One checks that the action of $\mathcal{L}_{\text{tot}}$ on the Pauli operators and the identity are

$$ \mathcal{L}_{\text{tot}}(1) = 0, \quad \mathcal{L}_{\text{tot}}(X) = \omega Y, \\ \mathcal{L}_{\text{tot}}(Y) = -\omega X - \gamma Y, \quad \mathcal{L}_{\text{tot}}(Z) = -\gamma Z. \qquad (\text{G42}) $$

Therefore, $\mathcal{L}_{\text{tot}}$ can be represented in the orthonormal basis $\{|1\rangle\rangle/\sqrt{2}, |X\rangle\rangle/\sqrt{2}, |Y\rangle\rangle/\sqrt{2}, |Z\rangle\rangle/\sqrt{2}\}$ of Pauli operators (denoted with subscript P) as

$$ [\mathcal{L}_{\text{tot}}]_P = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -\omega & 0 \\ 0 & \omega & -\gamma & 0 \\ 0 & 0 & 0 & -\gamma \end{pmatrix}_P . \qquad (\text{G43}) $$

One can verify that this matrix is diagonalized as

$$ [\mathcal{L}_{\text{tot}}]_P = S \begin{pmatrix} \lambda_+ & & & \\ & \lambda_- & & \\ & & -\gamma & \\ & & & \end{pmatrix} S^{-1}, \qquad (\text{G44}) $$

$$ \lambda_{\pm} = -\frac{\gamma}{2} \pm i\alpha, \quad \alpha = \frac{1}{2}\sqrt{4\omega^2 - \gamma^2}, \quad \lambda_{+}\lambda_{-} = \omega^2, \quad \lambda_{+} + \lambda_{-} = -\gamma, $$

$$ S = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -\frac{\lambda_{-}}{\omega} & -\frac{\lambda_{+}}{\omega} & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

$$ S^{-1} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{-i\omega}{2\alpha} & \frac{-i\lambda_{+}}{2\alpha} & 0 \\ 0 & \frac{i\omega}{2\alpha} & \frac{i\lambda_{-}}{2\alpha} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \qquad (\text{G45}) $$

We can solve the dynamics analytically using this diagonal representation to compute the matrix exponential as

$$ \mathcal{E}_t = [e^{t\mathcal{L}_{\text{tot}}}]_P = S \begin{pmatrix} 1 & & & \\ & e^{t\lambda_+} & & \\ & & e^{t\lambda_-} & \\ & & & e^{-\gamma t} \end{pmatrix} $$

$$ S^{-1} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & e_{xx} & e_{xy} & 0 \\ 0 & e_{yx} & e_{yy} & 0 \\ 0 & 0 & 0 & e^{-\gamma t} \end{pmatrix}_P, $$

$$ e_{xx} = e^{-\frac{\gamma t}{2}} \left[ \cos(\alpha t) + \frac{\gamma}{2\alpha} \sin(\alpha t) \right], \quad e_{xy} = -e_{yx}, \qquad (\text{G46}) $$

$$ e_{yx} = \frac{\omega}{\alpha} e^{-\frac{\gamma t}{2}} \sin(\alpha t), \quad e_{yy} = e_{xx}. $$

This gives us a useful expression of the linear operator $\mathcal{E}_t$ acting on the operator basis of Pauli operators. If we let the

<figure><img src="image_22.png" /><figcaption>FIG. 16. Trajectory on the equatorial slice of the Bloch sphere of the state of a qubit initialized in the state vector $|+\rangle$, evolving under the Hamiltonian $H = (\omega/2)Z$ and subject to continuous dephasing along the X axis. Here $\omega = 1$ and $\gamma = 0.1$.</figcaption></figure>

initial state $\psi_{\text{init}} = |+\rangle\langle +|$ evolve for a time $t$, we obtain

$$ \rho(t) = \mathcal{E}_t(\psi_{\text{init}}) = \mathcal{E}_t\left(\frac{1+X}{2}\right) = \frac{1}{2} + \frac{e^{-\frac{\gamma t}{2}}}{2} \left[ \left(\cos(\alpha t) + \frac{\gamma}{2\alpha} \sin(\alpha t)\right) X + \frac{\omega}{\alpha} \sin(\alpha t) Y \right]. \qquad (\text{G47}) $$

See Fig. 16 for a plot of the trajectory of the state $\rho(t)$ in the X-Y plane of the Bloch sphere.

We can compute the derivative $\partial_t \rho$ by directly differentiating the expression (G47) or by simply applying the Lindbladian since we have determined its action in the Pauli basis:

$$ \begin{aligned} \partial_t \rho = & \frac{e^{-\frac{\gamma t}{2}}}{2} \left[ \left( \cos(\alpha t) + \frac{\gamma}{2\alpha} \sin(\alpha t) \right) \omega Y \right. \\ & \left. + \frac{\omega}{\alpha} \sin(\alpha t) (-\omega X - \gamma Y) \right] \\ = & \frac{\omega}{2} e^{-\frac{\gamma t}{2}} \left[ -\frac{\omega}{\alpha} \sin(\alpha t) X + \left( \cos(\alpha t) - \frac{\gamma}{2\alpha} \sin(\alpha t) \right) Y \right]. \end{aligned} \qquad (\text{G48}) $$

The approximation we make to apply our uncertainty relation is to replace this expression for $\partial_t \rho$ by

$$ \begin{aligned} \mathcal{E}_t(-i[H, \psi_{\text{init}}]) &= \mathcal{E}_t\left(-i\left[\frac{\omega}{2}Z, \frac{1+X}{2}\right]\right) = \frac{\omega}{2}\mathcal{E}_t(Y) \\ &= \frac{\omega}{2}e^{-\frac{\gamma t}{2}}\left[-\frac{\omega}{\alpha}\sin(\alpha t)X + \left(\cos(\alpha t) + \frac{\gamma}{2\alpha}\sin(\alpha t)\right)Y\right]. \end{aligned} \qquad (\text{G49}) $$

We see that the two expressions (G48) and (G49) differ by a term $(2\alpha)^{-1}\gamma\omega e^{-\gamma t/2}\sin(\alpha t)Y$, which is small as long as $\gamma \ll \omega$.
The Fisher information $F_{\text{clock},t}$ given by Eq. (134) and $F_{\text{clock},U,t}$ given by Eq. (140) are plotted in Fig. 17 as a function of $t_0$. Our approximation matches the exact Fisher information well, except for an out-of-phase oscillation of relatively small amplitude. This error to the contribution of the phase damping is expected to be attributable to the difference in sign of the smaller terms in Eqs. (G48) and (G49).

## APPENDIX H: PERTURBING THE NOISY CHANNEL TO RESTORE EQUALITY IN THE UNCERTAINTY RELATION FOR METROLOGICAL CODES

In this Appendix, we study how to perturb a noisy channel $\mathcal{N}$ in order to restore uncertainty relation equality for a metrological code. We prove Proposition 2 of the main text, which shows that equality in the uncertainty relation can be restored by an infinitesimal perturbation of the Stinespring isometry, all while preserving the zero sensitivity loss conditions (148) (it might be necessary to enlarge Bob's system with an auxiliary qubit). The proposition is slightly reformulated to emphasize the fact that we can apply the same construction also without regards to the zero sensitivity loss condition.

**Proposition 22.**—let $V_{A \to BE}$ be an isometry, let $|\psi\rangle_A, |\xi\rangle_A$ with $\langle\psi|\xi\rangle_A = 0$ and let $\mathcal{N}(\cdot) = \text{tr}_E(V(\cdot) V^\dagger)$, $\tilde{\mathcal{N}}(\cdot) = \text{tr}_B(V(\cdot) V^\dagger)$. For any $\epsilon > 0$, there exists an isometry $V'_{A \to BE}$ with $\|V' - V\| \le \epsilon$ and such that $(P_{\rho'_B}^\perp \otimes P_{\rho'_E}^\perp) V' |\xi\rangle = 0$, where $\rho'_B = \text{tr}_E\{V' \psi V^\dagger\}$ and $\rho'_E = \text{tr}_B\{V' \psi V^\dagger\}$.

Furthermore, assume that $\tilde{\mathcal{N}}(|\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|) = 0$ and assume that there exists a unitary operator $G_B$ acting on the system $B$ with the properties that $P_{\rho_B} G_B P_{\rho_B} = 0$, $P_{\zeta_B} G_B P_{\zeta_B} = 0$, $P_{\rho_B} G_B P_{\zeta_B} = 0$, and $P_{\zeta_B} G_B P_{\rho_B} = 0$, where $\zeta_B = \mathcal{N}(|\xi\rangle\langle\xi|)$ and $\zeta_E = \tilde{\mathcal{N}}(|\xi\rangle\langle\xi|)$. Then the perturbed isometry $V''$ can be chosen to also satisfy $\tilde{\mathcal{N}}'(|\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|) = 0$, where $\tilde{\mathcal{N}}'(\cdot) = \text{tr}_B\{V'(\cdot) V^{\dagger}\}$.

**Proof.**—Write $\mathcal{N}(\cdot) = \text{tr}_E(V(\cdot) V^\dagger)$ and let

$$ \rho_B = \mathcal{N}(\psi); \quad \rho_E = \tilde{\mathcal{N}}(\psi). \qquad (H1) $$

The strategy to perturb $V$ is to include an infinitesimal rotation that rotates the state $V|\psi\rangle$ into the direction of another suitably chosen state $|\chi\rangle_{BE}$. We first compute some properties of a general such rotation, and then we will prove the stated claims.

Let $\epsilon > 0$. Let $\alpha > 0$ such that $4 \sin^2(\alpha/2) \le \epsilon$. Let $|\chi\rangle_{BE}$ be a state with the property that the reduced state on $B$ lies in a subspace that is orthogonal to the reduced state $\rho_B$ of $V|\psi\rangle$, i.e., $P_{\rho_B}|\chi\rangle = 0$, or equivalently, $|\chi\rangle_{BE}$ lies in the support of $P_{\rho_B}^\perp \otimes 1$. The state $|\chi\rangle_{BE}$ will be fixed later. Let $\{|\mu^{(j)}\rangle\}_j$ be a basis of $BE$ with $|\mu^{(1)}\rangle_{BE} = V|\psi\rangle$ and

$|\mu^{(2)}\rangle_{BE} = |\chi\rangle_{BE}$. Let

$$ W_{BE \to BE} = (\cos(\alpha)|\mu^{(1)}\rangle + \sin(\alpha)|\mu^{(2)}\rangle)\langle\mu^{(1)}| \\ + (\cos(\alpha)|\mu^{(2)}\rangle - \sin(\alpha)|\mu^{(1)}\rangle)\langle\mu^{(2)}| \\ + \sum_{j=3,...} |\mu^{(j)}\rangle\langle\mu^{(j)}|, \qquad (H2) $$

and note that $W_{BE \to BE}$ is a unitary close to the identity, effecting the rotation $W_0 = \begin{bmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{bmatrix}$ between $|\mu^{(1)}\rangle_{BE}$ and $|\mu^{(2)}\rangle_{BE}$. The eigenvalues $\theta_1, \theta_2$ of $W_0$ are determined from $\theta_1 + \theta_2 = \text{tr}(W_0) = 2 \cos(\alpha)$ and $\theta_1\theta_2 = \text{det}(W_0) = 1$ as $\theta_1 = \theta_2^* = e^{i\alpha}$. As the operator norm is the maximal singular value, we find $\|W_0 - 1\|_{\infty} = \max\{|e^{i\alpha} - 1|, |e^{-i\alpha} - 1|\} = (1 - \cos\alpha)^2 + (\sin\alpha)^2 = 2 - 2 \cos(\alpha) = 4 \sin^2(\alpha/2) \le \epsilon$, and $\|W - 1\|_{\infty} \le \epsilon$. Now let $V'' = W_{BE}V$, with

$$ \|V' - V\|_{\infty} \le \|W - 1\|_{\infty} \|V\|_{\infty} \le \epsilon. \qquad (H3) $$

We find

$$ \begin{aligned} \rho'_E &= \text{tr}_B(V' \psi V^{\dagger}) \\ &= \text{tr}_B[\cos^2(\alpha) |\mu_1\rangle\langle\mu_1| + \cos(\alpha) \sin(\alpha) \\ &\quad \times (|\mu_1\rangle\langle\mu_2| + |\mu_2\rangle\langle\mu_1|) \\ &\quad + \sin^2(\alpha) |\mu_2\rangle\langle\mu_2|] \\ &= \cos^2(\alpha) \rho_E + \cos(\alpha) \sin(\alpha) \text{tr}_B \\ &\quad \times [V|\psi\rangle\langle\chi| + |\chi\rangle\langle\psi|V^{\dagger}] + \sin^2(\alpha) \chi_E. \\ &= \cos^2(\alpha) \rho_E + \sin^2(\alpha) \chi_E. \end{aligned} \qquad (H4) $$

The last equality holds thanks to our assumption that $P_{\rho_B}|\chi\rangle = 0$.

We now prove the first part of the proposition. We can assume without loss of generality that $\text{rank}(P_{\rho_E}^\perp) \le \text{rank}(P_{\rho_B}^\perp)$, by exchanging the roles of the $B$ and $E$ systems if necessary. Let $\{|\chi_k\rangle_B\}_{k=1}^K$, $\{|\chi'_k\rangle_E\}_{k=1}^K$ be two orthonormal families of states lying in the support of $P_{\rho_B}^\perp$ and $P_{\rho_E}^\perp$, respectively, with $K = \min\{\text{rank}(P_{\rho_B}^\perp), \text{rank}(P_{\rho_E}^\perp)\} = \text{rank}(P_{\rho_E}^\perp)$. Define $|\chi\rangle_{BE} = (1/\sqrt{K}) \sum_k |\chi_k\rangle_B \otimes |\chi_k\rangle_E$. By construction, we have that $\chi_E = (1/K) \sum_{k=1}^K |k\rangle\langle k|_E = P_{\rho_E}^\perp / \text{tr}(P_{\rho_E}^\perp)$. It follows that the state (H4) has full rank, and therefore our conditions for our uncertainty relation equality are fulfilled.

Now we prove the second part of the proposition, and we assume that $\tilde{\mathcal{N}}(D_A^Z) = 0$, with $D_A^Z = |\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|$. The proof strategy is similar to above, to introduce a small “rotation” to fix the support of the state $\rho_E$ all while preserving the zero sensitivity loss conditions (148).
<figure><img src="image_23.png" /><figcaption>FIG. 17. Relative Fisher information with respect to time of a single qubit prepared in $|+\rangle$ and evolving according to the Hamiltonian $(\omega/2)Z$ and exposed to continuous dephasing along the X axis at a rate $\gamma$. The blue curve shows the sensitivity as a function of time $t_0$ of the probe to the signal if we turn off the noise. In orange, the exact Fisher information $F(\rho; \partial_{\omega}\rho)$ is computed directly. In green, an approximation to the desired Fisher information ignores the contribution of the time dependency $\partial_t N_t$ of the effective noisy channel. This approximation is the quantity that appears in our trade-off relation in the alternative setting where Alice sends a noiseless quantum clock over a noisy channel to Bob. Because the unitary and noise parts of the Lindbladian do not commute as superoperators, invoking our the trade-off relation requires the channel $N_t$ to be determined via Eq. (135). Here $\omega = 1$, $\gamma = 0.1$.</figcaption></figure>

Without loss of generality, we may assume that $\| |\xi\rangle \| = 1$. We define for later convenience

$$ Z_L = |\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|; \quad \Pi_L = |\psi\rangle\langle\psi| + |\xi\rangle\langle\xi|; \\ \tilde{Z}_L = Z_L + (1 - \Pi_L), \qquad (H5) $$

noting that $\tilde{Z}_L$ is the unitary operator that flips the normalized states $|\psi\rangle$ and $|\xi\rangle$ and acts as the identity on the subspace that is orthogonal to $|\psi\rangle$, $|\xi\rangle$.

As stated in the claim, we assume that there exists a unitary operator $G_B$ with the properties that $P_{\rho_B} G_B P_{\rho_B} = 0$, $P_{\xi_B} G_B P_{\xi_B} = 0$, $P_{\rho_B} G_B P_{\xi_B} = 0$, and $P_{\xi_B} G_B P_{\rho_B} = 0$.

Let $0 < \epsilon \le 1$. Let $\alpha = \epsilon/2$ with $0 < \alpha \le 1/2$ and let

$$ V' = (\cos(\alpha)V + \sin(\alpha)G_B V\tilde{Z}_L). \qquad (H6) $$

Then

$$ \begin{aligned} \|V' - V\| &= \|(\cos(\alpha) - 1)V + \sin(\alpha)G_B V\tilde{Z}\| \\ &\le (1 - \cos(\alpha))\|V\| + \sin(\alpha)\|G_B V\tilde{Z}\| \\ &\le 2\sin^2(\alpha/2) + \sin(\alpha) \le 2|\alpha|, \end{aligned} \qquad (H7) $$

using $\sin(\alpha) \le |\alpha|$ and with $|\alpha| \le 1/2$.

We first show that the perturbed isometry $V'$ also satisfies the zero sensitivity-loss conditions. Let $\tilde{N}'(\cdot) = \text{tr}_B\{V' \cdot V'^{\dagger}\}$ and we compute

$$ \begin{aligned} \tilde{N}'(Z_L) &= \text{tr}_B\{V' Z_L V'^{\dagger}\} \\ &= \text{tr}_B\{\cos^2(\alpha) V Z_L V'^{\dagger}\} \end{aligned} $$

$$ \begin{aligned} &+ \cos(\alpha) \sin(\alpha) [V Z_L \tilde{Z}_L V^{\dagger} G_B^{\dagger} + G_B V \tilde{Z}_L Z_L V^{\dagger}] \\ &+ \sin^2(\alpha) G_B V \tilde{Z}_L Z_L \tilde{Z}_L V^{\dagger} G_B^{\dagger} \} \\ &= 0, \end{aligned} \qquad (H8) $$

using $\text{tr}_B\{V Z_L V'^{\dagger}\} = \tilde{N}(Z_L) = 0$ and $\tilde{Z}_L Z_L \tilde{Z}_L = Z_L$, as well as the fact that

$$ \begin{aligned} \text{tr}_B[G_B V \tilde{Z}_L Z_L V^{\dagger}] &= \text{tr}_B[G_B V [ |\psi\rangle\langle\psi| + |\xi\rangle\langle\xi| ] V^{\dagger}] \\ &= \text{tr}_B[G_B P_{\rho_B} V \psi V^{\dagger} P_{\rho_B} \\ &\qquad + G_B P_{\xi_B} V \xi V^{\dagger} P_{\xi_B}] = 0, \end{aligned} \qquad (H9) $$

using the fact that $P_{\rho_B} G_B P_{\rho_B} = 0 = P_{\xi_B} G_B P_{\xi_B}$. We then have

$$ \begin{aligned} \rho'_E &= \text{tr}_B\{V' \psi V'^{\dagger}\} \\ &= \text{tr}_B\{\cos^2(\alpha) V \psi V^{\dagger} + \cos(\alpha) \sin(\alpha) \\ &\qquad \times [V |\psi\rangle\langle\xi| V^{\dagger} G_B^{\dagger} + G_B V |\xi\rangle\langle\psi| V^{\dagger}] \\ &\qquad + \sin^2(\alpha) G_B V \xi V G_B^{\dagger}\} \\ &= \cos^2(\alpha) \rho_E + \sin^2(\alpha) \zeta_E, \end{aligned} \qquad (H10) $$

where the two middle terms in the long expression vanish because $\text{tr}_B\{G_B V |\xi\rangle\langle\psi| V^{\dagger}\} = \text{tr}_B\{P_{\rho_B} G_B P_{\xi_B} V |\xi\rangle\langle\psi| V^{\dagger}\} = 0$.
Similarly,

$$
\begin{align}
\zeta'_{E} &= \operatorname{tr}_{B} \{ V' \xi V'^{\dagger} \} \nonumber \\
&= \operatorname{tr}_{B} \{ \cos^{2}(\alpha) V \xi V'^{\dagger} + \cos(\alpha) \sin(\alpha) \nonumber \\
&\qquad \times [V | \xi \rangle \langle \psi | V'^{\dagger} G_{B}^{\dagger} + G_{B} V | \psi \rangle \langle \xi | V'^{\dagger}] \nonumber \\
&\qquad + \sin^{2}(\alpha) G_{B} V \psi V'^{\dagger} G_{B}^{\dagger} \} \nonumber \\
&= \cos^{2}(\alpha) \zeta_{E} + \sin^{2}(\alpha) \rho_{E}. \tag{H11}
\end{align}
$$

Any state $|c\rangle_E$ that lies in the kernel of $\rho'_E$ must satisfy

$$0 = \langle c | \rho'_{E} | c \rangle = \cos^2(\alpha) \langle c | \rho_{E} | c \rangle + \sin^2(\alpha) \langle c | \zeta_{E} | c \rangle, \quad (H12)$$

which in turn implies $0 = \langle c | \rho_E | c \rangle = \langle c | \zeta_E | c \rangle$. We then find

$$
\begin{align}
\|(1_B \otimes \langle c|_E) V'|\xi\rangle\|^2 &= \langle c|_E \operatorname{tr}_B(V'\xi V'^{\dagger})|c\rangle_E \nonumber \\
&= \langle c|\zeta'_E|c\rangle \nonumber \\
&= \langle c|_E [\cos^2(\alpha) \zeta_E + \sin^2(\alpha) \rho_E]|c\rangle_E \nonumber \\
&= 0. \tag{H13}
\end{align}
$$

Therefore, $(1_B \otimes P_{\rho'_E}^{\perp}) V' |\xi\rangle = 0$, implying that $(P_{\rho'_B}^{\perp} \otimes P_{\rho'_E}^{\perp}) V' |\xi\rangle = 0$ and our uncertainty relation equality conditions are satisfied. ■

# APPENDIX I: BEHAVIOR OF METROLOGICAL CODES FOR WEAK IID NOISE; METROLOGICAL CODES, UNCERTAINTY RELATION EQUALITY, AND DISCONTINUITIES OF THE QUANTUM FISHER INFORMATION

In this Appendix, we consider a metrological code $(|\psi\rangle, |\xi\rangle)$ on $n$ qubits, with a metrological distance $d_m > 1$. For any noise channel that acts on fewer than $d_m$ qubits, we have seen in Sec. VII that $\Delta F_{\text{Bob},t} = 0$. Instead of noise acting on few qubits, we now consider examples of IID noise channels $[\mathcal{N}_1^{(p)}]^{\otimes n}$, where each channel $\mathcal{N}_1^{(p)}$ acts on a single qubit and depends on a noise parameter $p$ such that $\mathcal{N}_1^{(p=0)} = \text{id}$. We ask, for constant $n$, to what order in $p$ is the loss in quantum Fisher information $\Delta F_{\text{Bob},t}$ suppressed?

Let us first consider a similar question in the conventional setting of quantum error correction, where a logical state is encoded into a physical state, is exposed to a noise channel, and is subsequently decoded to attempt to recover the initial state. If a state $|\psi\rangle$, encoded with a distance-$d$ quantum error-correcting code, is exposed to a weak IID noise channel in which a single-site error happens with

probability $p$, then after a subsequent decoding operation, the fidelity of the state with respect to the original state differs with the ideal value one by at most $O(p^{d/2})$. In other words, the fidelity loss is suppressed by the quantum error-correction procedure to an order in the noise parameter that is proportional to the distance of the code. This suppressed fidelity loss is explained by a fundamental principle in quantum information: Two states (respectively, two channels) that are $\epsilon$ close in trace distance (respectively, diamond distance) may not be distinguished by any physical operation, except with probability of the order at most $O(\epsilon)$. In the case of weak IID noise, any error operator whose weight is larger than $(d-1)/2$ occurs only with probability at most $O(p^{d/2})$. Consequently, no experiment should be able to distinguish the weak IID noise from a noise operator with only weight-[(d-1)/2] operators with probability better than $O(p^{d/2})$, for which the quantum error-correction scheme enables perfect recovery.

By analogy, it is natural to expect that the quantum Fisher information loss $\Delta F_{\text{Bob},t}$ should scale as approximately $p^{cd_m}$, where $d_m$ is the metrological distance of the metrological code, and where $c$ is some constant. However, this is not the case, as we will see in the remainder of this Appendix. While $\Delta F_{\text{Bob},t}$ exhibits the expected behavior for certain examples of metrological codes, we can find counterexamples in which the quantum Fisher information loss scales as $\Delta F_{\text{Bob},t} \sim p$ despite the state forming a metrological code of an arbitrarily large, but fixed, metrological distance $d_m$. This counterexample shows that when measuring the accuracy of Bob's estimate to the time parameter in terms of the quantum Fisher information, the code distance is not necessarily related to the loss in sensitivity of the state. This might be worrying, since the metrological distance of the metrological code would not be related to the degree of protection offered by such codes in suppressing the sensitivity loss. We argue, however, that the quantum Fisher information might not be the relevant sensitivity measure to study in such regimes. More specifically, we know that there are regimes in which we should question the operational relevance of the quantum Fisher information, because infinitesimal perturbations in the state or the noise channel result in observable consequences in the purported sensitivity as reported by the quantum Fisher information. We attribute this behavior to the fact that it ignores the error associated with the estimation of the expectation value of the optimal sensing observable from a finite number of measurement repetitions. Based on our examples, we hypothesize that the settings where $\Delta F_{\text{Bob},t} \ll O(p^{d_m/2})$ fall into this regime. While confirming this hypothesis would invalidate known counterexamples in which a high metrological distance can still lead to a high accuracy loss, a full proof of the protection offered by metrological codes in the general setting remains elusive. Such a result would further require (a) establishing a measure of sensitivity that is robust to
perturbations of the physical setting by accounting for limits on the number of available measurement repetitions and (b) showing that its loss is suppressed as a function of the metrological distance of the metrological code.

In the following, we first compute the quantum Fisher information loss of some states that form metrological codes after exposure to weak IID noise. In order to explore the cause of the behavior of some examples that appear problematic, we study more closely some properties of the quantum Fisher information: we argue that there are regimes in which the quantum Fisher information, being discontinuous, cannot be a representative measure of sensitivity, and we attribute this problematic behavior to the failure to account for the number of finite available measurement repetitions. Finally, we consider a restricted setting with additional assumptions on the state and the noise channel, in which we prove the expected bound on the quantum Fisher information loss $\Delta F_{\text{Bob},t} \le O(p^{d_m/2})$.

# 1. Examples of metrological codes exposed to weak IID noise

We now consider three single-site noise channels: the amplitude-damping channel, the dephasing channel in the Z basis, and the bit-flip channel. In the basis $\{|\uparrow\rangle, |\downarrow\rangle\}$, the single-qubit amplitude-damping channel has Kraus operators

$$E_{\text{AD},0}^{(p)} = \begin{pmatrix} \sqrt{1-p} & 0 \\ 0 & 1 \end{pmatrix}; \quad E_{\text{AD},1}^{(p)} = \begin{pmatrix} 0 & 0 \\ \sqrt{p} & 0 \end{pmatrix}. \quad (11)$$

The second noise channel we consider is the dephasing channel in the Z basis, described by the Kraus operators

$$E_{\text{dephas.},0}^{(p)} = \sqrt{1 - \frac{p}{2}} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}; \quad E_{\text{dephas.},1}^{(p)} = \sqrt{\frac{p}{2}} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \quad (12)$$

Finally, the bit-flip channel is described by the Kraus operators

$$E_{\text{bit-flip},0}^{(p)} = \sqrt{1 - \frac{p}{2}} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}; \quad E_{\text{bit-flip},1}^{(p)} = \sqrt{\frac{p}{2}} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \quad (13)$$

## a. Four-qubit code state based on the [[4, 2, 2]] code

Consider the state vector introduced in Secs. VII E and VII F,

$$|\psi_{\text{code}}\rangle = \frac{1}{2} [ |\uparrow\uparrow\uparrow\rangle + |\downarrow\downarrow\downarrow\rangle + |\uparrow\downarrow\uparrow\rangle + |\downarrow\downarrow\downarrow\rangle ] . \quad (14)$$

Consider the Hamiltonian consisting of ZZ terms on the edges connecting the four qubits when they are arranged

<figure><img src="image_24.png" /><figcaption>FIG. 18. Quantum Fisher information loss $\Delta F_{\text{Bob},t}$ after exposure of $|\psi_{\text{code}}\rangle$ [cf. Eq. (14)] to IID amplitude-damping or dephasing noise in the Z basis, as a function of the noise parameter $p$. Based on our intuition of standard error-correcting codes, we might have expected that $\Delta F_{\text{Bob},t}$ depends only on an order in $p$ (for $p \to 0$) that is directly related to $d_m$ (or $d_m/2$). In the case of either noise model, we fit the data points where $p < 0.1$ to $\ln(y) = a \ln(p) + b$ (which corresponds to a power law $y \propto p^a$) to obtain the order in $p$ to which $\Delta F_{\text{Bob},t}$ is affected. We see that for amplitude-damping noise, the loss in quantum Fisher information is suppressed to depend only on $p$ to second order; for dephasing noise, the loss is affected to first order in $p$. The quantum Fisher information loss due to an IID bit-flip noise channel (not shown) behaves very similarly to the dephasing noise.</figcaption></figure>

in a square, as in Fig. 9(a); with a suitable normalization we obtain

$$|\xi_{\text{code}}\rangle = \frac{1}{2} [ |\uparrow\uparrow\uparrow\rangle + |\downarrow\downarrow\downarrow\rangle - |\uparrow\downarrow\uparrow\rangle - |\downarrow\downarrow\downarrow\rangle ] . \quad (15)$$

We have seen that $(|\psi_{\text{code}}\rangle, |\xi_{\text{code}}\rangle)$ forms a metrological code of metrological distance 2.

Let us consider how the quantum Fisher information of this state drops when exposed to IID amplitude-damping noise and to IID dephasing noise. The quantum Fisher information loss $\Delta F_{\text{Bob},t}$ is plotted in a log-log plot as a function of $p$ in Fig. 18. We fit the computed values for points with $p < 0.1$ to the model $\ln(y) = a \ln(p) + b$ in order to determine the quantum Fisher information loss order (as $y \propto p^a$). We observe that while the quantum Fisher information loss is indeed affected only to second order in $p$ for amplitude-damping noise, it is directly affected to first order for dephasing noise. The behavior of this small-scale example is not necessarily surprising, although it rules out an optimistic conjecture that states of the form (14) could have their loss in quantum Fisher information be protected to second order in $p$ against any IID noise channel, as could have been suggested from Fig. 13.

## b. Repetition code in the +/- basis

Now we investigate a larger example that shows that the metrological distance is not always indicative of the order
of quantum Fisher information loss in the noise parameter. On n qubits, let

$$|\psi\rangle = |+\rangle^{\otimes n}; \quad |\xi\rangle = |-\rangle^{\otimes n}. \qquad (16)$$

Here, the Hamiltonian corresponding to these states is the nonlocal operator $H = Z^{\otimes n}$. (Note that this example differs starkly from a standard ensemble of n spins where the Hamiltonian is as a sum of Z terms on each site. In that case, $|\xi\rangle$ would be a superposition of strings that consist of all $|+\rangle$ state vectors and a single $|-\rangle$ state vector.) The $(|\psi\rangle, |\xi\rangle)$ given above form a metrological code of distance $d_m = n$. Indeed, any operator $O$ with $wgt(O) < n$ cannot make $|\psi\rangle$ nonorthogonal to $|\xi\rangle$, and the conditions (155) are satisfied.

We show that if we expose this state to IID dephasing noise along the Z axis, the quantum Fisher information loss is indeed suppressed to the order $O(p^{n/2})$, as we would expect. On the other hand, if we expose the state to IID bit-flip noise, which can be seen as dephasing noise along the X axis, then the quantum Fisher information loss is not suppressed as expected and we find $\Delta F_{\text{Bob},t} \sim p$.

Let us first consider IID dephasing noise along the Z axis. We show that the quantum Fisher information loss is indeed suppressed to order $O(p^{n/2})$ for this noise channel. We now prove this statement. We may choose for the noise channel $\hat{\mathcal{N}}_{\text{dephas}}^{(p)}(\cdot) = (1-p/2)(\cdot) + (p/2)Z(\cdot)Z$ the Stinespring isometry

$$\begin{aligned} V_{A \to BE} &= \sqrt{1 - \frac{p}{2}} 1 \otimes |0\rangle_E + \sqrt{\frac{p}{2}} Z \otimes |1\rangle_E \\ &= |p_+\rangle_E \langle \uparrow|_A \otimes |\uparrow\rangle_B + |p_-\rangle_E \langle \downarrow|_A \otimes |\downarrow\rangle_B, \end{aligned} \qquad (17)$$

with respect to some basis $|0\rangle, |1\rangle$ on $E$, and with

$$|p_{\pm}\rangle = \sqrt{1 - \frac{p}{2}} |0\rangle \pm \sqrt{\frac{p}{2}} |1\rangle. \qquad (18)$$

This choice leads to the complementary channel

$$\hat{\mathcal{N}}_{\text{dephas}}^{(p)}(\cdot) = \langle \uparrow|\cdot|\uparrow\rangle_A |p_+\rangle \langle p_+|_E + \langle \downarrow|\cdot|\downarrow\rangle_A |p_-\rangle \langle p_-|_E. \qquad (19)$$

We find

$$\begin{aligned} \rho_{E_1} &= \hat{\mathcal{N}}_{\text{dephas}}^{(p)}(|+\rangle\langle+|) = \frac{1}{2} [ |p_+\rangle\langle p_+|_E + |p_-\rangle\langle p_-|_E ] = \begin{pmatrix} 1-p & 0 \\ 0 & p \end{pmatrix}; \\ \hat{\mathcal{N}}_{\text{dephas}}^{(p)}(|+\rangle\langle-|) &= \frac{1}{2} [ |p_+\rangle\langle p_+|_E - |p_-\rangle\langle p_-|_E ] = \sqrt{\frac{p}{2}} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{aligned} \qquad (110)$$

We can then compute

$$\begin{aligned} \Delta F_{\text{Bob},t} &= F([\hat{\mathcal{N}}_{\text{dephas}}^{(p)}(|+\rangle\langle+|)]^{\otimes n}; [\hat{\mathcal{N}}_{\text{dephas}}^{(p)}]^{\otimes n}([|+\rangle\langle-|]^{\otimes n} + [|-\rangle\langle+|]^{\otimes n})) \\ &= F\left(\begin{pmatrix} 1-p & 0 \\ 0 & p \end{pmatrix}^{\otimes n}; \left[\sqrt{\frac{p}{2}}\left(1-\frac{p}{2}\right)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right]^{\otimes n} + \text{h.c.}\right) \\ &= 4\left[\frac{p}{2}\left(1-\frac{p}{2}\right)\right]^n F\left(\begin{pmatrix} 1-p & 0 \\ 0 & p \end{pmatrix}^{\otimes n}; X^{\otimes n}\right) \\ &= 4p^n 2^{-n} \left(1-\frac{p}{2}\right)^n \sum_{\mathbf{x},\mathbf{x}'} \frac{2}{\lambda_{\mathbf{x}} + \lambda_{\mathbf{x}'}} |\langle\mathbf{x}|X^{\otimes n}|\mathbf{x}'\rangle|^2, \end{aligned} \qquad (111)$$

where $\mathbf{x}, \mathbf{x}'$ are bit strings and where $\lambda_{\mathbf{x}} = (1-p)^{|\mathbf{x}|}p^{n-|\mathbf{x}|}$ is the eigenvalue of $\rho_{E_1}^{\otimes n}$ associated with the eigenvector $|\mathbf{x}\rangle$. Observe that $\langle\mathbf{x}|X^{\otimes n}|\mathbf{x}'\rangle = \delta_{\mathbf{x},\mathbf{x}'}$, where $\tilde{\mathbf{x}}$ is the bit string obtained by flipping all the bits of $\mathbf{x}$. Then

noting that $\min(|\mathbf{x}|, n - |\mathbf{x}|) \le n/2$. Therefore,

$$(J11) = p^n \sum_{\mathbf{x}} O(p^{-n/2}) = O(p^{n/2}). \qquad (113)$$

$$\begin{aligned} \lambda_{\mathbf{x}} + \lambda_{\tilde{\mathbf{x}}} &= (1-p)^{|\mathbf{x}|}p^{n-|\mathbf{x}|} + (1-p)^{n-|\mathbf{x}|}p^{|\mathbf{x}|} \\ &= \Omega(p^{\min(|\mathbf{x}|,n-|\mathbf{x}|)}) = \Omega(p^{n/2}), \end{aligned} \qquad (112)$$

In other words, the quantum Fisher information on Bob's end after exposure of the state to IID dephasing noise along the Z axis is well protected, in that the loss is suppressed to
the order $O(p^{n/2})$. Observe that $\rho_E$ is full rank, and therefore our uncertainty relation holds with equality in this setting.

Consider now the IID bit-flip noise channel $[\mathcal{N}_{\text{bit-flip}}^{(p)}]^{\otimes n}$ determined by the single-site Kraus operators $E_{\text{bit-flip},0}^{(p)}$ and $E_{\text{bit-flip},1}^{(p)}$. We find

$$ \mathcal{N}_{\text{bit-flip}}^{(p)}(|+\rangle\langle+|) = |+\rangle\langle+|; \quad \mathcal{N}_{\text{bit-flip}}^{(p)}(|+\rangle\langle-|) \\ = (1-p)|+\rangle\langle-|. \qquad (114) $$

We would like to compute

$$ F_{\text{Bob},t} = F([\mathcal{N}_{\text{bit-flip}}^{(p)}]^{\otimes n}(|+\rangle\langle+|^{\otimes n}); [\mathcal{N}_{\text{bit-flip}}^{(p)}]^{\otimes n} \\ \times (-i[|-\rangle\langle+|]^{\otimes n} + i[|+\rangle\langle-|]^{\otimes n})) \\ = F(|+\rangle\langle+|^{\otimes n}; -i[(1-p)|-\rangle\langle+|]^{\otimes n} + \text{h.c.}) \\ = 4 \langle+^n|-i[(1-p)|-\rangle\langle+|]^{\otimes n} + \text{h.c.}]^2|+^n\rangle, \qquad (115) $$

where the last equality follows from Proposition 12. With

$$ \langle+^n|-i[(1-p)|-\rangle\langle+|]^{\otimes n} + \text{h.c.} = i(1-p)^n\langle-^n|, \qquad (116) $$

we find

$$ (115) = 4(1-p)^{2n} = 4 - 8np + O(p^2). \qquad (117) $$

Therefore, for bit-flip IID noise, we have

$$ \Delta F_{\text{Bob},t}[\text{bit-flip}] = 8np + O(p^2), \qquad (118) $$

meaning that the quantum Fisher information loss is linear in $p$ despite the high metrological distance $d_m$.

Note that, in the case of IID bit-flip noise, our uncertainty relation equality conditions are not satisfied, since the rank of $\rho_B$ changes locally as a function of time. In other words, we should not expect our uncertainty relation to hold with equality. This fact does not impact our calculation of the quantum Fisher information loss (118), since we determined this value by direct computation on Bob's side. However, based on this example, we are tempted to hypothesize that settings in which a high metrological distance does not inhibit a high accuracy loss under weak IID noise coincide with the settings in which our uncertainty relation does not hold with equality. In the remainder of this Appendix, we provide additional indications in favor of this hypothesis.

## 2. Discontinuities of the quantum Fisher and uncertainty relation equality conditions

We briefly return to study the behavior of the quantum Fisher information in a simple example in which our uncertainty relation equality conditions are not satisfied. In such cases, the state on Bob's side changes rank, and it is known that the quantum Fisher information can be discontinuous [33,34,110].

The definition of the quantum Fisher information that we use [Eq. 9], which can differ from the expression stemming from the second-order expansion of the Bures metric [33,34,110], directly expresses the accuracy to which one can sense an unknown parameter via an observable that reveals the true value of the parameter locally in expectation value (see Proposition 19 in Appendix D).

It is a fundamental principle in quantum information that a quantity that is measurable in a physical setting should be robust to infinitesimal perturbations of the quantum state. Yet, how is it possible that the quantum Fisher information is discontinuous, if it directly corresponds to the physically operational sensitivity to which one can estimate an unknown parameter locally? We attribute this discontinuity to the assumption, in Proposition 19 in Appendix D, that the sensing observable reveals the true parameter value in expectation value. An expectation value needs to be estimated using multiple rounds of measurements, and depending on the outcome distribution of the observable, an arbitrary large number of measurements might be required to accurately estimate its expectation value. In the following example, we study how the optimal sensing observable diverges close to discontinuity points of the quantum Fisher information; namely, the discontinuity can be associated with diverging eigenvalues of the observable associated with eigenstates that are outside the support of the state at the discontinuity point.

Overall, this example indicates that the operational relevance of the quantum Fisher information might break down in certain regimes where it is not possible to accurately estimate the expectation value of the optimal sensing observable.

The following example is based on Refs. [33,34,110]. Consider the example of Sec. IV B: a qubit state evolving along the equator of the Bloch sphere is collapsed by the noise channel along the X axis of the Bloch sphere. Bob's quantum Fisher information is constant and equal to $\omega^2$ almost all the time, except when the state is exactly a $\pm X$ eigenstate, in which case Bob's quantum Fisher information is equal to zero. The state on Bob's end is given by Eq. (89) as

$$ \begin{aligned} \rho_B &= p_+|+\rangle\langle+| + p_-|-\rangle\langle-|; \\ p_+ &= \cos^2\left(\frac{\omega t_0}{2}\right); \quad p_- = \sin^2\left(\frac{\omega t_0}{2}\right). \end{aligned} \qquad (119) $$
When Bob's quantum Fisher information $F_{\text{Bob},t}$ is nonzero, there is always an observable $O$ whose expectation value reveals the true parameter value locally, i.e., $\langle O \rangle_{\rho(t_0+dt)} = t_0 + dt + O(dt^2)$, and whose variance is $\langle O^2 \rangle - \langle O \rangle^2 = 1/\omega^2$ (cf. Appendix D). The optimal sensing observable is given by the suitably normalized symmetric logarithmic derivative (Proposition 19) and can be computed, when $\omega t_0$ is not a multiple of $\pi$, as follows:

$$ 
\begin{aligned}
O - t_0 1 &= \frac{1}{\omega^2} \mathcal{R}_{\rho t_0}^{-1}(\partial_t \rho) = \frac{1}{\omega^2} \\
&\quad \times \sum_{k,k'=\pm} \frac{2}{p_k + p_{k'}} \langle k | (\partial_t \rho) | k' \rangle |k\rangle \langle k'| \\
&= \frac{2}{2\omega^2 \cos^2(\frac{\omega t_0}{2})} \langle + | (\partial_t \rho) | + \rangle |+\rangle \langle +| \\
&\quad + \frac{2}{2\omega^2 \sin^2(\frac{\omega t_0}{2})} \langle - | (\partial_t \rho) | - \rangle |-\rangle \langle -| \\
&= -\frac{1}{\omega} \tan(\frac{\omega t_0}{2}) |+\rangle \langle +| + \frac{1}{\omega} \left[ \tan(\frac{\omega t_0}{2}) \right]^{-1} |-\rangle \langle -|,
\end{aligned} 
\quad (120) $$

using the relation $(\partial_t \rho) = -(\omega/2) \sin(\omega t_0) X$ [cf. Eq. (91)], which implies $\langle \pm | (\partial_t \rho) | \pm \rangle = \mp (\omega/2) \sin(\omega t_0) = \mp \omega \sin(\omega t_0/2) \cos(\omega t_0/2)$.

As a sanity check, we can verify that $O$ satisfies

$$ \langle O \rangle_{\rho(t_0+dt)} = t_0 + dt + O(dt^2), \quad (121) $$

as well as

$$ \sigma_O^2 = \langle O^2 \rangle_{\rho t_0} - \langle O \rangle_{\rho t_0}^2 = \frac{1}{\omega^2}. \quad (122) $$

As the state gets closer to a discontinuity (for instance, at $t_0 = 0$), this optimal sensing observable has one eigenvalue that diverges (for $t_0 = 0$, this eigenvalue is associated with the eigenvector $|−\rangle$). At the discontinuous point, the derivative is zero locally, so no observable will ever be able to correctly reveal the true value of the parameter to first order locally. The state does not change to first order in $t$ at all! We can attribute the discontinuity to the fact that an optimal sensing observable for one state might turn out to no longer be an acceptable sensing observable for a neighboring point. In other words, while the variance of an observable is continuous both as a function of the state and of the observable, the optimal variance in the local sensing scenario is discontinuous because the conditions of the optimization (D1) are discontinuous.

At the discontinuity $t_0 = 0$, the derivative $\partial_t \rho$ vanishes locally, and it is impossible to find an observable $O$ such that $\langle O \rangle_{\rho(t_0+dt)} = t_0 + dt + O(dt^2)$. By convention we set the corresponding quantum Fisher information to be zero; first, it is convenient because we do not have to modify the definition of the quantum Fisher information, and

second, it expresses the fact that we cannot have any sensitivity locally to first order in the parameter by measuring the expectation value of an observable. If the quantum Fisher information is defined starting from the Bures distance, a mismatch will be observed; this mismatch could be interpreted as a failure of the Cramér-Rao bound.

Operationally, even for $t_0$ not at one of the discontinuities, the use of the expectation value as the way of reading out the parameter in the estimation process might be problematic. Estimating the expectation value of $O$ to good accuracy, for $t_0 \approx 0$, requires that we observe sufficiently many times the $|−\rangle$ outcome, even though the latter only appears with the vanishing probability $\sin^2(\omega(t_0 + dt)/2)$. If we do not repeat the measurement on enough copies, we would only empirically observe $|+\rangle$ events and we would erroneously estimate the expectation value of $O$ to be equal to $-\left[\tan(\omega t_0/2)\right]/\omega$, and that its variance is zero. Not only this result would be wrong as it does not depend on the actual value $dt$ that we wanted to measure, but the variance is certainly incorrect since the optimal variance when an infinite number of measurements is available is $1/\omega^2$. There might be opportunities for defining and investigating refined measures of sensitivity that can account for the finite amount of measurement outcomes that can be collected in the estimation process.

The above example illustrates that the quantum Fisher information can be problematic to interpret in certain regimes close to points where the rank of the state can change. This type of regime can occur for metrological codes, if the noise happens to fix the state vector $|\psi\rangle$ while not fixing other states that are infinitesimally close to $|\psi\rangle$, resulting in a rank change for Bob and Eve's states. We observe that in the context of metrological codes exposed to weak IID noise, the quantum Fisher information is not actually discontinuous as a function of the noise parameter; rather, it is the order in $p$ of the Fisher information loss that can behave unexpectedly. That the quantum Fisher information loss must be suppressed at least to the order $O(p)$ follows from our continuity bound Proposition 18, noting that the weak IID noise channel is $O(p)$ close to the identity channel.

### 3. Suppression of quantum Fisher information loss in a restricted setting

Here we show that, when considering a metrological code exposed to weak IID noise in a restricted setting with additional assumptions, the quantum Fisher information loss $\Delta F_{\text{Bob},t}$ is suppressed to the expected order $O(p^{d_m/2})$, where $d_m$ is the metrological distance of the metrological code.

**Proposition 23.**—let $|\psi\rangle$, $|\xi\rangle$ define a metrological code of metrological distance $d_m$. Let $\mathcal{N}_1$ be a single-site noise operator with a Kraus representation $\{E_1^{(k)}\}_{k=1}^K$ that is such that $\|E_1^{(k)}\|_{\infty} = O(\sqrt{p})$ for $k' \neq 1$. Furthermore, if
x denotes a string of Kraus operator labels with $x_i \in \{1, \dots, K\}$, and if $E_x = (\bigotimes_{i=1}^n E_1^{(x_i)})$, we assume that the states $\{E_x|\psi\rangle\}_x$ are all nonzero and orthogonal, and that $\|E_x|\psi\rangle\| \ge \Omega(p^{|x|/2})$. Then $\Delta F_{\text{Bob},t} = O(p^{d_m/2})$.

This result follows fairly straightforwardly from Eq. (130) in Sec. V E.

*Proof.*—Using the notation in Eq. (130), with $\epsilon = p$, we have that $\Delta F_{\text{Bob},t} = O(p^m)$ with

$$m = \min_{\mathbf{x},\mathbf{x}'} \{2q_{\mathbf{x},\mathbf{x}'} - \min(r_{\mathbf{x}}, r_{\mathbf{x}'})\}, \quad (123)$$

where $r_x$ and $q_{x,x'}$ are defined via

$$\begin{gathered} \langle\psi|E_x^\dagger E_x|\psi\rangle = \Omega(p^{r_x}); \\ \text{tr}\{E_{x'}^\dagger E_x(|\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|)\} = O(p^{q_{x,x'}}) \end{gathered} \quad (124)$$

setting by convention $q_{x,x'} = \infty$ whenever we have $\text{tr}\{E_{x'}^\dagger E_x(|\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|)\} = 0$. From our assumption that $E_x|\psi\rangle \ne 0$, we see that $r_x$ is always finite.

We now consider different cases for $\mathbf{x}, \mathbf{x}'$. Suppose first that $|\mathbf{x}| + |\mathbf{x}'| < d_m$. Then, since $|\psi\rangle, |\xi\rangle$ form a metrological code of metrological distance $d_m$, we have $q_{\mathbf{x},\mathbf{x}'} = \infty$. Now suppose instead that $|\mathbf{x}| + |\mathbf{x}'| \ge d_m$, implying that either $|\mathbf{x}| \ge d_m/2$ or $|\mathbf{x}'| \ge d_m/2$. Then, since $\|E_x|\psi\rangle\| = \Omega(p^{|\mathbf{x}|/2})$, we find

$$\langle\psi|E_x^\dagger E_x|\psi\rangle = \|E_x|\psi\rangle\|^2 = \Omega(p^{|\mathbf{x}|}), \quad (125)$$

so we can pick $r_x = |\mathbf{x}|$. Since $E_1^{(x_i)} = O(\sqrt{p})$ for each $x_i \ne 0$, we have

$$\begin{aligned} \text{tr}\{E_{x'}^\dagger E_x(|\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|)\} \\ = \langle\xi|E_x^\dagger E_x|\psi\rangle + \langle\psi|E_x^\dagger E_x|\xi\rangle \\ = (\sqrt{p})^{|\mathbf{x}'|+|\mathbf{x}|} O(1) = O(p^{(|\mathbf{x}'|+|\mathbf{x}|)/2}), \end{aligned} \quad (126)$$

so we can pick $q_{\mathbf{x},\mathbf{x}'} = (|\mathbf{x}'| + |\mathbf{x}|)/2$. Then

$$\begin{aligned} 2q_{\mathbf{x},\mathbf{x}'} - \min\{r_{\mathbf{x}}, r_{\mathbf{x}'}\} &= |\mathbf{x}| + |\mathbf{x}'| - \min\{|\mathbf{x}|, |\mathbf{x}'|\} \\ &= \max\{|\mathbf{x}|, |\mathbf{x}'|\} \ge d_m/2. \end{aligned} \quad (127)$$

In all cases, we have $2q_{\mathbf{x},\mathbf{x}'} - \min\{r_{\mathbf{x}}, r_{\mathbf{x}'}\} \ge d_m/2$ and thus

$$\Delta F_{\text{Bob},t} \le O(p^{d_m/2}), \quad (128)$$

as claimed.

There are two strong assumptions made in the above proposition. First, we assume that the Kraus operator representation satisfies $\text{tr}\{E_x^\dagger E_x \psi\} \propto \delta_{\mathbf{x},\mathbf{x}'}$, or equivalently, that $\rho_E$ is diagonal; such a representation always exists but might be difficult to find. Second, the state on Eve must not be rank-deficient, or equivalently, there is no Kraus

operator $E_x$ that has zero probability of occurring when the channel is applied onto the state $\psi$. It is not immediately clear to us how to generalize the above proposition to weaken either of these assumptions.

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