the energy of Alice's state. This observation provides a simple explanation for why the GHZ state has a high Fisher information loss even for small values of $p$: when Alice exposes a GHZ state to the noise, then Eve can estimate the energy of Alice's GHZ state by noting whether or not she observes a photon. If Eve observes even a single photon, then she can safely guess the energy associated with the all-excited state, and if she observes no decay, she guesses the energy of the ground state. [Her guess is wrong with probability $(1-p)^n$, corresponding to the probability of the all-excited state suffering no decay.] Our trade-off relation thus tells us that we seek a state with a large energy spread, but for which a decay would not betray the value of the total energy of the state. As the effect of the decay becomes more significant with increasing $p$, some of the energy spread is sacrificed in order to make the state more resilient to Eve's probe.

Here we consider states that are invariant under permutations of the $n$ spins, motivated by the fact that the Hamiltonian is permutation invariant (see also Refs. [41,42]). These states live in the symmetric subspace and can be written in the basis of Dicke states of the symmetric subspace. A Dicke state is a permutation-invariant state with a fixed number of excitations. More specifically, for $q = 0, \dots, n$ we define

$$|h_q^n\rangle = \binom{n}{q}^{-1/2} \sum_{|\mathbf{x}|=q} |\mathbf{x}\rangle, \qquad (201)$$

where the sum ranges over all strings $\mathbf{x}$ with $x_i = \uparrow, \downarrow$, and where $|\mathbf{x}|$ denotes the number of sites $i$ where $x_i = \downarrow$. In the standard basis $\{|0\rangle = |\uparrow\rangle, |1\rangle = |\downarrow\rangle\}$ for spin-1/2 particles, the value $|\mathbf{x}|$ is the Hamming weight of the corresponding computational basis state $\mathbf{x}$.

A general pure symmetric state vector can therefore be written as

$$|\psi\rangle = \sum_{q=0}^{n} \psi_q |h_q^n\rangle. \qquad (202)$$

We consider symmetric states for convenience, although the optimal state in such settings need not be symmetric [67].

We can consider an even superposition of two Dicke states (as in Sec. IV E). Namely, for $0 \le q_1, q_2 \le n$ we consider the state vector

$$|\tilde{\psi}_{q_1;q_2}\rangle = \frac{1}{\sqrt{2}}[|h_{q_1}^n\rangle + |h_{q_2}^n\rangle]. \qquad (203)$$

Our upper bound on the sensitivity of the state vector $|\tilde{\psi}_{q_1;q_2}\rangle$ for all $q_1, q_2$ is depicted in Fig. 12 for $n = 50$ and the values of $p = 0.01, 0.05, 0.1, 0.25$. In contrast to the case of erasures (Sec. IV E), the states among this family where our bound is large have one of the terms being

close to the maximally excited state ($q_1 = 0$ or $q_2 = 0$). [The bound is not necessarily expected to be tight, in light of the gap that is apparent for $n = 12$ in Fig. 11(a) between our bound and the exact value of the quantum Fisher information. The discussion that follows aims to identify states that can potentially have high sensitivity, while ruling out states that are certain to have low sensitivity.] This property can again be understood from our trade-off relation. In the case of erasures, Eve receives the entire reduced state of the systems that have been lost. If $q_1 = 0$ or $q_2 = 0$, then the reduced state on each subsystem is the pure state vector $|\uparrow\rangle$; since it is a pure state, it is easier for Eve to distinguish it from the reduced state of the other Dicke state vector $|h_{q_2}^n\rangle$. In the case of amplitude damping, Eve knows only whether a decay happened or not on each site and she cannot access the full reduced state. An alternative phrasing of this argument is to express erasures as a random operation $X, Y, Z$ applied onto each site; equivalently, a random operation from the set $\{\sigma_+, \sigma_-, Z\}$ is applied on each site, where $\sigma_\pm = [X \pm iY]/\sqrt{2}$ are the creation and annihilation operators of the qubit excitation on a specific site. In the case of amplitude damping, the Kraus operators have no overlap with $\sigma_+$, meaning that physically, there is no event in which excitations are created in the system. Such events, however, happen in the case of erasures. If Eve receives the information that $k$ such events have occurred, she can safely assert that the energy of Alice's state could not have exceeded the energy of the state that can still accommodate $k$ further excitations. Thus, the state $q_1 = 0$ can easily be ruled out by Eve in the case of erasures if she receives a report of even a single $\sigma_+$ event.

Another interesting choice of state is the uniform superposition of all Dicke states, giving rise to

$$|\psi_{\text{unif}}\rangle = \frac{1}{\sqrt{n+1}} \sum_{q=0}^{n} |h_q^n\rangle. \qquad (204)$$

The intuitive reason we expect this state to achieve a good sensitivity after the noise is that if Eve observes emitted photons, she gains comparatively little information about the energy of Alice's state as opposed to if the state is a superposition of few spaced-out Dicke states. The Fisher information of this state after the application of the amplitude damping noisy channel is depicted in Fig. 11.

A more systematic, numerical optimization of $F_{\text{Bob},t}$ by varying Alice's state using different Ansätze for the coefficients $\{\psi_q\}$ indicate that the Fisher information obtained by states of the form (203) can be marginally exceeded for specific values of $p$ by states for which the amplitudes $\{\psi_q\}$ are concentrated around two values $q_1 = 0$ and some value $q_2$, but with some broadening to include some weight on neighboring Dicke states to $q_2$. Interestingly, there appears to be many states with very different profiles of $\{\psi_q\}$ that
<figure><img src="image_14.png" /><figcaption>FIG. 12. Upper bound (118) on the quantum Fisher information of a superposition of two Dicke states on $n = 50$ spin-1/2 particles after being exposed to IID amplitude-damping noise. A Dicke state with $q$ excitations is an even superposition of all states with exactly $q$ excitations. For each $q_1, q_2$, the considered state is an even superposition of the two Dicke states with a number $q_1$ and $q_2$ of downwards-pointing spins, respectively. Each plot corresponds to a different value of the single-site amplitude-damping noise parameter $p$. Our bound attains its maximum on this family of states (black crosses) for states that are an even superposition of the highest excited state and of a weakly excited Dicke state that is separated from the ground state. This separation hinders Eve from accurately guessing the energy of Alice's state, which via our trade-off relation improves the state's sensitivity after application of the noisy channel. The bound is computed with $k = n = 50$.</figcaption></figure>

achieve a very similar sensitivity after the application of the noisy channel.

One particular such state is the state vector $|\psi_{\text{half-Gauss}}\rangle$ of the form (202) where the $\psi_q$ coefficients are a half-Gaussian centered on the all-excited state as

$$|\psi_{\text{half-Gauss}}\rangle = \sum_q \psi_q |h_q^n\rangle, \quad \psi_q = \frac{1}{c} e^{-\frac{(q/n)^2}{2w^2}}, \quad (205)$$

where $c$ is determined from the normalization condition. Empirically, we find that this state with a value of $w = 0.4$ yields a sensitivity after application of the noisy channel that is competitive with respect to the other studied states. The half-Gaussian spreads over the entire Dicke basis. The amplitude of the ground state is $\psi_n = e^{-1/(2w^2)}\psi_0 \approx 0.04\psi_0$. Here again, the state vector $|\psi_{\text{half-Gauss}}\rangle$ balances a broad spread in energy values while still preventing Eve from easily finding out the energy of Alice's state.

For our numerical calculations, we employed the standard Python NumPy and SciPy toolboxes along with QuTip [68,69]. The permutation invariance of our setting greatly simplifies the calculation of terms of the form $\text{tr}(E_x^\dagger E_x \psi)$ and $\text{tr}(E_x^\dagger E_x \{\tilde{H}, \psi\})$ because $E_x$ is a tensor

product of single-site operators. Similarly, the reduced operator on a given number $k$ of sites of any operator acting on the symmetric subspace can be computed easily by combinatorial considerations in a basis of the symmetric subspace [41–43]. Even for $n = 50$, our pinched bound is easy to compute even for $k \sim n$ in the permutation-invariant setting: to determine the diagonal matrix elements associated with $\tilde{\mathcal{N}}(\psi)$ and $\tilde{\mathcal{N}}(\{\tilde{H}, \psi\})$, it suffices to compute terms of the form $\text{tr}(E_x^\dagger E_x \psi)$ and $\text{tr}(E_x^\dagger E_x \{\tilde{H}, \psi\})$ for operators $E_x$ of the form $E_1^{\otimes w} \otimes E_0^{\otimes (n-w)}$ (the other terms are determined by symmetry).

## B. Strongly interacting Ising Hamiltonian with a noisy channel

Consider a one-dimensional spin chain with nearest-neighbor ZZ couplings, with the Hamiltonian

$$H = \sum_{j=1}^{n-1} \frac{J}{2} Z_j Z_{j+1}. \quad (206)$$

Our upper bound on Bob's sensitivity to time, computed using the expression (118) for various states, is plotted in Fig. 13 for $n = 12$ and $n = 50$, with $J = 2$ in both plots. We first consider the state vector corresponding to an even superposition of a ferromagnetic all-zero state and an antiferromagnetic state vector

$$|\psi_{\text{F-AF}}\rangle = \frac{1}{\sqrt{2}} [ |\downarrow\downarrow\downarrow\downarrow\dots\rangle + |\downarrow\uparrow\downarrow\uparrow\dots\rangle ]. \quad (207)$$

Since $|\downarrow\downarrow\downarrow\downarrow\dots\rangle$ and $|\downarrow\uparrow\downarrow\uparrow\dots\rangle$ are energy eigenvectors of respective energies $\omega(n-1)$ and $-\omega(n-1)$, we see that the state vector $|\psi_{\text{F-AF}}\rangle$ has energy variance $\sigma_H^2(\psi_{\text{F-AF}}) = \omega^2(n-1)^2$. For comparison, we compute the true values of the Fisher information (for $n = 12$), plotted as dashed lines in Fig. 13(a), as well as an ad hoc lower bound, plotted as dotted lines. As can be seen in Fig. 13, our upper bound yields tight bounds on the time sensitivity of the many-body interacting probe, as witnessed by its proximity to the true value and to the ad hoc lower bound, provided $p$ is not too large and $k$ can be taken to be large enough.

The next probe state vector we consider is

$$|\psi_{\text{code-F-AF}}\rangle = \frac{1}{2} [ |\downarrow\downarrow\downarrow\downarrow\dots\rangle + |\uparrow\uparrow\uparrow\uparrow\dots\rangle \\ + |\downarrow\uparrow\downarrow\uparrow\dots\rangle + |\uparrow\downarrow\uparrow\downarrow\dots\rangle ]. \quad (208)$$

The state vector $|\psi_{\text{code-F-AF}}\rangle$ is the state (166) using the antiferromagnetic configuration as the bit string $\mathbf{x}$. Recall that this state satisfies our Knill-Laflamme-like conditions for a single located error. Here we study how this state's
<figure><img src="image_15.png" /><figcaption>FIG. 13. Numerical calculation of the upper bound on the sensitivity of a many-qubit state after IID amplitude-damping noise, for different states evolving according to the one-dimensional Ising Hamiltonian $H = \sum_j (J/2) Z_j Z_{j+1}$. (a) Upper bound on Bob's Fisher information $F_{Bob}(t)$ for $n = 12$ qubits as a function of amplitude-damping parameter $p$, for the state $|\psi_{F-AF}\rangle$, which is an even superposition of a ferromagnet state and an antiferromagnet state (in red), for $|\psi_{code-F-AF}\rangle$, which satisfies our Knill-Laflamme-like conditions for a single located error (in green), and for the spin-coherent state vector $|+\rangle^{\otimes n}$ where $|+\rangle$ is the $+1$ eigenvector of $X$ (in blue). Spin-coherent states are typically used in non-quantum-enhanced metrology. For each state, bounds are shown for various values of $k$, where $k$ is a parameter in the additional noisy channel acting on Eve's system, which is used to derive the upper bound; bounds with larger values of $k$ are harder to compute but are tighter. The upper bounds can increase again for $p \gtrsim 0.4$ because the bound accounts only for low-weight Kraus operators, and higher-weight errors cannot be ignored in this regime. Dashed curves indicate the true Fisher information values, determined by direct computation, and the dotted lines are lower bounds associated with $|\psi_{F-AF}\rangle$ and $|\psi_{code-F-AF}\rangle$. The red curves for $k = 6, 7, 8$, the associated true value, and the lower bound appear superimposed. (b) Upper bounds for the states $|\psi_{F-AF}\rangle$ and $|\psi_{code-F-AF}\rangle$ for $n = 50$ qubits, enlargement of low values of $p$. Dotted lines are lower bounds on the Fisher information for these states. Computing the true values of the Fisher information in this regime would require more advanced methods, such as tensor networks [28].</figcaption></figure>

sensitivity is affected when exposed to IID amplitude-damping noise. Our upper bound on Bob's Fisher information via Eq. (118) is plotted in Fig. 13 for $n = 12$ and $n = 50$, alongside that of $|\psi_{F-AF}\rangle$. The probe state remains almost maximally sensitive when $p$ is small, in contrast to the probe $|\psi_{F-AF}\rangle$, which immediately loses sensitivity at what appears to be a linear rate with $p$. This is a manifestation of the fact that the sensitivity of the probe state is unaffected by a single error, and only in the event that two simultaneous errors occur does the sensitivity decrease.

Finally, we consider for comparison the natural probe state given by an ensemble of independent spins, each pointing in the $X$ direction

$$|+^n\rangle = |+\rangle \otimes |+\rangle \otimes \cdots \otimes |+\rangle, \quad (209)$$

where $|+\rangle = [|\uparrow\rangle + |\downarrow\rangle]/\sqrt{2}$ is the $+1$ eigenvector of $X$. Our upper bound computed for the spin-coherent state vector $|+^n\rangle$ is plotted in blue in Fig. 13. We can see that this probe state performs significantly worse than the entangled probe states for $p \lesssim 0.4$. This is expected, since such a probe's noiseless sensitivity scales only linearly in $n$, as opposed to the quadratic scaling of the sensitivity of the

$|\psi_{F-AF}\rangle$ and $|\psi_{code-F-AF}\rangle$ probe states. However, the robustness of the spin-coherent state to the noise is significant. At larger values of the amplitude damping parameter ($p \sim 0.5$ for $n = 12$), the other probe states have all but lost their advantage in sensitivity.

For our numerical calculations, we employed the standard Python NumPy and SciPy toolboxes along with QuTip [68,69]. Our source code is published on Github [70]. To compute the trace terms in Eq. (118) we express $|\psi\rangle$ and $\tilde{H}|\psi\rangle$ as superpositions of a small number of computational basis vectors over the $n$ sites. The traces then factorize into tensor factors enabling their efficient computation. For the spin-coherent state we work with the local $X$ basis instead of the $Z$ basis, such that the spin-coherent state becomes a basis state in this picture. The direct computation of the Fisher information is performed via an eigenvalue decomposition of the full $n$-body noisy probe state $\rho_B$ to transform the anticommutator equation $1/2\{\rho_B, R\} = N(-i[H, \psi])$ in a basis where $\rho_B$ is diagonal, and then solving elementwise to determine $R$. The ad hoc lower bound is computed by numerically solving the symmetric logarithmic derivative in a restricted subspace consisting of the computational basis vectors that appear in the decomposition of the probe state and those bit
strings that are close by in Hamming distance. The result-
ing value is guaranteed to be a lower bound, because the
map that projects the state down to any subspace of the
state space is a trace-non-increasing, completely positive
map for which one can apply the data-processing inequal-
ity satisfied by the Fisher information [19] (see Proposition
13 in Appendix C for details). It is likely the quantum
Fisher information in this setting can also be computed
based on existing techniques, such as those introduced in
Refs. [21,31,53,71].

IX. CONCLUSIONS AND OUTLOOK

Our results present a new paradigm for characterizing
the sensitivity of a quantum clock or sensor when exposed
to noise, by establishing a quantitative trade-off between
the quantum Fisher information of the noisy system with
respect to the parameter of interest and the quantum Fisher
information that the environment acquires with respect to a
complementary parameter. Information trade-offs are inter-
esting because they reveal properties of the mathematical
structure of quantum theory, which in turn determine what
tasks can be accomplished within the laws of quantum
mechanics. Here, our results provide a guiding principle
for finding noise-resilient clock states: in order to avoid
sensitivity loss due to the application of a noise channel,
clock states should hide their energy from the environment.

Energy-time uncertainty relations have historically been
harder to formulate than position-momentum-type uncer-
tainty principles, because there is no global time observ-
able in quantum mechanics in the same sense as there is
a position observable. Our work contributes an additional
type of time-energy uncertainty relation, complement-
ing existing uncertainty relations such as Mandelstamm-
Tamm-type uncertainty relations [13], Fisher-based uncer-
tainty relations with a single system [12], and entropic
uncertainty relations [15]. Our relation exploits a type of
complementarity between the local optimal sensing oper-
ator for time and the Hamiltonian (Fig. 5), in the same
spirit as uncertainty relations derived in Refs. [1,12]. Our
relation furthermore connects the estimation capabilities
of two distinct parties (Bob and Eve); in this sense our
results can be seen as a Fisher information counterpart of
the entropic uncertainty relations for time and energy [15].

A. Summary and discussion

An overview of the results presented in this work can be
found in Fig. 3.

1. *Time-energy sensitivity trade-off*

Our main result is a quantitative time-energy sensitivity
trade-off relation in the setting of Fig. 1. If a quantum sys-
tem is subjected to an instantaneous noisy channel, then
the loss in sensitivity to time trades off exactly with the

environment's ability to sense the energy of the system as
laid out in Eq. (1).

The setting of our uncertainty relation (Fig. 1) is uncon-
ventional for quantum metrology: a quantum clock usually
accumulates noise continuously as time evolves, much
like a quantum probe usually accumulates noise continu-
ously while sensing an unknown parameter. Our setting is
instead the communication scenario studied in Ref. [29]:
Alice possesses a noiseless quantum clock that already
encodes some time value, and she sends it to Bob over
a noisy communication channel. In this alternative set-
ting one can analyze the quantum information that leaks
to the environment, which is more challenging to do if we
consider continuous noise.

An appealing feature of our trade-off relation is that
Bob’s time sensitivity and Eve’s sensitivity to energy are
related by an equality. Concretely, this feature means that
not only does a gain in energy sensitivity imply a time
sensitivity loss by Bob, but also a loss in time sensitivity
for Bob automatically implies a gain in energy sensitiv-
ity by Eve. In contrast, uncertainty relations in quantum
mechanics often relate two observable uncertainties or
two entropic quantities via an inequality. For instance, a
Schrödinger particle in one dimension that has a large vari-
ance in the momentum observable need not have a narrow
variance in the position observable.

Our results furthermore hold for an arbitrary pure probe
state vector $|\psi\rangle$ and Hamiltonian $H$. We evade the ques-
tion of formally optimizing over the probe state vec-
tor $|\psi\rangle$ itself—a central question in quantum metrology
that many contributions on using quantum error correc-
tion for metrology address [21,22,52,72,73]—by identify-
ing instead what features a probe state vector $|\psi\rangle$ must
exhibit to avoid being affected by the noise. The alterna-
tive expression of the Fisher information obtained by our
trade-off relation can potentially facilitate the computa-
tion of the Fisher information when optimizing the clock
or probe state, potentially improving state optimization
schemes such as those in Refs. [67,74] in the presence of
noise. Our trade-off relation also offers a guideline to seek
good clock states, especially in settings where it might not
be possible to reliably prepare the probe state that has the
absolute best sensitivity: noise-resilient clock states need
to hide their energy from the environment.

Importantly, it is not necessarily the probe state which
is least affected by the noise that is the most sensi-
tive. Another state might exhibit a better sensitivity after
the noisy channel, even if its sensitivity loss is greater,
by ensuring that it is initially sufficiently more sensi-
tive. As an extreme case, this point is illustrated by the
ground state of a qubit affected by amplitude-damping
noise; the ground state trivially remains unaffected by
the noise but has no sensitivity, whereas the $+X$ eigen-
state has a better sensitivity, even if it is affected by the
noise.
A key technique in our approach is the formulation of the quantum Fisher information as a semidefinite program [26,28] (see Appendix C). Semidefinite programming offers a versatile toolbox in which an alternate expression for an optimization (known as dual problem) can be derived and bounds on such optimizations can be proven more easily [75,76]. The technical proof of our main trade-off result (Appendix E 2) offers additional insight into the meaning of the dual problem associated with the semidefinite programming formulation of the quantum Fisher information.

## 2. The setting of continuous noise

In certain specific settings, the setup in Fig. 1 remains a good approximation of a quantum clock exposed to continuous noise described by a Lindbladian master equation (see Sec. VI). A sufficient condition that guarantees the accuracy of this approximation is to ensure, on one hand, that the Hamiltonian part $L_0$ of the evolution commutes (as a superoperator) with the noise part $L_1$ of the Lindbladian that contains all the noise operators, and, on the other hand, that the time derivative of the state is primarily driven by the Hamiltonian and not by the noise. More precisely, in the notation of Sec. VI, the sufficient condition consists in checking that $[L_0, L_1] = 0$ as well as ensuring that $\partial_t N$ contributes only a negligible part of the Fisher information $F(\rho; \partial_t \rho)$ [for which a rigorous bound can, for instance, be computed in Eq. (146)]. The second condition is rarely expected to be violated, as sensors are typically designed to have their signal imprinted on their state through their Hamiltonian evolution; noise is usually a degrading process and is typically not the mechanism by which the signal is acquired. If the Hamiltonian of a many-body system consists only of single-site Z terms, then both IID dephasing noise and IID amplitude-damping noise commute (as a superoperator) with the Hamiltonian part of the Lindbladian. Furthermore if the Hamiltonian $H$ commutes with the individual Lindblad jump operators, then the corresponding evolutions also commute as superoperators; this is the case, for instance, if $H$ consists of arbitrary-weight terms containing only Z operators and in the presence of IID dephasing noise. In the case where $[L_0, L_1] \neq 0$ the setting can still formally be mapped onto the setting of Fig. 1, by defining the effective noise as the full evolution map with a unitary applied on the input, as long as the time dependence of the effective noisy channel can be neglected. In this case, determining the effective noisy channel in general might be difficult.

## 3. Trade-off with generalized parameters

The trade-off relation for time and energy can be extended to other parameter evolutions. First of all, there is a choice in how $\psi$ evolves along the $t$ and $\eta$ parameters: any choice of $|\psi(t, \eta)\rangle$ such that Eq. (23) is satisfied

at $(t_0, \eta_0)$ (but not necessarily at other even neighboring points) leads to the same Fisher information quantities $F_{Alice,t}$, $F_{Alice,\eta}$, $F_{Bob,t}$, and $F_{Eve,\eta}$, so our trade-off relation directly applies. An alternative choice for the $\eta$ parameter is an evolution generated by the Lindbladian master equation $\partial_\eta \psi = \mathcal{L}[\psi]$ with $\mathcal{L}[\rho] = \sum_k [L_k \rho L_k^\dagger - \{L_k^\dagger L_k, \rho\}/2]$, where $L_k = \sigma_H^{-1} \sqrt{(e_k + c)} |\psi\rangle\langle e_k|$, where $\{|e_k\rangle\}$ are eigenvectors of the Hamiltonian, where $H = \sum e_k |e_k\rangle\langle e_k|$, and where $c \ge 0$ is chosen large enough such that $e_k + c \ge 0$ for all $k$. (We have the opposite sign for $\partial_\eta \psi$, but this can be corrected by redefining $\eta \mapsto -\eta$, and this does not impact the Fisher information.) We can check that this choice of $\partial_\eta \psi$ satisfies Eq. (26) at $(t_0, \eta_0)$, and therefore also Eq. (23). Another interesting choice for $\psi(t, \eta)$ is to set $|\psi(t_0, \eta)\rangle = e^{(\eta-\eta_0)\tilde{H}/(2\sigma_H^2)}|\psi(t_0, \eta_0)\rangle$ for $\eta$ in a neighborhood of $\eta_0$, recalling $\tilde{H} = H - \langle H \rangle_\psi$. Again, we see that Eq. (26) is satisfied. This evolution is nonunitary, but one can check that it does preserve the trace of $\psi$ locally to first order at $\eta_0$: we have $\partial_\eta \text{tr}(\psi)|_{\eta_0} = \text{tr}(\{\tilde{H}, \psi\}) = 0$. Either of these choices of evolution might be relevant depending on the specific application, though we expect the primary application of our trade-off relation is to help characterize Bob's Fisher information to time, in which case the specific choice of how the clock state is stated to evolve along $\eta$ might not be important.

The trade-off relation can further be extended to an inequality that is valid for any two arbitrary parameters (Sec. III C). The trade-off between the Fisher information that Alice and Bob, respectively, have with respect to either parameter is then quantified by a value that depends on the commutator of the generators of the two parameters (Theorem 3). The appearance of the commutator in this expression reinforces its central role in quantifying the incompatibility of physical observable quantities. Our main time-energy trade-off relation can be recovered from the more general Theorem 3 by plugging in the local generators for time and energy. While Theorem 3 appears to be tight whenever the Robertson-Weyl uncertainty relation (66) is saturated for the two generators, the bound can likely be improved when considering two generators that have a small commutator. We also present a sufficient condition under which a trade-off relation for any two parameters can be obtained in the form of an equality, mirroring the equality statement in our main trade-off relation for the time and energy parameters. One might have thought that equality in our general uncertainty relation would happen only if the parameters are complementary in the sense of Sec. II B and Fig. 5; in fact, it suffices that the parameters obey some suitable complementarity relation on the support of the complementary channel. Therefore, equality in our general uncertainty relation does not simply depend on the structure of the parameters a, b, but also on the noisy channel $N$.
## 4. Bounds on the quantum Fisher information

Computing the quantum Fisher information for general states involves the calculation of the symmetric logarithmic derivative in Eq. (9). This object is straightforward to determine for pure states, it is simple when represented in the diagonal basis of the state, and it can be computed using numerical methods such as the Bartels-Stewart algorithm [77]. However, in the absence of a simple diagonal representation of the state, it is in general difficult to characterize analytically the Fisher information or to derive useful bounds on the Fisher information that apply in general settings of mixed states, especially if the state is rank deficient or close to the boundary of state space. Our results provide an alternative expression for the Fisher information in the scenario of Fig. 1. Combined with the powerful semidefinite methods for the Fisher information reviewed in Appendix C, we provide a general toolbox to characterize the Fisher information for mixed states in a variety of situations. For instance, for an interacting many-body system subject to noise that acts locally, the noise process might be well approximated by an environment that is small relative to the full many-body system. In this case, the computation of the Fisher information on Eve's end happens on a smaller-dimensional system. This observation is, for instance, a main component of our bound (118).

By applying known Fisher information bounds on Eve's system, our trade-off relation enables us to straightforwardly obtain an opposite bound for the Fisher information of Bob's noisy state (see Sec. V). (Upper bounds on the Fisher information can be difficult to obtain; see, for instance, Refs. [31,78].) An example of such a bound to apply is the data-processing inequality for the Fisher information [19]: further processing of a state that has been exposed to the unknown parameter can only decrease the sensitivity with respect to that parameter. This procedure is useful when Eve obtains a state that is not diagonal in the computational basis, making the Fisher information harder to compute. In such cases, we can dephase Eve's state to set all the off-diagonal matrix elements to zero. The resulting Fisher information for Eve can only decrease; by our trade-off relation this immediately yields an upper bound on Bob's Fisher information. This bound, for instance, facilitates the computation of the sensitivity loss of a state exposed to weak amplitude-damping noise, as discussed in Sec. VIII.

## 5. Metrological codes

Our main uncertainty relation leads to necessary and sufficient conditions for when a clock state loses zero sensitivity when a given noisy channel is applied (Sec. VII). These conditions are a weaker version of the Knill-Laflamme conditions for quantum error correction. Given

a clock state vector $|\psi\rangle$ and a Hamiltonian $H$, we can consider the virtual qubit $L$ spanned by the vectors $|\psi\rangle$ and $H|\psi\rangle$. The clock state vector $|\psi\rangle$ loses no sensitivity under the application of a noisy channel with Kraus operators $\{E_k\}$ if and only if all operators of the form $E_k^\dagger E_k$, when projected onto the virtual qubit, do not have any overlap with the Pauli-Z operator on the virtual qubit. It would be, in principle, possible to prove these zero sensitivity-loss conditions directly on Alice's and Bob's systems, without invoking our trade-off relation; however, characterizing when Eve's Fisher information is zero provides an immediate proof whose simplicity we have not been able to match with alternative techniques.

The zero sensitivity-loss conditions (148) bear similarities with classical codes, where there is only a commutative algebra of observables that one wishes to reproduce [48,50]. Intuitively, the conditions simply ensure that there is a measurement on Bob's system that will reveal the time parameter as well as the local time sensing observable on Alice's system. In contrast to fully quantum error correction, however, there is in general no recovery operation that will restore the pure clock state accurately to first order in the parameter (see Sec. VII H 1 for a simple counterexample). An intriguing aspect of the zero sensitivity-loss conditions are that they do not appear to be formally equivalent to quantum error correction with respect to specific set of noise operators. (The results of Refs. [22,52] appear to indicate that it might be possible to implement certain metrological codes as an error-correcting code involving ancillary systems.) In some cases, such as the qubit example of Sec. VII H 1, the clock state can be thought of as an error-correcting code that corrects only a certain type of error (X or Y Pauli errors). But this is not generally the case—there are examples of a clock state and a (highly nonlocal) Hamiltonian that fulfill the metrological code condition for low-weight errors, but that are not quantum error-correcting codes with respect to neither low-weight X errors nor low-weight Z errors (Sec. VII H 2).

The conditions for zero sensitivity loss are closely related to the recent series of works detailing how to use quantum error correction for metrology in the presence of noise [21, 22, 52, 72, 73, 79, 80]. The main difference with our results is the setting that is being considered. We ask which initial clock states one can prepare on the clock system such that no sensitivity is lost when a noisy channel is applied (and what the associated optimal sensing measurement after the application of the noisy channel is), whereas the mentioned references consider the setting where, during the time a probe system is exposed to the signal and the continuous noise, one can control the probe [81] to repeatedly apply the recovery procedure associated with the quantum error-correcting code.

Metrological codes might be useful for ancillary measurements of error syndromes [82]. Consider an ancillary qubit (of dimension greater than two) which extracts a
bit-valued error syndrome via an entangling gate, correlating a pair of states $|\psi\rangle, |\xi\rangle$ with respective binary syndrome values 0, 1. If the ancillary subspace participating in syndrome extraction satisfies the zero sensitivity-loss conditions (148) against physical noise, then, by definition, a measurement in the $\{|\psi\rangle, |\xi\rangle\}$ basis will not be affected by such noise. In other words, the loss conditions ensure protection against X-type logical noise, yielding more robust syndrome extraction using a Z-type measurement. However, such conditions do not preclude any Y-type logical noise. They also do not guarantee fault tolerance, which would require that ancilla errors not spread to any logical encoding via backaction.

## 6. Numerics for many-body systems

Characterizing the quantum Fisher information of a state exposed to a noise channel using the bound presented in Sec. V A is convenient in the setting of a many-body system subject to noise that acts locally. In the case of $n$ qubits prepared in a permutation-invariant state and exposed to an IID amplitude-damping noise channel, we empirically find that the bound (118) with $k=n$ appears reasonably tight for the states that we investigated and for small values of the noise parameter $p$; furthermore, for a selection of states including the GHZ state, the bound appears to remain tight even in the regime of high values of $p$. Our bound can be computed for systems of size $n \gtrsim 50$ on a standard desktop computer.

If instead of on-site terms we consider only Ising-type nearest-neighbor interactions, we can study the robustness of the example “metrological code” introduced in Sec. VII F to an IID amplitude-damping channel with local noise parameter $p$. This state retains its sensitivity after a single located error. Our numerics show that in the presence of IID noise, the decrease in the quantum Fisher information scales only as $p^2$, and not linearly in $p$ as for the other studied states with similar sensitivity. We observe that if we expose the interacting system to continuous amplitude-damping noise, then the noise part and the unitary part of the Lindblad evolution do not commute as superoperators (i.e., the setting is not that of phase-covariant noise); it is then possible that the advantages of the metrological code state might not persist in the setting of continuous noise.

## B. Outlook

Our trade-off relation is perhaps most relevant in an intermediate regime where the clock is exposed to a signal without the possibility for intermittent quantum control. In such cases, the noise is expected to spoil any Heisenberg scaling that could be achieved using quantum error-correcting schemes due to the lack of recovery operations during the evolution (see, e.g., Refs. [22,31]). Provided the setting can be modeled with a single noisy channel, our

results present an alternative expression for the sensitivity of the noisy probe in this regime where the sensitivity is not yet dominated by the asymptotic scaling. Our results might therefore help identify which states present sufficient robustness to the noise to present an advantage in sensitivity with respect to commonly used states (such as a GHZ state or a spin-coherent state).

In the situation where the clock evolves according to a Lindbladian master equation, and the Hamiltonian and noise parts of the Lindbladian fail to commute as superoperators, one might expect in certain cases to still be able to consider time-dependent noise using the following trick. Let us identify the system $A$ as a full copy of the bipartite system $B \otimes E$, and let the unitary evolution of $|\psi(t)\rangle$ cover both systems. A time-dependent channel $\mathcal{E}_t(\cdot)$ can be written as $\mathcal{E}_t(\cdot) = \text{tr}_E[U(t)\cdot U^\dagger(t)]$ where all the time dependence is encoded in the unitary $U(t)$ and where the environment system $E$ is chosen suitably. We then select the noisy channel $\mathcal{N}_{A \to B} = \text{tr}_E$ that simply performs the partial trace over $E$; the complementary channel is correspondingly $\mathcal{N}_{E \to A} = \text{tr}_B$. While writing a Markovian master equation in this form might require a huge environment system $E$ with rapidly mixing internal dynamics, we expect that our formalism can still account for simple time dependence in the noisy channel in this way. Note also that the unitary $U(t)$ only has to approximate on $B$ the noisy channel $\mathcal{E}_t$ locally to first order around a fixed value of the parameter (e.g., $t=0$) in order to determine the Fisher information.

A potential domain of application of our main uncertainty relation is for quantum thermometry [83], where the goal is to estimate the temperature of a quantum system. In the simple setting of quantum thermometry where the temperature of the system is known to some approximation, and a measurement is performed in order to refine that knowledge, the optimal measurement to carry out is an energy measurement [83]. Since our main result (1) involves the sensitivity of a party with respect to a parameter representing the energy, which is optimally measured using the Hamiltonian of the noiseless system, we expect that one can leverage our main results to yield new sensitivity bounds for quantum thermometry.

Our results are also likely to be relevant in situations where only a restricted set of operators can be measured on a system. Such a restriction could be imposed by limitations in control for a given experimental platform. Suppose we prepare a clock state vector $|\psi\rangle$ evolving noiselessly according to a Hamiltonian $H$. We would like to measure the clock at time $t \approx t_0$, but we are only permitted to use a measurement from a given set of measurements. What is the optimal local sensitivity that we can achieve? Should the set of allowed measurement operators form an algebra, then the problem is equivalent to sending the clock through a channel that represents the projection onto that algebra. Our results then imply that the
resulting sensitivity trades off exactly with the sensitiv-
ity that one can achieve with the set of measurements
in the commutant of that algebra, with respect to the
complementary parameter $\eta$.

It might be possible to extend our results to the multipa-
parameter metrology regime where more than one parameter
is estimated by Bob. There are known uncertainty rela-
tions that determine trade-offs between the precision to
which individual parameters can be simultaneously esti-
mated by a single party [18,72,84–87]. In fact, the *t* and *η*
parameters form a so-called *D*-invariant model [86,88,89],
the latter referring to a multiparameter quantum statistical
model in which the tangent space is invariant under taking
symmetric logarithmic derivatives of the possible gen-
erated state-evolution directions. *D*-invariant models are
interesting in multiparameter quantum metrology, because
different sensitivity bounds, which in general are difficult
to relate, can be shown to coincide [86]. It seems plau-
sible that known multiparameter uncertainty relations can
be extended to the present bipartite setting, either where all
parameters are simultaneously estimated by Bob while Eve
simultaneously estimates a set of complementary parame-
ters, or where a number of parties estimate each individual
parameter, where each party might be part of the output or
the environment.

Our main uncertainty relation might offer a connection
between the setting of quantum metrology hindered by a
noisy quantum channel and the setting of multiparame-
ter, noiseless quantum metrology. It appears that a key
ingredient for our main uncertainty relation is that the *t*
and *η* parameters form a *D*-invariant model, in the sense
of the preceding paragraph. By construction, our uncer-
tainty relation applies to any pure state *D*-invariant model
consisting of two complementary generators related by
Eq. (20), given that we made no specific assumptions about
the parameter *t* or its local Hermitian generator *H*.

However, it remains unclear whether our results extend
to general multiparameter *D*-invariant models, as our proof
seems to utilize the fact that the space is spanned by
only two complementary generators *T* and *H*. *D*-invariant
models have a rich geometric structure [88], which might
prove an essential conceptual component of our results;
such connections nevertheless remain to be better under-
stood. A further connection to *D*-invariant models appears
in Eq. (69), which appears to be a *D*-invariance con-
dition restricted onto the support of the complementary
channel $\hat{\mathcal{N}}$. In fact, one can view Bob's and Eve's mea-
surements $T_B$ and *E* as measurement operators $\hat{\mathcal{N}}^\dagger(T_B)$
and $\hat{\mathcal{N}}^\dagger(E)$ on Alice's system through the action of the
adjoint channels $\mathcal{N}^\dagger$ and $\hat{\mathcal{N}}^\dagger$. We could ask whether our
uncertainty relation translates into a trade-off in how the
two parameters *t* and *η* can be estimated by Alice, if the
estimation of *t* (respectively, *η*) is required to employ an
observable in the set of operators that is specified as the
image of $\mathcal{N}^\dagger$ (respectively, of $\hat{\mathcal{N}}^\dagger$). It is not clear if this

is the case, as the quantum Fisher information attained
by the observable $\mathcal{N}^{\dagger}(T_B)$ [respectively, $\hat{\mathcal{N}}^{\dagger}(E)$] on the
state $\psi$ is not necessarily expected to match the corre-
sponding value of the quantum Fisher information of $T_B$
on $\mathcal{N}[\psi]$ (respectively, of $E$ on $\hat{\mathcal{N}}[\psi]$). It is thus unclear
if or how our main uncertainty relation is connected with
general bounds that hold in the multiparameter regime,
such as multiparameter versions of the quantum Cramér-
Rao bound [86] or the Gill-Massar inequality [90,91]. We
might expect that deeper connections can be developed
between the setting of parameter estimation after the appli-
cation of a noisy channel and noiseless multiparameter
estimation.

Also, our results apply locally to first order around
a given fixed value of the unknown parameter; whether
similar results can be derived in the global parameter
estimation regime [89,92–95] is unknown. Global param-
eter estimation might be more relevant for applications to
atomic quantum clocks [96,97]. We also anticipate exten-
sions of our results to the finite-sample regime where the
quantum Fisher information might no longer accurately
quantify the sensitivity of a quantum state to an unknown
parameter [98].

Along a similar vein, there are settings where one
seeks to compute different variants of the quantum Fisher
information. For instance, the so-called right-logarithmic
derivative (see, e.g., Ref. [87]) is often used to bound the
standard quantum Fisher information. Alternative sensitiv-
ity measures include the truncated Fisher information [54],
which not only give useful bounds on the standard quan-
tum Fisher information but can be more relevant in the
regime of limited measurement data. An interesting ques-
tion would be to study whether our results extend to such
generalized sensitivity measures.

Entropic uncertainty relations play a central role in
quantum cryptography [11, 15, 16, 99], and cryptographic
schemes have been studied for quantum metrology [100].
It is possible that our parameter-estimation trade-off can
similarly form the basis of cryptographic schemes in which
a parameter encoded in a quantum state is to be shielded
from a malevolent eavesdropper. Furthermore, the Fisher
information is closely related to relative entropy mea-
sures [87, 101, 102]; we might expect our trade-off rela-
tion to translate into a statement about Rényi relative
entropies.

The development of quantum atomic clocks as ultraprecise time references [103] makes it all the more important to achieve a thorough understanding of how noise can be prevented from spoiling sensitivity. We also anticipate that our results will be relevant for recently developed atomic clocks built with a lattice of interacting atoms [9] and correlated many-body sensing probes [6, 104], as these platforms will offer new possibilities for metrology by exploiting the strong interactions between the particles.
## ACKNOWLEDGMENTS

The authors are grateful to Fernando Brandão, Jonathan Conrad, Rafał Demkowicz-Dobrzański, Richard Küng, Johannes Meyer, Yingkai Ouyang, Renato Renner, Ralph Silva, Ryan Sweke, and Nathan Walk for discussions. We warmly thank Gian Michele Graf for his invaluable input for our proofs in infinite-dimensional spaces. M. W. acknowledges support from the Swiss National Science Foundation (SNSF) via an Ambizione Fellowship (PZ00P2_179914). J.M.R. acknowledges support from the Swiss National Science Foundation Sinergia grant CRSII5 186364. M.W. and J.M.R. acknowledge the National Centre of Competence in Research QSIT. Ph.F. and J.E. acknowledge support from the DFG (FOR 2724, CRC 183, EI 519/21-1), the FQXi, the QuantERA (HQCC), the BMBF (RealistiQ, Hybrid, MuniQC-Atoms) the ERC (DebugQC) and the Einstein Research Unit on quantum devices. This research is also part of the Munich Quantum Valley (K8), which is supported by the Bavarian state government with funds from the Hightech Agenda Bayern Plus. V.V.A. acknowledges funding from NSF QLCI Award No. OMA-2120757. Contributions to this work by NIST, an agency of the U.S. government, are not subject to U.S. copyright. Any mention of commercial products does not indicate endorsement by NIST. J.P. acknowledges funding from the U.S. Department of Energy Office of Science (DE-NA0003525, DE-SC0020290, DE-AC02-07CH11359, DE-SC0018407), the Simons Foundation It from Qubit Collaboration, the Air Force Office of Scientific Research (FA9550-19-1-0360), and the National Science Foundation (PHY-1733907). The Institute for Quantum Information and Matter is an NSF Physics Frontiers Center.

## APPENDIX A: NOTATION, PRELIMINARIES, AND AUXILIARY LEMMAS

We first introduce some preliminaries and notation that will be used throughout the Appendices. All Hilbert spaces are finite dimensional unless otherwise indicated, and all projectors are Hermitian. A pure quantum state is a vector $|ψ⟩$ in the Hilbert space that is normalized to unit norm, and the terminology pure quantum state is also used by extension for the associated density operator $|ψ⟩⟨ψ|$. Quantum states are positive semidefinite operators $\rho$ with unit trace, $\text{tr}(\rho) = 1$. A subnormalized quantum state is a positive semidefinite operator $\rho$ satisfying $\text{tr}(\rho) \le 1$. States are normalized to unit trace unless explicitly specified as being subnormalized.

For any Hermitian operator $O$, we denote by $P_O$ the projector onto the support of $O$, and by $P_O^\perp = 1 - P_O$ its complement. For any positive semidefinite operator $A$, we denote by $A^{-1}$ its Moore-Penrose pseudoinverse, i.e., the operator obtained by taking the inverse on the support of $A$.

We denote by $||A||$ the maximal singular value of an operator $A$. We also define the Schatten one-norm as $||A||_1 = \text{tr}\sqrt{A^\dagger A}$.

It will prove convenient to “vectorize” operators by viewing them as vectors in Hilbert-Schmidt space using the following representation. The vector space of operators acting on a Hilbert space $\mathcal{H}$ is isomorphic to $\mathcal{H} \otimes \mathcal{H}$. Let $|1⟩⟩$ denote the element

$$ \sum_{i=1}^{d} |i⟩ \otimes |i⟩ \in \mathcal{H} \otimes \mathcal{H}, \qquad (A1) $$

where $\{|i⟩\}_{i=1}^d$ is a fixed basis of $\mathcal{H}$. We define the “vectorized” representation of any operator $A$ acting on $\mathcal{H}$ as $|A⟩ = (A \otimes 1)|1⟩$. Similarly, we define $⟨1| = \sum_{i=1}^d ⟨i| \otimes ⟨i|$ and $⟨A| = ⟨1|(A^\dagger \otimes 1)$. We recall the useful identity

$$ (X \otimes 1)|1⟩ = (1 \otimes X^T)|1⟩, \qquad (A2) $$

and note that $|1⟩ = |1⟩$ is the vectorized operator representation of the identity matrix $1$. We denote a rank-one operator $|φ⟩⟨ψ|$ in this representation as $|φ, ψ⟩$, with $|φ, ψ⟩ = |φ⟩ \otimes (|ψ⟩)^*$ and $⟨φ, ψ| = ⟨φ| \otimes (⟨ψ|)^*$. The Hilbert-Schmidt inner product in this notation is simply $\text{tr}(A^\dagger B) = ⟨A|B⟩$. The matrix elements of $A$ in any basis $\{|ℓ⟩\}$ are also simply given by $⟨ℓ|A|ℓ'⟩ = ⟨ℓ, ℓ'|A⟩$. A superoperator $\mathcal{E}$ acting on an operator $M$ is denoted by $\mathcal{E}|M⟩$. The superoperator consisting of a left multiplication by $A$ and a right multiplication by $B$, i.e., $M \mapsto AMB$, is represented by $|M⟩ \mapsto (A \otimes B^T)|M⟩$. The identity superoperator id is represented by $1 \otimes 1$. Also, $⟨A|\mathcal{E}|B⟩ = ⟨B|\mathcal{E}^\dagger|A⟩^*$, where $\mathcal{E}^\dagger$ is the usual superoperator adjoint defined by $\text{tr}(M\mathcal{E}^\dagger(N)) = \text{tr}(\mathcal{E}(M)N)$. In the following and unless otherwise stated, superoperators are expressed in this representation, unless they are explicitly applied onto an operator with the notation $\mathcal{E}(\cdot)$.

We now compute a few quantities that often recur throughout these appendices. Let $|ψ⟩$ be a state vector and consider the evolution $\partial_t ψ = -i[H, ψ]$, where $H$ is any Hermitian operator. Let $M$ be any Hermitian operator. We have

$$ \left\langle \left( \frac{d\psi}{dt} \right)^2 \right\rangle = \langle (-i[H, \psi])^2 \rangle = -\text{tr}[\psi(H\psi H - H\psi H - \psi H^2\psi + \psi H\psi H)] = \text{tr}(\psi H^2) - [\text{tr}(\psi H)]^2; \qquad (A3) $$

$$ \begin{aligned} -i[i[M, \psi], \psi] &= (M\psi - \psi M)\psi - \psi(M\psi - \psi M) \\ &= \{M, \psi\} - 2\langle M \rangle \psi \\ &= \{M - \langle M \rangle, \psi\}. \end{aligned} \qquad (A4) $$

The notion of Schur complement will serve multiple times in these appendices, so we state it here.
Theorem 5 (Positive semidefiniteness via Schur complement).—let $A \in \mathbb{C}^{n \times n}$, $B \in \mathbb{C}^{m \times m}$ be positive semidefinite matrices. Let $W \in \mathbb{C}^{n \times m}$ be an arbitrary complex matrix. The following statements are equivalent:

(i) $$ \begin{bmatrix} A & W \\ W^\dagger & B \end{bmatrix} \ge 0, $$

(ii) $WP_B^\perp = 0$ and $A \ge WB^{-1}W^\dagger$,

(iii) $P_A^\perp W = 0$ and $B \ge W^\dagger A^{-1} W$.

Moreover, (ii) implies $P_A^\perp W = 0$ and (iii) implies $WP_B^\perp = 0$.

For a proof, see, e.g., Ref. [105]. With respect to the proof of similar statements in standard textbooks, we can see that (ii) implies $P_A^\perp W = 0$ as follows: hitting the inequality with $P_A^\perp(\cdot)P_A^\perp$ and noting that $WB^{-1}W^\dagger \ge 0$, we see that $P_A^\perp WB^{-1}W^\dagger P_A^\perp = 0$, which implies $P_A^\perp WB^{-1/2} = 0$. Therefore, $P_A^\perp W = 0$ using the fact that $WP_B^\perp = 0$. Similarly, (iii) implies $WP_B^\perp = 0$.

Now we present a simple method to relate operator inequalities before and after the application of a completely positive map.

Lemma 1 (Positive semidefiniteness of block matrices under completely positive maps).—let $A, B, W \in \mathbb{C}^{n \times n}$ be complex matrices and assume that

$$ \begin{bmatrix} A & W \\ W^\dagger & B \end{bmatrix} \ge 0. \qquad (A5) $$

Let $\Phi$ be any completely positive map that maps operators on $\mathbb{C}^n$ to operators on $\mathbb{C}^m$. Then

$$ \begin{bmatrix} \Phi(A) & \Phi(W) \\ \Phi(W^\dagger) & \Phi(B) \end{bmatrix} \ge 0. \qquad (A6) $$

*Proof*.—The matrix (A6) is obtained by applying the completely positive map $id_2 \otimes \Phi$ onto Eq. (A5). ■

While the above lemma is fairly trivial, paired with Theorem 5 it enables us to show less obvious inequalities such as the following.

Corollary 2 (Image of matrix squared under a subunitary, completely positive map).—let $M$ be a Hermitian operator and let $\Phi$ be any completely positive map that satisfies $\Phi(1) \le 1$. Then $\Phi(M^2) \ge [\Phi(M)]^2$.

*Proof*.—Observe first that $\begin{bmatrix} M^2 & M \\ M & 1 \end{bmatrix} = [M \quad 1]^\dagger$ [$M \quad 1] \ge 0$. With the subunitarity condition on $\Phi$ and Lemma 1, we have

$$ \begin{bmatrix} \Phi(M^2) & \Phi(M) \\ \Phi(M) & 1 \end{bmatrix} \ge \begin{bmatrix} \Phi(M^2) & \Phi(M) \\ \Phi(M) & \Phi(1) \end{bmatrix} \ge 0. \qquad (A7) $$

Thanks to Theorem 5, this implies $\Phi(M^2) \ge \Phi(M)$ [$1^{-1} \Phi(M) = [\Phi(M)]^2$]. ■

## APPENDIX B: SOLUTIONS OF THE ANTICOMMUTATOR EQUATION

In this Appendix, we briefly review the solutions of the anticommutator equation

$$ \frac{1}{2}\{\rho, M\} = N, \qquad (B1) $$

where $M$ is the unknown operator, $N$ is a fixed operator, $\rho$ is a subnormalized quantum state, and $\{A, B\} := AB + BA$ denotes the anticommutator.

For any subnormalized state $\rho$, it is convenient to define the Hermiticity-preserving superoperator $\mathcal{R}_\rho$ as

$$ \mathcal{R}_\rho(\cdot) = \frac{1}{2}\{\rho, (\cdot)\}. \qquad (B2) $$

Note that $\mathcal{R}_\rho$ is neither completely positive nor trace preserving. The operator $\mathcal{R}_\rho$ is self-adjoint, since $\text{tr}(N\mathcal{R}_\rho(M)) = \frac{1}{2}\text{tr}(N\{\rho, M\}) = \frac{1}{2}\text{tr}(\{\rho, N\}M) = \text{tr}(\mathcal{R}_\rho(N)M)$. It is interesting to study the superoperator $\mathcal{R}_\rho$ as a linear operator in Hilbert-Schmidt space. In vectorized operator space, it is represented as

$$ \mathcal{R}_{\rho} = \frac{1}{2}(\rho \otimes 1 + 1 \otimes \rho^T). \qquad (B3) $$

This matrix is Hermitian and positive, and it is positive definite if and only if $\rho$ has full rank. The fact that the vectorized matrix representing $\mathcal{R}_\rho$ is positive is not to be confused with the usual notion of a superoperator being positive, which means preserving the positivity of its argument. Here, $\mathcal{R}_\rho$ has a positive semidefinite vectorized representation, which means that $\langle\langle M | \mathcal{R}_\rho | M \rangle\rangle \ge 0$ for all operators $|M\rangle\rangle$.

Suppose for a moment that $\rho$ has full rank. Then the superoperator $\mathcal{R}_\rho$ can be inverted, because its vectorized operator matrix representation has full rank, and we denote the inverse by $\mathcal{R}_\rho^{-1}$. The operator $M = \mathcal{R}_\rho^{-1}(N)$ is then the unique solution to the anticommutator equation $\frac{1}{2}\{\rho, M\} = N$. If $\{k\}$ is a basis of the Hilbert space that diagonalizes $\rho$ as $\rho = \sum_k p_k |k\rangle\langle k|$, then Eq. (B3) provides a diagonal representation of $\mathcal{R}_\rho$, and we obtain the familiar expression of $\mathcal{R}_\rho^{-1}$ as

$$ \begin{align} \mathcal{R}_{\rho}^{-1}|N\rangle\rangle &= \sum_{k,k'} \frac{2}{p_k + p_{k'}} |k, k'\rangle\langle\langle k, k'|N\rangle\rangle, \quad \text{i.e.,} \nonumber \\ \mathcal{R}_{\rho}^{-1}(N) &= \sum_{k,k'} \frac{2}{p_k + p_{k'}} \langle k|N|k' \rangle |k\rangle\langle k'|. \tag{B4} \end{align} $$

If $\rho$ is not full rank, then we define $\mathcal{R}_\rho^{-1}$ as the Moore-Penrose inverse of the superoperator $\mathcal{R}_\rho$, i.e., we take the
inverse on its support. From Eq. (B3) we can identify the kernel $\mathcal{R}_\rho$ of the superoperator $\mathcal{R}_\rho$ as the space spanned by operators of the form $|\phi, \psi\rangle$ where $P_\rho|\phi\rangle = P_\rho|\psi\rangle = 0$, where $P_\rho$ is the projector onto the support of $\rho$. If $\{|k\rangle\}$ is a basis of the Hilbert space that diagonalizes $\rho$ as $\rho = \sum_k p_k |k\rangle\langle k|$, then Eq. (B3) is diagonal in the basis $\{|k, k'\rangle\}$ and we see that the expression (B4) remains the correct expression for $\mathcal{R}_\rho^{-1}$, provided we only keep those terms in the sum for which $p_k + p_{k'} \neq 0$.

We may now state the following useful proposition that characterizes the full solution set of the anticommutator equation $1/2\{\rho, M\} = N$ for $M$.

**Proposition 3 (Solutions to the anticommutator equation).**—let $\rho$ be any subnormalized quantum state, let $N$ be any operator, and let $\mathcal{R}_\rho$ be given by Eq. (B2). Let $P_\rho$ denote the projector onto the support of $\rho$ and let $P_\rho^\perp = 1 - P_\rho$. Then the set $\mathcal{S}$ of solutions of the equation $Y = \mathcal{R}_\rho(M)$ for the operator $M$ is

$$ S = \begin{cases} \emptyset & \text{if } P_{\rho}^{\perp} N P_{\rho}^{\perp} \neq 0; \\ \{\mathcal{R}_{\rho}^{-1}(N) + P_{\rho}^{\perp} M' P_{\rho}^{\perp} : M' \text{ any operator}\} & \text{if } P_{\rho}^{\perp} N P_{\rho}^{\perp} = 0, \end{cases} \quad (B5) $$

where $\mathcal{R}_\rho^{-1}$ denotes as above the Moore-Penrose pseudoinverse of the superoperator $\mathcal{R}_\rho$. Furthermore, if $Y$ is Hermitian, then the set $S_H$ of Hermitian solutions of the equation $N = \mathcal{R}_\rho(M)$ for the operator $M$ is

$$ S_H = \begin{cases} \emptyset & \text{if } P_{\rho}^{\perp} N P_{\rho}^{\perp} \neq 0; \\ \{\mathcal{R}_{\rho}^{-1}(N) + P_{\rho}^{\perp} M'' P_{\rho}^{\perp} : M'' \text{ any Hermitian operator}\} & \text{if } P_{\rho}^{\perp} N P_{\rho}^{\perp} = 0, \end{cases} \quad (B6) $$

where $\mathcal{R}_\rho^{-1}(Y)$ is always a Hermitian operator.

This proposition is essentially obvious if we think of superoperators as linear operators in Hilbert-Schmidt space. Indeed, it is well known that the general solution to a system of equations given in matrix form can be expressed by the matrix pseudoinverse, plus anything that is in the matrix kernel.

*Proof.*—Let $\mathcal{P}^\perp$ be the superoperator projector onto the kernel of $\mathcal{R}_\rho$. In the vectorized-operator representation, we have $\mathcal{P}^\perp := P_\rho^\perp \otimes (P_\rho^\perp)^T$ as can be seen from Eq. (B3). Let $\mathcal{P} = \text{id} - \mathcal{P}^\perp$ be the superoperator projector onto the complementary operator subspace, which is the support of $\mathcal{R}_\rho$. Observe that $\mathcal{R}_\rho \mathcal{R}_\rho^{-1} = \mathcal{R}_\rho^{-1} \mathcal{R}_\rho = \mathcal{P}$ and that $\mathcal{R}_\rho \mathcal{P}^\perp = 0$.

The claim we want to show is that if $\mathcal{P}^\perp|N\rangle\rangle \neq 0$, then there is no solution to the equation $|N\rangle\rangle = \mathcal{R}_\rho|M\rangle\rangle$; otherwise, then the equation is satisfied if and only if

$$ |M\rangle\rangle = \mathcal{R}_{\rho}^{-1}|N\rangle\rangle + \mathcal{P}^{\perp}|M'\rangle\rangle \quad (B7) $$

for some operator $|M'\rangle\rangle$. The condition $\mathcal{P}^\perp|N\rangle\rangle = 0$ is necessary for any solution to the equation $|N\rangle\rangle = \mathcal{R}_\rho|M\rangle\rangle$ to exist, as otherwise $|N\rangle\rangle$ would not be in the range of $\mathcal{R}_\rho$. We can therefore assume for the rest of this proof that $\mathcal{P}^\perp|N\rangle\rangle = 0$.

Suppose $M$ solves $\mathcal{R}_\rho|M\rangle\rangle = |N\rangle\rangle$. Applying $\mathcal{R}_\rho^{-1}$ on both sides, we have $\mathcal{P}|M\rangle\rangle = \mathcal{R}_\rho^{-1}|N\rangle\rangle$, which determines $|M\rangle\rangle$ on the operator space projected onto by $\mathcal{P}$. On the complementary space (associated with $\mathcal{P}^\perp$), the operator $|M\rangle\rangle$ can be arbitrary because this subspace is the kernel of

$\mathcal{R}_\rho$. A general operator in this subspace can be written as $\mathcal{P}^\perp|M'\rangle\rangle$ for some operator $M'$. This proves that the solution $M$ must have the form given in the claim. Conversely, if

$$ |M\rangle\rangle = \mathcal{R}_{\rho}^{-1}|N\rangle\rangle + \mathcal{P}^{\perp}|M'\rangle\rangle \quad (B8) $$

for some operator $|M'\rangle\rangle$, then we see that $\mathcal{R}_\rho|M\rangle\rangle = \mathcal{R}_\rho(\mathcal{R}_\rho^{-1}|N\rangle\rangle + \mathcal{P}^\perp|X'\rangle\rangle) = \mathcal{P}|N\rangle\rangle = |N\rangle\rangle$, thus proving the claim.

If $N$ is Hermitian, then $\mathcal{R}_\rho^{-1}(N)$ is Hermitian because $\mathcal{R}_\rho$, and hence $\mathcal{R}_\rho^{-1}$, is Hermiticity preserving. Any two Hermitian solutions $M_0, M_1$, as seen above, must differ by a term $P_\rho^\perp M' P_\rho^\perp$ for some arbitrary $M'$; because the difference $M_0 - M_1$ is Hermitian, $M'$ can be chosen to be Hermitian as well [specifically, one can set $M'' = (M' + M'^\perp)/2$].

We now compute the map $\mathcal{R}_\psi^{-1}(\cdot)$ in the case the reference state is a pure (normalized) state vector $|\psi\rangle$.

**Proposition 4 (Computing $\mathcal{R}_\rho^{-1}$ when $\rho$ is a pure state).**—let $|\psi\rangle$ be a (normalized) state vector and let $P_\psi^\perp = 1 - |\psi\rangle\langle\psi|$. Then for any Hermitian $O$ we have

$$ \mathcal{R}_{\psi}^{-1}(O) = 2(O - P_{\psi}^{\perp} O P_{\psi}^{\perp}) - \langle O \rangle_{\psi} \psi. \quad (B9) $$

*Proof.*—Define $\bar{O} := O - P_\psi^\perp O P_\psi^\perp - \langle O \rangle_\psi \psi$. By linearity, we have

$$ \mathcal{R}_{\psi}^{-1}(\bar{O}) = \mathcal{R}_{\psi}^{-1}(O) - \langle O \rangle_{\psi} \psi, \quad (B10) $$
noting that $1/2\{\psi, \psi\} = \psi$ and therefore $\mathcal{R}_{\psi}^{-1}(\psi) = \psi$. The operator $\tilde{O}$ satisfies $P_{\psi}^{\perp}\tilde{O}P_{\psi}^{\perp} = 0$ and $\langle\tilde{O}\rangle_{\psi} = 0$, the latter implying that $\tilde{O}\psi = P_{\psi}^{\perp}\tilde{O}\psi$. Then

$\tilde{O} = (\psi + P_{\psi}^{\perp})\tilde{O}(\psi + P_{\psi}^{\perp}) = P_{\psi}^{\perp}\tilde{O}\psi + \psi\tilde{O}P_{\psi}^{\perp} = \{\psi, \tilde{O}\}$, and therefore $\mathcal{R}_{\psi}^{-1}(\tilde{O}) = 2\tilde{O}$. From Eq. (B10) we then find

$$ \mathcal{R}_{\psi}^{-1}(O) = 2\tilde{O} + \langle O \rangle \psi = 2(Z - P_{\psi}^{\perp}OP_{\psi}^{\perp}) - \langle O \rangle \psi. $$

<figure><img src="image_16.png" /></figure>

## APPENDIX C: SEMIDEFINITE PROGRAMMING METHODS FOR THE FISHER INFORMATION

In this Appendix, we review some methods based on semidefinite programming [75,76] for computing the Fisher information, and review some elementary properties of the Fisher information. Let $\rho$ be any subnormalized quantum state, and let $D$ be any Hermitian operator that satisfies $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$ (recall $P_{\rho}^{\perp}$ is the projector onto the kernel of $\rho$). Define the quantity

$$ F(\rho; D) := \mathrm{tr}(\rho R^2), \qquad (C1) $$

where $R$ is any solution to $(1/2)\{\rho, R\} = D$. For a normalized state $\rho$ and for traceless $D$, the quantity $F(\rho; D)$ corresponds to the Fisher information associated with a one-parameter family of states $\lambda \mapsto \rho_{\lambda}$ taken at a value of $\lambda$ where $\rho_{\lambda} = \rho$ and $d\rho_{\lambda}/d\lambda = D$. We allow subnormalized states $\rho$ and operators $D$ with nonzero trace in the definition (C1) for later technical convenience. We

require that $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$ as otherwise the anticommutator equation $(1/2)\{\rho, R\} = D$ has no solution for $R$.

The definition of $F(\rho; D)$ does not depend on the choice of $R$ that solves $(1/2)\{\rho, R\} = D$. Indeed, Proposition 3 guarantees that any two solutions differ only by a term $P_{\rho}^{\perp}M^{\prime}P_{\rho}^{\perp}$; such a term does not contribute to the trace in Eq. (C1). We may therefore write, using the notation of Appendix B,

$$ F(\rho; D) = \mathrm{tr}(\rho [\mathcal{R}_{\rho}^{-1}(D)]^2). \qquad (C2) $$

We now write this expression as a pair of convex optimizations. These expressions have been derived in Refs. [26,28]; we provide a proof using our notation for self-consistency.

**Proposition 5 (Fisher information in terms of convex optimization problems).**—let $\rho$ be a subnormalized quantum state and $D$ be a Hermitian operator that satisfies $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. The quantity $F(\rho; D)$ defined in Eq. (C1) is equivalently expressed as the following optimizations:

$$ F(\rho; D) = \max_{S=S^{\dagger}} 4[\mathrm{tr}(DS) - \mathrm{tr}(\rho S^2)] \qquad (C3a) $$

$$ = \min \{4 \mathrm{tr}(L^{\dagger}L) : \rho^{1/2}L + L^{\dagger}\rho^{1/2} = D\}, \qquad (C3b) $$

where the first optimization ranges over all Hermitian operators $S$ and where in the second optimization $L$ is an arbitrary complex matrix. Optimal choices for the variables are $S = (1/2)\mathcal{R}_{\rho}^{-1}(D)$ and $L = \rho^{1/2}S$, noting that $\{\rho, S\} = D$. Furthermore, alternative forms for the minimization are

$$ F(\rho; D) = 4 \min \left\{ \mathrm{tr}(N) : \begin{bmatrix} \rho & 0 \\ 0^{\dagger} & N \end{bmatrix} \ge 0 \quad \text{with} \quad O + O^{\dagger} = D, N \ge 0 \right\} \qquad (C3c) $$

$$ = \min \left\{ \mathrm{tr}(J) : \begin{bmatrix} \rho & D+iK \\ D-iK & J \end{bmatrix} \ge 0 \quad \text{with} \quad K = K^{\dagger}, J \ge 0 \right\}, \qquad (C3d) $$

in which optimal choices are $O = \rho S$, $N = S \rho S$, $K = -i[\rho, S]$, and $J = \mathcal{R}_{\rho}^{-1}(D) \rho \mathcal{R}_{\rho}^{-1}(D)$.

Note that the condition in the optimization (C3d) implicitly enforces the fact that $P_{\rho}^{\perp}(D+iK) = 0$ (see Theorem 5); this can make it more complicated to guess a candidate for $K$ in Eq. (C3d) if $\rho$ does not have full rank, especially if $P_{\rho}^{\perp}D \ne 0$. Also, note that if there is any feasible choice of candidates in Eq. (C3c), then automatically

$P_{\rho}^{\perp}O = 0$ and $O^{\dagger}P_{\rho}^{\perp} = 0$, such that $P_{\rho}^{\perp}DP_{\rho}^{\perp} = P_{\rho}^{\perp}(O + O^{\dagger})P_{\rho}^{\perp} = 0$. Therefore, finding feasible candidates automatically enforces the condition in the definition (C1). A similar argument holds if feasible candidates are found in Eqs. (C3b) or (C3d).

**Proof.**—The maximization (C3a) is a quadratic optimization can be cast into a semidefinite program using Schur complements (Theorem 5). We stick closely to the
formalism of Watrous [76,106]. We introduce a variable $Q \ge 0$ with the constraint $Q \ge S^2$ expressed as a Schur complement condition:

$$ 
\begin{aligned}
\frac{1}{4} \{ & \text{maximization in Eq. (C3a)} \\
= \text{maximize} & \quad [\text{tr}(DS) - \text{tr}(\rho Q)] \quad \text{(C4)} \\
\text{over variables} & \quad S = S^{\dagger}, Q \ge 0 \\
\text{subject to} & \quad \begin{bmatrix} Q & -S \\ -S & 1 \end{bmatrix} \ge 0.
\end{aligned}
$$

(The sign of $-S$ in the last constraint is for later convenience.) We now determine the corresponding dual problem. Let $[\begin{matrix} M & O \\ O^{\dagger} & N \end{matrix}] \ge 0$ be the Lagrange dual variable corresponding to the primal constraint, with $M, N \ge 0$ and $O$ arbitrary. The primal constraint can be written as

$$ \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} \otimes Q + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \otimes S \le \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \otimes 1. \quad \text{(C5)} $$

The dual objective is obtained by collecting the constant terms of the constraints and taking the inner product with the corresponding dual variable. Here we only have the right-hand side of Eq. (C5) and we obtain the objective that is simply to minimize $\text{tr}(N)$. There are two dual constraints, one for each primal variable $S$ and $Q$; to a Hermitian variable corresponds an equality constraint and to a positive semidefinite variable corresponds a positive semidefinite constraint. The primal objective gives the constant terms for each constraint, which are $(...)=D$ and $(...)>-\rho$. For the left-hand side of the first constraint we obtain the term $\text{tr}_1([\begin{matrix} M & O \\ O^{\dagger} & N \end{matrix}][\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]) = O + O^{\dagger}$. For the left-hand side of the second constraint, we find $-\text{tr}_1([\begin{matrix} M & O \\ O^{\dagger} & N \end{matrix}][\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]) = -M$. We thus obtain the following dual problem:

$$ 
\begin{aligned}
(C4) = & \quad \text{minimize} \quad \text{tr}(N) \quad \text{(C6)} \\
\text{over variables} & \quad M \ge 0, N \ge 0, O, \\
\text{subject to} & \quad O + O^{\dagger} = D, \\
& \quad M \le \rho, \\
& \quad \begin{bmatrix} M & O \\ O^{\dagger} & N \end{bmatrix} \ge 0.
\end{aligned}
$$

Equality with the primal optimization problem holds thanks to strong duality, which is ensured by the Slater conditions [76,106]. We can further simplify the dual problem. First, the choice $M = \rho$ is optimal: indeed, for any optimal choices of variables with $M \le \rho$, we can replace $M$ by $\rho$

while still achieving the same value. Therefore,

$$ 
\begin{aligned}
(C6) = & \quad \text{minimize} \quad \text{tr}(N) \quad \text{(C7)} \\
\text{over variables} & \quad N \ge 0, O, \\
\text{subject to} & \quad O + O^{\dagger} = D, \\
& \quad \begin{bmatrix} \rho & O \\ O^{\dagger} & N \end{bmatrix} \ge 0.
\end{aligned}
$$

Using the Schur complement argument again (Theorem 5), we find that $n \ge O^{\dagger}\rho^{-1}O$, and for the same reason as above, there is an optimal choice of variables with $Y = O^{\dagger}\rho^{-1}O$. Hence

$$ 
\begin{aligned}
(C7) = & \quad \text{minimize} \quad \text{tr}(O^{\dagger}\rho^{-1}O) \quad \text{(C8)} \\
\text{over variables} & \quad O \text{ arb.} \\
\text{subject to} & \quad O + O^{\dagger} = D \\
& \quad P_{\rho}^{\perp}O = 0.
\end{aligned}
$$

We may introduce the variable $L = \rho^{-1/2}O$, which yields

$$ 
\begin{aligned}
(C8) = & \quad \text{minimize} \quad \text{tr}(L^{\dagger}L) \quad \text{(C9)} \\
\text{over variables} & \quad L \text{ arb.} \\
\text{subject to} & \quad \rho^{1/2}L + L^{\dagger}\rho^{1/2} = D.
\end{aligned}
$$

We recognize the optimization in Eq. (C3b). At this point we have shown that both optimizations in the claim, Eqs. (C3a) and (C3b), are equal thanks to semidefinite programming duality. It remains to show that the common optimal value is $F(\rho; D)$ as given by Eq. (C2).

To find optimal variables, we examine the complementary slackness conditions [106] corresponding to the primal-dual problem pair (C4) and (C6). Namely, taking the product of an inequality constraint with the corresponding dual variable turns the inequality into an equality for optimal primal and dual choices of variables. For the primal constraint this gives us the equalities

$$ 
\begin{aligned}
& QM - SO^{\dagger} = 0; \quad QO - SN = 0; \\
& -SM + O^{\dagger} = 0; \quad -SO + N = 0.
\end{aligned}
\quad \text{(C10)} $$

From $-SM + O^{\dagger} = 0$ along with the optimal $M = \rho$ we deduce that $\rho S = O$ and thus $\rho^{1/2}S = \rho^{-1/2}O = L$. Plugging this into the constraint in Eq. (C9) we find

$$ \rho S + S\rho = D. \quad \text{(C11)} $$

The solutions of this anticommutator equation have been studied in Appendix B, leading us to the primal candidate

$$ S = \frac{1}{2} R_{\rho}^{-1}(D). \quad \text{(C12)} $$
Plugging this choice into Eq. (C4), along with the choice
$Q = S^2$, we obtain

$$
\begin{equation}
\begin{split}
(C4) \ge & \frac{1}{2} \operatorname{tr}(DR_{\rho}^{-1}(D)) \\
& - \frac{1}{4} \operatorname{tr}(\rho[\mathcal{R}_{\rho}^{-1}(D)]^2) = \frac{1}{4} F(\rho; D),
\end{split}
\tag{C13}
\end{equation}
$$

where we have used the fact that tr($DR_{\rho}^{-1}$(D)) =
$\langle\langle D|\mathcal{R}_{\rho}^{-1}|D\rangle\rangle = \langle\langle D|\mathcal{R}_{\rho}^{-1}\mathcal{R}_{\rho}\mathcal{R}_{\rho}^{-1}|D\rangle\rangle = \langle\langle \mathcal{R}_{\rho}^{-1}(D)|\mathcal{R}_{\rho}$
$|\mathcal{R}_{\rho}^{-1}(D)\rangle\rangle = \operatorname{tr}(\rho[\mathcal{R}_{\rho}^{-1}(D)]^2)$.

By construction, $L = \rho^{1/2}S$ satisfies the constraint in Eq. (C9), noting that we have used the assumption that $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$ as in the proposition statement. The corresponding value attained in the dual problem is

$$
(C9) \le \operatorname{tr}(L^\dagger L) = \frac{1}{4} \operatorname{tr}(\rho[\mathcal{R}_\rho^{-1}(D)]^2) = \frac{1}{4} F(\rho; D), \quad (C14)
$$

noting that tr(L†L) = tr(ρS²). Combining Eqs. (C13) and (C14) with the above statement that (C4) = (C9) proves the first part of the claim.

The alternative form (C3c) is nothing else than Eq. (C7).
Now we show the alternative form (C3d). Consider the
optimization (C7). Decompose $O = O_R + iO_I$ into Her-
mitian and anti-Hermitian parts with $O_R = (O + O^\dagger)/2 =$
$O_R^\dagger$, $O_I = -i(O - O^\dagger)/2 = O_I^\dagger$. The constraint on O indi-
cates that the Hermitian part $O_R$ of O must satisfy $2O_R =$
$D$. The second constraint then becomes

$$
\begin{bmatrix}
\rho & D/2 + iO_I \\
D/2 - iO_I & N
\end{bmatrix} \ge 0. \quad (C15)
$$

Conjugating by $\begin{bmatrix} 1 & 0 \\ 0 & 21 \end{bmatrix}$, we see that this condition is equivalent to

$$
\begin{bmatrix}
\rho & D + 2iO_I \\
D - 2iO_I & 4N
\end{bmatrix} \ge 0. \quad (C16)
$$

Now we set $K = 2O_I$ and $J = 4N$, showing that the optimization (C3d) is equivalent to Eq. (C7) (up to a factor of 4), and therefore equal to $F(\rho; D)$.

For completeness, we exhibit optimal choices for K, J.
Choose K to be the anti-Hermitian part of ρR̄<sub>ρ</sub><sup>-1</sup>(D), i.e.,
K = (ρR̄<sub>ρ</sub><sup>-1</sup>(D) - R̄<sub>ρ</sub><sup>-1</sup>(D)ρ)/(2i). Then

$$
\begin{equation}
\begin{split}
D + iK = {}& \frac{1}{2} (\rho \mathcal{R}_{\rho}^{-1}(D) + \mathcal{R}_{\rho}^{-1}(D) \rho) \\
            & + \frac{1}{2} (\rho \mathcal{R}_{\rho}^{-1}(D) - \mathcal{R}_{\rho}^{-1}(D) \rho) = \rho \mathcal{R}_{\rho}^{-1}(D),
\end{split}
\tag{C17}
\end{equation}
$$

and its Hermitian conjugate is $D - iK = \mathcal{R}_{\rho}^{-1}(D)\rho$. Now choose $J = (D - iK)\rho^{-1}(D + iK) = \mathcal{R}_{\rho}^{-1}(D)\rho\mathcal{R}_{\rho}^{-1}(D)$;

the constraint in Eq. (C3d) is satisfied thanks to Theorem
5. The value reached by this choice of candidates is then
the optimal value tr(J) = F(ρ; D). ■

The expressions in Proposition 5 lead to simple proofs
of elementary properties of the Fisher information.

Proposition 6 (Simple bounds for the Fisher information).—let $\rho$ be a subnormalized quantum state and $D$ be a Hermitian operator that satisfies $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. Then we have

$$
\|D\|^2 \le F(\rho; D) \le \operatorname{tr}(\rho^{-1} D^2), \quad (C18)
$$

where $D' = 2D - P_{\rho}DP_{\rho}$.

*Proof*.—First we show the lower bound. Let $|\phi\rangle$ be a (normalized) eigenvector associated with the largest eigenvalue of $D$ (in magnitude), such that $\langle\phi|D|\phi\rangle = \|D\|$. For some $s \in \mathbb{R}$ to be determined later, we choose the optimization candidate $S = s|\phi\rangle\langle\phi|$ in Eq. (C3a). Then the corresponding objective value is

$$
F(\rho; D) \ge 4 \operatorname{tr}(DS) - 4 \operatorname{tr}(\rho S^2) = 4s\|D\| - 4s^2 \langle\phi|\rho|\phi\rangle. \quad (C19)
$$

The latter expression is maximal when $0 = (d/ds)(\cdots) = 4\|D\| - 8s\langle\phi|\rho|\phi\rangle$, i.e., when $s = \|D\|/(2\langle\phi|\rho|\phi\rangle)$. We obtain the bound

$$
F(\rho; D) \ge 2 \frac{\|D\|^2}{\langle\phi|\rho|\phi\rangle} - \frac{\|D\|^2}{\langle\phi|\rho|\phi\rangle} = \frac{\|D\|^2}{\langle\phi|\rho|\phi\rangle} \ge \|D\|^2, \quad (C20)
$$

recalling furthermore that $\langle\phi|\rho|\phi\rangle \le 1$.

For the upper bound, consider the optimization problem
(C3b) and choose the candidate $L = \rho^{-1/2}D'/2$. This is a
feasible candidate because

$$
\begin{align*}
\rho^{1/2}L + L^{\dagger}\rho^{1/2} &= P_{\rho}D' + (\text{h.c.}) \\
&= 2P_{\rho}D - P_{\rho}DP_{\rho} + (\text{h.c.}) \\
&= 2P_{\rho}D(P_{\rho} + P_{\rho}^{\perp}) - P_{\rho}DP_{\rho} + (\text{h.c.}) \\
&= P_{\rho}DP_{\rho} + 2P_{\rho}DP_{\rho}^{\perp} + (\text{h.c.}) \\
&= 2(P_{\rho}DP_{\rho} + P_{\rho}DP_{\rho}^{\perp} + P_{\rho}^{\perp}DP_{\rho}) = 2D, \tag{C21}
\end{align*}
$$

where h.c. stands for the Hermitian conjugate of the entire
preceding expression, and where we furthermore recall that
$P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. The objective value attained by this choice of
candidate is $F(\rho; D) \le 4 \operatorname{tr}(L^{\dagger}L) = \operatorname{tr}(\rho^{-1}D^{2})$. ■

Proposition 7 (Right logarithmic derivative (RLD) bound [89]).—let $\rho$ be a subnormalized quantum state and let $D$ be a Hermitian operator satisfying $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. Let $G$ be any operator (possibly non-Hermitian) that satisfies $(\rho G + G^{\dagger}\rho)/2 = D$. Then

$$
F(\rho; D) \le \operatorname{tr}(\rho GG^{\dagger}). \quad (C22)
$$
Proof.—Use $L = \rho^{1/2}G/2$ in Eq. (C3b). ■

**Proposition 8 (Fisher information under parameter rescaling).**—let $\rho$ be a sub-normalized quantum state and $D$ be a Hermitian operator that satisfies $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. Then for any $\alpha \le 1$, $\beta \in \mathbb{R}$,

$$F(\alpha\rho; \beta D) = \frac{\beta^2}{\alpha} F(\rho; D). \quad (C23)$$

Proof.—Let $S, L$ be optimal variables in Eq. (C3a) and (C3b) for $F(\alpha\rho; \beta D)$. Let $S' = (\alpha/\beta)S$ and $L' = (\sqrt{\alpha}/\beta)L$. Then

$$\begin{aligned} \frac{1}{4} F(\alpha\rho; \beta D) &= \operatorname{tr}(\beta DS) - \operatorname{tr}(\alpha\rho S^2) \\ &= \frac{\beta^2}{\alpha} [\operatorname{tr}(DS') - \operatorname{tr}(\rho S'^2)] \le \frac{\beta^2}{\alpha} \frac{1}{4} F(\rho; D); \end{aligned} \quad (C24a)$$

$$\frac{1}{4} F(\alpha\rho; \beta D) = \operatorname{tr}(LL^{\dagger}) = \frac{\beta^2}{\alpha} \operatorname{tr}(L'L^{\dagger}) \ge \frac{\beta^2}{\alpha} \frac{1}{4} F(\rho; D), \quad (C24b)$$

noting that $L'$ is a valid choice of optimization candidate in (C3b) for $F(\rho; D)$ because $\rho^{1/2}L' + L^{\dagger}\rho^{1/2} = (1/\beta)((\alpha\rho)^{1/2}L + L^{\dagger}(\alpha\rho)^{1/2}) = D$. ■

**Proposition 9 (Fisher information bound for trace-decreasing maps).**—let $|\psi\rangle$ be a (normalized) state vector and let $|\xi\rangle$ be any vector such that $\langle\psi|\xi\rangle = 0$. Let $\mathcal{N}$ be any completely positive, trace-non-increasing map and let $0 \le \alpha \le 1$ such that $\mathcal{N}^{\dagger}(1) \le \alpha 1$. Then

$$F(\mathcal{N}(|\psi\rangle\langle\psi|); \mathcal{N}(|\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|)) \le 4\alpha|\xi|\xi. \quad (C25)$$

Proof.—Let $O = \mathcal{N}(|\psi\rangle\langle\xi|)$ and $N = \mathcal{N}(|\xi\rangle\langle\xi|)$. These choices are feasible in Eq. (C3c) because applying the completely positive map $\mathcal{N} \otimes \text{id}_2$ onto the positive semidefinite matrix

$$\begin{bmatrix} |\psi\rangle\langle\psi| & |\psi\rangle\langle\xi| \\ |\xi\rangle\langle\psi| & |\xi\rangle\langle\xi| \end{bmatrix} = \begin{bmatrix} |\psi\rangle \\ |\xi\rangle \end{bmatrix} [\langle\psi| \quad \langle\xi|] \ge 0 \quad (C26)$$

gives again a positive semidefinite matrix. This choice of variable yields the objective value $\operatorname{tr}(\mathcal{N}(|\xi\rangle\langle\xi|)) = \operatorname{tr}(\mathcal{N}^{\dagger}(1)|\xi\rangle\langle\xi|) \le \alpha|\xi|\xi$, proving the claim. ■

**Proposition 10 (Joint convexity of the Fisher information).**—let $\{\rho_k\}$ be a set of subnormalized states and $\{D_k\}$ be a set of Hermitian operators such that $P_{\rho_k}^{\perp}D_kP_{\rho_k}^{\perp} = 0$. Let $\{\alpha_k\}$ be a real positive coefficients such that $\sum_k \alpha_k \operatorname{tr}(\rho_k) \le 1$. Then

$$F\left(\sum_k \alpha_k \rho_k; \sum_k \alpha_k D_k\right) \le \sum_k \alpha_k F(\rho_k; D_k). \quad (C27)$$

Proof.—For each $k$, let $K_k, J_k$ be optimal choices in Eq. (C3d) for $F(\rho_k; D_k)$. Set $K = \sum_k \alpha_k K_k$ and $J = \sum_k \alpha_k J_k$. Then

$$\begin{bmatrix} \rho & D+iK \\ D-iK & J \end{bmatrix} = \sum_k \alpha_k \begin{bmatrix} \rho_k & D_k+iK_k \\ D_k-iK_k & J_k \end{bmatrix} \ge 0, \quad (C28)$$

and so $K, J$ are feasible candidates in the problem (C3d) for $F(\rho; D)$. The objective value achieved for this choice of variables gives the bound $F(\rho; D) \le \operatorname{tr}(J) = \sum_k \alpha_k \operatorname{tr}(J_k) = \sum_k \alpha_k F(\rho_k; D_k)$. ■

**Proposition 11 (Additivity of independent probes).**—let $\rho_A, \rho'_B$ be two subnormalized quantum states on two systems $A, B$, and let $D_A, D'_B$ be two traceless Hermitian operators such that $P_{\rho_A}^{\perp}D_A P_{\rho_A}^{\perp} = 0$ and $P_{\rho'_B}^{\perp}D'_B P_{\rho'_B}^{\perp} = 0$. Then

$$\begin{aligned} F(\rho_A \otimes \rho'_B; D_A \otimes \rho'_B + \rho_A \otimes D'_B) \\ = F(\rho_A; D_A) + F(\rho'_B; D'_B). \end{aligned} \quad (C29)$$

Observe that the second argument on the left-hand side corresponds to the derivative of the state of a composite system that remains in a tensor product, $(d/dt)(\rho_A \otimes \rho'_B) = (d\rho_A/dt) \otimes \rho'_B + \rho_A \otimes (d\rho'_B/dt)$.

Proof.—Here we may directly guess a solution $R$ to $\mathcal{R}_{\rho_A \otimes \rho'_B}(R) = D_A \otimes \rho'_B + \rho_A \otimes D'_B$. Compute first

$$\begin{aligned} \mathcal{R}_{\rho_A \otimes \rho'_B}(1_A \otimes M_B) &= \frac{1}{2} \{\rho_A \otimes \rho'_B, 1_A \otimes M_B\} \\ &= \frac{1}{2} \rho_A \otimes \{\rho'_B, M_B\}, \end{aligned} \quad (C30)$$

so we see that, setting

$$R = 1_A \otimes \mathcal{R}_{\rho'_B}^{-1}(D'_B) + \mathcal{R}_{\rho_A}^{-1}(D_A) \otimes 1_B, \quad (C31)$$

we have $\mathcal{R}_{\rho_A \otimes \rho'_B}(R) = \rho_A \otimes D'_B + D_A \otimes \rho'_B$. Then

$$\begin{aligned} F(\rho_A \otimes \rho'_B; D_A \otimes \rho'_B + \rho_A \otimes D'_B) &= \operatorname{tr}((\rho_A \otimes \rho'_B) R^2) \\ &= \operatorname{tr}((\rho_A \otimes \rho'_B)(1_A \otimes [\mathcal{R}_{\rho_A}^{-1}(D'_B)]^2 + [\mathcal{R}_{\rho_A}^{-1}(D_A)]^2 \otimes 1_B \\ &\quad + 2 \mathcal{R}_{\rho_A}^{-1}(D_A) \otimes \mathcal{R}_{\rho'_B}^{-1}(D'_B))) \\ &= F(\rho_A; D_A) + F(\rho'_B; D'_B), \end{aligned} \quad (C32)$$

where in the last line we have used $\operatorname{tr}(\rho_A \mathcal{R}_{\rho_A}^{-1}(D_A)) = \operatorname{tr}(D_A - P_{\rho_A}^{\perp}D_A P_{\rho_A}^{\perp}) = \operatorname{tr}(D_A) = 0$. ■

**Proposition 12 (Fisher information for pure states).**—let $|\psi\rangle$ be a subnormalized state vector and let $D$ be a Hermitian operator satisfying $\operatorname{tr}(D) = 0$ and $P_{\psi}^{\perp}DP_{\psi}^{\perp} = 0$. Then
$\langle D \rangle_{\psi} = 0$ and

$$F(\psi; D) = \frac{1}{(\text{tr} \psi)^2} [4 \text{tr}(\psi D^2)]. \quad (\text{C33})$$

Furthermore, if $\text{tr}(\psi) = 1$ and $D = -i[H, \psi]$ for some Hermitian operator $H$, then

$$F(\psi; D) = 4\sigma_H^2 = 4(\langle H^2 \rangle_{\psi} - \langle H \rangle_{\psi}^2). \quad (\text{C34})$$

*Proof*.—First of all thanks to Proposition 8 we assume without loss of generality that $\text{tr}(\psi) = 1$. Then, to see that $\langle D \rangle = 0$ we write

$$0 = \text{tr}(D) = \text{tr}[(\psi + P_{\psi}^{\perp})D] = \langle D \rangle + \text{tr}[P_{\psi}^{\perp}DP_{\psi}^{\perp}] = \langle D \rangle. \quad (\text{C35})$$

Using Eq. (C2) and Proposition 4, we then find

$$F(\psi; D) = \text{tr}(\psi (2D)^2) = 4 \text{tr}(\psi D^2). \quad (\text{C36})$$

If furthermore $D = -i[H, \psi]$ for some Hermitian $H$, then we use Eq. (A3) to see that $\text{tr}(\psi D^2) = \langle H^2 \rangle_{\psi} - \langle H \rangle_{\psi}^2$. ■

*Proposition 13 (Data-processing inequality for the Fisher information [19]).*—let $\rho$ be a subnormalized quantum state and $D$ be a Hermitian operator that satisfies $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. Let $\mathcal{E}$ be any completely positive, trace-nonincreasing map. Then

$$F(\rho; D) \ge F(\mathcal{E}(\rho); \mathcal{E}(D)). \quad (\text{C37})$$

*Proof*.—First we show that $P_{\rho}^{\perp}\mathcal{E}(D)P_{\rho}^{\perp} = 0$, ensuring that the right-hand side in Eq. (C37) is well defined. Decompose $D = P_{\rho}DP_{\rho} + P_{\rho}^{\perp}DP_{\rho} + P_{\rho}DP_{\rho}^{\perp} = D_0 + D_0^{\dagger}$, defining $D_0 = (P_{\rho}DP_{\rho})/2 + P_{\rho}DP_{\rho}^{\perp}$ such that $P_{\rho}^{\perp}D_0 = 0$. For $c > 0$ large enough, we have $\begin{bmatrix} \rho & D_0 \\ D_0^{\dagger} & c1 \end{bmatrix} \ge 0$ thanks to Theorem 5. Applying the completely positive map $\text{id}_2 \otimes \mathcal{E}$ we obtain $\begin{bmatrix} \mathcal{E}(\rho) & \mathcal{E}(D_0) \\ \mathcal{E}(D_0^{\dagger}) & c\mathcal{E}(1) \end{bmatrix} \ge 0$, and therefore thanks to Theorem 5, $P_{\rho}^{\perp}\mathcal{E}(D_0) = 0$. Then $P_{\rho}^{\perp}\mathcal{E}(D)P_{\rho}^{\perp} = 0$ recalling $D = D_0 + D_0^{\dagger}$.

Let $S$ be optimal in Eq. (C3a) for $F(\mathcal{E}(\rho); \mathcal{E}(D))$, thus satisfying $F(\mathcal{E}(\rho); \mathcal{E}(D)) = 4[\text{tr}(S\mathcal{E}(D)) - \text{tr}(\mathcal{E}(\rho)S^2)]$. Choosing the candidate $\mathcal{E}^{\dagger}(S)$ in Eq. (C3a) for $F(\rho; D)$ we obtain

$$\begin{aligned} F(\rho; D) &\ge 4[\text{tr}(D \mathcal{E}^{\dagger}(S)) - \text{tr}(\rho [\mathcal{E}^{\dagger}(S)]^2)] \\ &\ge 4[\text{tr}(\mathcal{E}(D)S) - \text{tr}(\rho \mathcal{E}^{\dagger}(S^2))] \\ &\ge 4[\text{tr}(\mathcal{E}(D)S) - \text{tr}(\mathcal{E}(\rho)S^2)] \\ &\ge F(\mathcal{E}(\rho); \mathcal{E}(D)), \end{aligned} \quad (\text{C38})$$

where we have used Corollary 2 in the second inequality. ■

In the case of commuting state and differential, the symmetric logarithmic derivative reduces to a matrix inverse as described by the following proposition.

*Proposition 14 (Fisher information for commuting state and derivative).*—let $\rho$ be any subnormalized quantum state and let $D$ be a Hermitian operator that satisfies $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. Suppose that $\rho$ and $D$ commute. Then

$$F(\rho; D) = \text{tr}(\rho^{-1}D^2). \quad (\text{C39})$$

*Proof*.—This can be shown from the properties of the symmetric logarithmic derivative, but we give a simple alternative proof using our convex optimizations for fun. Choose $S = \rho^{-1}D/2$ in Eq. (C3a), which we note is a Hermitian operator because $\rho$ and $D$ commute. This gives $F(\rho; D) \ge (D39)(C39)$. Similarly, the choice $L = \rho^{-1/2}D/2$ in Eq. (C3b) provides the opposite bound. ■

The following proposition interprets the Fisher information for subnormalized states according to the definition (C1) as the Fisher information of a normalized state that was projected onto a smaller subspace. This interpretation works as long as the subnormalized state does not change trace along its evolution, meaning that the derivative $D$ has zero trace.

*Proposition 15 (Fisher information for subnormalized and normalized states).*—let $\rho$ be any subnormalized quantum state and let $D$ be any Hermitian operator that satisfies both $\text{tr}(D) = 0$ and $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0$. Define $\rho', D'$, with an additional new Hilbert space dimension, as

$$\rho' = \begin{pmatrix} \rho & 0 \\ 0 & 1 - \text{tr}(\rho) \end{pmatrix}; \quad D' = \begin{pmatrix} D & 0 \\ 0 & 0 \end{pmatrix}. \quad (\text{C40})$$

Then

$$F(\rho; D) = F(\rho'; D'). \quad (\text{C41})$$

*Proof*.—Let $P$ denote the projector onto the subspace of the Hilbert space on which the upper left block of $\rho', D'$ acts. Let $R = \mathcal{R}_{\rho}^{-1}(D)$, and define

$$R' = \begin{pmatrix} R & 0 \\ 0 & 0 \end{pmatrix}. \quad (\text{C42})$$

Multiplying together block-diagonal matrices preserves the block-diagonal structure, hence

$$\frac{1}{2}\{\rho', R'\} = \begin{pmatrix} \frac{1}{2}\{\rho, R\} & 0 \\ 0 & 0 \end{pmatrix} = D'. \quad (\text{C43})$$

Then with the definition (C1),

$$F(\rho'; D') = \text{tr}(\rho'R^2) = \text{tr}(\rho R^2) = F(\rho; D).$$
We can furthermore prove a relation between the Fisher information of two different directions in state space that might be associated with two different parametrized evolutions.

**Proposition 16 (Relation between the Fisher information of two directions).**—let $\rho$ be a subnormalized quantum state and let $D, D'$ be two Hermitian operators that satisfy $P_{\rho}^{\perp}DP_{\rho}^{\perp} = P_{\rho}^{\perp}D'P_{\rho}^{\perp} = 0$. Then

$$F(\rho; D) \le F(\rho; D') + [F(\rho; D+D')F(\rho; D-D')]^{1/2}. \quad (C44)$$

Consequently,

$$|F(\rho; D) - F(\rho; D')| \le [F(\rho; D+D')F(\rho; D-D')]^{1/2}. \quad (C45)$$

Furthermore, equality holds in Eq. (C45) if and only if $D, D'$ are linearly dependent.

**Proof.**—Define the shorthand $\Delta_{\pm} = D \pm D'$. We compute

$$\begin{aligned} F(\rho; D) &= \operatorname{tr}(\rho [\mathcal{R}_{\rho}^{-1}(D') + \mathcal{R}_{\rho}^{-1}(\Delta_{-})]^2) \\ &= F(\rho; D') + \operatorname{tr}(\rho [(\mathcal{R}_{\rho}^{-1}(\Delta_{-}))^2 \\ &\quad + \{\mathcal{R}_{\rho}^{-1}(D'), \mathcal{R}_{\rho}^{-1}(\Delta_{-})\}]) \\ &= F(\rho; D') + \operatorname{tr}(\rho \{\frac{1}{2}\mathcal{R}_{\rho}^{-1}(\Delta_{-}) \\ &\quad + \mathcal{R}_{\rho}^{-1}(D'), \mathcal{R}_{\rho}^{-1}(\Delta_{-})\}), \quad (C46) \end{aligned}$$

where in the last equality we have used $M^2 = \{\frac{1}{2}M, M\}$ for any operator $M$ along with the linearity of the anti-commutator in the first argument. Furthermore, we see from the definition of $\Delta_{-}$ that

$$D' + \frac{1}{2}\Delta_{-} = \frac{1}{2}(D + D') = \frac{1}{2}\Delta_{+}. \quad (C47)$$

Then

$$\begin{aligned} (C46) &= F(\rho; D') + \frac{1}{2} \operatorname{tr}(\rho \{\mathcal{R}_{\rho}^{-1}(\Delta_{+}), \mathcal{R}_{\rho}^{-1}(\Delta_{-})\}) \\ &= F(\rho; D') + \operatorname{Re} \operatorname{tr}(\rho \mathcal{R}_{\rho}^{-1}(\Delta_{+}) \mathcal{R}_{\rho}^{-1}(\Delta_{-})) \\ &=: F(\rho; D') + C_{\rho}(\Delta_{+}, \Delta_{-}), \end{aligned} \quad (C48)$$

where $C_{\rho}(\Delta_{+}, \Delta_{-})$ is defined as the second term in the above expression. From the Cauchy-Schwarz inequality,

$$|C_{\rho}(\Delta_{+}, \Delta_{-})|^2 \le \operatorname{tr}(\rho[\mathcal{R}_{\rho}^{-1}(\Delta_{+})]^2) \operatorname{tr}(\rho[\mathcal{R}_{\rho}^{-1}(\Delta_{-})]^2). \quad (C49)$$

Hence,

$$(C48) \le F(\rho; D) + [F(\rho; \Delta_{+}) F(\rho; \Delta_{-})]^{1/2}. \quad (C50)$$

Equation (C45) follows by repeating the argument while inverting the roles of $D$ and $D'$.

Equality in Eq. (C45) is equivalent to the Cauchy-Schwarz inequality being tight. In turn is equivalent to the operators $\rho^{1/2}\mathcal{R}_{\rho}^{-1}(\Delta_{+})$ and $\rho^{1/2}\mathcal{R}_{\rho}^{-1}(\Delta_{-})$ being linearly dependent, i.e., there exist $\alpha_1, \alpha_2 \in \mathbb{R}$, $(\alpha_1, \alpha_2) \ne (0, 0)$, such that

$$\alpha_1 \rho^{1/2} \mathcal{R}_{\rho}^{-1}(\Delta_{+}) + \alpha_2 \rho^{1/2} \mathcal{R}_{\rho}^{-1}(\Delta_{-}) = 0. \quad (C51)$$

Since the operator $\mathcal{R}_{\rho}^{-1}(\Delta_{\pm})$ vanishes on the operator subspace spanned by $P_{\rho}^{\perp}(\cdot)P_{\rho}^{\perp}$, we have that Eq. (C51) is equivalent to

$$\alpha_1 \mathcal{R}_{\rho}^{-1}(\Delta_{+}) + \alpha_2 \mathcal{R}_{\rho}^{-1}(\Delta_{-}) = 0, \quad (C52)$$

and therefore to

$$\mathcal{R}_{\rho}^{-1}[\alpha_1 \Delta_{+} + \alpha_2 \Delta_{-}] = 0. \quad (C53)$$

Because the kernel of the superoperator $\mathcal{R}_{\rho}^{-1}$ is spanned by $P_{\rho}^{\perp}(\cdot)P_{\rho}^{\perp}$, onto which $\Delta_{\pm}$ have no support by assumption, then Eq. (C53) is further equivalent to

$$\alpha_1 \Delta_{+} + \alpha_2 \Delta_{-} = 0. \quad (C54)$$

Therefore, equality in Eq. (C45) is achieved if and only if $\Delta_{\pm}$ are linearly dependent, which is equivalent to the linear dependence of $D$ with $D'$. ■

Using a similar idea, we can also prove a continuity bound on the Fisher information with respect to its second argument.

**Proposition 17 (A continuity bound of the Fisher information in its second argument).**—let $\rho$ be any subnormalized quantum state and let $D, \Delta$ be any Hermitian operators such that $P_{\rho}^{\perp}DP_{\rho}^{\perp} = 0 = P_{\rho}^{\perp}\Delta P_{\rho}^{\perp}$. Then

$$\begin{aligned} &|F(\rho; D + \Delta) - F(\rho; D) - F(\rho; \Delta)| \\ &\le 2[F(\rho; D)F(\rho; \Delta)]^{1/2}. \end{aligned} \quad (C55)$$

As a consequence,

$$\begin{aligned} &|F(\rho; D + \Delta) - F(\rho; D)| \\ &\le F(\rho; \Delta) + 2[F(\rho; D)F(\rho; \Delta)]^{1/2}. \end{aligned} \quad (C56)$$
*Proof.*—Using the formula $F(\rho; D') = \text{tr}(D' \mathcal{R}_{\rho}^{-1}(D'))$ for the Fisher information, we write

$$F(\rho; D + \Delta) = \text{tr}((D + \Delta) \mathcal{R}_{\rho}^{-1}(D + \Delta)) \\ = F(\rho; D) + F(\rho; \Delta) + 2 \text{tr}(D \mathcal{R}_{\rho}^{-1}(\Delta)), \quad (C57)$$

recalling that $\mathcal{R}_{\rho}^{-1}$ is superoperator self-adjoint. The claim follows by bounding the last term in the above expression using the Cauchy-Schwarz inequality, to get

$$|\text{tr}(D \mathcal{R}_{\rho}^{-1}(\Delta))| \le \sqrt{\text{tr}(D \mathcal{R}_{\rho}^{-1}(D)) \text{tr}(\Delta \mathcal{R}_{\rho}^{-1}(\Delta))} \\ = \sqrt{F(\rho; D)F(\rho; \Delta)}.$$

We can consider more precisely how $F_{\text{Bob},t}$ behaves when seen as a function of the noise channel $\mathcal{N}$, for channels $\mathcal{N}$ that are close to the identity channel id. More

specifically, we prove a continuity bound for the quantum Fisher information $F(\mathcal{N}(\psi); \mathcal{N}(\partial_t\psi))$ at the point $\mathcal{N} = \text{id}$, when that quantity is seen as a function of $\mathcal{N}$.

*Proposition 18.*—Let $|\psi\rangle$ be a pure state and let $D$ be a Hermitian operator such that $\langle D|\psi = 0$ and $P_{\psi}^{\perp}DP_{\psi}^{\perp} = 0$. Let $\epsilon > 0$ and let $\mathcal{N}$ be a channel with $\|\mathcal{N} - \text{id}\|_{\diamond} \le \epsilon$. Then

$$F(\psi, D) \ge F(\mathcal{N}(\psi), \mathcal{N}(D)) \ge F(\psi, D) \\ - 8\epsilon \|D\|_1 \|D\|_{\infty}. \quad (C58)$$

Observe that the stated conditions on $D$ are satisfied if $D = -i[H, \psi]$ for some Hermitian operator $H$.

*Proof.*—Let $\epsilon > 0$ and let $\mathcal{N} = \text{id} + \Delta$ where $\Delta$ is a Hermiticity-preserving superoperator with $\|\Delta\|_{\diamond} \le \epsilon$. The first claimed inequality immediately follows from the data processing inequality. We now prove the second inequality. Using Proposition 4, let $S = \frac{1}{2}\mathcal{R}_{\psi}^{-1}(D) = D$. Since this $S$ is known to be optimal in Eq. (C3a) for $F(\psi; D)$, we can compute

$$\begin{aligned} F(\psi; D) &= 4\{\text{tr}[(D)S] - \text{tr}[\psi S^2]\} \\ &= 4\{\text{tr}[\mathcal{N}(D)S] - \text{tr}[\mathcal{N}(\psi)S^2]\} - 4\{\text{tr}[\Delta(D)S] - \text{tr}[\Delta(\psi)S^2]\} \\ &\le F(\mathcal{N}(\psi), \mathcal{N}(D)) + 4\|\Delta(D)\|_1 \|S\|_{\infty} + 4\|\Delta(\psi)\|_1 \|S\|_{\infty}^2 \\ &\le F(\mathcal{N}(\psi), \mathcal{N}(D)) + 8\epsilon \|D\|_1 \|D\|_{\infty}, \end{aligned} \quad (C59)$$

using $\|D\|_{\infty} \le \|D\|_1$, and thus proving the claim.

## APPENDIX D: OPTIMAL LOCAL SENSING AND THE CRAMÉR-RAO BOUND

Here we review which operators achieve the optimal variance in estimating an unknown parameter [1, 12, 23, 24, 89]. An unknown parameter $t$ of an evolution $\rho_t$ of a (normalized) quantum state is estimated locally around $t_0$ using an observable $T$, whose measurement outcomes are the estimates of the parameter. We ask for the observable to have the correct average and first order deviation, $\langle T \rangle_{\rho_{t_0+dt}} = t_0 + dt + O(dt^2)$; except in edge cases, this condition can be enforced by a suitable scaling factor and a suitable shift by the identity. The conditions then become $\langle T \rangle_{\rho_{t_0}} = t_0$ and $\text{tr}\{(\partial_t \rho_t|_{t_0}) T\} = 1$. We seek to minimize the operator $T$'s variance $\langle T^2 \rangle_{\rho_{t_0}} - \langle T \rangle_{\rho_{t_0}}^2$. We call such an operator with minimal variance an *optimal local sensing operator*, and the square root of the minimal variance is the *optimal estimation error* $\Delta t_{\text{unc}}(t_0)$ locally at $t_0$. That is, the

optimal estimation error locally at $t_0$, along with an optimal local sensing operator at $t_0$, are given by the following optimization problem:

$$\Delta t_{\text{unc}}^2(t_0) = \min_{T=T^{\dagger}} \text{tr}\{\rho_{t_0}(T - t_0 \mathbb{1})^2\}, \\ \text{such that} \\ \text{tr}\{\rho_{t_0} T\} = t_0, \text{tr}\{(\partial_t \rho_t|_{t_0}) T\} = 1. \quad (D1)$$

In the event that $\partial_t \rho_t|_{t_0} = 0$, there is no operator $T$ that satisfies the given conditions. We conventionally set $\Delta t_{\text{unc}} = \infty$, since the state is locally stationary and no observable is able to detect a first-order deviation in the parameter $t$.

A more general scheme would enable an agent to use a generalized measurement given by a POVM instead of an observable $T$. However, as shown in, e.g., Ref. [1], the optimal POVM can in fact be chosen to be a projective measurement. Therefore, one cannot sense a parameter more accurately using a POVM instead of an observable.
The following proposition fully characterizes the locally optimal sensing observables (cf., e.g., Ref. [1]). In the following, we write as a shorthand $\rho$ and $\partial_t\rho$ instead of $\rho_{t_0}$ and $\partial_t\rho_{t|_{t_0}}$.

**Proposition 19 (Locally optimal sensing).**—assume $\partial_t\rho \neq 0$. Then any operator $T$ that is optimal in Eq. (D1) is of the form

$$T = t\mathbb{1} + (\Delta t_{\text{unc}}^2) \mathcal{R}_{\rho}^{-1}(\partial_t \rho) + P_{\rho}^{\perp}MP_{\rho}^{\perp}, \quad (\text{D2})$$

for some Hermitian operator $M$.

If $P_{\rho}^{\perp}(\partial_t\rho)P_{\rho}^{\perp} = 0$, then $\Delta t_{\text{unc}}^2 = [F(\rho; \partial_t\rho)]^{-1}$ with the Fisher information defined in Eq. (C1), and $M$ can be arbitrary.

If $P_{\rho}^{\perp}(\partial_t\rho)P_{\rho}^{\perp} \neq 0$, then $\Delta t_{\text{unc}}^2 = 0$ and $M$ satisfies $\text{tr}(MP_{\rho}^{\perp}\partial_t\rho P_{\rho}^{\perp}) = 1$.

Let us further note that if $\partial_t\rho = 0$, we have $F(\rho; \partial_t\rho) = 0$. Therefore, provided that $P_{\rho}^{\perp}\partial_t\rho P_{\rho}^{\perp} = 0$, we can in full generality write

$$\Delta t_{\text{unc}}^2 = \frac{1}{F(\rho; \partial_t \rho)}, \quad (\text{D3})$$

along with the convention that $\Delta t_{\text{unc}} = \infty$ if $F(\rho; \partial_t\rho) = 0$. In our setting, the optimal sensing scheme always achieves the value of the Cramér-Rao bound.

**Proof of Proposition 19.**—Without loss of generality, we assume $t_0 = 0$ throughout this proof; this is achieved by shifting the parameter to center it at zero, implying the corresponding shift $T \to T' = T - t_0\mathbb{1}$. We thus consider the optimization problem

$$\Delta t_{\text{unc}}^2 = \min_{T=T^{\dagger}} \begin{cases} \text{tr}\{\rho_{t_0} T^2\}, \\ \text{tr}\{\rho_{t_0} T\} = 0, \text{ tr}\{(\partial_t \rho_{t|_{t_0}}) T\} = 1. \end{cases} \quad (\text{D4})$$

First of all we observe that the first condition, $\text{tr}(\rho T) = 0$, can be ignored without changing the optimal value of the problem. Indeed, for any $T$ that satisfies $\text{tr}((\partial_t\rho)T) = 1$ but with $\text{tr}(\rho T) \neq 0$, we can define $T' = T - \text{tr}(\rho T)\mathbb{1}$, with $\text{tr}(\rho T') = 0$ and $\text{tr}((\partial_t\rho)T) = \text{tr}((\partial_t\rho)T')$ since $\text{tr}(\partial_t\rho) = \partial_t \text{tr}(\rho) = 0$; then $\text{tr}(\rho T^{\prime 2}) = \text{tr}(\rho T^2) - [\text{tr}(\rho T)]^2 \leq \text{tr}(\rho T^2)$, meaning that $T'$ not only satisfies $\text{tr}(\rho T') = 0$ in addition to the other condition, but it achieves a better objective function value.

We can recast this optimization as semidefinite problem, following Refs. [89,107], by using Schur complements (Theorem 5):

$$\Delta t_{\text{unc}}^2 = \min_{Q \ge 0, T=T^{\dagger}} \begin{cases} \text{tr}(\rho Q) & (\text{D5}) \\ \text{tr}((\partial_t \rho) T) = 1; \\ \begin{bmatrix} Q & -T \\ -T & \mathbb{1} \end{bmatrix} \ge 0. \end{cases}$$

The associated dual problem takes the following form, noting that strong duality holds thanks to Slater's conditions

[76,106].

$$\begin{align*}
\Delta t_{\text{unc}}^2 &= \max_{A,C \ge 0, B \text{ arb., } \mu \in \mathbb{R}} && \mu - \text{tr}(C) \\
&\text{such that} && A \le \rho \\
& && B + B^{\dagger} = \mu \partial_t \rho \\
& && \begin{bmatrix} A & B \\ B^{\dagger} & C \end{bmatrix} \ge 0 \\[1em]
&= \max_{B \text{ arb., } \mu \in \mathbb{R}} && \mu - \text{tr}(B^{\dagger} \rho^{-1} B) \quad (\text{D6}) \\
&\text{such that} && B + B^{\dagger} = \mu \partial_t \rho \\
& && P_{\rho} B = B
\end{align*}$$

$$= \max_{L \text{ arb., } \mu \in \mathbb{R}} \mu - \text{tr}(L^{\dagger}L) \quad (\text{D7}) \\
\text{such that} \quad \rho^{1/2}L + L^{\dagger}\rho^{1/2} = \mu \partial_t \rho,$$

using again Schur complements and where we introduced the variable $L$ via $B = \rho^{1/2}L$, and where $P_{\rho} = \mathbb{1} - P_{\rho}^{\perp}$ is the projector onto the support of $\rho$.

A powerful characterization of the whole family of optimal solutions to a semidefinite problem with strong duality are the complementary slackness relations. An inequality constraint multiplied by the corresponding dual variable becomes an equality for any choice of primal and dual optimal solutions [76,106]. Here, this means that

$$\begin{bmatrix} Q & -T \\ -T & \mathbb{1} \end{bmatrix} \begin{bmatrix} A & B \\ B^{\dagger} & C \end{bmatrix} = 0. \quad (\text{D8})$$

This gives us the following relations that must be satisfied for any choice of optimal variables:

$$Q\rho = TB^{\dagger}; \quad QB = TC; \quad B^{\dagger} = T\rho; \quad C = TB. \quad (\text{D9})$$

The third equality ($B^{\dagger} = T\rho$) along with the dual constraint in Eq. (D6) implies that $\rho T + T\rho = \mu \partial_t\rho$. Proposition 3 asserts that the solutions are necessarily of the form $T = (\mu/2)\mathcal{R}_{\rho}^{-1}(\partial_t\rho) + P_{\rho}^{\perp}MP_{\rho}^{\perp}$ for some Hermitian $M$.

Now first suppose that $P_{\rho}^{\perp}(\partial_t\rho)P_{\rho}^{\perp} = 0$. The primal value achieved for a $T$ of this form, and for any $\mu$ and $M$, is

$$\text{primal achieved} = \text{tr}(\rho T^2) = \frac{\mu^2}{4} F(t), \quad (\text{D10})$$

with $F(t)$ as in Eq. (9). From complementary slackness we have $B^{\dagger} = T\rho$ and hence $\text{tr}(B^{\dagger}\rho^{-1}B) = \text{tr}(\rho T^2) = \mu^2 F(t)/4$. The dual problem therefore reaches the value

$$\text{dual achieved} = \mu - \mu^2 F(t)/4. \quad (\text{D11})$$

Optimality implies that the primal and dual values are equal, $\mu^2 F(t)/4 = \mu - \mu^2 F(t)/4$ and therefore $\mu =$
$2/F(t)$ [note $\mu = 0$ is ruled out because the primal constraint $\mathrm{tr}((\partial_t \rho) T) = 1$ would be impossible to satisfy]. Therefore, the optimal solution to the problem is

$$ \Delta t_{\text{unc}}^2 = \frac{1}{F(t)}. \qquad (\text{D12}) $$

Now suppose that $P_{\rho}^{\perp}(\partial_t \rho)P_{\rho}^{\perp} \neq 0$. Then there cannot be any solution for $L$ in the constraint in Eq. (D7) unless $\mu = 0$ [the left-hand side vanishes entirely if we hit it with $P_{\rho}^{\perp}(\cdot)P_{\rho}^{\perp}$ but not the right-hand side if $\mu \neq 0$]. Then $T = P_{\rho}^{\perp}XP_{\rho}^{\perp}$, which implies $\mathrm{tr}(\rho T^2) = 0$, and furthermore $M$ must satisfy $\mathrm{tr}((\partial_t \rho)P_{\rho}^{\perp}MP_{\rho}^{\perp}) = 1$ from the primal constraint. The dual candidate $L = 0$ yields objective value of zero in the dual problem, and therefore the optimal value of the optimization problem is zero, $\Delta t_{\text{unc}}^2 = 0$. ■

## APPENDIX E: PROOF OF THE SENSITIVITY UNCERTAINTY RELATION

The goal of this section is to prove the statements made in Sec. III. The setting is the one introduced in Sec. II. We provide two independent proofs of the uncertainty relation. The first proof is more intuitive and straightforward. The second proof is slightly more general and provides greater insight into some technicalities that underpin the uncertainty relation. The second proof directly relates the semidefinite characterizations of the quantities $F_{\text{Bob},t}$ and $F_{\text{Eve},\eta}$, making it easier to analyze edge cases, to gain insight on what choices of semidefinite variables are optimal, and to consider the more general situation where $\mathcal{N}$ is a trace-non-increasing map.

### 1. Proof via the second-order expansion of the fidelity

The strategy of our first proof of our uncertainty relation is to provide a direct proof of the statement presented as Corollary 1; we have already seen in the main text that the statement in Theorem 1 is equivalent to Corollary 1.

First observe that without loss of generality, we can assume that the Hamiltonian is time independent. This is because the Fisher information depends only on the state and its local time derivative at $t$, which is given by Eq. (42) and depends only on the value of the Hamiltonian at the fixed value $t$ of interest.

Our proof proceeds in a similar fashion to that of the channel-extension bound developed in Refs. [30,31,78]. While our uncertainty relation could also be derived from the results in those references, we provide a self-contained proof for completeness and consistency of notation.

A remarkable property of the Fisher information is that it is directly related to the Bures distance and the fidelity of quantum states [1,12,25,108] according to

$$ F_{\text{Bob},t} = -4 \left. \frac{d^2}{dt'^2} \right|_{t'=t} F(\rho_B(t), \rho_B(t')), \qquad (\text{E1}) $$

where $F(\rho, \rho') = \|\rho^{1/2} \rho'^{1/2}\|_1 = \mathrm{tr}[(\rho^{1/2} \rho' \rho^{1/2})^{1/2}]$ is the root fidelity between two quantum states [63], where $\|A\|_1$ denotes trace norm, i.e., the sum of the singular values of $A$. Note that at $t' = t$, the fidelity reaches its maximum value 1. We assume that $\rho(t)$ is does not change rank at $t' = t$, avoiding edge cases where the expression (E1) is incomplete [32–34].

By Uhlmann’s theorem, and writing $|\rho(t)\rangle_{BE} = V_{A \to BE}|\psi(t)\rangle_A$ in terms of the Stinespring dilation $V_{A \to BE}$ of $\mathcal{N}$ given in Eq. (43), we have that

$$ F(\rho_B(t), \rho_B(t')) = \max_{W_E \text{ unitary}} |\langle \rho(t')|_{BE} W_E |\rho(t)\rangle_{BE}|, \quad (\text{E2}) $$

where $W_E$ is a unitary operation on $E$. We therefore have the following equivalent expressions:

$$ F(\rho_B(t), \rho_B(t')) = \max_{W_E} \mathrm{Re} \langle \psi(t')|_A V^\dagger W_E V_{A \to BE} |\psi(t)\rangle_A \qquad (\text{E3a}) $$

$$ = \max_{W_E} \mathrm{Re} \langle \psi(t')|_A \hat{\mathcal{N}}^\dagger (W_E) |\psi(t)\rangle_A \qquad (\text{E3b}) $$

$$ = \max_{W_E} \mathrm{Re} \langle \psi(t)|_A e^{iH(t'-t)} \hat{\mathcal{N}}^\dagger (W_E) |\psi(t)\rangle_A \qquad (\text{E3c}) $$

$$ = \max_{W_E} \mathrm{Re} \mathrm{tr}(\hat{\mathcal{N}}(\psi_A(t) e^{iH(t'-t)}) W_E) \qquad (\text{E3d}) $$

$$ = \|\hat{\mathcal{N}}(\psi e^{iH(t'-t)})\|_1, \qquad (\text{E3e}) $$

where the complementary channel $\hat{\mathcal{N}}$ is given by Eq. (44). In the above expressions, the maximization can be taken over operators $W_E$ that are unitary, or equivalently, it can be relaxed to all operators $W_E$ satisfying $\|W_E\| \le 1$.

The optimal unitary $W_E$ is given by the polar decomposition of the operator $\hat{\mathcal{N}}(\psi e^{iH(t'-t)})$. For $t' = t + dt$ with a small $dt$, we have that the optimal $W_E$ is close to the identity, which is the optimal for $t' = t$. Let us expand $W_E = 1 - idtS - (1/2)dt^2S_2 + O(dt^3)$ for general matrices $S$ and $S_2$ to be determined. The unitary constraint $W_E^\dagger W_E = 1$ for all $dt$ implies that $S = S^\dagger$ and that $S_2 + S_2^\dagger = 2S^\dagger S = 2S^2$. Starting from Eq. (E3c) and expanding up to order $dt^2$ we find
$$
\begin{align}
F(\rho_B(t), \rho_B(t')) &= \max_{S=S^\dagger, S_2} \mathrm{Re} \operatorname{tr} \left\{ \psi \left( 1 + i dt H - \frac{H^2}{2} dt^2 \right) \hat{\mathcal{N}}^\dagger \left( 1 - i dt S - \frac{S_2}{2} dt^2 \right) \right\} + O(dt^3) \nonumber \\
&= 1 + \max_{S=S^\dagger, S_2} \left\{ dt \mathrm{Re} \operatorname{tr}[i \psi H - i \psi \hat{\mathcal{N}}^\dagger(S)] + dt^2 \mathrm{Re} \operatorname{tr} \left[ -\frac{1}{2} \psi H^2 - \frac{1}{2} \psi \hat{\mathcal{N}}^\dagger(S_2) + \psi H \hat{\mathcal{N}}^\dagger(S) \right] + O(dt^3) \right\} \nonumber \\
&= 1 + \frac{dt^2}{2} \max_{S=S^\dagger, S_2} \mathrm{Re} \operatorname{tr} \left\{ -\psi H^2 - \psi \hat{\mathcal{N}}^\dagger(S_2) + 2 \psi H \hat{\mathcal{N}}^\dagger(S) \right\} + O(dt^3), \tag{E4}
\end{align}
$$

recalling that $\hat{\mathcal{N}}^\dagger(1) = 1$, and where the first-order term vanishes because a product of two Hermitian operators has a real trace; with the factor $i$ the term is killed by taking the real part. Continuing with only the second-order term we find

$$
\begin{align}
\left. \frac{d^2}{dt'^2} \right|_{t'=t} F(\rho_B(t), \rho_B(t')) &= \max_{S=S^\dagger, S_2} \left\{ -\mathrm{tr}(\psi H^2) - \frac{1}{2} \mathrm{tr}(\hat{\mathcal{N}}(\psi)(S_2 + S_2^\dagger)) + \mathrm{tr}[\{\psi, H\} \hat{\mathcal{N}}^\dagger(S)] \right\} \nonumber \\
&= \max_{S=S^\dagger} \left\{ -\mathrm{tr}(\psi H^2) - \mathrm{tr}(\hat{\mathcal{N}}(\psi) S^2) + \mathrm{tr}[\hat{\mathcal{N}}(\{\psi, H\} S)] \right\}, \tag{E5}
\end{align}
$$

where we have used the identity $2 \mathrm{Re} \operatorname{tr}(AO) = \mathrm{tr}(A(O + O^\dagger))$ for Hermitian $A$, the identity $2 \mathrm{Re} \operatorname{tr}(ABC) = \mathrm{tr}(\{A, B\} C)$ for Hermitian $A, B, C$, as well as the condition $S_2 + S_2^\dagger = 2S^2$ that came from enforcing the unitarity of $W_E$.

It is instructive to briefly comment on the situation of a time-dependent Hamiltonian. The derivation of the above expression, especially Eq. (E3c) and the expansion of the time-evolution operator leading up to Eq. (E4), looks like it necessitated the assumption of time independence of the Hamiltonian and that a time-dependent Hamiltonian might have led to a different result. In fact, we obtain the same result with a time-dependent Hamiltonian, which can be seen as follows. Write

$$H(t) = H + tH' + O(t^2) \quad (E6)$$

and expand the time-evolution operator via the time-ordered exponential as $U^\dagger(t' - t) = T e^{i \int_t^{t'} dt'' H(t'')} = 1 + i \int_t^{t'} dt'' H(t'') - \int_t^{t'} dt'' H(t'') \int_t^{t''} dt''' H(t''') + O(t^3) = 1 + i dt H + (dt^2/2)(iH' - H^2) + O(dt^3)$, then we see that the only difference in the expressions leading up to Eq. (E4) is an additional term $\mathrm{Re} \operatorname{tr}\{\psi i H' dt^2\}$, which is equal to zero.

Now, we proceed to prove the uncertainty relation. With the definition $\Delta F_{\mathrm{Bob},t} = F_{\mathrm{Alice},t} - F_{\mathrm{Bob},t}$, we have

$$
\begin{align}
\Delta F_{\mathrm{Bob},t} ={}& 4(\mathrm{tr}(\psi H^2) - (\mathrm{tr}(\psi H))^2) \nonumber \\
& + 4 \left. \frac{d^2}{dt'^2} \right|_{t'=t} F(\rho_B(t), \rho_B(t')) \nonumber \\
={}& \max_{S=S^\dagger} \left\{ 4 \mathrm{tr}[\hat{\mathcal{N}}(\{\psi, H\} S)] - 4 \mathrm{tr}[\hat{\mathcal{N}}(\psi) S^2] \right. \nonumber \\
& \qquad \left. - 4(\mathrm{tr}(\psi H))^2 \right\}, \tag{E7}
\end{align}
$$

recalling that $F_{\mathrm{Alice},t} = 4\sigma_H^2 = 4(\langle H^2 \rangle - \langle H \rangle^2)$ and using the expression (E5). Observe that $\Delta F_{\mathrm{Bob},t}$ is necessarily invariant under a constant shift of the Hamiltonian $H \mapsto H + c1$, because such a shift does not influence the evolution $\psi(t)$ and therefore both $F_{\mathrm{Alice},t}$ and $F_{\mathrm{Bob},t}$ are invariant under such shifts. [This invariance can also be checked explicitly by carrying out the corresponding transformations $H \mapsto H + c1$ and $S \to S + c1$ in Eq. (E7).] Applying the shift $H \mapsto H - \langle H \rangle_\psi$ yields

$$(F7) = \max_{S=S^{\dagger}} \left\{ 4 \operatorname{tr}[\hat{\mathcal{N}}(\{\psi, \bar{H}\} S)] - 4 \operatorname{tr}[\hat{\mathcal{N}}(\psi) S^2] \right\}, \quad (E8)$$

using the shorthand $\bar{H} := H - \langle H \rangle_\psi$. At this point we recognize the expression of the Fisher information from Proposition 5, with $\rho = \hat{\mathcal{N}}(\psi)$ and $D = \hat{\mathcal{N}}(\{\psi, \bar{H}\})$. Let us briefly check that the requirement $P_\rho^\perp D P_\rho^\perp = 0$ in Proposition 5 and in the definition of the Fisher information (C1) is satisfied. Thanks to Theorem 5, we have $[\psi \quad \psi \bar{H}] \ge 0$, and furthermore, by Lemma 1, $[\hat{\mathcal{N}}(\psi) \quad \hat{\mathcal{N}}(\psi \bar{H})] \ge 0$; by Theorem 5 again, this implies that $P_\rho^\perp \hat{\mathcal{N}}(\psi \bar{H}) = 0$. Therefore, $P_\rho^\perp \hat{\mathcal{N}}(\{\psi, \bar{H}\}) P_\rho^\perp = 0$. It follows that

$$\Delta F_{\mathrm{Bob},t} = (F8) = F(\hat{\mathcal{N}}(\psi); \hat{\mathcal{N}}(\{\psi, \bar{H}\})), \quad (E9)$$

as claimed.

## 2. Direct proof using the semidefinite characterization of the Fisher information

For this section, we fix $|\psi\rangle, |\xi\rangle$ be such that $\langle\psi|\psi\rangle = 1$ and $\langle\psi|\xi\rangle = 0$, and let $\mathcal{N}$ be a completely positive, trace-non-increasing map. Let $V_{A\to BE}$ be a Stinespring
dilation of $\mathcal{N}$, i.e., $\mathcal{N}(\cdot) = \text{tr}_E(V(\cdot) V^\dagger)$, and let $\hat{\mathcal{N}}(\cdot) = \text{tr}_B(V(\cdot) V^\dagger)$. Let

$$D_A^Y = -i(|\xi\rangle\langle\psi| - |\psi\rangle\langle\xi|); \quad D_A^Z = |\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|. \qquad (E10)$$

Suppose that $|\Phi_{B:E}\rangle$ is a maximally entangled ket between two suitable subspaces of $B$ and $E$ that are sufficiently large to ensure that there exist $M$, $\Lambda$ matrices on $B$ satisfying

$$V|\phi\rangle = (\Lambda \otimes 1)|\Phi_{B:E}\rangle; \quad V|\xi\rangle = (M \otimes 1)|\Phi_{B:E}\rangle. \qquad (E11)$$

(Alternatively, one can embed both $B$ and $E$ into larger systems $B'$, $E'$ with $B' \simeq E'$, on which one can consider the canonical maximally entangled ket $|\Phi'_{B'E'}\rangle = \sum |k\rangle_{B'}|k\rangle_{E'}$ with respect to the canonical bases of $B'$, $E'$. We then define $|\Phi_{B:E}\rangle$ by projecting down $|\Phi'_{B'E'}\rangle$ onto $B \otimes E$. Through-out the following, we only ever consider operators that are in the support of the reduced operators of $\Phi_{B:E}$ on $B$ and $E$.

We define the operation $t_{B \to E}(\cdot) := \text{tr}_B\{\Phi_{B:E}[(\cdot) \otimes 1_E]\}$, which is the partial transpose operation with respect to the bases used to define $|\Phi_{B:E}\rangle$. Equivalently, a defining property of this operation is that for any operator $X_B$, we have $(X_B \otimes 1_E)|\Phi_{B:E}\rangle = (1_B \otimes t_{B \to E}(X_B))|\Phi_{B:E}\rangle$. Furthermore, for any $M$, we have $t_{B \to E}(M^\dagger) = [t_{B \to E}(M)]^\dagger$ and for any $X$, $Y$ we have $t_{B \to E}(MN) = t_{B \to E}(N) t_{B \to E}(M)$. Similarly, we define the inverse operation $t_{E \to B}(\cdot) = \text{tr}\{\Phi_{B:E}[1_B \otimes (\cdot)]\}$, which has the same properties.

Observe that $\Lambda \Lambda^\dagger = \mathcal{N}(|\psi\rangle\langle\psi|) = \rho_B$ and $MM^\dagger = \mathcal{N}(|\xi\rangle\langle\xi|)$. Furthermore, we define $W$ via the polar decomposition of $\Lambda = \rho^{1/2}W$, with

$$\begin{aligned} \Lambda &= \rho^{1/2}W; \quad \Lambda^\dagger = W^\dagger \rho^{1/2}; \\ \Lambda^{-1} &= W^\dagger \rho^{-1/2}; \quad \Lambda^{-\dagger} = \rho^{-1/2}W. \end{aligned} \qquad (E12)$$

The operators $\Lambda^{-1}$ and $\Lambda^{-\dagger}$ are the Moore-Penrose pseudoinverses of $\Lambda$ and $\Lambda^\dagger$, respectively, as can be seen by computing $\Lambda\Lambda^{-1} = P_\rho$ and $\Lambda^{-1}\Lambda = W^\dagger P_\rho W$ as well as $\Lambda^\dagger\Lambda^{-\dagger} = W^\dagger P_\rho W$ and $\Lambda^{-\dagger}\Lambda^\dagger = P_\rho$. Furthermore, we have

$$\mathcal{N}(D_A^Y) = -i(M\Lambda^\dagger - \Lambda M^\dagger), \qquad (E13a)$$

$$\begin{aligned} \hat{\mathcal{N}}(D_A^Z) &= \text{tr}_B\{M_B \Phi_{B:E} \Lambda_B^\dagger + \Lambda_B \Phi_{B:E} M_B^\dagger\} \\ &= t_{B \to E}[\Lambda^\dagger M + M^\dagger \Lambda]. \end{aligned} \qquad (E13b)$$

We may also relate these objects to the state on Eve's system, via the partial transpose operation $t_{B \to E}$. Observe that $V|\psi\rangle = (\Lambda \otimes 1)|\Phi_{B:E}\rangle = (1 \otimes t_{B \to E}(\Lambda))|\Phi_{B:E}\rangle$, and therefore $\rho_E = \text{tr}_B(V\psi V^\dagger) = [t_{B \to E}(\Lambda)][t_{B \to E}(\Lambda)]^\dagger = t_{B \to E}(\Lambda^\dagger \Lambda) = t_{B \to E}(W^\dagger \rho_B W)$. Then $P_{\rho E} = t_{B \to E}(W^\dagger P_{\rho B} W)$ and $P_{\rho E}^\perp = t_{B \to E}(W^\dagger P_{\rho B}^\perp W)$. We begin with a characterization of when our uncertainty relation holds with equality.

**Proposition 20 (Conditions for uncertainty relation equality).**—the following statements are equivalent:

(i) $\langle P_{\rho B}^\perp \otimes P_{\rho E}^\perp \rangle V|\xi\rangle = 0$.

(ii) We have $P_{\rho B}^\perp M W^\dagger P_{\rho E}^\perp = 0$.

(iii) We have $P_{\rho B}^\perp \mathcal{N}(|\xi\rangle\langle\xi|) P_{\rho B}^\perp = P_{\rho B}^\perp \mathcal{N}(|\xi\rangle\langle\psi|) P_{\rho B}^{-1} \mathcal{N}(|\psi\rangle\langle\xi|) P_{\rho B}^\perp$.

(iv) We have $P_{\rho E}^\perp \hat{\mathcal{N}}(|\xi\rangle\langle\xi|) P_{\rho E}^\perp = P_{\rho E}^\perp \hat{\mathcal{N}}(|\xi\rangle\langle\psi|) P_{\rho E}^{-1} \hat{\mathcal{N}}(|\psi\rangle\langle\xi|) P_{\rho E}^\perp$.

(v) Let $\{E_k\}$ are Kraus operators for $\mathcal{N}$. For any linear combination $E = \sum_k c_k E_k$ with $c_k \in \mathbb{C}$ and such that $E|\psi\rangle = 0$, we have $P_{\rho B}^\perp E|\xi\rangle = 0$.

Furthermore, consider the setting of Theorem 1 and suppose that $|\xi\rangle$ is defined as $|\xi\rangle = (H - \langle H \rangle)|\psi\rangle$. Then (i)–(v) are furthermore equivalent to the following.

(vi) For any eigenvalue $p_k(t)$ of $\mathcal{N}(\psi(t))$ such that $p_k(t_0) = 0$, we have $\partial_t^2 p_k(t_0) = 0$.

Observe that all the conditions above do not depend on the choice of Stinespring dilation and/or on the choice of the Kraus operator representation, as all such choices differ by a partial isometry on the $E$ system. In other words, if the conditions above hold for particular choices of $V$, $\hat{\mathcal{N}}$, and $\{E_k\}$, they hold for all other choices as well.

**Proof of Proposition 20.**—We have the following implications. (i) $\Leftrightarrow$ (ii): consider

$$\begin{aligned} (P_{\rho B}^\perp \otimes P_{\rho E}^\perp) V|\xi\rangle &= (\langle P_{\rho B}^\perp M \rangle \otimes P_{\rho E}^\perp)|\Phi\rangle_{BE} \\ &= (\langle P_{\rho B}^\perp M t_{E \to B} (P_{\rho E}^\perp) \rangle \otimes 1)|\Phi\rangle_{BE}. \end{aligned} \qquad (E14)$$

Since $V|\psi\rangle = (\Lambda \otimes 1)|\Phi\rangle = (1 \otimes t_{B \to E}(\Lambda))|\Phi\rangle$, we have $\rho_E = t_{B \to E}(\Lambda) t_{B \to E}(\Lambda)^\dagger = t_{B \to E}(\Lambda^\dagger \Lambda) = t_{B \to E}(W^\dagger \rho_B W)$. Then $P_{\rho E} = t_{B \to E}(W^\dagger P_{\rho B} W)$ and $P_{\rho E}^\perp = t_{B \to E}(W^\dagger P_{\rho B}^\perp W) = t_{B \to E}(\tilde{P}_{\rho B}^\perp)$, and

$$(F14) = (\langle P_{\rho B}^\perp M \tilde{P}_{\rho B}^\perp \rangle \otimes 1)|\Phi\rangle_{BE}. \qquad (E15)$$

Therefore, we have that $(P_{\rho B}^\perp \otimes P_{\rho E}^\perp) V|\xi\rangle = 0$ is equivalent to $0 = P_{\rho B}^\perp M \tilde{P}_{\rho B}^\perp$.

(ii) $\Rightarrow$ (iii): let $K = P_\rho^\perp \mathcal{N}(|\xi\rangle\langle\psi|) P_B^{-1/2} = P_\rho^\perp \text{tr}_E(M\Phi_{BE}\Lambda^\dagger) P_B^{-1/2} = P_\rho^\perp M W^\dagger P_\rho$. Now assume that (ii) holds; then $K = P_\rho^\perp MW^\dagger$ and we have $KK^\dagger = P_\rho^\perp MM^\dagger P_\rho^\perp = P_\rho^\perp \mathcal{N}(|\xi\rangle\langle\xi|) P_\rho^\perp$, showing (iii).

(iii) $\Rightarrow$ (ii): conversely, assuming (iii) and if $K = P_{\rho B}^\perp \mathcal{N}(|\xi\rangle\langle\psi|) P_B^{-1/2} = P_\rho^\perp M W^\dagger P_\rho$, we have by assumption that $KK^\dagger = P_\rho^\perp MM^\dagger P_\rho^\perp = (P_\rho^\perp M W^\dagger P_\rho)(P_\rho M W^\dagger P_\rho)^\dagger + P_\rho^\perp M W^\dagger P_\rho^\perp W M^\dagger P_\rho^\perp$. This means that $0 = (P_\rho^\perp M W^\dagger P_\rho^\perp)(P_\rho^\perp W M^\dagger P_\rho^\perp)$. The latter equation can only hold if $P_\rho^\perp M W^\dagger P_\rho^\perp = 0$, showing (ii).
(i) ⇔ (iv): condition (i) is symmetric if we replace $B \leftrightarrow E$ (and correspondingly $\mathcal{N} \leftrightarrow \hat{\mathcal{N}}$), meaning that the condition holds if and only if the condition with $B$ and $E$ swapped also holds. Therefore, we can swap $B \leftrightarrow E$ in the other conditions and those will also hold if and only if (i) holds. Condition (iv) is obtained by performing this transformation on (iii).

(i) ⇒ (v): we choose the representation $V = \sum E_k \otimes |k\rangle_E$ and assume (i), i.e., that we have

$$ (P_{\rho B}^{\perp} \otimes P_{\rho E}^{\perp}) V |\xi\rangle = 0. \qquad (E16) $$

Let $\{c_k\}$ with $c_k \in \mathbb{C}$ such that $\sum_k c_k E_k |\psi\rangle = 0$ and let $E = \sum_k c_k E_k$. Define $|e\rangle_E = \sum_k c_k^* |k\rangle_E$. We have

$$ \begin{aligned} \langle e | \hat{\mathcal{N}}(\psi) | e \rangle &= \sum_{k,k'} \langle e | k \rangle \operatorname{tr}(E_k \psi E_{k'}^{\dagger}) \langle k' | e \rangle \\ &= \operatorname{tr} \left[ \left( \sum c_k E_k \right) \psi \left( \sum c_k E_k \right)^{\dagger} \right] \\ &= \| E | \psi \rangle \|^2 = 0, \end{aligned} \qquad (E17) $$

which implies that $|e\rangle_E \in \ker \hat{\mathcal{N}}(\psi)$, i.e., $P_{\rho E}^{\perp} |e\rangle_E = |e\rangle_E$. Applying $(1 \otimes \langle e|)$ onto Eq. (E16) we find

$$ 0 = (P_{\rho B}^{\perp} \otimes \langle e | P_{\rho E}^{\perp} | V \rangle) \langle e | \xi \rangle = \sum_k (P_{\rho B}^{\perp} E_k | \xi \rangle) \langle e | k \rangle = P_{\rho B}^{\perp} E | \xi \rangle, \qquad (E18) $$

showing that (v) holds.

(i) ⇔ (v): we now suppose that condition (v) holds. Let $|\chi_j\rangle_E$ be a set of orthonormal states that span the support of $P_{\rho E}^{\perp}$, i.e., $P_{\rho E}^{\perp} = \sum_j |\chi_j\rangle \langle \chi_j|$. Fix any such $|\chi_j\rangle$ and define $E^{(j)} = \sum_j \langle \chi_j | k \rangle E_k$. We repeat Eq. (E17) by replacing $|e\rangle \rightarrow |\chi_j\rangle$, $c_k \rightarrow \langle \chi_j | k \rangle$ to find

$$ 0 = \langle \chi_j | \hat{\mathcal{N}}(\psi) | \chi_j \rangle = \dots = \| E^{(j)} | \psi \rangle \|^2, \qquad (E19) $$

which implies that $E^{(j)}|\psi\rangle = 0$. We use the assumption that (v) holds to deduce that $P_{\rho B}^{\perp} E^{(j)} |\xi\rangle = 0$; we note the latter expression holds for all $j$ by repeating this argument for each $j$ individually. Then

$$ \begin{aligned} (P_{\rho B}^{\perp} \otimes P_{\rho E}^{\perp}) V |\xi\rangle &= (P_{\rho B}^{\perp} \otimes \sum |\chi_j\rangle \langle \chi_j|) V |\xi\rangle \\ &= \sum_{k,j} (P_{\rho B}^{\perp} E_k |\xi\rangle) \otimes (|\chi_j\rangle \langle \chi_j| k) \\ &= \sum_j (P_{\rho B}^{\perp} E^{(j)} |\xi\rangle) \otimes |\chi_j\rangle = 0, \end{aligned} \qquad (E20) $$

showing that (i) holds.

(iii) ⇔ (vi): now consider the setting of Theorem 1 and suppose that $|\xi\rangle$ is defined as $|\xi\rangle = (H - \langle H \rangle)|\psi\rangle$. We

invoke Ref. [33, Eq. (B15)], which in the present context reads

$$ \operatorname{tr}(P_{\rho}^{\perp} \partial_{t}^{2} \rho) = \sum_{k: p_{k}=0} \partial_{t}^{2} p_{k} + 2 \sum_{\substack{k, \ell: \\ p_{k}>0 \\ p_{\ell}=0}} \frac{|\langle \lambda_{k} | \partial_{t} \rho | \lambda_{\ell} \rangle|^{2}}{p_{k}}, \qquad (E21) $$

where $\{|\lambda_k\rangle\}$ is a complete eigenbasis of $\rho$ with eigenvalues $p_k$. Using Eq. (11) one can check that the second term on the right-hand side satisfies

$$ \begin{aligned} 2 \sum_{\substack{k, \ell: \\ p_k>0 \\ p_\ell=0}} \frac{|\langle \lambda_k | \partial_t \rho | \lambda_\ell \rangle|^2}{p_k} &= 2 \operatorname{tr}\{\rho^{-1} (\partial_t \rho) P_\rho^\perp (\partial_t \rho)\} \\ &= 2 \operatorname{tr}\{\rho^{-1} \mathcal{N}(|\psi\rangle\langle\xi|) P_\rho^\perp \mathcal{N}(|\xi\rangle\langle\psi|)\}, \end{aligned} \qquad (E22) $$

using the fact that $\partial_t \rho = \mathcal{N}(-i[H, \psi]) = \mathcal{N}(-i|\xi\rangle\langle\psi| + i|\psi\rangle\langle\xi|)$ and that $\mathcal{N}(X\psi)P_\rho^\perp = 0$ for any $X$. On the other hand, we can see that $\partial_t^2 \rho = \partial_t \mathcal{N}(-i[H, \psi]) = \mathcal{N}(-i[\partial_t H, \psi] - [H, [H, \psi]])$, and recalling that $\mathcal{N}(X\psi)P_\rho^\perp = 0$ for any $X$ we obtain

$$ \begin{aligned} \operatorname{tr}(P_{\rho}^{\perp} \partial_{t}^{2} \rho) &= \operatorname{tr}(P_{\rho}^{\perp} \mathcal{N}(2H\psi H)) = 2 \operatorname{tr}(P_{\rho}^{\perp} \mathcal{N}(\bar{H}\psi \bar{H})) \\ &= 2 \operatorname{tr}(P_{\rho}^{\perp} \mathcal{N}(|\xi\rangle\langle\xi|)), \end{aligned} \qquad (E23) $$

writing $\bar{H} = H - \langle H \rangle_\psi$ and where $|\xi\rangle = \bar{H}|\psi\rangle$. Now suppose that (iii) holds. Then

$$ (E23) = 2 \operatorname{tr}(P_{\rho}^{\perp} \mathcal{N}(|\xi\rangle\langle\psi|) \rho^{-1} \mathcal{N}(|\psi\rangle\langle\xi|)) = (E22), \qquad (E24) $$

and therefore the first term on the right-hand side of Eq. (E21) must vanish, and since $\partial_t^2 p_k \ge 0$ for all $k$ for which $p_k = 0$ as $p_k$ reaches a minimum at that point, we must necessarily have that $\partial_t^2 p_k = 0$ for all those $k$.

Conversely, if the first term on the right-hand side of Eq. (E21) vanishes, then we have

$$ \operatorname{tr}(P_{\rho}^{\perp} \mathcal{N}(|\xi\rangle\langle\psi|) \rho^{-1} \mathcal{N}(|\psi\rangle\langle\xi|)) = \operatorname{tr}(P_{\rho}^{\perp} \mathcal{N}(|\xi\rangle\langle\xi|)). \qquad (E25) $$

By applying the completely positive map $\mathrm{id}_2 \otimes \mathcal{N}$ onto the matrix $\begin{pmatrix} |\psi\rangle\langle\psi| & |\psi\rangle\langle\xi| \\ |\xi\rangle\langle\psi| & |\xi\rangle\langle\xi| \end{pmatrix}$ and further conjugating by
$\begin{pmatrix} 1 \\ P_{\rho}^{\perp} \end{pmatrix}$ we find that

$$ \begin{bmatrix} \rho & \mathcal{N}(|\psi\rangle\langle\xi|)P_{\rho}^{\perp} \\ P_{\rho}^{\perp}\mathcal{N}(|\xi\rangle\langle\psi|) & P_{\rho}^{\perp}\mathcal{N}(|\xi\rangle\langle\xi|)P_{\rho}^{\perp} \end{bmatrix} \ge 0. \quad (E26) $$

From the Schur complement (Theorem 5) we find that

$$ P_{\rho}^{\perp}[\mathcal{N}(|\xi\rangle\langle\xi|) - \mathcal{N}(|\xi\rangle\langle\psi|) \rho^{-1} \mathcal{N}(|\psi\rangle\langle\xi|)]P_{\rho}^{\perp} \ge 0. \quad (E27) $$

But a positive semidefinite operator has trace zero if and only if it is identically equal to zero, so with Eq. (E25) we find that $P_{\rho}^{\perp}[\mathcal{N}(|\xi\rangle\langle\xi|) - \mathcal{N}(|\xi\rangle\langle\psi|) \rho^{-1} \mathcal{N}(|\psi\rangle\langle\xi|)]P_{\rho}^{\perp} = 0$, showing that (iii) holds. ■

Our main technical theorem is the following.

**Theorem 6 (Time-energy uncertainty relation in the virtual metrological qubit picture).**—let A, B, and E be finite-dimensional quantum systems. Let $\mathcal{N}_{A \to B}$ be a completely positive, trace-non-increasing map. Let $V_{A \to BE}$ be such that $\mathcal{N}_{A \to B}(\cdot) = \text{tr}_E(V(\cdot)V^\dagger)$ and $V^\dagger V \le 1$, i.e., V is

a Stinespring dilation of $\mathcal{N}$. Let $\tilde{\mathcal{N}}_{A \to E}(\cdot) = \text{tr}_B(V(\cdot)V^\dagger)$. Let $|\psi\rangle$ be any subnormalized state on A, and let $|\xi\rangle$ be any vector on A such that $\langle\psi|\xi\rangle = 0$. Define $D_A^Y = -i(|\xi\rangle\langle\psi| - |\psi\rangle\langle\xi|)$ and $D_A^Z = |\xi\rangle\langle\psi| + |\psi\rangle\langle\xi|$. Then

$$ \begin{aligned} & F(\mathcal{N}(\psi); \mathcal{N}(D_A^Y)) + F(\tilde{\mathcal{N}}(\psi); \tilde{\mathcal{N}}(D_A^Z)) \\ & \le 4|\xi|\mathcal{N}^{\dagger}(1)|\xi|. \end{aligned} \quad (E28) $$

Furthermore, if $(P_{\rho B}^{\perp} \otimes P_{\rho E}^{\perp})V|\xi\rangle = 0$, then equality holds.

First, we remark that both Fisher information expressions in Eq. (E28) are well defined in that we always have $P_{\tilde{\mathcal{N}}(\psi)}^{\perp}\mathcal{N}(D_A^Y)P_{\tilde{\mathcal{N}}(\psi)}^{\perp} = 0$ and $P_{\tilde{\mathcal{N}}(\psi)}^{\perp}\tilde{\mathcal{N}}(D_A^Z)P_{\tilde{\mathcal{N}}(\psi)}^{\perp} = 0$ as required in the definition (C1). These conditions can be verified by first noting that the following matrix is positive semidefinite:

$$ \begin{bmatrix} |\psi\rangle\langle\psi| & |\psi\rangle\langle\xi| \\ |\xi\rangle\langle\psi| & |\xi\rangle\langle\xi| \end{bmatrix} = \begin{bmatrix} |\psi\rangle \\ |\xi\rangle \end{bmatrix} [\langle\psi| \quad |\xi|] \ge 0, \quad (E29) $$

and applying either completely positive map $\text{id}_2 \otimes \mathcal{N}$ or $\text{id}_2 \otimes \tilde{\mathcal{N}}$ to obtain

$$ \begin{bmatrix} \mathcal{N}(|\psi\rangle\langle\psi|) & \mathcal{N}(|\psi\rangle\langle\xi|) \\ \mathcal{N}(|\xi\rangle\langle\psi|) & \mathcal{N}(|\xi\rangle\langle\xi|) \end{bmatrix} \ge 0; \quad \begin{bmatrix} \tilde{\mathcal{N}}(|\psi\rangle\langle\psi|) & \tilde{\mathcal{N}}(|\psi\rangle\langle\xi|) \\ \tilde{\mathcal{N}}(|\xi\rangle\langle\psi|) & \tilde{\mathcal{N}}(|\xi\rangle\langle\xi|) \end{bmatrix} \ge 0. \quad (E30) $$

Then, Theorem 5 ensures that $P_{\tilde{\mathcal{N}}(\psi)}^{\perp}\mathcal{N}(|\psi\rangle\langle\xi|) = 0$ and therefore $P_{\tilde{\mathcal{N}}(\psi)}^{\perp}\mathcal{N}(D_A^Y)P_{\tilde{\mathcal{N}}(\psi)}^{\perp} = 0$; likewise $P_{\tilde{\mathcal{N}}(\psi)}^{\perp}\tilde{\mathcal{N}}(D_A^Z)P_{\tilde{\mathcal{N}}(\psi)}^{\perp} = 0$.

**Proof of Theorem 6.**—Let $\Lambda, M$ be operators acting on B such that $V|\psi\rangle = (\Lambda \otimes 1)|\Phi\rangle$ and $V|\xi\rangle = (M \otimes 1)|\Phi\rangle$. We can write

$$ \begin{aligned} D_B &= \mathcal{N}(D_A^Y) = \text{tr}_E(-i(V|\xi\rangle\langle\psi|V^\dagger - V|\psi\rangle\langle\xi|V^\dagger)) \\ &= -i(M\Lambda^\dagger - \Lambda M^\dagger), \end{aligned} \quad (E31a) $$

$$ \begin{aligned} \hat{D}_E &= \tilde{\mathcal{N}}(D_A^Z) = \text{tr}_B(V|\xi\rangle\langle\psi|V^\dagger + V|\psi\rangle\langle\xi|V^\dagger) \\ &= \text{tr}_B(M\Phi\Lambda^\dagger + \Lambda\Phi M^\dagger); \end{aligned} \quad (E31b) $$

where in Eq. (E31b) the operators M, $\Lambda$ act only on B with a tensor product with the identity on E implied but $\Phi = \Phi_{BE} = |\Phi\rangle\langle\Phi|_{BE}$. Now consider

$$ \frac{1}{4} \left\{ \begin{aligned} & 4\langle\xi|\mathcal{N}^{\dagger}(1)|\xi\rangle - F(\rho_E; \hat{D}_E) \\ & - \max_{S_E=S_E^{\dagger}} \left\{ \text{tr}(\hat{D}_E S_E) - \text{tr}(\rho_E S_E^2) \right\} \end{aligned} \right\} = \text{tr}(MM^{\dagger}) \quad (E32) $$

using Eq. (C3a) and noting that $\langle\xi|\mathcal{N}^{\dagger}(1)|\xi\rangle = \text{tr}(\mathcal{N}(|\xi\rangle\langle\xi|)) = \text{tr}(MM^{\dagger})$. Then, using Eq. (E31b), and writing $t(\cdot) = t_{E \to B}(\cdot)$ as a shorthand,

$$ \begin{aligned} (E32) &= \min_{S_E=S_E^{\dagger}} \left\{ \text{tr}(MM^{\dagger}) - \text{tr}\left((M\Phi\Lambda^{\dagger} + \Lambda\Phi M^{\dagger})S_E\right) + \text{tr}\left(S_E\Lambda\Phi\Lambda^{\dagger}S_E\right) \right\} \\ &= \min_{S_E=S_E^{\dagger}} \left\{ \text{tr}(MM^{\dagger}) - \text{tr}\left(M(t(S_E)\Lambda^{\dagger} + \Lambda t(S_E)M^{\dagger}) + \text{tr}\left(\Lambda(t(S_E))^2\Lambda^{\dagger}\right) \right) \right\} \\ &= \min_{S'=S^{\dagger}} \left\{ \text{tr}(MM^{\dagger}) - \text{tr}\left(MS'\Lambda^{\dagger} + \Lambda S'M^{\dagger}\right) + \text{tr}\left(\Lambda S'^2\Lambda^{\dagger}\right) \right\}, \\ &= \min_{S'=S^{\dagger}} \text{tr}\left((M - \Lambda S')(M - \Lambda S')^{\dagger}\right), \end{aligned} \quad (E33) $$
where the optimization now ranges over all Hermitian operators $S'$ acting on $B$. On the other hand, using Eq. (C3b),

$$ \frac{1}{4} F(\rho; D) = \min \left\{ \text{tr}(L^\dagger L) : \rho^{1/2} L + L^\dagger \rho^{1/2} = D \right\}, \quad (E34) $$

where $\rho, D$ refer to operators on $B$. To prove the inequality (E28), which is the first part of our main theorem claim, our strategy is to show that for any candidate $S'$ in Eq. (E33), there is a valid candidate $L$ in Eq. (E34) that achieves the same value. This statement then implies that (E34) $\le$ (E33) as desired.

Recall that $\Lambda\Lambda^\dagger = \rho$ (where $\rho \equiv \rho_B$ for short in this proof), and therefore the polar decomposition of $\Lambda$ can be written as $\Lambda = \rho^{1/2}W$ for some unitary matrix $W$. Let $S'$ be any Hermitian operator that is candidate in the optimization (E33), and let $L = iW(M^\dagger - S'\Lambda^\dagger)$. Then one can verify that

$$ \begin{aligned} \rho^{1/2} L + L^\dagger \rho^{1/2} &= i\Lambda(M^\dagger - S'\Lambda^\dagger) - i(M - \Lambda S')\Lambda^\dagger \\ &= -i(M\Lambda^\dagger - \Lambda M^\dagger) = D, \end{aligned} \quad (E35) $$

and thus $L$ is a feasible candidate in Eq. (E34). Furthermore it holds that $\text{tr}(L^\dagger L) = \text{tr}((M - \Lambda S')(M - \Lambda S')^\dagger)$, thus proving the inequality (E28).

We now show that, assuming $(P_{\rho_B}^{\perp} \otimes P_{\rho_E}^{\perp})V|\xi\rangle = 0$, the inequality becomes an equality. The proof strategy is to go in reverse direction above, starting with an optimal candidate $L$ in Eq. (E34), and constructing a candidate $S'$ in Eq. (E33) that achieves the same value. From Proposition 20 we see that $(P_{\rho_B}^{\perp} \otimes P_{\rho_E}^{\perp})V|\xi\rangle = 0$ is equivalent to

$$ P_{\rho}^{\perp} M W^{\dagger} P_{\rho}^{\perp} = 0. \quad (E36) $$

Let $L$ be an optimal candidate in Eq. (E34), i.e., such that $\rho^{1/2}L + L^\dagger\rho^{1/2} = D$ and $F(\rho; D) = 4\text{tr}(L^\dagger L)$. Without loss of generality, we may assume that $P_\rho L = L$, since otherwise $P_\rho L$ would yield a better optimization candidate in Eq. (E34). Denoting by $P_X^{\text{supp}}$ and $P_X^{\text{mg}}$ the projectors onto the support and the range of an operator $X$, and defining $\tilde{P}_\rho = W^\dagger P_\rho W$, we have

$$ \begin{aligned} P_{\Lambda}^{\text{mg}} &= P_{\Lambda^\dagger}^{\text{supp}} = P_\rho, & P_{\Lambda}^{\text{supp}} &= P_{\Lambda}^{\text{mg}} = W^\dagger P_\rho W = \tilde{P}_\rho, \\ P_{\rho}^{\perp} &= 1 - P_\rho, & \tilde{P}_{\rho}^{\perp} &= 1 - \tilde{P}_\rho. \end{aligned} \quad (E37) $$

Let us compute the object $LP_\rho^\perp$:

$$ \begin{aligned} LP_\rho^\perp &= P_\rho LP_\rho^\perp = \rho^{-1/2}(\rho^{1/2}L + L^\dagger\rho^{1/2})P_\rho^\perp \\ &= \rho^{-1/2}(-i(M\Lambda^\dagger - \Lambda M^\dagger))P_\rho^\perp \\ &= i\rho^{-1/2}(\Lambda M^\dagger)P_\rho^\perp = iP_\rho WM^\dagger P_\rho^\perp = iW\tilde{P}_\rho M^\dagger P_\rho^\perp \\ &= iWM^\dagger P_\rho^\perp, \end{aligned} \quad (E38) $$

where we have employed Eq. (E36) in the last equality.

Now let us get started with constructing $S'$. Our goal is to find a Hermitian matrix $S'$ such that

$$ L \stackrel{!}{=} iW(M^\dagger - S'\Lambda^\dagger). \quad (E39) $$

Indeed, this would ensure a valid candidate in Eq. (E33) reaching the same value as $\text{tr}(L^\dagger L)$. The equality (E39) is equivalent to both simultaneous conditions

$$ LP_\rho \stackrel{!}{=} iW(M^\dagger - S'\Lambda^\dagger)P_\rho; \quad LP_\rho^\perp \stackrel{!}{=} iW(M^\dagger - S'\Lambda^\dagger)P_\rho^\perp. \quad (E40) $$

The latter follows immediately from Eq. (E38), noting that $\Lambda^\dagger P_\rho^\perp = 0$. It suffices, therefore, to find a Hermitian matrix $S'$ such that the first equality in Eq. (E40) is satisfied.

Let $\Lambda^{-1} = W^\dagger\rho^{-1/2}$ noting that $\Lambda^{-1}\Lambda = \tilde{P}_\rho$ and $\Lambda\Lambda^{-1} = P_\rho$. Define

$$ S' = \Lambda^{-1} \left[ \Lambda M^\dagger + i \Lambda W^\dagger L \right] (\Lambda^{-1})^\dagger + \tilde{P}_\rho^\perp M^\dagger (\Lambda^{-1})^\dagger \\ + \Lambda^{-1} M \tilde{P}_\rho^\perp. \quad (E41) $$

First we show that $S'$ is Hermitian by proving that the term in brackets in the first term above is, in fact, Hermitian. Using $\Lambda W^\dagger = \rho^{1/2}$ we can compute

$$ \begin{aligned} & \left[ \Lambda M^\dagger + i \Lambda W^\dagger L \right] - \left[ \Lambda M^\dagger + i \Lambda W^\dagger L \right]^\dagger \\ &= (\Lambda M^\dagger - M \Lambda^\dagger) + i(\rho^{1/2} L + L^\dagger \rho^{1/2}) \\ &= -iD + iD = 0, \end{aligned} \quad (E42) $$

using properties of $L$ noted above and using Eq. (E31a). Therefore, $S'$ is Hermitian. Then

$$ \begin{aligned} iW(M^\dagger - S'\Lambda^\dagger)P_\rho = iWM^\dagger P_\rho - iW\tilde{P}_\rho M^\dagger P_\rho \\ + P_\rho LP_\rho - iW\tilde{P}_\rho^\perp M^\dagger P_\rho = LP_\rho, \end{aligned} \quad (E43) $$

noting that $\tilde{P}_\rho^\perp \Lambda^\dagger = 0$, $(\Lambda^{-1})^\dagger \Lambda^\dagger = (\Lambda\Lambda^{-1})^\dagger = P_\rho$, and recalling that $P_\rho L = L$. With this choice of $S'$, the first equality in Eq. (E40) is thus also satisfied, thereby completing the proof. ■
### 3. Additional equivalent conditions for zero sensitivity loss

The following theorem provides additional conditions under which zero sensitivity loss is achieved (see Sec. VII), leading to an explicit form of Bob's optimal sensing observable whenever these conditions are satisfied.

**Theorem 7.**—we use the notation of Appendix E 2. Suppose that the conditions for our uncertainty relation equality (Proposition 20) hold. Then the following statements are equivalent:

(i) We have $F(\mathcal{N}(\psi); \mathcal{N}(D_A^Y)) = 4\langle\xi|\mathcal{N}^\dagger(1)|\xi\rangle$.

(ii) We have $\text{tr}(E_k^\dagger E_k D_A^Z) = 0$ for all $k, k'$, where $\{E_k\}$ is any set of Kraus operators for $\mathcal{N}$.

(iii) We have $\hat{\mathcal{N}}(D_A^Z) = 0$.

(iv) We have $\Lambda^\dagger M + M^\dagger \Lambda = 0$.

(v) The operator $i\rho^{1/2}MW^\dagger$ is Hermitian.

(vi) The operator $i\rho \mathcal{N}(|\xi\rangle\langle\psi|)$ is Hermitian and $\mathcal{N}(|\xi\rangle\langle\xi|) = \mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|)$.

(vii) The operator $i\rho \mathcal{N}(|\xi\rangle\langle\psi|)$ is Hermitian and $\langle\xi|\mathcal{N}^\dagger(1)|\xi\rangle = \text{tr}[\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|)]$.

Furthermore, if these conditions are satisfied then

$$ \mathcal{R}_{\rho B}^{-1}(\mathcal{N}(D_A^Y)) = -2i\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1} + 2i\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|)P_{\rho}^{\perp}. \quad (E44) $$

*Proof.*—The proof of (i) $\Leftrightarrow$ (ii) $\Leftrightarrow$ (iii) is presented in the main text (Sec. VII).

(iii) $\Rightarrow$ (iv) : write $0 = \hat{\mathcal{N}}(D_A^Z) = \text{tr}_B\{\Phi_{B:E}[\Lambda^\dagger M + M^\dagger \Lambda]\}$. Observe that $\text{tr}_B\{\Phi_{B:E}(\cdot)\}$ is the partial transpose map with respect to the bases used to define $\Phi_{B:E}$; therefore, $\Lambda^\dagger M + M^\dagger \Lambda = 0$.

(iv) $\Leftrightarrow$ (v) : we compute

$$ i\rho^{1/2}MW^\dagger - (i\rho^{1/2}MW^\dagger)^\dagger = iW(\Lambda^\dagger M + M^\dagger \Lambda)W^\dagger, \quad (E45) $$

which vanishes thanks to the assumption that (iv) holds. Conversely, because $W$ is unitary we may only have Eq. (E45) $= 0$ if $\Lambda^\dagger M + M^\dagger \Lambda = 0$.

(iv) $\Rightarrow$ (vi) : recall that $\rho = \Lambda\Lambda^\dagger$ and $\mathcal{N}(|\xi\rangle\langle\psi|) = \text{tr}_E(|\xi\rangle\langle\psi|V^\dagger) = M\Lambda^\dagger$. Then $i\rho\mathcal{N}(|\xi\rangle\langle\psi|) = i\Lambda\Lambda^\dagger M\Lambda^\dagger$. To check that $i\rho\mathcal{N}(|\xi\rangle\langle\psi|)$ is Hermitian we compute

$$ \begin{aligned} & i\rho\mathcal{N}(|\xi\rangle\langle\psi|) - (i\rho\mathcal{N}(|\xi\rangle\langle\psi|))^\dagger \\ & = i\Lambda(\Lambda^\dagger M + M^\dagger \Lambda)\Lambda^\dagger = 0, \end{aligned} \quad (E46) $$

using our assumption that (iv) holds. Furthermore, we have

$$ 0 = \rho^{-1/2}W(\Lambda^\dagger M + M^\dagger \Lambda)W^\dagger P_\rho^\perp = P_\rho MW^\dagger P_\rho^\perp; \quad (E47) $$

recalling point (ii) of Proposition 20, we find that

$$ MW^\dagger P_\rho^\perp = 0. \quad (E48) $$

Then

$$ \begin{aligned} \mathcal{N}(|\xi\rangle\langle\xi|) &= MM^\dagger = MW^\dagger(P_\rho + P_\rho^\perp)WM^\dagger \\ &= \mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|). \end{aligned} \quad (E49) $$

(vi) $\Rightarrow$ (vii) : this implication follows immediately from $\langle\xi|\mathcal{N}^\dagger(1)|\xi\rangle = \text{tr}(\mathcal{N}(|\xi\rangle\langle\xi|))$.

(vii) $\Rightarrow$ (i) : our proof strategy for this implication is to show that the expression of the symmetric logarithmic derivative in Eq. (E44) is correct, and that the corresponding Fisher information at Bob's end has no sensitivity loss. Let

$$ R = -2i\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1} + 2i\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|)P_\rho^\perp. \quad (E50) $$

We can see that $R$ is Hermitian by writing

$$ \begin{aligned} R &= -2i(P_\rho + P_\rho^\perp)\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1} + 2i\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|)P_\rho^\perp \\ &= -2i\rho^{-1}[\rho\mathcal{N}(|\xi\rangle\langle\psi|)]\rho^{-1} + (-2iP_\rho^\perp\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1} \\ &\qquad + \text{h.c.}). \end{aligned} \quad (E51) $$

The first term is Hermitian by assumption and the second term is manifestly Hermitian. We note for convenience that $RP_\rho = -2i\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1}$ and $P_\rho R = 2i\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|)$. We can compute

$$ \frac{1}{2}(\rho R + R\rho) = i\mathcal{N}(|\psi\rangle\langle\xi|) - i\mathcal{N}(|\xi\rangle\langle\psi|) = \mathcal{N}(D_A^Y). \quad (E52) $$

Combining with the fact that $P_\rho^\perp RP_\rho^\perp = 0$ we have that $\mathcal{R}_\rho^{-1}(\mathcal{N}(D_Y)) = R$ (see also Proposition 3), thus proving Eq. (E44). The Fisher information at the output of the mapping $\mathcal{N}$ is therefore

$$ \begin{aligned} F(\mathcal{N}(\psi); \mathcal{N}(D_A^Y)) &= \text{tr}(\rho R^2) \\ &= \text{tr}[\rho (2i\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|))(-2i\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1})] \\ &= 4\text{tr}(\mathcal{N}(|\xi\rangle\langle\psi|)\rho^{-1}\mathcal{N}(|\psi\rangle\langle\xi|)) \\ &= 4\langle\xi|\mathcal{N}^\dagger(1)|\xi\rangle. \end{aligned} \quad (E53) $$

We conclude that (i) holds.

<figure><img src="image_17.png" /></figure>

### 4. Proof of the generalized bipartite Fisher information uncertainty relation for any two parameters

In this Appendix, we prove the generalized uncertainty relation (68) that applies to any two parameters generated by unitary evolutions.

**Proposition 21 (Uncertainty relation for any two parameters with associated generators).**—let $|\psi\rangle$ be a state vector on Alice's system, and let $A, B$ be two Hermitian
operators. The latter generate two respective parametrized evolutions

$$ \partial_a \psi = -i[A, \psi]; \quad \partial_b \psi = -i[B, \psi]. \qquad (E54) $$

Consider the setting depicted in Fig. 1, where $\mathcal{N}$ can be any completely positive, trace-non-increasing map. Then

$$ \frac{F_{\text{Bob},a}}{F_{\text{Alice},a}} + \frac{F_{\text{Eve},b}}{F_{\text{Alice},b}} \le 1 + 2 \sqrt{1 - \frac{\langle i[A, B] \rangle^2}{4 \sigma_A^2 \sigma_B^2}}. \quad (E55) $$

Furthermore, assume that $\mathcal{N}[\psi(a)]$ does not change rank locally as a function of $a$ and that there exists $\beta \in \mathbb{R}, \beta \ne 0$ such that

$$ \tilde{\mathcal{N}}\left(-i\left[\frac{B}{\sigma_B}, \psi\right]\right) = \beta \tilde{\mathcal{N}}\left(\left\{\frac{A - \langle A \rangle}{\sigma_A}, \psi\right\}\right). \quad (E56) $$

Then

$$ \frac{F_{\text{Bob},a}}{F_{\text{Alice},a}} + \frac{1}{\beta^2} \frac{F_{\text{Eve},b}}{F_{\text{Alice},b}} = 1. \quad (E57) $$

Corollary 3 (Uncertainty relation for any two parameters).—let $|\psi(a, b)\rangle$ be any state vector depending on parameters $a, b$. Then

$$ \frac{F_{\text{Bob},a}}{F_{\text{Alice},a}} + \frac{F_{\text{Eve},b}}{F_{\text{Alice},b}} \le 1 + 2 \sqrt{1 - \frac{\langle i[\partial_a \psi, \partial_b \psi] \rangle^2}{4 \langle (\partial_a \psi)^2 \rangle \langle (\partial_b \psi)^2 \rangle}}. \quad (E58) $$

We first prove the following lemma.

Lemma 2.—let $|\psi\rangle$ be any state vector and let $\mathcal{M}$ be any completely positive, trace-non-increasing map. Consider two Hermitian operators $C, B$ generating respective evolutions

$$ \partial_c \psi = -i[C, \psi], \quad \partial_b \psi = -i[B, \psi]. \quad (E59) $$

We write $\rho = \mathcal{M}(\psi)$, $\partial_c\rho = \mathcal{M}(-i[C, \psi])$ and $\partial_b\rho = \mathcal{M}(-i[B, \psi])$. Then for any $x, y > 0$,

$$ \frac{y}{\sigma_B^2} F(\rho; \partial_b \rho) \le \frac{x}{\sigma_C^2} F(\rho; \partial_c \rho) + 4(x+y) \sqrt{1 - \frac{xy}{(x+y)^2} \frac{4[\text{Re}\langle \tilde{C}\tilde{B} \rangle]^2}{\sigma_C^2 \sigma_B^2}}, \quad (E60) $$

where $\tilde{C} = C - \langle C \rangle \mathbf{1}$ and $\tilde{B} = B - \langle B \rangle \mathbf{1}$. In addition, suppose that $C$ can be written as $C = i\alpha[A, \psi]$ for some

Hermitian operator $A$ and some $\alpha \in \mathbb{R}$. Then the above inequality takes the form

$$ \frac{y}{\sigma_B^2} F(\rho; \partial_b \rho) \le \frac{x}{\alpha^2 \sigma_A^2} F(\rho; \partial_c \rho) + 4(x+y) \sqrt{1 - \frac{xy}{(x+y)^2} \frac{\langle i[A, B] \rangle^2}{\sigma_A^2 \sigma_B^2}}. \quad (E61) $$

Furthermore, let $x, y > 0$. If there exists $s \in \{+1, -1\}$ such that

$$ \mathcal{M}\left(-i\left[\frac{\sqrt{y}B}{\sigma_B} + s\frac{\sqrt{x}C}{\sigma_C}, \psi\right]\right) = 0, \quad (E62) $$

then

$$ \frac{y}{\sigma_B^2} F(\rho; \partial_b \rho) = \frac{x}{\sigma_C^2} F(\rho; \partial_c \rho). \quad (E63) $$

Proof of Lemma 2.—For any $x, y > 0$, define the short-hands

$$ \tilde{C} = \frac{\sqrt{x}}{\sigma_C} (C - \langle C \rangle), \quad \tilde{B} = \frac{\sqrt{y}}{\sigma_B} (B - \langle B \rangle). \quad (E64) $$

Observe that $\sigma_C^2 = x$ and $\sigma_B^2 = y$. Furthermore, we define for convenience $D_c = \mathcal{M}(-i[\cdot, \psi])$, observing that $D_C = \partial_c\rho$ and $D_B = \partial_b\rho$. Then using Proposition 8 we see that

$$ F(\rho; D_{\tilde{C}}) = \frac{x}{\sigma_C^2} F(\rho; D_C); \quad F(\rho; D_{\tilde{B}}) = \frac{y}{\sigma_B^2} F(\rho; D_B). \quad (E65) $$

Invoking Proposition 16,

$$ F(\rho; D_{\tilde{B}}) \le F(\rho; D_{\tilde{C}}) + [F(\rho; \Delta_+)F(\rho; \Delta_-)]^{1/2}, \quad (E66) $$

where $\Delta_{\pm} = D_{\tilde{C}} \pm D_{\tilde{B}} = D_{\tilde{C}\pm\tilde{B}}$. We proceed to compute the second term on the right-hand side of this inequality. The data-processing inequality (Proposition 13), along with Proposition 12, gives us

$$ F(\rho; \Delta_{\pm}) \le F(\psi; -i[\tilde{C} \pm \tilde{B}, \psi]) = 4\text{Var}_{\psi}(\tilde{C} \pm \tilde{B}), \quad (E67) $$

where we write $\text{Var}_{\rho}(X) = \langle X^2 \rangle_{\rho} - (\langle X \rangle_{\rho})^2$. We find

$$ \begin{aligned} 4\text{Var}_{\psi}(\tilde{C} \pm \tilde{B}) &= 4 \langle (\tilde{C} \pm \tilde{B})^2 \rangle = 4 \langle \tilde{C}^2 + \tilde{B}^2 \pm \{\tilde{C}, \tilde{B}\} \rangle \\ &= 4(x+y) \pm 8 \text{Re} \langle \tilde{C}\tilde{B} \rangle. \end{aligned} \quad (E68) $$
Then

$$4^2 \mathrm{Var}_{\psi}(\tilde{C} + \tilde{B}) \mathrm{Var}_{\psi}(\tilde{C} - \tilde{B}) = 4^2 (x+y)^2 - 8^2 \frac{xy}{\sigma_C^2 \sigma_B^2} [\mathrm{Re} \langle \tilde{C}\tilde{B} \rangle]^2, \quad (\text{E69})$$

where $\tilde{C} = C - \langle C \rangle$ and $\tilde{B} = B - \langle B \rangle$. Combining the above,

$$\begin{aligned} & \left[ F(\rho; \Delta_+) F(\rho; \Delta_-) \right]^{1/2} \\ & \le 4(x+y) \sqrt{1 - \frac{xy}{(x+y)^2} \frac{4[\mathrm{Re} \langle \tilde{C}\tilde{B} \rangle]^2}{\sigma_C^2 \sigma_B^2}}. \end{aligned} \quad (\text{E70})$$

Plugging this expression back into Eq. (E66), along with Eq. (E65), proves Eq. (E60). Now suppose that $C = i\alpha[A, \psi]$ for some Hermitian operator $A$ and for a real number $\alpha$. Then $\langle C \rangle = 0$ so $\tilde{C} = C$ and

$$\begin{aligned} \mathrm{Re} \langle \tilde{C}\tilde{B} \rangle &= \alpha \mathrm{Re} \langle i[A, \psi]\tilde{B} \rangle = \alpha \mathrm{Re} (\langle \psi | iH | \psi \rangle \langle \psi | \tilde{B} | \psi \rangle \\ & \quad - \langle \psi | iA\tilde{B} | \psi \rangle) \\ &= \alpha \mathrm{Re}(-i\langle A\tilde{B} \rangle) = \frac{\alpha}{2} (-i\langle A\tilde{B} \rangle + i\langle \tilde{B}A \rangle) \\ &= \frac{\alpha}{2} \langle i[A, \tilde{B}] \rangle = \frac{\alpha}{2} \langle i[A, B] \rangle. \end{aligned} \quad (\text{E71})$$

Equation (E61) follows from this and using the fact that $\sigma_C^2 = \langle \tilde{C}^2 \rangle = \alpha^2 \langle -(A\psi - \psi A)^2 \rangle = \alpha^2 \langle A^2 \rangle - \langle A \rangle^2 = \alpha^2 \sigma_A^2$.

Now assume that Eq. (E62) is satisfied. Recalling that $\Delta_{\pm} = D_{\tilde{C}} \pm D_{\tilde{B}} = M(-i[\tilde{C} \pm \tilde{B}, \psi])$, we find that condition (E62) immediately implies that either $\Delta_+ = 0$ or $\Delta_- = 0$ and therefore either $F(\rho; \Delta_+) = 0$ or $F(\rho; \Delta_-) = 0$. In this case, Proposition 16 immediately implies that $F(\rho; D_{\tilde{C}}) = F(\rho; D_{\tilde{B}})$. We conclude that Eq. (E63) holds, recalling Eq., (E65). ■

*Proof of Proposition 21.*—Consider the evolution $\psi$ $(a, c)$, where the parameter $a$ is generated by the first given Hermitian operator $A$ and where the parameter $c$ is generated by the complementary generator $C$ (as per Fig. 5 in the main text) given by

$$\partial_c \psi = i[C, \psi], \quad C = \frac{1}{2\mathrm{Var}_{\psi}(A)} (-i[A, \psi]). \quad (\text{E72})$$

Recall $F_{\mathrm{Alice},c} = 4\sigma_C^2 = \sigma_A^{-2}$ from Eq. (27) with $H \to A$ and $T \to C$. Our time-energy uncertainty relation, in its form of Theorem 6, asserts that

$$\frac{1}{4\sigma_A^2} F_{\mathrm{Bob},a} + \sigma_A^2 F_{\mathrm{Eve},c} \le 1. \quad (\text{E73})$$

Now we invoke Lemma 2, with $M = \tilde{\mathcal{N}}$, $c$, $b$, $C = i\alpha[A, \psi]$, $\alpha = -(2\sigma_A^2)^{-1}$, $B$, and $x = y = 1/4$. From

Eq. (E61) we find

$$\frac{1}{4\sigma_B^2} F_{\mathrm{Eve},b} \le \sigma_A^2 F_{\mathrm{Eve},c} + 2 \sqrt{1 - \frac{\langle i[A, B] \rangle^2}{4\sigma_A^2 \sigma_B^2}}. \quad (\text{E74})$$

We find, applying Eqs. (E73) and (E74) in succession,

$$\begin{aligned} \frac{1}{4\sigma_A^2} F_{\mathrm{Bob},a} + \frac{1}{4\sigma_B^2} F_{\mathrm{Eve},b} &\le 1 - \sigma_A^2 F_{\mathrm{Eve},c} + \frac{1}{4\sigma_B^2} F_{\mathrm{Eve},b} \\ &\le 1 + 2 \sqrt{1 - \frac{\langle i[A, B] \rangle^2}{4\sigma_A^2 \sigma_B^2}}. \end{aligned} \quad (\text{E75})$$

This shows the desired uncertainty relation.

Now assume that $\mathcal{N}[\psi]$ does not change rank locally as a function of $a$ and that Eq. (E56) holds. Let $M = \tilde{\mathcal{N}}$, $\rho_E = \tilde{\mathcal{N}}[\psi]$, $C = i\alpha[A, \psi]$, and $\alpha = -(2\sigma_A^2)^{-1}$. Then as computed above $\sigma_C = |\alpha|\sigma_A = 1/(2\sigma_A)$. Let us compute now

$$\begin{aligned} -i \left[ \sqrt{x} \frac{C}{\sigma_C}, \psi \right] &= -i \left[ \sqrt{x} \frac{-i[A, \psi]}{\sigma_A}, \psi \right] \\ &= -\sqrt{x} \left\{ \frac{A - \langle A \rangle}{\sigma_A}, \psi \right\}, \end{aligned} \quad (\text{E76})$$

recalling that $[[A, \psi], \psi] = \{A - \langle A \rangle, \psi\}$. Let $y = 1$, $x = |\beta|^2$ and $s = \mathrm{sign}(\beta)$ such that $s\sqrt{\tilde{x}}/\sqrt{\tilde{y}} = \beta$. We then have

$$\begin{aligned} \mathcal{M} \left( -i \left[ \sqrt{\tilde{y}} \frac{B}{\sigma_B} + s \sqrt{x} \frac{C}{\sigma_C}, \psi \right] \right) \\ &= \hat{\mathcal{N}} \left( -i \left[ \sqrt{\tilde{y}} \frac{B}{\sigma_B}, \psi \right] \right) + s \hat{\mathcal{N}} \left( -i \left[ \sqrt{x} \frac{C}{\sigma_C}, \psi \right] \right) \\ &= \frac{\sqrt{\tilde{y}}}{2} \left\{ \hat{\mathcal{N}} \left( -i \left[ \frac{B}{\sigma_B}, \psi \right] \right) - \beta \hat{\mathcal{N}} \left( \left\{ \frac{A - \langle A \rangle}{\sigma_A}, \psi \right\} \right) \right\} \\ &= 0. \end{aligned} \quad (\text{E77})$$

The latter expression then vanishes thanks to our assumption that Eq. (E56) holds. Thanks to Lemma 2 we find

$$\begin{aligned} \frac{F_{\mathrm{Eve},b}}{F_{\mathrm{Alice},b}} &= \frac{1}{4\sigma_B^2} F(\rho_E; \partial_b \rho_E) = \frac{x}{4\sigma_C^2} F(\rho_E; \partial_c \rho_E) \\ &= \beta^2 \frac{F_{\mathrm{Eve},c}}{F_{\mathrm{Alice},c}}. \end{aligned} \quad (\text{E78})$$

Thanks to our assumption that $\mathcal{N}[\psi]$ does not change rank locally as a function of $a$, we know that our main
uncertainty relation (Theorem 6) holds with equality:

$$ \frac{1}{4\sigma_A^2} F_{\text{Bob},a} + \sigma_A^2 F_{\text{Eve},c} = 1. \qquad (\text{E79}) $$

We therefore find, recalling $\sigma_A^2 = 1/(4\sigma_C^2)$,

$$ \begin{aligned} \frac{F_{\text{Bob},a}}{F_{\text{Alice},a}} + \frac{1}{\beta^2} \frac{F_{\text{Eve},b}}{F_{\text{Alice},b}} &= \left[ 1 - \sigma_A^2 F_{\text{Eve},c} \right] + \frac{1}{\beta^2} \frac{F_{\text{Eve},b}}{4\sigma_B^2} \\ &= 1 - \frac{1}{\beta^2} \frac{F_{\text{Eve},c}}{4\sigma_C^2} + \frac{1}{\beta^2} \frac{F_{\text{Eve},b}}{4\sigma_B^2} = 1, \end{aligned} \qquad (\text{E80}) $$

using Eq. (E78), thus proving the claim.

<figure><img src="image_18.png" /></figure>

*Proof of Corollary 3.*—The main idea of this corollary is to note that the Fisher information depends only on the state and its first derivative with respect to the parameter, and that any derivative $\partial_a \psi$ can be written in the form $\partial_a \psi = -i[A, \psi]$ for some Hermitian generator $A$. Therefore, we seek Hermitian operators $A, B$ such that $\partial_a \psi = -i[A, \psi]$ and $\partial_b \psi = -i[B, \psi]$, such that we can apply Proposition 21. We let $A = i[\partial_a \psi, \psi]$ and $B = i[\partial_b \psi, \psi]$, and we compute

$$ \begin{aligned} -i[A, \psi] &= -i[i[\partial_a \psi, \psi], \psi] = \{\partial_a \psi, \psi\} \\ &= \partial_a(\psi^2) = \partial_a \psi, \end{aligned} \qquad (\text{E81}) $$

using Eq. (A4) and the fact that $\langle \partial_a \psi \rangle = \text{tr}[\partial_a \psi] = \partial_a \text{tr}(\psi) = 0$. Similarly,

$$ -i[B, \psi] = \partial_b \psi. \qquad (\text{E82}) $$

We can therefore apply Proposition 21. It remains to compute the quantities appearing in the right-hand side of Eq. (E55). We have

$$ \begin{aligned} \langle i[A, B] \rangle &= i \operatorname{tr} \{ \psi [i[\partial_a \psi, \psi], i[\partial_b \psi, \psi]] \} \\ &= i \operatorname{tr} \{ [\psi, i[\partial_a \psi, \psi]] (i[\partial_b \psi, \psi]) \} \\ &= \operatorname{tr} \{ -i[i[\partial_a \psi, \psi], \psi] (i[\partial_b \psi, \psi]) \} \\ &= \operatorname{tr} \{ (\partial_a \psi) (i[\partial_b \psi, \psi]) \} \\ &= \langle i[\partial_a \psi, \partial_b \psi] \rangle, \end{aligned} \qquad (\text{E83}) $$

using the cyclicity of the trace and invoking Eq. (E81) for the fourth equality. Furthermore,

$$ \begin{aligned} \sigma_A^2 &= \langle A^2 \rangle - \langle A \rangle^2 = \langle (i[\partial_a \psi, \psi])^2 \rangle - \langle i[\partial_a \psi, \psi] \rangle^2 \\ &= -\langle ((\partial_a \psi) \psi - \psi (\partial_a \psi))((\partial_a \psi) \psi - \psi (\partial_a \psi)) \rangle \\ &= \langle \psi (\partial_a \psi) (\partial_a \psi) \psi \rangle = \langle (\partial_a \psi)^2 \rangle, \end{aligned} \qquad (\text{E84}) $$

where we have made use of $\psi(\partial_a \psi) \psi = 0$. Similarly $\sigma_B^2 = \langle (\partial_b \psi)^2 \rangle$, which ends the proof.

<figure><img src="image_19.png" /></figure>

# APPENDIX F: GENERALIZATIONS TO INFINITE-DIMENSIONAL HILBERT SPACES

While the main text has put an emphasis on discussing notions of quantum metrology making use of finite-dimensional quantum systems, in this section, we generalize the above findings to the setting of infinite-dimensional Hilbert spaces. A specific attention is given to unbounded operators, as many physical systems of practical use fall under this category.

## 1. Uncertainty relation for any two parameters

We start with a generalisation of Theorem 3 to infinite dimensions (cf. Proposition 21).

*Theorem 8 (Uncertainty relation for infinite-dimensional systems).*—let $A, B$ be two self-adjoint operators (possibly unbounded) on a separable Hilbert space $\mathcal{H}_A$ with domains $D(A)$ and $D(B)$, respectively. Let $|\psi\rangle \in D(A) \cap D(B)$ and $|\psi(a)\rangle \in D(A)$, $|\psi(b)\rangle \in D(B)$ for some $b, a \in \mathbb{R}$ where $|\psi(a)\rangle := e^{-iaA}|\psi\rangle$, $|\psi(b)\rangle := e^{-ibB}|\psi\rangle$. Let $V_{A \to BE}$ be any isometry $\mathcal{H}_A \to \mathcal{H}_B \otimes \mathcal{H}_E$, where the Hilbert spaces $\mathcal{H}_A, \mathcal{H}_B, \mathcal{H}_E$ associated with Bob and Eve are also separable and possibly of infinite dimensions. Consider the two pure state evolutions given by Eq. (E54). Then Eq. (E55) holds, with the following quantities defined by

$$ \langle i[A, B] \rangle := i\langle A\psi, B\psi \rangle - i\langle B\psi, A\psi \rangle, \qquad (\text{F1}) $$

$$ \langle A^2 \rangle := \langle A\psi, A\psi \rangle \qquad (\text{F2}) $$

and

$$ F_M(y) := \liminf_{l \to \infty} \operatorname{tr} \left[ \rho_M^{(l)}(y) R^2 \right] \in \mathbb{R}, \qquad (\text{F3}) $$

where $M \in \{B, E\}$, $y \in \{a, b\}$ and $\rho_M^{(l)}$ is an $l$-dimensional subnormalised density operator and $R = R(l)$ is defined in Eq. (11) on an $l$-dimensional Hilbert space for $\rho_M^{(l)}$. Specifically,

$$ \rho_M^{(l)}(y) := \operatorname{tr}_{\setminus M} \left[ P_{BE}^{(l)} V_{A \to BE} \rho_A(y) V_{A \to BE}^{\dagger} P_{BE}^{(l)} \right], \qquad (\text{F4}) $$

where $\setminus E := B, \setminus B := E$, and $P_{BE}^{(l)}$ is the orthogonal projection onto the first $l$ basis elements of a basis for $\mathcal{H}_B \otimes \mathcal{H}_E$. Furthermore, the derivative of $\rho_M^{(l)}(y)$ is defined via

$$ \frac{d}{dy} \rho_M^{(l)}(y) := \operatorname{tr}_{\setminus M} \left[ P_{BE}^{(l)} V_{A \to BE} \frac{d}{dy} \rho_A(y) V_{A \to BE}^{\dagger} P_{BE}^{(l)} \right], \qquad (\text{F5}) $$

where

$$ \frac{d}{da} \rho_A(a) := i|\psi(a)\rangle\langle\psi(a)|A - i\bar{A}|\psi(a)\rangle\langle\psi(a)|, \qquad (\text{F6}) $$

$$ \frac{d}{db} \rho_A(b) := i|\psi(b)\rangle\langle\psi(b)|B - i\bar{B}|\psi(b)\rangle\langle\psi(b)|. \qquad (\text{F7}) $$
*Proof of Theorem 8.*—The proof will proceed in two steps. First we will approximate B and A by bounded operators (if they are already bounded, then this first step is not necessary, although the approximation will nevertheless be well defined). Second, we will approximate these bounded operators by finite dimensional operators. Then we will apply Eq. (E55) before taking a sequence of limits in which the approximations vanish. We start with a few elementary definitions and results, which will be necessary for our proof.

Let $A$, $(A_n)_n$, be bounded operators on a Hilbert space $\mathcal{H}$. We define all bounded operators we consider to have domain equal to the entire Hilbert space. We say that $A_n$ converges (as $n \to \infty$) to $A$ in the strong limit if $A_n\Psi \to A\Psi$ as $n \to \infty$ for any $\Psi \in \mathcal{H}$. We denote this as $A_n \xrightarrow{s} A$. Some properties are the following.

(i) Let $A$, $(A_n)_n$, $B$, $(B_n)_n$, $C$, $(C_n)_n$, be bounded operators on a Hilbert space $\mathcal{H}$. $A_n \xrightarrow{s} A$, $B_n \xrightarrow{s} B$ and $C_n \xrightarrow{s} C$ imply $A_n B_n \xrightarrow{s} AB$ and $A_n B_n C_n \xrightarrow{s} ABC$. 

*Proof.* $(A_n B_n - AB)\Psi = A_n(B_n - B)\Psi + (A_n - A)B\Psi$. By the uniform boundedness principle, $A_n \xrightarrow{s} A$ implies $\|A_n\| \le c$ for some $c \in \mathbb{R}$ for all $n$. Therefore,

$$ \|(A_n B_n - AB)\Psi\| \le c \|(B_n - B)\Psi\| + \|(A_n - A)B\Psi\|, \quad (F8) $$

where the rhs tends to zero as $n \to \infty$. This proves the first claim. For the second, simply define $\tilde{A}_n := A_n B_n$. Hence $\tilde{A}_n \xrightarrow{s} AB$ and thus $\tilde{A}_n C_n \xrightarrow{s} (AB)C$, hence proving the second claim.

(ii) $A_n \xrightarrow{s} A$ implies $e^{-iA_n t} \xrightarrow{s} e^{-iAt}$ for $t \in \mathbb{R}$.

*Proof.* $e^{-iA_n t} - e^{-iAt} = e^{-iA_n s}e^{-iA(t-s)} \Big|_{s=0}^{s=t} = -i \int_0^t ds \ e^{-iA_n s}(A_n - A)e^{-iA(t-s)}$. But we have $(A_n - A)e^{-iA(t-s)} \xrightarrow{s} \tilde{0}$ pointwise in $s$, where $\tilde{0}$ is the bounded operator mapping all vectors in $\mathcal{H}$ to the zero vector in $\mathcal{H}$. Thus via (i), $e^{-iAs}(A_n - A)e^{-iA(t-s)} \xrightarrow{s} \tilde{0}$ pointwise in $s$ and the result follows by dominated convergence.

(iii) Let $A$ be self-adjoint and possibly unbounded. Let $f$, $(f_n)_n : \mathbb{R} \to \mathbb{C}$ be uniformly bounded functions with $f_n \to f$ as $n \to \infty$ point-wise. Then $f_n(A) \xrightarrow{s} f(A)$.

*Proof.* See Ref. [109].

We can now prove the theorem. Let $(P_N^{(n)})_n$ be the orthogonal projections onto the span of the first $n$ basis elements of a separable Hilbert space $\mathcal{H}_N$. Consider two bounded operators $\tilde{A}$ and $\tilde{B}$ on $\mathcal{H}_A$ and define $\tilde{A}_n, \tilde{B}_n$ by

$$ \tilde{A}_n := P_A^{(n)} \tilde{A} P_A^{(n)}, \quad \tilde{B}_n := P_A^{(n)} \tilde{B} P_A^{(n)}. \quad (F9) $$

Furthermore, consider the sequence of states $(\rho_B^{(n,l)})_{n,l}$ on $\mathcal{H}_B$, and $(\rho_E^{(n,l)})_{n,l}$ on $\mathcal{H}_E$, where

$$ \begin{aligned} \rho_B^{(n,l)}(a) &:= \operatorname{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (|\psi_n(a)\rangle\langle\psi_n(a)|) P_A^{(n)} V_{A \to BE}^\dagger P_{BE}^{(l)} \right], \\ \rho_E^{(n,l)}(b) &:= \operatorname{tr}_B \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (|\psi_n(b)\rangle\langle\psi_n(b)|) P_A^{(n)} V_{A \to BE}^\dagger P_{BE}^{(l)} \right], \end{aligned} \quad (F10) $$

where $|\psi_n(a)\rangle := e^{-ia\tilde{A}_n}|\psi\rangle$, $|\psi_n(b)\rangle := e^{-ib\tilde{B}_n}|\psi\rangle$ and the sequences of derivatives, $(\frac{d}{da}\rho_B^{(n,l)}(t))_{n,l}$ on $\mathcal{H}_B$, and $(\frac{d}{da}\rho_E^{(n,l)}(a))_{n,l}$ on $\mathcal{H}_E$ are

$$ \begin{aligned} \frac{d}{da} \rho_B^{(n,l)}(a) &= \operatorname{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (i |\psi_n(a)\rangle\langle\psi_n(a)| \tilde{A}_n - i \tilde{A}_n |\psi_n(a)\rangle\langle\psi_n(a)|) P_A^{(n)} V_{A \to BE}^\dagger P_{BE}^{(l)} \right], \\ \frac{d}{db} \rho_E^{(n,l)}(b) &= \operatorname{tr}_B \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (i |\psi_n(b)\rangle\langle\psi_n(b)| \tilde{B}_n - i \tilde{B}_n |\psi_n(b)\rangle\langle\psi_n(b)|) P_A^{(n)} V_{A \to BE}^\dagger P_{BE}^{(l)} \right]. \end{aligned} \quad (F11) $$

We can use Eqs. (F10) and (F11) to construct the Fisher information for these states. Since $V_{A \to BE} V_{A \to BE}^\dagger = 1_{BE}$, where $1_{BE}$ is the identity operator on $\mathcal{H}_{BE}$, it follows that

$$ P_A^{(n)} - \left( P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} \right)^\dagger \left( P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} \right) \ge 0 \quad (F12) $$

for all $l, n$. Hence, by Kraus' theorem, Eq. (F10) are completely positive and trace nonincreasing maps evaluated on inputs $|\psi_n(a)\rangle\langle\psi_n(a)|$. Since Proposition 21 holds

for any completely positive, trace nonincreasing map, we can apply it to our setup. This yields

$$ \frac{F(\rho_B^{(n,l)}(a))}{F(\rho_A^{(n)}(a))} + \frac{F(\rho_E^{(n,l)}(b))}{F(\rho_A^{(n)}(b))} \le 1 + 2 \sqrt{1 - \frac{\langle i[\tilde{A}_n, \tilde{B}_n] \rangle^2}{4 \tilde{\sigma}_{\tilde{A},n}^2 \tilde{\sigma}_{\tilde{B},n}^2}}, \quad (F13) $$
recalling that the uncertainty relation also applies to sub-normalized positive operators, and where

$$F(\rho_A^{(n)}(a)) = 4 \tilde{\sigma}_{\tilde{A},n}^2, \qquad (\text{F14})$$

$$F(\rho_A^{(n)}(b)) = 4 \tilde{\sigma}_{\tilde{B},n}^2, \qquad (\text{F15})$$

$$\tilde{\sigma}_{\tilde{A},n} := \left( (\tilde{A}_n \psi, \tilde{A}_n \psi) - \langle \psi, \tilde{A}_n \psi \rangle^2 \right)^{1/2}, \qquad (\text{F16})$$

$$\tilde{\sigma}_{\tilde{B},n} := \left( (\tilde{B}_n \psi, \tilde{B}_n \psi) - \langle \psi, \tilde{B}_n \psi \rangle^2 \right)^{1/2}. \qquad (\text{F17})$$

We can now take the limit $n \to \infty$ on both sides of Eq. (F13). Due to property (i), it follows

$$\lim_{n \to \infty} F(\rho_B^{(n,l)}(a)) + \lim_{n \to \infty} F(\rho_E^{(n,l)}(b)) \\ \le 1 + 2\sqrt{1 - \frac{\langle i[\tilde{A}, \tilde{B}] \rangle^2}{4 \tilde{\sigma}_{\tilde{A}}^2 \tilde{\sigma}_{\tilde{B}}^2}}, \qquad (\text{F18})$$

where

$$F(\rho_A^{(\infty)}(a)) := 4 \tilde{\sigma}_A^2, \qquad (\text{F19})$$

$$F(\rho_A^{(\infty)}(b)) := 4 \tilde{\sigma}_B^2, \qquad (\text{F20})$$

$$\tilde{\sigma}_A := \left( (\tilde{A}\psi, \tilde{A}\psi) - \langle \psi, \tilde{A}\psi \rangle^2 \right)^{1/2}, \qquad (\text{F21})$$

$$\tilde{\sigma}_B := \left( (\tilde{B}\psi, \tilde{B}\psi) - \langle \psi, \tilde{B}\psi \rangle^2 \right)^{1/2}. \qquad (\text{F22})$$

Observe that the quantities $\lim_{n \to \infty} F(\rho_B^{(n,l)}(a))$, $\lim_{n \to \infty} F(\rho_E^{(n,l)}(b))$ cannot diverge, since it would contradict the inequality (since the Fisher information is non-negative). This observation follows alternatively from applying the data processing inequality (110) to bound Bob's Fisher information in terms of Alice's, followed by talking the $n \to \infty$ limit. Similarly for Eve's Fisher information. By direct calculation, we observe that the Fisher information $F$ of a state $\rho$ on a $d$-dimensional Hilbert space, according to Eqs. (9) and (11), is given by

$$F = \sum_{\substack{k,k'=1 \\ \text{s.t. } p_k+p_{k'}>0}}^{d} \frac{p_k}{(p_k+p_{k'})^2} \left|\left\langle k \left| \frac{d\rho}{da} \right| k' \right\rangle\right|^2, \qquad (\text{F23})$$

where $\rho = \sum_{k=1}^{d} p_k |k\rangle\langle k|$. Hence

$$\lim_{n \to \infty} F(\rho_B^{(n,l)}(a)) = \lim_{n \to \infty} \sum_{\substack{k,k'=1 \\ \text{s.t. } p_k^{(n,l)} + p_{k'}^{(n,l)} > 0}}^{d_B(l)} \frac{p_k^{(n,l)}}{(p_k^{(n,l)} + p_{k'}^{(n,l)})^2} \\ \left| \langle k, n, l | \frac{d\rho_B^{(n,l)}(a)}{da} | k', n, l \rangle \right|^2, \qquad (\text{F24})$$

where $d_B(l)$ is the dimension of Bob's reduced system, which is $l$ independent, and $\rho_B^{(n,l)}(a) = \sum_{k=1}^{d_B(l)} p_k^{(n,l)} |k, n, l\rangle\langle k, n, l|$. Observe that all terms in the summation must be finite in the limit, since they are all non-negative and we are guaranteed that the rhs of Eq. (F24) does not diverge. Observe that for terms in the summation for which $\lim_{n \to \infty} p_k^{(n,l)} + p_{k'}^{(n,l)} > 0$, the summation can be interchanged with the limit. However, while for terms such that $p_k^{(n,l)} + p_{k'}^{(n,l)} > 0$ for all $n$, but $\lim_{n \to \infty} p_k^{(n,l)} + p_{k'}^{(n,l)} = 0$, the summation and integration cannot be interchanged, the interchange of the limit and summation will result in the lower bound

$$\lim_{n \to \infty} F(\rho_B^{(n,l)}(a)) \ge \sum_{\substack{k,k'=1 \\ \text{s.t. } p_k^{(\infty,l)} + p_{k'}^{(\infty,l)} > 0}}^{d_B(l)} \frac{p_k^{(\infty,l)}}{(p_k^{(\infty,l)} + p_{k'}^{(\infty,l)})^2} \\ \left| \langle k, \infty, l | \frac{d\rho_B^{(\infty,l)}(a)}{da} | k', \infty, l \rangle \right|^2, \qquad (\text{F25})$$

where $\rho_B^{(\infty,l)}(a) = \sum_{k=1}^{d_B(l)} p_k^{(\infty,l)} |k, \infty, l\rangle\langle k, \infty, l|$, with

$$\rho_B^{(\infty,l)}(a) := \lim_{n \to \infty} \text{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (|\psi_n(a)\rangle\langle\psi_n(a)|) \right. \\ \left. P_A^{(n)} V_{A \to BE}^\dagger P_{BE}^{(l)} \right] \qquad (\text{F26})$$

$$= \text{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} \left( |\tilde{\psi}(a)\rangle\langle\tilde{\psi}(a)| \right) V_{A \to BE}^\dagger P_{BE}^{(l)} \right], \qquad (\text{F27})$$

where $|\tilde{\psi}(a)\rangle := e^{-i\tilde{A}a}|\psi\rangle$ and using properties (i) and (ii). Similarly, use properties (i) and (ii) again to obtain

$$\frac{d}{da} \rho_B^{(\infty,l)}(a) := \lim_{n \to \infty} \text{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (i|\psi_n(a)\rangle\langle\psi_n(a)|\tilde{A}_n - i\tilde{A}_n|\psi_n(a)\rangle\langle\psi_n(a)|) \right. \\ \left. P_A^{(n)} V_{A \to BE}^\dagger P_{BE}^{(l)} \right] \qquad (\text{F28})$$

$$= \text{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} \left( i|\tilde{\psi}(a)\rangle\langle\tilde{\psi}(a)|\tilde{A} - i\tilde{A}|\tilde{\psi}(a)\rangle\langle\tilde{\psi}(a)| \right) V_{A \to BE}^\dagger P_{BE}^{(l)} \right]. \qquad (\text{F29})$$
Likewise, we obtain the same expression for $\lim_{n \to \infty} F(\rho_E^{(n,l)}(b))$ that we have achieved for

$$ \lim_{n \to \infty} F(\rho_B^{(n,l)}(a)), \qquad (F30) $$

but interchanging $a \mapsto b$, $\tilde{A} \mapsto \tilde{B}$ and $\mathrm{tr}_E \mapsto \mathrm{tr}_B$.

Now that we have an expression for the bound, which holds for bounded operators $\tilde{A}$ and $\tilde{B}$, our next step is to move to unbounded operators. For this task, we define sequences of bounded operators $(A_m)_m$ and $(B_m)_m$ as

$$ A_m := \frac{A}{1 + A^2/m}, \quad B_m := \frac{B}{1 + B^2/m}. \qquad (F31) $$

We now evaluate Eq. (F18) choosing $\tilde{A}$ equal to $\tilde{A}_m$ and $\tilde{A}$ equal to $\tilde{A}_m$, followed by taking the limit $m \to \infty$ on both sides of the equation. Since by (iii), it follows that $1/(1+A^2/m) \xrightarrow{s} 1_A$ and $1/(1+B^2/m) \xrightarrow{s} 1_A$, where $1_A$ is the identity operator on $\mathcal{H}_A$, we have that $A_m\psi \to A\psi$ and $B_m\psi \to B\psi$ for all $\psi \in \mathcal{D}(A) \cap \mathcal{D}(B)$, and we find

$$ \lim_{m \to \infty} \lim_{n \to \infty} F(\rho_B^{(n,l)}(a)) \overline{F(\rho_A^{(\infty,\infty)}(a))} + \lim_{m \to \infty} \lim_{n \to \infty} F(\rho_E^{(n,l)}(b)) \overline{F(\rho_A^{(\infty,\infty)}(b))} \le 1 + 2\sqrt{1 - \frac{\langle i[A, B] \rangle^2}{4 \sigma_A^2 \sigma_B^2}}, \qquad (F32) $$

where

$$ F(\rho_A^{(\infty,\infty)}(a)) := 4\sigma_A^2, \qquad (F33) $$

$$ F(\rho_A^{(\infty,\infty)}(b)) := 4\sigma_O^2, \qquad (F34) $$

$$ \sigma_A := \left( \langle A\psi, A\psi \rangle - \langle \psi, A\psi \rangle^2 \right)^{1/2}, \qquad (F35) $$

$$ \sigma_B := \left( \langle B\psi, B\psi \rangle - \langle \psi, B\psi \rangle^2 \right)^{1/2}. \qquad (F36) $$

The rhs of this inequality is now of the form in the corollary statement. We continue with the rhs. First observe that

$$ \lim_{m \to \infty} \lim_{n \to \infty} F(\rho_B^{(n,l)}(a)) \ge \qquad (F37) $$

$$ \sum_{\substack{k,k'=1 \\ \text{s.t. } p_k^{(\infty,\infty,l)} + p_{k'}^{(\infty,\infty,l)} > 0}}^{d_B(l)} \frac{p_k^{(\infty,\infty,l)}}{(p_k^{(\infty,\infty,l)} + p_{k'}^{(\infty,\infty,l)})^2} \left| \langle k, \infty, \infty, l | \frac{d\rho_B^{(\infty,\infty,l)}(a)}{da} | k', \infty, \infty, l \rangle \right|^2, \qquad (F38) $$

$$ \text{where} \quad \rho_B^{(\infty,\infty,l)}(a) = \sum_{k=1}^{d_B(l)} p_k^{(\infty,\infty,l)} |k, \infty, \infty, l\rangle \langle k, \infty, \infty, l|, \text{ with} $$

$$ \rho_B^{(\infty,\infty,l)}(a) := \lim_{m \to \infty} \mathrm{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} \left( |\tilde{\psi}_m(a)\rangle \langle \tilde{\psi}_m(a)| \right) V_{A \to BE}^{\dagger} P_{BE}^{(l)} \right], \qquad (F39) $$

$$ \frac{d}{da} \rho_B^{(\infty,\infty,l)}(a) := \lim_{m \to \infty} \mathrm{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} \left( i |\tilde{\psi}_m(a)\rangle \langle \tilde{\psi}_m(a)| A_m - i A_m |\tilde{\psi}_m(a)\rangle \langle \tilde{\psi}_m(a)| \right) V_{A \to BE}^{\dagger} P_{BE}^{(l)} \right], \qquad (F40) $$

and $|\tilde{\psi}_m(a)\rangle = e^{-ia\lambda_m}|\psi\rangle$. To see that Eq. (F38) holds, observe that the same reasoning to why the limit and sum- mation could be interchanged going from Eq. (F23) to Eq. (F24), holds for the limit $m \to \infty$ also. Now define $f_m(x) = e^{-iax/(1+x^2/m)}$ and $f(x) = e^{-iax}$. Assumptions in (iii) hold, thus $e^{-ia\lambda_m} \xrightarrow{s} e^{-ia\lambda}$, hence using (ii) $1/(1+A^2/m)e^{-ia\lambda_m} \xrightarrow{s} e^{-ia\lambda}$. Furthermore, since, by definition $e^{-ia\lambda}|\psi\rangle \in \mathcal{D}(A)$, we have $H|\psi(a)\rangle \in \mathcal{H}_A$. Taking all these things into account, we conclude that

$$ \rho_B^{(\infty,\infty,l)}(a) = \mathrm{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} (|\psi(a)\rangle \langle \psi(a)|) V_{A \to BE}^{\dagger} P_{BE}^{(l)} \right], \qquad (F41) $$

$$ \frac{d}{da} \rho_B^{(\infty,\infty,l)}(a) = \mathrm{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} (i |\psi(a)\rangle \langle \psi(a)| A - i A |\psi(a)\rangle \langle \psi(a)|) V_{A \to BE}^{\dagger} P_{BE}^{(l)} \right]. \qquad (F42) $$

Likewise, we obtain the same expression for $\lim_{m \to \infty} \lim_{n \to \infty} F(\rho_E^{(n,l)}(b))$ that we have achieved for $\lim_{m \to \infty} \lim_{n \to \infty} F(\rho_B^{(n,l)}(a))$, but interchanging $a \mapsto b$, $A \mapsto B$ and $\mathrm{tr}_E \mapsto \mathrm{tr}_B$. Lastly, by comparing the rhs of Eq. (F38) with the rhs of Eq. (F23), one sees that $\lim_{m \to \infty} \lim_{n \to \infty} F(\rho_B^{(n,l)}(a))$ is given by evaluating the Fisher information for $\rho_B^{(\infty,\infty,l)}(a)$ [defined by Eq. (F41)] with derivative $\frac{d}{da} \rho_B^{(\infty,\infty,l)}(a)$ [defined by Eq. (F41)] according to Eqs. (9) and (11). The same observation holds for Eve's Fisher information. Hence to conclude the proof, we take $\lim_{l \to \infty}$ on both sides of the equation. ■

## 2. Time-energy uncertainty equality in infinite dimensions

In fact, building on the previous result, we get the fol- lowing statement in the case where the commutator in the previous theorem vanishes. This can be viewed as a generalization of Theorem 1 to the unbounded operator case.
**Theorem 9 (Time-energy uncertainty relation for infinite-dimensional systems).**—let $|\psi\rangle$ be a state vector in a separable Hilbert space $\mathcal{H}_A$ of possibly infinite dimensions, let $H, X$ be self-adjoint operators (possibly unbounded) with domains $\mathcal{D}(H)$ and $\mathcal{D}(X)$, respectively, so that $|\psi\rangle \in \mathcal{D}(H) \cap \mathcal{D}(X)$. Define $\sigma_H := [\langle H\psi, H\psi \rangle - \langle \psi, H\psi \rangle^2]^{1/2}$, which is finite due to $|\psi\rangle \in \mathcal{D}(H)$, and $P_\rho^\perp := 1 - P_\rho$, where $P_\rho$ denotes the projector onto the support of $\rho$. Define analogously as before

$$T := t_0 - \frac{i[H, \psi]}{2\sigma_H^2} + P_\psi^\perp X P_\psi^\perp, \quad (\text{F43})$$

where $X$ captures the freedom left when defining the optimal local time sensing observable, and consider for real $t, \eta$ and $t_0, \eta_0$ the two-parameter family $|\psi(t, \eta)\rangle$ with $|\psi(t_0, \eta_0)\rangle = |\psi\rangle$, again $\psi = |\psi\rangle\langle\psi|$ and

$$|\psi(t, \eta)\rangle = \exp\{-i[(t - t_0)H - (\eta - \eta_0)T]\}|\psi\rangle. \quad (\text{F44})$$

Let, as in Theorem 8, $V_{A \to BE}$ be any isometry $\mathcal{H}_A \to \mathcal{H}_B \otimes \mathcal{H}_E$, where the Hilbert spaces $\mathcal{H}_B, \mathcal{H}_E$ associated with Bob and Eve are also separable and possibly of infinite dimensions, and define $F_A, F_B, F_E$ analogously as in Theorem 8. Then the uncertainty principle

$$\frac{F_B(t)}{F_A(t)} + \frac{F_E(\eta)}{F_A(\eta)} \le 1 \quad (\text{F45})$$

holds.

Indeed, even in the infinite-dimensional setting for unbounded operators, the uncertainty principle can be attained with equality, so that

$$\frac{F_B(t)}{F_A(t)} + \frac{F_E(\eta)}{F_A(\eta)} = 1 \quad (\text{F46})$$

still holds true.

**Proof of Theorem 9.**—The proof follows the same line of thought as that of Theorem 8, with some differences. To start with, consider the bounded operators $\tilde{H}$ and $\tilde{X}$ on $\mathcal{H}_A$ and define for a positive integer $n$ the truncated operators $\tilde{H}_n$ as

$$\tilde{H}_n := P_A^{(n)} \tilde{H} P_A^{(n)} \quad (\text{F47})$$

and

$$\tilde{T}_n := P_A^{(n)} \tilde{T} P_A^{(n)}, \quad (\text{F48})$$

with $\tilde{T}$ being defined as in Eq. (F43) with $T$ being replaced by $\tilde{T}$ and $X$ by $\tilde{X}$. As above, one can define the time-evolved states as

$$|\psi_n(t, \eta)\rangle = \exp\{-i[(t - t_0)\tilde{H}_n - (\eta - \eta_0)\tilde{T}_n]\}|\psi\rangle, \quad (\text{F49})$$

with $|\psi_n\rangle := |\psi_n(t_0, \eta_0)\rangle$. In the same way as before, for positive integers $l$ (and $n$), we can consider the sequence of

positive operators $(\rho_B^{(n,l)})_{n,l}$ on $\mathcal{H}_B$ defined as

$$\begin{aligned} \rho_B^{(n,l)}(t, \eta) &:= \text{tr}_E \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (|\psi_n(t, \eta)\rangle\langle\psi_n(t, \eta)|) \right. \\ & \qquad \left. P_A^{(n)} V_{A \to BE} P_{BE}^{(l)} \right], \quad (\text{F50}) \end{aligned}$$

and

$$\begin{aligned} \rho_E^{(n,l)}(t, \eta) &:= \text{tr}_B \left[ P_{BE}^{(l)} V_{A \to BE} P_A^{(n)} (|\psi_n(t, \eta)\rangle\langle\psi_n(t, \eta)|) \right. \\ & \qquad \left. P_A^{(n)} V_{A \to BE} P_{BE}^{(l)} \right]. \quad (\text{F51}) \end{aligned}$$

Using these quantities, and proceeding as in the proof of Theorem 8, since this is a valid finite-dimensional setting in which the above proof in terms of a semidefinite program holds true, one has

$$\frac{F_{\text{Bob},t}}{4\langle\tilde{H}\psi_n, \tilde{H}_n\psi_n\rangle - \langle\psi_n, \tilde{H}_n\psi_n\rangle^2} + \langle\tilde{H}_n\psi_n, \tilde{H}_n\psi_n\rangle - \langle\psi_n, \tilde{H}_n\psi_n\rangle^2 F_{\text{Eve},\eta} = 1, \quad (\text{F52})$$

with equality, since $|\psi\rangle \in \mathcal{D}(H) \cap \mathcal{D}(X)$ and hence the state vector is in the domains of $H$ and $X$. Here,

$$F_{\text{Bob},t} := F(\rho_B^{(n,l)}(t_0); \partial_t \rho_B^{(n,l)}(t_0)), \quad (\text{F53})$$

with

$$\rho_B^{(n,l)}(\cdot) := \rho_B^{(n,l)}(\cdot, \eta_0), \quad (\text{F54})$$

and $F_{\text{Eve},\eta}$ defined analogously based on $\rho_E^{(n,l)}(\eta)$ with $\rho_E^{(n,l)}(\cdot) := \rho_E^{(n,l)}(t_0, \cdot)$. The limit to the infinite-dimensional setting involving the suitable limit of $n \to \infty$ and $l \to \infty$ can be performed as in Theorem 8, while maintaining equality for each $n$ and $l$. ■

# APPENDIX G: CALCULATIONS FOR THE CASE OF CONTINUOUS LINDBLADIAN NOISE

## 1. Sensing an unknown parameter in the Hamiltonian

Consider a probe initialized in the state vector $|\psi_{\text{init}}\rangle$ and subject to the Lindblad dynamics

$$\dot{\rho} = \mathcal{L}_{\text{tot}}^{(\omega)}(\rho), \quad (\text{G1})$$

with

$$\begin{aligned} \mathcal{L}_{\text{tot}}^{(\omega)} &= \mathcal{L}_{\text{sig}}^{(\omega)} + \mathcal{L}_{\text{rest}}; \quad \mathcal{L}_{\text{sig}}^{(\omega)}(\rho) = -i[\omega G, \rho]; \\ \mathcal{L}_{\text{rest}}(\rho) &= -i[H_{\text{rest}}, \rho] + \sum_j \left[ L_j \rho L_j^\dagger - \frac{1}{2} \{L_j^\dagger L_j, \rho\} \right]. \end{aligned} \quad (\text{G2})$$
Here, $\omega$ is the unknown parameter to be estimated. The overall evolution up to some total time $T$ is given by

$$ \mathcal{E}_T^{(\omega)} = e^{T[\mathcal{L}_{\text{sig}}^{(\omega)} + \mathcal{L}_{\text{rest}}]} . \quad (G3) $$

As we did earlier, we can decompose the overall evolution into the unitary evolution driven by the signal (which depends on the unknown parameter $\omega$), followed by an effective instantaneous noisy channel $\mathcal{N}_{T,\omega}$:

$$ \mathcal{E}_T^{(\omega)} = \mathcal{N}_{T,\omega} e^{T \mathcal{L}_{\text{sig}}^{(\omega)}}, \quad (G4) $$

where $\mathcal{N}_{T,\omega}$ is given by

$$ \mathcal{N}_{T,\omega} = \mathcal{E}_T^{(\omega)} e^{-T \mathcal{L}_{\text{sig}}^{(\omega)}}. \quad (G5) $$

We are interested in the sensitivity of the probe to the parameter $\omega$, locally around $\omega_0$, after letting the probe evolve for some fixed time $T$. The sensitivity is given in terms of the Fisher information

$$ F_{T,\omega}(\omega_0) = F(\rho_{T,\omega_0}; (\partial_{\omega}\rho_{T,\omega})(\omega_0)). \quad (G6) $$

Defining the (fictitious) family of states

$$ \begin{aligned} \psi_{T,\omega} &= e^{-iT\omega G} \psi_{\text{init}} e^{iT\omega G}; \\ \partial_{\omega}\psi_{T,\omega} &= -iT[G, \psi_{T,\omega}], \end{aligned} \quad (G7) $$

we may write

$$ F_{T,\omega}(\omega_0) = F(\rho_T; \mathcal{N}_T(\partial_\omega \psi_{T,\omega}) + (\partial_\omega \mathcal{N}_{T,\omega})(\psi_T)), \quad (G8) $$

where we omit the subscript $(\cdot)_{\omega_0}$ on all objects, which are ultimately evaluated at $\omega = \omega_0$.

Again as earlier we assume that we can neglect the second term in the derivative in Eq. (G8), and carry on with the approximation

$$ F_{T,\omega} \approx F(\mathcal{N}_{T,\omega_0}(\psi_{T,\omega_0}); \mathcal{N}_{T,\omega_0}(\partial_{\omega}\psi_{T,\omega})) =: F_{T,\omega}^{\text{unit}}. \quad (G9) $$

As above we are now in the setting of our main uncertainty relation; we can identify the above quantity with $F_{\text{Bob},t}$ in Theorem 1, where now the relevant evolution generator is TG. Theorem 1 then implies that

$$ \begin{aligned} F_{T,\omega}^{\text{unit.}} &= 4T^2 \sigma_G^2 - \Delta F_{T,\omega}^{\text{unit.}}; \\ \Delta F_{T,\omega}^{\text{unit.}} &= T^2 F(\hat{\mathcal{N}}_{T,\omega_0}(\psi_{T,\omega_0}); \hat{\mathcal{N}}_{T,\omega_0}(\{\bar{G}, \psi_{T,\omega_0}\})), \end{aligned} \quad (G10) $$

where $\hat{\mathcal{N}}_{T,\omega_0}$ is a channel that is complementary to $\mathcal{N}_{T,\omega_0}$, and where $\bar{G} = G - \langle G \rangle$ with $\langle G \rangle = \text{tr}[G \psi_{T,\omega_0}]$. As earlier, the complementary channel can be written $\hat{\mathcal{N}}_{T,\omega_0} = \hat{\mathcal{E}}_{T,\omega_0} e^{-T \mathcal{L}_{\text{sig}}^{(\omega)}}$.

The absolute error $\delta$ in the approximation (G9) can be bounded as earlier using Proposition 17 in Appendix C as

$$ \begin{aligned} |\delta| \le F(\rho; (\partial_{\omega}\mathcal{N}_{T,\omega})(\psi_{T,\omega_0})) \\ + [F(\rho; (\partial_{\omega}\mathcal{N}_{T,\omega})(\psi_{T,\omega_0})) F_{T,\omega}^{\text{unit.}}]^{1/2}. \end{aligned} \quad (G11) $$

Similar arguments to those presented earlier apply when computing $\partial_{\omega}\mathcal{N}_{T,\omega}$ in order to bound $\delta$; we have

$$ (\partial_{\omega}\mathcal{N}_{T,\omega})(\psi_{T,\omega}) = \partial_{\omega}\rho - \mathcal{E}_{T}^{(\omega)}(-iT[G, \psi_{0}]). \quad (G12) $$

Any numerical or analytical upper bound on $F(\rho; (\partial_{\omega}\mathcal{N}_{T,\omega})(\psi_{T,\omega_0}))$ then directly gives an upper bound to $|\delta|$ in Eq. (G11).

## 2. Example: continuous dephasing noise along the Z axis

A qubit is initialized in the state vector

$$ |\psi_{\text{init}}\rangle = |+\rangle = \frac{1}{\sqrt{2}}[|\uparrow\rangle + |\downarrow\rangle], \quad (G13) $$

and evolves according to the Hamiltonian $H = \omega Z/2$. Suppose that the qubit is subject to continuous dephasing along the Z axis. This noise is represented by the Lindbladian jump operators

$$ L_0 = \sqrt{\gamma} |\uparrow\rangle\langle\uparrow|, \quad L_1 = \sqrt{\gamma} |\downarrow\rangle\langle\downarrow|. \quad (G14) $$

In vectorized operator notation (same conventions as in the appendices of our work, i.e., row-major convention), we have

$$ \begin{aligned} \mathcal{L}_1 &= \sum \left[ L_j \otimes L_j^T - \frac{1}{2} [L_j^\dagger L_j \otimes 1 + 1 \otimes (L_j^\dagger L_j)^T] \right] \\ &= \begin{bmatrix} 0 & -\gamma \\ -\gamma & 0 \end{bmatrix}; \end{aligned} $$

$$ \begin{aligned} \mathcal{L}_0 = (\dots) &= \begin{bmatrix} 0 & -i\omega \\ i\omega & 0 \end{bmatrix}, \quad \mathcal{E}_t = e^{t(\mathcal{L}_0 + \mathcal{L}_1)} \\ &= \begin{bmatrix} 1 & e^{-\gamma t - it\omega} \\ e^{-\gamma t + it\omega} & 1 \end{bmatrix}. \end{aligned} $$

The full evolution map, represented as an operator in terms of matrix elements $\rho_{ij} = \langle i|\rho|j\rangle$, is

$$ \mathcal{E}_t(\rho) = \begin{bmatrix} \rho_{00} & \rho_{01}e^{-it\omega-\gamma t} \\ \rho_{10}e^{it\omega-\gamma t} & \rho_{11} \end{bmatrix}. \quad (G15) $$

The next steps for this example are (a) a direct computation of Bob's sensitivity; (b) a calculation of Eve's sensitivity
to energy via our effective picture; and (c) an assessment of the error made in the approximation (140).

a. Direct computation of the sensitivity of the noisy
probe

At a time $t$, the state is

$$
\begin{align}
\rho(t) &= \frac{1}{2} \begin{bmatrix} 1 & e^{-it\omega-\gamma t} \\ e^{it\omega-\gamma t} & 1 \end{bmatrix} \\
&= U_t \begin{bmatrix} 1 & e^{-\gamma t} \\ e^{-\gamma t} & 1 \end{bmatrix} U_t^\dagger = U_t \frac{1+e^{-\gamma t}X}{2} U_t^\dagger \\
&= \frac{1+e^{-\gamma t}}{2} U_t|+\rangle\langle+| U_t^\dagger + \frac{1-e^{-\gamma t}}{2} U_t|-\rangle\langle-| U_t^\dagger, \tag{G16}
\end{align}
$$

where we use the shorthand $U_t = e^{-iHt}$. The last expres-
sion in Eq. (G16) provides a diagonal form for $\rho$, which
will serve in the calculation of the Fisher information. The
derivative of the state is

$$
\begin{align}
\mathcal{L}_{\text{tot}}[\rho(t)] = \dot{\rho}(t) &= \frac{1}{2} \nonumber \\
&\quad \begin{bmatrix} 0 & (-i\omega - \gamma)e^{-it\omega - \gamma t} \\ (i\omega - \gamma)e^{it\omega - \gamma t} & 0 \end{bmatrix} \nonumber \\
&= \frac{e^{-\gamma t}}{2} \left[ -\gamma U_t X U_t^\dagger + \omega U_t Y U_t^\dagger \right], \tag{G17}
\end{align}
$$

noting that $U_t X U_t^\dagger = \begin{bmatrix} 0 & e^{-it\omega} \\ e^{it\omega} & 0 \end{bmatrix}$ and $U_t Y U_t^\dagger =$
$\begin{bmatrix} 0 & -ie^{-it\omega} \\ ie^{it\omega} & 0 \end{bmatrix}$. We can interpret this derivative in terms

of two different dynamics: one $\propto \omega U_t Y U_t^\dagger$, which drives
the rotation around the Bloch sphere and one $\propto -\gamma U_t X U_t^\dagger$,
which drives decoherence (Fig. 14). The matrix elements
of the derivative in the eigenbasis $\{U_t|\pm\rangle\}$ of $\rho$ are

$$
\begin{align*}
\langle +|U_t^\dagger \dot{\rho} U_t|+\rangle &= -\gamma \frac{e^{-\gamma t}}{2}, & \langle +|U_t^\dagger \dot{\rho} U_t|-\rangle &= i\omega \frac{e^{-\gamma t}}{2}, \\
\langle -|U_t^\dagger \dot{\rho} U_t|+\rangle &= -i\omega \frac{e^{-\gamma t}}{2}, & \langle -|U_t^\dagger \dot{\rho} U_t|-\rangle &= \gamma \frac{e^{-\gamma t}}{2}, \tag{G18}
\end{align*}
$$

where $\langle +|Y|-\rangle = \langle +|YZ|+\rangle = i\langle +|X|+\rangle = i$. Now we compute the Fisher information using Eq. (12) as

$$
F_{\text{clock},t} = F(\rho(t_0); \dot{\rho}(t_0)) = \frac{2}{1 + e^{-\gamma t_0}} \left|\gamma \frac{e^{-\gamma t_0}}{2}\right|^2 + 2 \left|i\omega \frac{e^{-\gamma t_0}}{2}\right|^2 + 2 \left|i\omega \frac{e^{-\gamma t_0}}{2}\right|^2
$$

<figure><img src="image_20.png" /><figcaption>FIG. 14. Top view of the Bloch sphere for a single qubit pre-
pared in the +X eigenstate, evolving under the Hamiltonian
H = ωZ/2 and subject to continuous dephasing along the Z axis.
The derivative of the state can be decomposed into a “longitudi-
nal part” along σY (t) associated with the Hamiltonian dynamics,
and a “radial part” along −X (t) associated with the noise terms.
The assumption that enables the mapping from the Lindblad set-
ting to our bipartite uncertainty relation setting is that the noise
component (“radial” component) contributes negligibly to the
overall time sensitivity of the clock.</figcaption></figure>

$$
\begin{align}
&+ \frac{2}{1 - e^{-\gamma t_0}} \left| \gamma \frac{e^{-\gamma t_0}}{2} \right|^2 \nonumber \\
&= \omega^2 e^{-2\gamma t_0} + \gamma^2 \frac{2e^{-2\gamma t_0}}{1 - e^{-2\gamma t_0}}. \tag{G19}
\end{align}
$$

b. Eve's Fisher information with respect to energy

Now we turn to using the methods of our paper to char-
acterize the sensitivity of the noisy probe. As described in
Sec. VI A, we turn to computing

$$
F_{\text{clock,U,t}} = F(\rho(t_0); \mathcal{N}_{t_0}(\partial_t \psi(t_0))), \quad (\text{G20})
$$

for the instantaneous effective noisy channel $\mathcal{N}_t$ and fictitious unitary evolution $\psi(t)$ defined in Sec. VI A. We will then later discuss how good of an approximation $F_{\text{clock,U,t}}$ is to the original desired quantity $F_{\text{clock,t}}$.

We decompose the full evolution $\mathcal{E}_t$ as in Eq. (136).
Since $[H, L_j] = 0$, we have

$$
\begin{align}
\mathcal{N}_t &= e^{t\mathcal{L}_1} = \begin{bmatrix} 1 & e^{-\gamma t} \\ e^{-\gamma t} & 1 \end{bmatrix} \rightarrow \nonumber \\
\mathcal{N}_t(\rho) &= \begin{bmatrix} \rho_{00} & \rho_{01} e^{-\gamma t} \\ \rho_{10} e^{-\gamma t} & \rho_{11} \end{bmatrix}. \tag{G21}
\end{align}
$$

This channel can be described by the two Kraus operators

$$
E_0^{(t)} = \sqrt{\frac{1 + e^{-t\gamma}}{2}} \mathbf{1}; \quad E_1^{(t)} = \sqrt{\frac{1 - e^{-t\gamma}}{2}} \mathbf{Z}. \quad (G22)
$$
The (fictitious) pure unitary evolution of the initial state vector $|\psi_{\text{init}}\rangle = |+\rangle$ is

$$ \psi(t) = U_t \psi_{\text{init}} U_t^\dagger = \frac{1}{2} \begin{bmatrix} 1 & e^{-it\omega} \\ e^{it\omega} & 1 \end{bmatrix}. \quad (G23) $$

We compute Eve's Fisher information with respect to energy, which characterizes the sensitivity loss of the noisy probe. For any $t$, a complementary channel to Eq. (G21) is given by

$$ \hat{\mathcal{N}}_t(\rho) = \begin{bmatrix} \frac{1+e^{-\gamma t}}{2} \operatorname{tr}(\rho) & \frac{\sqrt{1-e^{-2\gamma t}}}{2} \operatorname{tr}(Z\rho) \\ \frac{\sqrt{1-e^{-2\gamma t}}}{2} \operatorname{tr}(Z\rho) & \frac{1-e^{-\gamma t}}{2} \operatorname{tr}(\rho) \end{bmatrix}. \quad (G24) $$

We would like to compute

$$ F(\hat{\mathcal{N}}_t(\psi); \hat{\mathcal{N}}_t(\{H - \langle H \rangle, \psi\})). \quad (G25) $$

Noting that $\langle H \rangle_{\psi(t)} = 0$ for all $t$ and that $\psi_{\text{init}} = (1 + X)/2$, we can compute

$$ \begin{aligned} \{H - \langle H \rangle, \psi\} &= \left\{ \frac{\omega}{2} Z, U_t \psi_{\text{init}} U_t^\dagger \right\} \\ &= \frac{\omega}{2} U_t \left\{ Z, \frac{1+X}{2} \right\} U_t^\dagger = \frac{\omega}{2} Z. \end{aligned} \quad (G26) $$

We then see that

$$ \begin{aligned} \hat{\mathcal{N}}_t(\psi) &= \begin{bmatrix} \frac{1+e^{-\gamma t}}{2} & 0 \\ 0 & \frac{1-e^{-\gamma t}}{2} \end{bmatrix}; \quad \hat{\mathcal{N}}_t\left(\frac{\omega}{2} Z\right) \\ &= \frac{\omega}{2} \begin{bmatrix} 0 & \sqrt{1-e^{-2\gamma t}} \\ \sqrt{1-e^{-2\gamma t}} & 0 \end{bmatrix}. \end{aligned} \quad (G27) $$

Then using Eq. (12) we find

$$ \begin{aligned} F(\hat{\mathcal{N}}(\psi); \hat{\mathcal{N}}(\{H - \langle H \rangle, \psi\})) \\ &= 0 + 2 \left[ \frac{\omega^2}{4} (1 - e^{-2\gamma t_0}) \right] + (\text{same term}) + 0 \\ &= \omega^2 (1 - e^{-2\gamma t_0}). \end{aligned} \quad (G28) $$

In the present picture of the effective noisy channel being applied instantly after unitary evolution of duration $t_0$, we see that Eve obtains no information about the energy direction for $t \approx 0$. However, for large $t$ Eve obtains near-perfect information, which hinders Bob's sensitivity. Since the noiseless Fisher information is $\omega^2$, we have via our uncertainty relation that

$$ F_{\text{Bob},t} = \omega^2 e^{-2\gamma t_0}. \quad (G29) $$

Our method therefore correctly gives us the first term in Eq. (G19). We can also check by direct calculation that the

first term in Eq. (G19) is indeed the Fisher information of the noisy clock state if we neglect the term in the derivative that is associated with the time derivative of the effective noise channel itself. First observe that

$$ \partial_t \psi = \frac{1}{2} \begin{bmatrix} 0 & -i\omega e^{-it\omega} \\ i\omega e^{it\omega} & 0 \end{bmatrix}, \quad (G30) $$

$$ \mathcal{N}(\partial_t \psi) = \frac{1}{2} \begin{bmatrix} 0 & -i\omega e^{-it\omega - \gamma t} \\ i\omega e^{it\omega - \gamma t} & 0 \end{bmatrix} = \frac{\omega e^{-\gamma t}}{2} U_t Y U_t^\dagger. \quad (G31) $$

We see that the object $\mathcal{N}(\partial_t \psi)$ is exactly the part of the derivative $\dot{\rho}$ with respect to the full dynamics that is associated with the Hamiltonian evolution of $\rho$, i.e., it is the “longitudinal” component of the derivative depicted in Fig. 14.

We use again Eq. (12) of our manuscript, recalling the diagonal form for $\rho$ given in Eq. (G16):

$$ F_{\text{clock,U},t} = F(\rho; \mathcal{N}(\partial_t \psi)) = 0 + 2 \left| \frac{\omega e^{-\gamma t_0}}{2} \right| + 2 \left| \frac{\omega e^{-\gamma t_0}}{2} \right| + 0 = \omega^2 e^{-2\gamma t_0}. \quad (G32) $$

The difference between $F_{\text{clock,U},t}$ and $F_{\text{clock},t}$ is

$$ \delta = F_{\text{clock},t} - F_{\text{clock,U},t} = \gamma^2 \frac{2e^{-2\gamma t_0}}{1 - e^{-2\gamma t_0}}. \quad (G33) $$

The relative error of the approximation is

$$ \frac{\delta}{F_{\text{clock,U},t}} = \frac{\gamma^2}{\omega^2} \frac{1}{1 - e^{-2\gamma t_0}}. \quad (G34) $$

(We computed the relative error with respect to $F_{\text{clock,U},t}$ because it is simpler.) We can see that $\delta$ is small relative to $F_{\text{clock,U},t}$ if the ratio $\gamma/\omega$ of the loss rate to the qubit's energy gap is small.

Numerical plots for $\omega = 1, \gamma = 0.1$ are presented in Fig. 15.

### c. Error bound for the mapping from the Lindblad master equation to our setting

As a sanity check we compute the error bound (146). We have

$$ \partial_t \mathcal{N} = \begin{bmatrix} 0 & -\gamma e^{-\gamma t} \\ -\gamma e^{-\gamma t} & 0 \end{bmatrix}, \quad (G35) $$
<figure><img src="image_21.png" /><figcaption>FIG. 15. Fisher information (FI) of a single qubit prepared in a +X eigenstate evolving according to the Hamiltonian $H = ωZ/2$ and subject to continuous dephasing along the Z axis. The horizontal axis represents the time $t_0$ at which we consider the clock sensitivity, and the vertical axis is the value of the different versions of the Fisher information (top plot) and relative Fisher information (bottom plot). The relative Fisher information is the Fisher information times $t_0^2$, which is relevant if we are interested in the relative sensitivity to time. In these plots we have set $ω = 1$ and $γ = 0.1$ (see main text). We verify from these plots that the time dependency $∂_t N_t$ of the effective Lnoisy channel contributes negligibly to the overall Fisher information; this example in the setting of continuous noise can therefore be reduced to a setting as in Fig. 1.</figcaption></figure>

and thus

$$ (\partial_t \mathcal{N})(\psi(t)) = \frac{1}{2} \begin{bmatrix} 0 & -\gamma e^{-\gamma t} e^{-i\omega t} \\ -\gamma e^{-\gamma t} e^{i\omega t} & 0 \end{bmatrix} = -\gamma e^{-\gamma t} U_t X U_t^\dagger. \quad (G36) $$

The matrix elements in the state's eigenbasis are

$$ \begin{aligned} \langle +|U_t^\dagger(\partial_t \mathcal{N})U_t|+\rangle &= -\gamma e^{-\gamma t}, & \langle +|U_t^\dagger(\partial_t \mathcal{N})U_t|-\rangle &= 0, \\ \langle -|U_t^\dagger(\partial_t \mathcal{N})U_t|+\rangle &= 0, & \langle -|U_t^\dagger(\partial_t \mathcal{N})U_t|-\rangle &= \gamma e^{-\gamma t}. \end{aligned} \quad (G37) $$

Then we can compute

$$ F(\rho; \partial_t \mathcal{N}(\psi)) = \frac{2}{1 + e^{-\gamma t_0}} |\gamma e^{-\gamma t_0}|^2 + \frac{2}{1 - e^{-\gamma t_0}} |\gamma e^{-\gamma t_0}|^2 = 4\gamma^2 \frac{e^{-2\gamma t_0}}{1 - e^{-2\gamma t_0}} \quad (G38) $$

Our bound (146) on the error $δ$ becomes

$$ \delta \le 4\gamma^2 \frac{e^{-2\gamma t_0}}{1 - e^{-2\gamma t_0}} + 2\gamma\omega \frac{e^{-2\gamma t_0}}{\sqrt{1 - e^{-2\gamma t_0}}}. \quad (G39) $$

The bound is consistent with our computed value of $δ$. However, in this case our bound is loose: The second term in our bound would suggest that the relative error with respect to $F_{\text{clock,U},t_0}$ behaves only as $γ/ω$ (if $γ \ll ω$), whereas we know from our explicit calculation of $δ$ that the behavior of this relative error is $γ^2/ω^2$.

### 3. Example: continuous dephasing noise along the transversal X axis

Consider the qubit state vector

$$ |\psi\rangle = |+\rangle = \frac{1}{\sqrt{2}} [|\uparrow\rangle + |\downarrow\rangle]. \quad (G40) $$

Suppose that the evolution of the qubit is given by the Lindbladian (132) with

$$ H = \frac{\omega}{2} Z, \quad L_0 = \sqrt{\gamma}|+\rangle\langle +|, \quad L_1 = \sqrt{\gamma}|-\rangle\langle -|. \quad (G41) $$