**A**

Direct calculations show that *H*TOTAL = ( *π* − *<sup>α</sup>*1)118 + *M*, where 118 is the 3-qubit identity operator, and *M* ≡ *M*0 0 0 *M*1 is a block diagonal matrix with *M*0 and *M*1 given by:

$$M\_0 \equiv \begin{bmatrix} \xi\_1 & 0 & 0 & 0 \\ 0 & \xi\_2 & 0 & 0 \\ 0 & 0 & \xi\_2 & 0 \\ 0 & 0 & 0 & \xi\_3 \end{bmatrix} \tag{A1a}$$

$$M\_1 \equiv \begin{bmatrix} -y & -\mathbf{x}/2 & -\mathbf{x}/2 & 0\\ -\mathbf{x}/2 & 0 & 0 & -\mathbf{x}/2\\ -\mathbf{x}/2 & 0 & 0 & -\mathbf{x}/2\\ 0 & -\mathbf{x}/2 & -\mathbf{x}/2 & y \end{bmatrix} \prime \tag{A1b}$$

where we denote:

$$\mathbf{j\_1} \equiv -\pi + 2\alpha\mathbf{1} + 2\alpha\mathbf{2} + 4\alpha\mathbf{3},\tag{A2a}$$

$$\mathbf{g}\_2^\* \equiv -\pi + 2\mathbf{a}\_1 - 2\mathbf{a}\_3,\tag{A2b}$$

$$
\xi\_3 \equiv -\pi + 2\mathfrak{a}\_1 - 2\mathfrak{a}\_2,\tag{A2c}
$$

$$x \equiv \pi \cos(\theta),\tag{A2d}$$

$$y \equiv \pi \sin(\theta) - 2\alpha\_2 + 2\alpha\_3. \tag{A2e}$$

We often use the following term:

$$w \equiv \sqrt{x^2 + y^2}.\tag{A3}$$

Calculating the eigendecomposition of (A1b), we find that *M*1 = *U* · *D* · *U*†, where *D* is the diagonal matrix with elements (0, 0, *w*, <sup>−</sup>*<sup>w</sup>*), and unitary *U* is given by:

$$M \equiv (1/2) \begin{bmatrix} 0 & \sqrt{2} \mathbf{x}/w & (w-y)/w & (w+y)/w \\ \sqrt{2} & -\sqrt{2} y/w & -\mathbf{x}/w & \mathbf{x}/w \\ -\sqrt{2} & -\sqrt{2} y/w & -\mathbf{x}/w & \mathbf{x}/w \\ 0 & -\sqrt{2} \mathbf{x}/w & (w+y)/w & (w-y)/w \end{bmatrix}. \tag{A4}$$

Using this formula, we can calculate *V* ≡ exp(−*itM*) to be block diagonal with blocks *V*0 and *V*1, where *V*0 is the diagonal matrix with elements (*<sup>u</sup>*1, *u*2, *u*2, *<sup>u</sup>*3),

$$
u\_i \equiv \exp(-it\xi\_i),\tag{A5}$$

and *V*1 = *R* + *iQ*, with *R* and *Q* defined as follows:

$$R \equiv \begin{bmatrix} r\_1 & r\_2 & r\_2 & r\_3 \\ r\_2 & r\_4 & r\_3 & -r\_2 \\ r\_2 & r\_3 & r\_4 & -r\_2 \\ r\_3 & -r\_2 & -r\_2 & r\_1 \end{bmatrix} \tag{A6a} \tag{A6a}$$

$$\mathbf{Q} \equiv \begin{bmatrix} -q\_1 & q\_2 & q\_2 & 0 \\ q\_2 & 0 & 0 & q\_2 \\ q\_2 & 0 & 0 & q\_2 \\ 0 & q\_2 & q\_2 & q\_1 \end{bmatrix}' \tag{A6b}$$

where

$$\begin{aligned} r\_1 &\equiv 0.5(x^2 + (w^2 + y^2)\cos(tw))/w^2, \\ r\_2 &\equiv -0.5xy(1 - \cos(tw))/w^2, \\ r\_3 &\equiv -0.5x^2(1 - \cos(tw))/w^2, \\ r\_4 &\equiv 0.5(x^2\cos(tw) + w^2 + y^2)/w^2, \end{aligned} \tag{A7}$$

and

$$\begin{aligned} q\_1 &\equiv -y \sin(tw) / w, \\ q\_2 &\equiv 0.5x \sin(tw) / w. \end{aligned} \tag{A8}$$

One can check by direct calculations that the following identities hold:

$$\begin{aligned} r\_1^2 + q\_1^2 - r\_4^2 &= 0, \\ q\_2^2 + r\_2^2 + r\_3^2 + r\_3 &= 0, \\ r\_1 r\_2 + r\_2 r\_3 - q\_1 q\_2 + r\_2 &= 0, \\ q\_1 r\_2 + q\_2 r\_1 - q\_2 r\_3 - q\_2 &= 0, \\ r\_4 - r\_3 - 1 &= 0, \end{aligned} \tag{A9}$$

and that *r*3 ∈ [−1, 0] and *r*4 ∈ [0, 1].
