r}_i,\vec{r}_j)}d\vec{r}'\,\cdot\vec{\mathcal{A}}(\vec{r}')]`$
where $`L(\vec{r}_i,\vec{r}_j)`$ is straight line from $`\vec{r}_j`$
to $`\vec{r}_i`$ and $`\mathrm{Pexp}`$ is the path-ordered integral.
Since the gauge field $`\vec{\mathcal{A}}`$ is fixed, the link
matrices can be computed ahead of time.
We want a discretized theory that is invariant under the discrete gauge transformation
```math
Q(j) \mapsto u(j) Q u(j)^{-1} \,,
```
and
```math
U_{ij}
\mapsto
u(i)U_{ij}u(j)^{-1}
\,,
```
and in the continuum limit reduces to the original theory. Such
discretization has the advantage of being insensitive to the gauge
choice, and having exact discrete conservation laws associated with
the symmetry.
## Derivative term
We discretize the derivatives a
```math
D_{ij}
:=
\frac{1}{h}[Q(i) U_{ij} - U_{ij} Q(j)]
\,,
\qquad
\Rightarrow
Q(i) D_{ij} = - D_{ij} Q(j)
\,,
```
which satisfies an anticommutation relation analogous to continuum derivative.
We then discretize
```math
S_2
=
\mathrm{Tr} D(\hat{\nabla} Q)^2
\mapsto
-D \frac{2d}{|\mathrm{neigh}|} h^d \sum_{\mathrm{neigh}(i,j)}\mathrm{Tr} D_{ij} D_{ji}
```
where $`d`$ is the space dimension and `|neigh|` the number of neighbors on the grid.
Expanding in small $`h`$ we have
```math
S_2
\simeq
\frac{h^{d}d}{|\mathrm{neigh}|}
\sum_{i}\sum_{j\in{}\mathrm{neigh}(i)}\mathrm{Tr}\Bigl(\frac{Q(i) - Q(j)}{h} - [i\mathcal{A}(\frac{\vec{r}_i+\vec{r}_j}{2}), Q(j)]\Bigr)^2
\,,
```
which verifies the reduction to the continuum limit.
## Field strength
The link matrices have the usual continuum limit expansion
```math
U_{ji}
=
\sum_{n=0}^\infty i^n |\vec{r}_i-\vec{r}_j|^n \int_{-1/2}^{1/2}ds_1\int_{-1/2}^{s_1}ds_2\ldots\int_{-1/2}^{s_{n-1}}ds_n\, \mathcal{A}(s_1)\cdots\mathcal{A}(s_n)
```
where $`\mathcal{A}(s)=\mathcal{A}[(\frac{1}{2} - s)\vec{r}_i + (\frac{1}{2} + s)\vec{r}_j]`$.
As well known in lattice QCD, the field strength can be expressed in
terms of the link matrices $`U`$ via a plaquette loop. Consider then
the plaquette (see Fig. 1), with edges $`j\leftarrow{}i`$ and $`k\leftarrow{}l`$ in
direction $`\mu`$ and $`l\leftarrow{}i`$, $`k\leftarrow{}j`$ in direction $`\nu`$, and expand around
$`\vec{r}_0=\frac{\vec{r}_i+\vec{r}_j+\vec{r}_k+\vec{r}_l}{4}`$:
```math
P_{lkji}
=
U_{il}U_{lk}U_{kj}U_{ji}
\,,
```
```math
\Rightarrow
P_{lkji}
\simeq
1
- i h^2
\bigl(
\partial_\mu \mathcal{A}_\nu - \partial_\nu \mathcal{A}_\mu - i[\mathcal{A}_\mu,\mathcal{A}_\nu]
\bigr)
+
\mathcal{O}(h^3)
=
1
+ i h^2 F_{\mu \nu}(\vec{r}_0)
+
\mathcal{O}(h^3)
```
which then allows expressing $`F_{ij}`$ in terms of the link matrices.
## Hall term
In the plaquette of Fig. 1 we have
```math
\mathrm{tr}
F_{\mu\nu}Q \hat{\nabla}_\mu Q \hat{\nabla}_\nu Q
\rvert_{\mu=ji, \nu=li}
\simeq
\frac{i}{h^4}
\mathrm{tr}
(1 - P_{lkji}) U_{ij} D_{ji} Q(i) D_{il} U_{li}
```
```math
=
\frac{-i}{h^4}
\mathrm{tr}
(U_{lk} U_{kj} - U_{li} U_{ij}) D_{ji} Q(i) D_{il}
```
So the structure is $`\frac{-i}{h^4}`$ $`\times`$ (gauge factors for
counterclockwise loop $`-`$ return back clockwise) $`\times`$ $`DQD`$ for
three sites in counterclockwise order. Since only $`Q(i)`$, $`Q(j)`$,
$`Q(l)`$ appear here, one can also think of this as a plaquette corner
with $`Q(i)`$ being the corner.
To do the sum over the $`\mu`$, $`\nu`$ indices, one can average over
corners of the plaquettes as follows, which gives the final discretization
of the Hall term:
```math
S_H
=
\eta
\mathrm{Tr} F_{\mu\nu}Q\hat{\nabla}_\mu Q\hat{\nabla}_\nu Q
```
```math
S_H
\mapsto
\eta
\frac{h^d d}{4ih^4}
\sum_{\mathrm{plaqc}(ijkl; i)}
\mathrm{Tr}
\Bigl(
(U_{lk} U_{kj} - U_{li} U_{ij}) D_{ji} Q(i) D_{il}
\Bigr)
```
```math
S_H
=
\frac{\eta h^{d-4} d}{4i}
\sum_{\mathrm{plaqc}(i; jkl)}
\mathrm{Tr}
\Bigl(
(P_{lkji} - 1)(Q(i) - U_{ij} Q(j) U_{ji} Q(i) U_{il} Q(l) U_{li})
\Bigr)
\,,
```
where $`\mathrm{plaqc}(i;jkl)`$ are the plaquette corners. The last line
follows with direct calculation, and considering the cyclic
permutations.
## Matrix current
We can also calculate the matrix current $`j\mapsto{}i`$ between neighboring cells:
```math
J_{ij}^a
=
\frac{\partial}{\partial\xi}
S[
U_{ij}
\mapsto
U_{ij}
+
T^a U_{ij}\xi
\,,
\;
U_{ji}
\mapsto
U_{ji}
-
U_{ji} T^a \xi
]\rvert_{\xi=0}
```
where $`T^a`$ is an appropriate matrix generator, and the transformation
is inserted to only one of the links.
The local gauge invariance now implies that there is an exact discrete
continuity equation
```math
\sum_{j\in{}N_i} J_{ij}^a
=
\frac{\partial}{\partial\xi}
S[Q(i)\mapsto{}e^{\xi T^a}Q(i)e^{-\xi T^a}]\rvert_{\xi=0}
=
\frac{\partial}{\partial\xi}
S[\Omega(i)\mapsto{}e^{-\xi T^a}\Omega(i)e^{\xi T^a}]\rvert_{\xi=0}
=
R_{i}^a
\,.
```
This is one main advantage of the procedure, in addition to its
invariance vs. gauge fixing.
However, the definition of the matrix current is *asymmetric*, so
that $`J_{ij}\ne{}-J_{ji}`$ as they differ by a quantity of order
$`\mathcal{O}(h^2)`$. The asymmetry arises when $`T^a`$ does not commute
with $`U_{ij}`$, so it is an issue only for the SU(2) component. The
problem is only in the interpretation.
The currents measured at sites *i* and *j* are different: since
the parallel transport of spin between these two locations can imply
rotation due to the gauge field, there's no reason why we should have
$`J_{ij}=-J_{ji}`$, except in the continuum limit where the two points
become close to each other.
## Implementation
The above is sufficient for numerical implementation: it is not
necessary to find out the saddle point equation symbolically. It can
be deduced by parametrizing first so that $`Q^2=1`$ and then using
automatic differentiation (AD). Only a routine evaluating the value of
the action is then needed, in terms of the AD dual variables.
When looking for the saddle point in equilibrium, the problem becomes
even simpler; *Q* has then additional restrictions, and we are looking
for the minimum of the free energy *S*.