This document contains derivations I did on when, depending on initial state, I should expect coherent errors to add constructively or destructively in a way that a detector error model that pick up on. We find that even when coherent errors are overrotated in the same direction for all data qubits, the coherent errors can cancel out in ancilla measurements for boundary DEM edges where two data qubits contribute to the same DEM edge if the initial state is in the -1 eigenspace of the $Z$ stabilizer for these two data qubits.
The paper claims that if I have a weight-4 $X$ check, there will be a case where two data qubits contribute to the same boundary spacelike DEM edge for this weight 4 check, and there I'll see an addition of angles of the coherent errors.
These two data qubits are also going to be connected by a weight 2 $Z$ stabilizer check. If during the $\ket{+}_L$ state, I measured $Z_1Z_2$ for these qubits to be +1, then I can prove that the angles add. If $Z_1Z_2=-1$ during the initial state prep syndrome cycle, and I dont correct this before moving on to the next syndrome cycle where I inject noise, the coherent errors cancel out completely.
If I look at the case where I just don't take the $Z_1Z_2$ stabilizer checks into account at all, and just lump $Z_1Z_2=\pm1$ cases together, then essentially I'm working with a 50/50 incoherent mixture of these two cases. In this case, I am taking an average of cases where either the coherent errors add completely or cancel completely, so I just get the Pauli twirl stochastic answer.
The claim is that if $q_1$ and $q_2$ contribute to one spacelike DEM edge corresponding to an $X$ check, the angles add.
Notationally, I write $\rho_L=\prod_{s} \pi_{s}\ket{+}^{\oplus n}\bra{+}^{\oplus n}\pi_s$ for $n=d^2=9$ .
$\pi_s$ is essentially the projector for a given $Z$ stabilizer measurement $s$. $\prod_s$ iterates over all $Z$ checks except for one: the stabilizer check for $Z_1Z_2$ . We can see visually from the $d=3$ diagram that $Z_1Z_2$ exists as a weight 2 stabilizer check (e.g. if $q_1,q_2$ are the qubits contributing to the DEM edge $p_{22}$ in my diagram above).
The stabilizer check for $Z_1Z_2$ is measured in the first round of syndrome extraction when preparing the quiescent state, and thus defines the Pauli frame. We denote the projector for this mreasurement as $\Pi_\pm$ in all subsequent notation. $\Pi_\pm$ projects out the case where we measured the weight 2 $Z_1Z_2=\pm1$ when preparing the quiescent state. Our initial state then is $\rho(t=0)=\Pi_\pm\rho_L\Pi_\pm$. The other stabilizer checks and their projectors are included in the definition of $\rho_L$.
We now look at $\rho =U_2U_1\rho(t=0)U_1^\dagger U_2^\dagger = U_2U_1\Pi_\pm \rho_L\Pi_\pm U_1^\dagger U_2^\dagger$ where $U_i=e^{-iZ_i\theta_i}$ performs a coherent error.
$$\begin{align}
\rho = &\left[\cos\theta_1\cos\theta_2 -\sin\theta_1\sin\theta_2 Z_1Z_2-i(Z_1\sin\theta_1\cos\theta_2+Z_2\sin\theta_2\cos\theta_1)\right]\Pi_\pm\rho_L\Pi_\pm\ &\left[\cos\theta_1\cos\theta_2 -\sin\theta_1\sin\theta_2 Z_1Z_2+i(Z_1\sin\theta_1\cos\theta_2+Z_2\sin\theta_2\cos\theta_1)\right]
\end{align}
$$
Now we use the fact that $\Pi_\pm$ tells us whether $Z_1Z_2=\pm 1$
$$\begin{align}
\rho = &\left[\cos(\theta_1\pm\theta_2)-i(Z_1\sin\theta_1\cos\theta_2+Z_2\sin\theta_2\cos\theta_1)\right]\Pi_\pm\rho_L\Pi_\pm\ &\left[\cos(\theta_1\pm\theta_2)+i(Z_1\sin\theta_1\cos\theta_2+Z_2\sin\theta_2\cos\theta_1)\right]
\end{align}
$$
$$\begin{align}
\rho = &\left[\cos(\theta_1\pm\theta_2)-iZ_1(\sin\theta_1\cos\theta_2+Z_1Z_2\sin\theta_2\cos\theta_1)\right]\Pi\rho_L\Pi\ &\left[\cos(\theta_1\pm\theta_2)+iZ_1(\sin\theta_1\cos\theta_2+Z_1Z_2\sin\theta_2\cos\theta_1)\right]
\end{align}
$$
$$
\begin{align}
\rho = &\left[\cos(\theta_1\pm\theta_2)-iZ_1(\sin\theta_1\cos\theta_2\pm\sin\theta_2\cos\theta_1)\right]\Pi_\pm\rho_L\Pi_\pm\ &\left[\cos(\theta_1\pm\theta_2)+iZ_1(\sin\theta_1\cos\theta_2\pm\sin\theta_2\cos\theta_1)\right]
\end{align}
$$
$$
\begin{align}
\rho = &\left[\cos(\theta_1\pm\theta_2)-iZ_1(\sin(\theta_1\pm\theta_2))\right]\Pi_\pm\rho_L\Pi_\pm\ &\left[\cos(\theta_1\pm\theta_2)+iZ_1\sin(\theta_1\pm\theta_2)\right]
\end{align}
$$
$$
\rho = \cos^2(\theta_1\pm\theta_2)\Pi_\pm\rho_L\Pi_\pm+\sin^2(\theta_1\pm\theta_2)Z_1\Pi_\pm\rho_L\Pi_\pm Z_1
$$
So probability of $X$ detector firing $P(a_0=-)=\sin^2(\theta_1\pm\theta_2)$.
If we have $Z_1Z_2=1$, we get that $p=\sin^2(2\theta)$. If we have $Z_1Z_2=-1$, we should get $P(a_0=-)=0$ , thus washing out the coherent error.
if we had not taken the stabilizer measurement into account, we would have a 50 / 50 mixture of the two cases:
$$\rho_L = \frac{1}{2}\left(\Pi_+\rho_L\Pi_++\Pi_-\rho_L\Pi_-\right)$$
Which would give us $P=0.5(\sin^2(2\theta)+0)=\sin^2 (2\theta)/2$ which is the Pauli twirl stochastic answer.