**Appendix C**

The state of an environmental qubit is given as the average of (A12), and thus it equals

$$\begin{bmatrix} 0.5 - r\_4(0.5 - p) & 0.5(1 - 2p)(r\_2 + iq\_2) \\ 0.5(1 - 2p)(r\_2 - iq\_2) & 0.5 + r\_4(0.5 - p) \end{bmatrix} \text{.} \tag{A17}$$

Now, let us consider how close is this state to the Gibbs state, in particular directly after the short-term C-INOT interaction, i.e., for *t* = 1?

Since the Hamiltonians (7) are diagonal in computational basis and proportional to *σZ* = 1 0 0 −1, the Gibbs state will also be diagonal, with the second diagonal value greater orequalthefirst(for*σZ*|1isthegroundstate).

Recall that from the form of the thermal environment (9), we have *p* ∈ [0, 0.5]. One can check that *r*4 ∈ [0, 1], and thus *<sup>r</sup>*4(0.5 − *p*) ≥ 0. Thereby, the second diagonal term of (A17) is greater or equal to the first, and thus the trace distance of the state (A17) from the closest thermal state is given by

$$\begin{array}{c} \left| \begin{bmatrix} 0 & 0.5(1-2p)(r\_2+iq\_2) \\ 0.5(1-2p)(r\_2-iq\_2) & 0 \end{bmatrix} \right| \right|\_{\text{Tr}} \end{array} \tag{A18}$$

where ||· ||Trdenotes the trace norm. This is equal to

$$|1 - 2p|\sqrt{r\_2^2 + q\_2^2} = |(1 - 2p)|\sqrt{-r\_4^2 + r\_4}. \tag{A19}$$

Direct calculations using (A9) show that either of the sides of (A19) can be rewritten also as

$$|0.5 - p| \ge \sqrt{1 - \cos\left(t\sqrt{\mathbf{x}^2 + y^2}\right)} \sqrt{\left(1 + \cos\left(t\sqrt{\mathbf{x}^2 + y^2}\right)\right)} \mathbf{x}^2 + 2y^2/(\mathbf{x}^2 + y^2). \tag{A20}$$

For fixed *p* and *θ*, the value of (A19) is a function of *r*4 that depends on the difference *α*2 − *α*3. The same holds for F(0, <sup>1</sup>), cf. (A16), as *μ* and *ν* are also functions of *p*, *θ* and *r*4, and thus there is a direct interplay between those two phenomena, the orthogonalization of observables and thermalization.
