simply set $\phi_z$ to be zero when we fed our fit parameters into the integrator program, the only change we had to make to the existing integrator was in accounting for the constant offset $b_y$. To demonstrate the changes we made to the Bmad symplectic integrator, we first present the Hamiltonian of a particle traveling through an LGB (and also a wiggler magnet) [@bmad]

$H = H_x + H_y + H_z$,

with\

$ H_x = \frac{(p_x - a_x)^2}{2(1+p_z)}$, $H_y = \frac{(p_y - a_y)^2}{2(1+p_z)}$, $H_z = -a_z$,\

in which\
$ a_x = 0, \hspace{10mm} a_y =  \frac{q}{P_0c} \int\limits_0^x B_z dx', \hspace{4mm}    a_z =  -\frac{q}{P_0c} \int\limits_0^x B_y dx' ,$\
are the components of the vector potential, normalized by the particle’s charge, $q$, the speed of light, $c$, and the reference momentum, $P_0$. The form of our symplectic integrator is

$$\begin{aligned}
\mathcal{M} = T_{s/2} \hspace{1mm} I_{x/2} \hspace{1mm} I_{y/2} \hspace{1mm} I_z \hspace{1mm} I_{y/2} \hspace{1mm} I_{x/2} \hspace{1mm} T_{s/2} +  O(\Delta s^3)\end{aligned}$$

in which\

$T_{s/2}: s\rightarrow s+\frac{\Delta s}{2}$,\

$I_{x/2} = \exp(:-\frac{\Delta s}{2} H_x:)$,\

$I_{y/2} = \exp(:-\frac{\Delta s}{2} H_y:)$,\

$I_{z} = \exp(:-\Delta s H_z:)$.

To evaluate $I_{x/2}$ and $I_{y/2}$, one would need to solve Hamilton’s equations corresponding to the Hamiltonians $H_x$ and $H_y$. This is difficult to 