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Exponential enhancement of quantum metrology using continuous variables

Li Sun Institute of Fundamental and Frontier Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 610051, China    Xi He Institute of Fundamental and Frontier Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 610051, China    Chenglong You Quantum Photonics Laboratory, Department of Physics & Astronomy, Louisiana State University, Baton Rouge, LA 70803, USA    Chufan Lyu Institute of Fundamental and Frontier Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 610051, China    Bo Li The Key Laboratory of Mathematics Mechanization, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China    Seth Lloyd slloyd@mit.edu Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge, Massachusetts 02139, USA    Xiaoting Wang xiaoting@uestc.edu.cn Institute of Fundamental and Frontier Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 610051, China
(August 18, 2025)
Abstract

Coherence time is an important resource to generate enhancement in quantum metrology. In this work, based on continuous-variable models, we propose a new design of the signal-probe Hamiltonian which generates an exponential enhancement of measurement sensitivity. The key idea is to include into the system an ancilla that does not couple directly to the signal. An immediate benefit of such design is one can expand quantum Fisher information(QFI) into a power series in time, making it possible to achieve a higher-order time scaling in QFI. Specifically, one can design the interaction for a qubit-oscillator Ramsey interferometer to achieve a quartic time scaling, based on which, one can further design a chain of coupled harmonic oscillators to achieve an exponential time scaling in QFI. Our results show that linear scaling in both time and the number of coupling terms is sufficient to obtain exponential enhancement. Such exponential advantage is closely related to the characteristic commutation relations of quadratures.

Quantum metrology aims to study the limitation of the measurement accuracy governed by quantum mechanics and to explore how to achieve better measurement sensitivity with quantum resourcesCaves (1981); Giovannetti et al. (2006). In recent years, quantum metrology has been long pursued due to its vital importance in applied physics, such as gravitational wave detection Schnabel et al. (2010); Danilishin and Khalili (2012); Adhikari (2014), atomic clocks Derevianko and Katori (2011); Schioppo et al. (2017), quantum imaging Kolobov (1999); Dowling and Seshadreesan (2015); Magaña-Loaiza and Boyd (2019). A quantum metrology process includes three steps: preparing probes in a designed initial state; probes evolve under a parameter-dependent Hamiltonian HfH_{f} for time TT; the measurement of the final state. A well-studied example of parameter estimation is to estimate a parameter ff introduced in a Hamiltonian with the form of Hf=fH0H_{f}=fH_{0}, where H0H_{0} is a known Hamiltonian. The sensitivity of estimating this type of parameter scales as 1/(TΔH0)1/(T\Delta H_{0}), where ΔH0\Delta H_{0} is the standard deviation of H0H_{0}. There are two ways to improve measurement accuracy: either by increasing the coherent interaction time TT, or by maximizing the standard deviation of H0H_{0} through preparing the probe in a special entangled state. Specifically, with quantum entanglement, one can maximize ΔH0\Delta H_{0} scaling as ΔH0N\Delta H_{0}\propto N, where NN is the number of probes Giovannetti et al. (2006). Accordingly, the estimation sensitivity 1/(TN){1}/{(TN)} is considered as the Heisenberg limit, where NN can be perceived as the quantum parallel resource, and TT as the quantum serial resource Huelga et al. (1997); Shaji and Caves (2007).

Quantum metrology has been studied for a wide range of systems with quantum resource in parallel schemeGiovannetti et al. (2006, 2004); Boixo et al. (2007, 2008); Woolley et al. (2008); Anisimov et al. (2010); Giovannetti et al. (2011); Thomas-Peter et al. (2011); Hall and Wiseman (2012); Demkowicz-Dobrzański et al. (2012); You et al. (2019); Arvidsson-Shukur et al. (2020); You et al. (2020). For instance, with kk-body interactions between the probes, a sensitivity limit that scales as 1/Nk1/N^{k} can be obtained Boixo et al. (2007), while an exponential scaling can be achieved by introducing an exponentially large number of coupling terms Roy and Braunstein (2008). For quantum resource in serial scheme, in terms of the coherence time TT, many questions remain open. It has been shown that the minimum sensitivity Δf\Delta f scales as 1/T1/T with a time-independent Hamiltonian Huelga et al. (1997); Yuan and Fung (2015); Jones et al. (2020), while 1/T21/T^{2} scaling can be realized with a time-dependent Hamiltonian Pang and Jordan (2017); Naghiloo et al. (2017). Interesting open questions include: what is the ultimate limit of such enhancement, whether one can achieve sensitivity scaling 1/Tk1/T^{k} for arbitrary kk, or even the exponential scaling with the amount of other physical resources polynomial in TT?

Here we show that exponential sensitivity can be efficiently achieved with the number of coupling terms scaling linear with time. Specifically, we study a special type of models with the Hamiltonian in the form Hf=fH0+H1H_{f}=fH_{0}+H_{1}, where H0H_{0} is coupled with the signal ff and H1H_{1} is an auxiliary Hamiltonian not coupled directly to ff. Such model utilizes the non-commutativity of H0H_{0} and H1H_{1} to expand QFI into a power series in TT, permitting us to obtain higher-order time scaling. In the model of a qubit-harmonic-oscillator(HO) Ramsey interferometer, sensitivity characterized by QFI with time scaling of T4T^{4} can be achieved; in the second model with a chain of coupled harmonic resonators, the QFI can obtain an exponential improvement in the measurement accuracy. Remarkably, the second model only requires a polynomial (linear) scaling of coupling terms, which is crucial to justify the efficiency and the effectiveness of exponential enhancement. After all, it is of no surprise to realize exponential enhancement with an exponential amount of physical resource.

Time resource for quantum enhancement. — In quantum metrology, one aims to estimate a parameter ff encoded in the quantum dynamics from ν\nu repeated quantum measurements. The variance of the estimation or the measurement sensitivity Δf\Delta f is bounded by the quantum Cramér-Rao bound (QCRB): Δf1/νFQ\Delta f\geq 1/\sqrt{\nu F_{Q}}, where the QFI FQ=Tr(ρfLf2){F_{Q}}=\operatorname{Tr}(\rho_{f}L_{f}^{2}) corresponds to the minimum quantum measurement sensitivityBraunstein and Caves (1994); Braunstein et al. (1996). The parameter ff is first encoded into the Hamiltonian HfH_{f}, and then into the final state ρf\rho_{f} after an evolution time TT under HfH_{f}, with ρf=Ufρ0Uf\rho_{f}=U_{f}\rho_{0}U_{f}^{\dagger} and Uf=eiHfTU_{f}=e^{-iH_{f}T}. Here, LfL_{f} is called the symmetric logarithmic derivative (SLD), defined through the relation fρf=12(ρfLf+Lfρf)\partial_{f}{\rho_{f}}=\frac{1}{2}(\rho_{f}L_{f}+L_{f}\rho_{f}). Alternatively, QFI can be reformulated in terms of UfU_{f} and ρ0\rho_{0}. Specifically, for pure initial state ρ0=|ψ0ψ0|\rho_{0}=|\psi_{0}\rangle\langle\psi_{0}|, we have FQ=4Δ2h4ψ0|h2|ψ0|ψ0|h|ψ0|2F_{Q}=4\Delta^{2}h\equiv 4\langle\psi_{0}|h^{2}|\psi_{0}\rangle-|\langle\psi_{0}|h|\psi_{0}\rangle|^{2}, where hi(fUf)Uf=0TeiHftfHfeiHftdth\equiv i(\partial_{f}U_{f}^{{\dagger}})U_{f}=-\int_{0}^{T}e^{iH_{f}t}{\partial_{f}H_{f}}e^{-iH_{f}t}dt Holevo (2011); Liu et al. (2015). It turns out that such reformulation of FQI is very useful to understand the scheme of quantum enhancement. For example, in the standard setting of quantum metrology where Hf=fH0H_{f}=fH_{0}, the above formula gives FQ=4T2Δ2H0F_{Q}=4T^{2}\Delta^{2}H_{0}. Hence, one way to improve QFI is to increase the coherent interaction time TT. However, due to the decoherence and dephasing effects, TT cannot be extended arbitrarily long to improve QFI. In comparison, achieving higher-order terms of TT in QFI is a more efficient and practical method to improve the estimation precision.

In order to find the relationship between FQF_{Q} and TT, we consider a general form of HfH_{f} on ff and rewrite hh into the following polynomial expansion in TT:

h=j=0ijTj+1(j+1)!Cj,\displaystyle h=-\sum_{j=0}^{\infty}\frac{i^{j}T^{j+1}}{(j+1)!}C_{j}, (1)

where Cj[Hf,Cj1]C_{j}\equiv[H_{f},C_{j-1}] is the jjth-order commutator of HfH_{f} and fHf\partial_{f}H_{f} with C0=fHfC_{0}=\partial_{f}H_{f} Kumar (1965); Wilcox (1967); Pang and Brun (2014); Liu et al. (2015). Thus, we can obtain the expression of QFI:

FQ=4Δ2(j=0ijTj+1(j+1)!Cj).\displaystyle F_{Q}=4\Delta^{2}\Big{(}\sum_{j=0}^{\infty}\frac{i^{j}T^{j+1}}{(j+1)!}C_{j}\Big{)}. (2)

Notice that the series in hh could be of finite length if Cj=0C_{j}=0 for some jj. If CjC_{j} are local operators on different subsystems, then for a separable initial state |ψ|\psi\rangle, we have the covariance Cov(Ck,Cj)ψ|CkCj|ψψ|Ck|ψψ|Cj|ψ=0{\rm Cov}(C_{k},C_{j})\equiv\langle\psi|C_{k}C_{j}|\psi\rangle-\langle\psi|C_{k}|\psi\rangle\langle\psi|C_{j}|\psi\rangle=0, and the QFI can be further simplified into

FQ=4j=0(1)jT2j+2[(j+1)!]2Δ2Cj.\displaystyle F_{Q}=4\sum_{j=0}^{\infty}\frac{(-1)^{j}T^{2j+2}}{[(j+1)!]^{2}}\Delta^{2}C_{j}. (3)

As shown in Eq. (3), T2j+2T^{2j+2} appears in the expression of QFI together with the variance of the jjth-order commutator CjC_{j} in initial state, which provides a possibility to realize higher-order time scaling in QFI.

In the following, we will design the Hamiltonian into the form Hf=fH0+H1H_{f}=fH_{0}+H_{1} such that an ancilla probe is introduced to the system and does not couple directly with the signal ff. We will explore the role played by the commutator [H0,H1][H_{0},H_{1}] in generating higher-order term in coherence time in QFI. In particular, we propose two specific models, the qubit-HO Ramsey interferometer and a chain of coupled harmonic oscillator to investigate the sensitivity enhancement.

Refer to caption
Figure 1: The qubit-HO Ramsey interferometer: the system is composed of a qubit coupled with a HO through the force ff acting on the HO. The qubit rotates π/2\pi/2 along the xx-axis firstly. Then ff is encoded on the HO and the system evolves under HfH_{f}, which makes the qubit rotate ϕ\phi along the zz-axis. Lastly, the qubit rotates π/2\pi/2 along the yy-axis.

Achieving quartic time scaling in QFI. — We start our analysis with a qubit-HO Ramsey interferometry model. As shown in Fig. 1, the initial state of the qubit and the HO is prepared into a separable state |φ0=|ψ0|0|\varphi_{0}\rangle=|\psi_{0}\rangle\otimes|0\rangle, where |ψ0|\psi_{0}\rangle is the initial state of HO and |0|0\rangle (|1|1\rangle) is the ground (excited) state of the qubit. The qubit is subsequently subjected to a π/2\pi/2 pulse along yy-axis, transforming it into 12(|0+|1)\frac{1}{{\sqrt{2}}}(|0\rangle+|1\rangle). Afterwards, the system evolves under the Hamiltonian Hf=fX+gPσzH_{f}=fX+gP\sigma_{z} for time TT, where X=12(a+a)X=\frac{1}{\sqrt{2}}(a^{\dagger}+a), P=i2(aa)P=\frac{i}{\sqrt{2}}({a^{\dagger}}-a), σz\sigma_{z} is the Pauli-Z gate, aa is the annihilation operator and gg is the qubit-HO coupling strength. Finally, another π/2\pi/2-pulse around the xx-axis is applied to the ancilla qubit.

In order to obtain QFI, we calculate commutators as: C0=XC_{0}=X, C1=[(fX+gPσz),X]=igσzC_{1}=[(fX+gP\sigma_{z}),X]=-ig\sigma_{z} and Cj=0C_{j}=0 for j2j\geq 2. Since the initial state of the system is |φ0=12(|0+|1)|ψ0|\varphi_{0}\rangle=\frac{1}{{\sqrt{2}}}(|0\rangle+|1\rangle)|\psi_{0}\rangle, thus Cov(Ck,Cj)=0{\rm Cov}(C_{k},C_{j})=0. According to Eq. (3) we calculate QFI as:

FQ=4Δ2(12gT2σz+TX)=g2T4Δ2σz+4T2Δ2X\displaystyle F_{Q}=4\Delta^{2}(\frac{1}{2}gT^{2}\sigma_{z}+TX)=g^{2}T^{4}\Delta^{2}\sigma_{z}+4T^{2}\Delta^{2}X (4)

for a separable |φ0|\varphi_{0}\rangle. In particular, if |ψ0=|0|\psi_{0}\rangle=|0\rangle is the vacuum state, then FQ=g2T4+4T2F_{Q}=g^{2}T^{4}+4T^{2}; if |ψ0=S(r)|0|\psi_{0}\rangle=S(r)|0\rangle, where S(r)S(r) is the squeeze operator and r>0r>0, then FQg2T4F_{Q}\approx g^{2}T^{4} for sufficiently large rr. In both cases, a T4T^{4}-scaling of QFI can be achieved. With a strong coupling gg and relatively long coherent interaction time TT, a T2T^{2} enhancement of Δf\Delta f can be obtained. As shown in Eq. (4), the higher-order term g2T4g^{2}T^{4} originates from the non-commutativity between XX and PP. Therefore, the non-commutativity in the Hamiltonian could be a useful quantum resource to enhance estimation precision.

The QCRB only gives the optimal lower bound for Δf\Delta f. We still need to show there exits an optimal quantum measurement to saturate this bound Zhou et al. (2020). One such optimal measurement strategy is given by the measurement observable M=ΠσxM=\Pi\otimes\sigma_{x} where Π=(1)aa\Pi=(-1)^{a^{\dagger}a} is the parity operator on the quantum HO, satisfying Πψ(x)=ψ(x)\Pi\psi(x)=\psi(-x). By preparing the HO in the initial state |ψ0=S(r)|0|\psi_{0}\rangle=S(r)|0\rangle, we can calculate Δf\Delta f via the error propagation formula:

Δf=ΔM|Mf|1gT2=1FQ, for sufficiently large r,\displaystyle\Delta f=\frac{{\Delta M}}{{|{\frac{{\partial\langle M\rangle}}{{\partial f}}}|}}\approx\frac{1}{gT^{2}}=\frac{1}{\sqrt{F_{Q}}},\text{ for sufficiently large }r,

where ΔM\Delta M and M\langle M\rangle are the variance and the expectation value of MM in the final state |φf|\varphi_{f}\rangle (See details in Supplementary Materials). Thus, a T2T^{2} enhancement of Δf\Delta f can be obtained under such measurement design.

Furthermore, the non-commutativity and entanglement can be used simultaneously to increase estimation precision. We design a system of nn non-interacting HOs, each of which is coupled with the global force ff, and an ancila qubit, under the Hamiltonian Hf=fk=1nXk+gk=1nPkσz(k)H_{f}=f\sum_{k=1}^{n}X_{k}+g\sum_{k=1}^{n}P_{k}\sigma_{z}^{(k)}. By analogy with Eq. (4), if the nn qubits are in the GHZ state, we can calculate the QFI according to Eq. (2):

FQ\displaystyle F_{Q} =g2T4Δ2(knσz(k))+4T2Δ2(knXk)\displaystyle=g^{2}T^{4}\Delta^{2}(\sum_{k}^{n}\sigma_{z}^{(k)})+4T^{2}\Delta^{2}(\sum_{k}^{n}X_{k})
=n2g2T4+4T2Δ2(knXk).\displaystyle=n^{2}g^{2}T^{4}+4T^{2}\Delta^{2}(\sum_{k}^{n}X_{k}). (5)

Hence, a quadratic improvement with respect to nn is obtained in QFI, multiplied by the quartic time scaling, compared with the scheme where the nn qubits are in a separable state.

Achieving T2n+2T^{2n+2} time scaling in QFI. — Next, we continue the exploration of the choice of the Hamiltonian HfH_{f} in order to generate higher-order terms of TT in QFI, based on the intuition gained from the qubit-HO Ramsey interferometry model. The first attempt is to choose Hf=fXn+gPσzH_{f}=fX^{n}+gP\sigma_{z}, which gives:

Cj=(ig)jn!(nj)!Xnjσzj.\displaystyle C_{j}=\frac{(-ig)^{j}n!}{(n-j)!}X^{n-j}\sigma_{z}^{j}. (6)

in Eq. (3). For odd nn, and for initial state |φ0=12|ψ0(|0+|1)|\varphi_{0}\rangle=\frac{1}{\sqrt{2}}|\psi_{0}\rangle(|0\rangle+|1\rangle), where |ψ0=|0|\psi_{0}\rangle=|0\rangle is the vacuum state, QFI is further simplified into:

FQ=\displaystyle F_{Q}= 4Δ2(j=0n(gT)j+1(nj)g(j+1)Xnjσzj)=4(gT)2n+2g2(n+1)2\displaystyle 4\Delta^{2}\big{(}\sum_{j=0}^{n}\frac{(gT)^{j+1}\binom{n}{j}}{g(j+1)}X^{n-j}\sigma_{z}^{j}\big{)}=\frac{4(gT)^{2n+2}}{g^{2}(n+1)^{2}}
+j,k=0n14(gT)j+k+2(nk)(nj)g2(j+1)(k+1)(2njk1)!!(2)2njk,\displaystyle+\sum_{j,k=0}^{n-1}\frac{4(gT)^{j+k+2}\tbinom{n}{k}\tbinom{n}{j}}{g^{2}(j+1)(k+1)}\frac{(2n-j-k-1)!!}{(\sqrt{2})^{2n-j-k}},

where j+kj+k is even in the summation. Thus we can achieve T2n+2T^{2n+2} time scaling in QFI. For instance, for n=3n=3, we have

FQ=152T2+514g2T4+72g4T6+14g6T8.\displaystyle F_{Q}=\frac{15}{2}T^{2}+\frac{51}{4}g^{2}T^{4}+\frac{7}{2}g^{4}T^{6}+\frac{1}{4}g^{6}T^{8}.

Nevertheless, it is difficult to experimentally implement such Hamiltonian XnX^{n}. Alternatively, we can design HfH_{f} to be a chain of nn coupled HOs with common HO-HO interactions to reach higher-order TT scaling in QFI.

Refer to caption
Figure 2: A model containing a chain of nn coupled HOs, where the signal ff couples to the first HO via the interaction fX1fX_{1}, and each HO interacts with nearest neighbors via the interaction gjPjXj+1g_{j}P_{j}X_{j+1}.
Refer to caption
Refer to caption
Figure 3: Optimal averaged QFI also demonstrates exponential enhancement. In (a), for gT=4gT=4, the averaged QFI versus nn is plotted, with the maximal value achieved at n=4n=4. In (b), the curves for FQ¯(n0(gT),gT)\overline{F_{Q}}(n_{0}(gT),gT) and FQ¯(3gT,gT)\overline{F_{Q}}(\lfloor 3gT\rfloor,gT) grow exponentially with gTgT, while the curve for FQ¯(n=2,gT)\overline{F_{Q}}(n=2,gT) grows polynomially with gTgT.

Achieving exponential enhancement of QFI. — As shown in Fig. 2, we design a chain of nn coupled HOs, where the parameter ff is coupled with the first HO, and each HO interacts only with its adjacent neighbors, characterized by the total Hamiltonian Hf=fH0+H1=fX1+j=1n1gjPjXj+1H_{f}=fH_{0}+H_{1}=fX_{1}+\sum^{n-1}_{j=1}g_{j}P_{j}X_{j+1}, with gjg_{j} as the coupling constant, and XjX_{j} and PjP_{j} as the quadratures of the jjth HO. The signal ff to be estimated is coupled to the first HO through X1X_{1}, and we introduce n1n-1 additional HOs as an ancila for better measurement precision. This model is also known as the bosonic Kitaev-Majorana chain McDonald et al. (2018). Analogous to the previous setup, after the evolution Uf=ei(fH0+H1)TU_{f}=e^{-i(fH_{0}+H_{1})T}, we can define hh and calculate the commutators:

Cj=[Hf,Cj1]=(ig)jXj+1.\displaystyle C_{j}=[H_{f},C_{j-1}]=(-ig)^{j}X_{j+1}. (7)

which are local operators on different subsystems. Hence, if the entire system is prepared in a separable initial state, we have Cov(Ck,Cj)=0{\rm Cov}(C_{k},C_{j})=0. In order to analyze the expression of FQF_{Q}, we further assume all HOs are prepared in the same initial state, which gives Δ2Xj+1=Δ2X\Delta^{2}X_{j+1}=\Delta^{2}X. In this case Eq. (3) is reduced to

FQ=4Δ2Xg2j=1n((gT)jj!)2,\displaystyle F_{Q}=\frac{4\Delta^{2}X}{g^{2}}\sum\limits_{j=1}^{n}\Big{(}\frac{(gT)^{j}}{j!}\Big{)}^{2}, (8)

One can further show the following inequality for gT2gT\geq 2 and n=3gTn=\lfloor 3gT\rfloor:

FQ\displaystyle{F_{Q}} =4Δ2Xg2j=13gT((gT)jj!)2>2Δ2Xg2egT.\displaystyle=\frac{4\Delta^{2}X}{g^{2}}\sum\limits_{j=1}^{\lfloor 3gT\rfloor}\Big{(}\frac{(gT)^{j}}{j!}\Big{)}^{2}>\frac{2\Delta^{2}X}{g^{2}}e^{gT}. (9)

Detailed proof of Eq. (9) can be found in the appendix. The intuition behind the proof is, the expression of FQF_{Q} in Eq. (9) is very similar to the first nn terms in the Taylor expansion of egTe^{gT}. The key point is, given the value of gTgT, one needs to determine how large nn has to be so that FQF_{Q} can scale as egTe^{gT} up to a constant. It turns out that n=3gTn=\lfloor 3gT\rfloor is sufficient to make QFI scale exponentially with respect to gTgT, as long as gT2gT\geq 2. Notice that such exponential enhancement of QFI only requires n1n-1 number of HO-HO nearest-neighbor interactions among nn HOs, which is crucial to justify the efficiency and the effectiveness of the proposal for exponential enhancement. In addition, we can also design an optimal measurement to saturate the QCRB for such exponential scaling of QFI. Specifically, we can choose the parity operator M=k=1nΠkM=\otimes_{k=1}^{n}\Pi_{k} as the measurement observable to reach the exponential enhancement of Δf\Delta f when nn and TT grows linearly under the condition n=3gTn=\lfloor 3gT\rfloor Li et al. (2021).

Achieving the optimal averaged QFI for exponential enhancement — It turns out that n=3gTn=\lfloor 3gT\rfloor is only sufficient but not necessary or efficient to obtain exponential enhancement of QFI in the previous model. In fact, we can define the averaged QFI per HO as FQ¯(n,gT)=FQ/n\overline{F_{Q}}(n,gT)=F_{Q}/n, which is a function of both nn and gTgT. For a fixed gTgT, FQ¯\overline{F_{Q}} is found to first increase and then decrease as nn grows, as shown in Fig. 3(a); in other words, there exists an optimal value of n=n0(gT)n=n_{0}(gT) to reach to the maximum of FQ¯\overline{F_{Q}} for fixed gTgT. One can then numerically find the value n0(gT)n_{0}(gT) for different values of gTgT. Hence, FQ¯(n0(gT),gT)\overline{F_{Q}}(n_{0}(gT),gT) can be plotted as a curve against gTgT in Fig. 3(b), and in the plot it seems to grow exponentially with gTgT. For comparison, we also plot FQ¯(3gT,gT)\overline{F_{Q}}(\lfloor 3gT\rfloor,gT) and FQ¯(n=2,gT)\overline{F_{Q}}(n=2,gT) in the same figure. Analogous to the previous discussion, one can rigorously prove that for gT2gT\geq 2 and n=3gTn=\lfloor 3gT\rfloor, FQ¯\overline{F_{Q}} grows exponentially with gTgT, as demonstrated in Fig. 3(b); nevertheless, the corresponding FQ¯\overline{F_{Q}} is far from being optimal, compared to the FQ¯\overline{F_{Q}} curve for n=n0(gT)n=n_{0}(gT). Hence, the n=3gTn=\lfloor 3gT\rfloor condition is only helpful to construct the rigorous proof for exponential enhancement, but not necessary. In practice, n=n0(gT)n=n_{0}(gT) should be enough to achieve the exponential behavior. Moreover, keeping nn growing with gTgT is crucial to obtain the exponential enhancement; for a fixed value of nn, FQ¯\overline{F_{Q}} only grows polynomially against gTgT, as illustrated in Fig. 3(b).

Conclusion.—As a type of quantum resource, coherent interaction time plays a crucial role in quantum precise measurement. By introducing an auxiliary system which do not couple directly to the signal to be estimated, we can express QFI as a power series in coherence time. For the qubit-oscillator Ramsey interferometer model, QFI has a quartic time scaling; for a chain of coupled harmonic oscillators, with the number of coupling terms growing linearly with time, the corresponding QFI is shown to have an exponential time scaling. Our results suggest that linear scaling in both time and the number of coupling terms is sufficient to obtain exponential enhancement in continuous-variable quantum metrology.

This research was supported by the National Key R&\&D Program of China, Grant No. 2018YFA0306703.

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Measurement design to achieve 1/T21/T^{2} scaling in measurement uncertainty in the qubit-harmonic-oscillator model

In this work, we use the following definition of the quadratures,

X\displaystyle X 12(a+a)\displaystyle\equiv\frac{1}{\sqrt{2}}(a+a^{\dagger})
P\displaystyle P i2(aa)\displaystyle\equiv\frac{i}{\sqrt{2}}(a^{\dagger}-a)

satisfying [X,P]=i[X,P]=i.

Let’s consider a quantum harmonic oscillator(HO) system coupled with a signal ff, under the Hamiltonian H0=fXH_{0}=fX. Our aim is to measure ff as accurately as possible. We introduce a probing qubit to interact with the HO through the Hamiltonian H1=gσzPH_{1}=g\sigma_{z}\otimes P. The total Hamiltonian of this qubit-HO model is Hf=H0+H1=fX+gσzPH_{f}=H_{0}+H_{1}=fX+g\sigma_{z}P. Let the initial state of the qubit-HO system to be |φ0|\varphi_{0}\rangle, its unitary evolution after time TT is described by:

Uf=eiT(fX+gPσz)\displaystyle U_{f}=e^{-iT(fX+gP\sigma_{z})} =eiTgPσzeiTfXei2fgT2σz=eiTgPσzi2fgT2σzeiTfX\displaystyle=e^{-iTgP\sigma_{z}}e^{-iTfX}e^{-\frac{i}{2}fgT^{2}\sigma_{z}}=e^{-iTgP\sigma_{z}-\frac{i}{2}fgT^{2}\sigma_{z}}e^{-iTfX}

Thus, the information about ff is encoded into UfU_{f}. If we choose the initial state as a separable state |φ0=12(|0+|1)|ψ0|\varphi_{0}\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|\psi_{0}\rangle, then the final state after UfU_{f} becomes:

|φf=Uf|φ0\displaystyle|\varphi_{f}\rangle=U_{f}|\varphi_{0}\rangle =12eiTgPσzi2fgT2σz(|0+|1)eiTfX|ψ0\displaystyle=\frac{1}{\sqrt{2}}e^{-iTgP\sigma_{z}-\frac{i}{2}fgT^{2}\sigma_{z}}(|0\rangle+|1\rangle)e^{-iTfX}|\psi_{0}\rangle
=12(eiTgPi2fgT2|0+eiTgP+i2fgT2|1)eiTfX|ψ0\displaystyle=\frac{1}{\sqrt{2}}\Big{(}e^{-iTgP-\frac{i}{2}fgT^{2}}|0\rangle+e^{iTgP+\frac{i}{2}fgT^{2}}|1\rangle\Big{)}e^{-iTfX}|\psi_{0}\rangle
=ei2fgT22(|0eiTgPeiTfX|ψ0+eifgT2|1eiTgPeiTfX|ψ0)\displaystyle=\frac{e^{-\frac{i}{2}fgT^{2}}}{\sqrt{2}}\Big{(}|0\rangle e^{-iTgP}e^{-iTfX}|\psi_{0}\rangle+e^{ifgT^{2}}|1\rangle e^{iTgP}e^{-iTfX}|\psi_{0}\rangle\Big{)}

The corresponding QFI is:

FQ\displaystyle F_{Q} =4Δ2(j=0ijTj+1(j+1)!Cj)=4Δ2(TX+12T2gσz)=4T2Δ2X+g2T4Δ2σz\displaystyle=4\Delta^{2}(\sum_{j=0}^{\infty}\frac{i^{j}T^{j+1}}{(j+1)!}C_{j})=4\Delta^{2}(TX+\frac{1}{2}T^{2}g\sigma_{z})=4T^{2}\Delta^{2}X+g^{2}T^{4}\Delta^{2}\sigma_{z}

where C0=XC_{0}=X and C1=igσzC_{1}=-ig\sigma_{z}, and Cj=0C_{j}=0 for j2j\geq 2. In particular, if we further choose the initial state of the HO to be |ψ0=|x=0|\psi_{0}\rangle=|x=0\rangle, an eigenstate of XX at position x=0x=0, then |φ0=12(|0+|1)|x=0|\varphi_{0}\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|x=0\rangle, Δ2X=0\Delta^{2}X=0, Δ2σz=1\Delta^{2}\sigma_{z}=1, and

FQ\displaystyle F_{Q} =4T2Δ2X+g2T4Δ2σz=g2T4\displaystyle=4T^{2}\Delta^{2}X+g^{2}T^{4}\Delta^{2}\sigma_{z}=g^{2}T^{4}

Hence the minimum measurement uncertainty for such |ψ0|\psi_{0}\rangle is minΔf=1FQ=1gT2\min\Delta f=\frac{1}{\sqrt{F_{Q}}}=\frac{1}{gT^{2}}.

Next, we show there does exist a quantum measurement that can achieve such minimum measurement accuracy. Let’s choose M=σxΠM=\sigma_{x}\otimes\Pi where Π\Pi is the parity operator on the quantum HO satisfying Πψ(x)=ψ(x)\Pi\psi(x)=\psi(-x). It turns out that Π=eiπaa=(1)aa\Pi=e^{i\pi a^{\dagger}a}=(-1)^{a^{\dagger}a}.

For the initial state |φ0=12(|0+|1)|x=0|\varphi_{0}\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|x=0\rangle, the final state becomes:

|φf=Uf|φ0=ei2fgT22(|0|gT+eifgT2|1|gT)\displaystyle|\varphi_{f}\rangle=U_{f}|\varphi_{0}\rangle=\frac{e^{-\frac{i}{2}fgT^{2}}}{\sqrt{2}}(|0\rangle|gT\rangle+e^{ifgT^{2}}|1\rangle|-gT\rangle)

Hence, the error propagation formula gives:

M\displaystyle\langle M\rangle =12(0|gT|+eifgT21|gT|)(σxΠ)(|0|gT+eifgT2|1|gT)\displaystyle=\frac{1}{2}(\langle 0|\langle gT|+e^{-ifgT^{2}}\langle 1|\langle-gT|)(\sigma_{x}\otimes\Pi)(|0\rangle|gT\rangle+e^{ifgT^{2}}|1\rangle|-gT\rangle)
=12(0|gT|+eifgT21|gT|)(|1|gT+eifgT2|0|gT)\displaystyle=\frac{1}{2}(\langle 0|\langle gT|+e^{-ifgT^{2}}\langle 1|\langle-gT|)(|1\rangle|-gT\rangle+e^{ifgT^{2}}|0\rangle|gT\rangle)
=12(eifgT2+eifgT2)=cos(fgT2)\displaystyle=\frac{1}{2}(e^{-ifgT^{2}}+e^{ifgT^{2}})=\cos(fgT^{2})
M2\displaystyle\langle M^{2}\rangle =1\displaystyle=1
Δ2M\displaystyle\Delta^{2}M =sin(2fgT2)\displaystyle=\sin(2fgT^{2})
Δf\displaystyle\Delta_{f} =ΔM|fM|=|sin(fgT2)|gT2sin(fgT2)|=1gT2=1FQ\displaystyle=\frac{\Delta M}{|\partial_{f}\langle M\rangle|}=\frac{|\sin(fgT^{2})|}{gT^{2}\sin(fgT^{2})|}=\frac{1}{gT^{2}}=\frac{1}{\sqrt{F_{Q}}}

In practice, we cannot exactly prepare the HO in the unphysical state |x=0|x=0\rangle; instead, we can prepare the HO in a squeezed state centered as the zero position to approximate |x=0|x=0\rangle. In the case, the final state after UfU_{f} becomes

|φf\displaystyle|\varphi_{f}\rangle =ei2fgT22(|0eigPT|α0+eifgT2|1eigPT|α0)\displaystyle=\frac{e^{-\frac{i}{2}fgT^{2}}}{\sqrt{2}}\Big{(}|0\rangle e^{-igPT}|\alpha_{0}\rangle+e^{ifgT^{2}}|1\rangle e^{igPT}|\alpha_{0}\rangle\Big{)}
=12(|0|gT+eifgT2|1|gT)\displaystyle=\frac{1}{\sqrt{2}}(|0\rangle|gT\rangle+e^{ifgT^{2}}|1\rangle|-gT\rangle)

For Fock state |n|n\rangle, in the position representation, ψn(x)n(x)=x|n\psi_{n}(x)\equiv n(x)=\langle x|n\rangle. ψn\psi_{n} are real functions: \mathbb{R}\to\mathbb{R}. ψn\psi_{n} is an even-function for an even nn, and an odd-function for an odd nn. Hence, Π^ψn(x)=ψn(x)=(1)nψn(x)\hat{\Pi}\psi_{n}(x)=\psi_{n}(-x)=(-1)^{n}\psi_{n}(x).

α(x)\displaystyle\alpha(x) =x|α=e|α|22nαnn!ψn(x)\displaystyle=\langle x|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}\psi_{n}(x)
Π^α(x)\displaystyle\hat{\Pi}\alpha(x) =e|α|22nαnn!Π^ψn(x)=e|α|22n(α)nn!ψn(x)=(α)(x)\displaystyle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}\hat{\Pi}\psi_{n}(x)=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{(-\alpha)^{n}}{\sqrt{n!}}\psi_{n}(x)=(-\alpha)(x)

Hence,

Π^|α\displaystyle\hat{\Pi}|\alpha\rangle =|α\displaystyle=|-\alpha\rangle

The Displacement operator is given by

D(α)=eαa^αa^\displaystyle D(\alpha)=e^{\alpha\hat{a}^{\dagger}-\alpha^{\ast}\hat{a}}

A few important facts:

eigTP\displaystyle e^{-igTP} =e12gT(aa)=D(12gT)\displaystyle=e^{\frac{1}{\sqrt{2}}gT(a^{\dagger}-a)}=D(\frac{1}{\sqrt{2}}gT)
eifTX\displaystyle e^{-ifTX} =ei2fT(a+a)=D(i2fT)\displaystyle=e^{-\frac{i}{\sqrt{2}}fT(a^{\dagger}+a)}=D(-\frac{i}{\sqrt{2}}fT)
D(α)D(β)\displaystyle D(\alpha)D(\beta) =e(αβαβ)D(α+β)\displaystyle=e(\alpha\beta^{\ast}-\alpha^{\ast}\beta)D(\alpha+\beta)
D(12gT)D(i2fT)\displaystyle D(\frac{1}{\sqrt{2}}gT)D(-\frac{i}{\sqrt{2}}fT) =D(12(gTifT))ei2fgT2\displaystyle=D(\frac{1}{\sqrt{2}}(gT-ifT))e^{\frac{i}{2}fgT^{2}}
D(12gT)D(i2fT)\displaystyle D(-\frac{1}{\sqrt{2}}gT)D(-\frac{i}{\sqrt{2}}fT) =D(12(gT+ifT))ei2fgT2\displaystyle=D(-\frac{1}{\sqrt{2}}(gT+ifT))e^{-\frac{i}{2}fgT^{2}}

Hence, the final state:

|φf=\displaystyle|\varphi_{f}\rangle= 12(|0ei2fgT2eigPTeifTX|ϕ0+|1ei2fgT2eigPTeifTX|ϕ0)\displaystyle\frac{1}{\sqrt{2}}\Big{(}|0\rangle e^{-\frac{i}{2}fgT^{2}}e^{-igPT}e^{-ifTX}|\phi_{0}\rangle+|1\rangle e^{\frac{i}{2}fgT^{2}}e^{igPT}e^{-ifTX}|\phi_{0}\rangle\Big{)}
=\displaystyle= 12(|0ei2fgT2D(12gT)D(i2fT)|ϕ0+|1ei2fgT2D(12gT)D(i2fT)|ϕ0)\displaystyle\frac{1}{\sqrt{2}}\Big{(}|0\rangle e^{-\frac{i}{2}fgT^{2}}D(\frac{1}{\sqrt{2}}gT)D(-\frac{i}{\sqrt{2}}fT)|\phi_{0}\rangle+|1\rangle e^{\frac{i}{2}fgT^{2}}D(-\frac{1}{\sqrt{2}}gT)D(-\frac{i}{\sqrt{2}}fT)|\phi_{0}\rangle\Big{)}
=\displaystyle= 12(|0D(12(gTifT))|ϕ0+|1D(12(gTifT))|ϕ0)\displaystyle\frac{1}{\sqrt{2}}\Big{(}|0\rangle D(\frac{1}{\sqrt{2}}(gT-ifT))|\phi_{0}\rangle+|1\rangle D(\frac{1}{\sqrt{2}}(-gT-ifT))|\phi_{0}\rangle\Big{)}

If we choose the initial state of HO to be the squeezed vacuum state |ϕ0=S|0|\phi_{0}\rangle=S|0\rangle, then the final state becomes:

|φf=12(|0D(12(gTifT))|ϕ0+|1D(12(gTifT))|ϕ0)\displaystyle|\varphi_{f}\rangle=\frac{1}{\sqrt{2}}\Big{(}|0\rangle D(\frac{1}{\sqrt{2}}(gT-ifT))|\phi_{0}\rangle+|1\rangle D(\frac{1}{\sqrt{2}}(-gT-ifT))|\phi_{0}\rangle\Big{)}

For M=σxΠM=\sigma_{x}\otimes\Pi, we have:

M|ψf\displaystyle M|\psi_{f}\rangle =12(|1D(12(gT+ifT))|ϕ0+|0D(12(gT+ifT))|ϕ0)\displaystyle=\frac{1}{\sqrt{2}}\Big{(}|1\rangle D(\frac{1}{\sqrt{2}}(-gT+ifT))|\phi_{0}\rangle+|0\rangle D(\frac{1}{\sqrt{2}}(gT+ifT))|\phi_{0}\rangle\Big{)}
ψf|\displaystyle\langle\psi_{f}| =12(0|ϕ0|D(12(gT+ifT))+1|ϕ0|D(12(gT+ifT)))\displaystyle=\frac{1}{\sqrt{2}}\Big{(}\langle 0|\langle\phi_{0}|D(\frac{1}{\sqrt{2}}(-gT+ifT))+\langle 1|\langle\phi_{0}|D(\frac{1}{\sqrt{2}}(gT+ifT))\Big{)}
M\displaystyle\langle M\rangle =12(0|ϕ0|D(12(gT+ifT))+1|ϕ0|D(12(gT+ifT)))\displaystyle=\frac{1}{2}\Big{(}\langle 0|\langle\phi_{0}|D(\frac{1}{\sqrt{2}}(-gT+ifT))+\langle 1|\langle\phi_{0}|D(\frac{1}{\sqrt{2}}(gT+ifT))\Big{)}
×(|1D(12(gT+ifT))|ϕ0+|0D(12(gT+ifT))|ϕ0)\displaystyle\times\Big{(}|1\rangle D(\frac{1}{\sqrt{2}}(-gT+ifT))|\phi_{0}\rangle+|0\rangle D(\frac{1}{\sqrt{2}}(gT+ifT))|\phi_{0}\rangle\Big{)}
=12(ϕ0|D(12(gT+ifT))D(12(gT+ifT))|ϕ0+ϕ0|D(12(gT+ifT)))D(12(gT+ifT))|ϕ0\displaystyle=\frac{1}{2}(\langle\phi_{0}|D(\frac{1}{\sqrt{2}}(-gT+ifT))D(\frac{1}{\sqrt{2}}(gT+ifT))|\phi_{0}\rangle+\langle\phi_{0}|D(\frac{1}{\sqrt{2}}(gT+ifT)))D(\frac{1}{\sqrt{2}}(-gT+ifT))|\phi_{0}\rangle
=12(eifgT2+eifgT2)ϕ0|D(2ifT)|ϕ0\displaystyle=\frac{1}{2}(e^{-ifgT^{2}}+e^{ifgT^{2}})\langle\phi_{0}|D(\sqrt{2}ifT)|\phi_{0}\rangle
=cos(fgT2)ϕ0|D(2ifT)|ϕ0=cos(fgT2)e2f2T2er\displaystyle=\cos(fgT^{2})\langle\phi_{0}|D(\sqrt{2}ifT)|\phi_{0}\rangle=\cos(fgT^{2})e^{-\sqrt{2}f^{2}T^{2}e^{-r}}

where we have used:

K\displaystyle K ϕ0|D(2ifT)|ϕ0=0|Se2ifT(a+a)S|0=0|e2ifTS(a+a)S|0\displaystyle\equiv\langle\phi_{0}|D(\sqrt{2}ifT)|\phi_{0}\rangle=\langle 0|S^{\dagger}e^{\sqrt{2}ifT(a^{\dagger}+a)}S|0\rangle=\langle 0|e^{\sqrt{2}ifTS^{\dagger}(a^{\dagger}+a)S}|0\rangle
=0|e2ifTer(a+a)|0=0|α=2ifTer=e2f2T2er\displaystyle=\langle 0|e^{\sqrt{2}ifTe^{-r}(a^{\dagger}+a)}|0\rangle=\langle{0}|\alpha=\sqrt{2}ifTe^{-r}\rangle=e^{-\sqrt{2}f^{2}T^{2}e^{-r}}
|α\displaystyle|\alpha\rangle =e|α|22n=0+αnn!|n\displaystyle=e^{\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{+\infty}\frac{\alpha^{n}}{\sqrt{n!}}|n\rangle
K\displaystyle K =0|α=e|α|22=e2f2T2er\displaystyle=\langle 0|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}=e^{-\sqrt{2}f^{2}T^{2}e^{-r}}

Also we have:

M2\displaystyle M^{2} =I\displaystyle=I
M2\displaystyle\langle M^{2}\rangle =1\displaystyle=1
Δ2M\displaystyle\Delta^{2}M =1cos2(fgT2)e2f4T4e2r\displaystyle=1-\cos^{2}(fgT^{2})e^{-2f^{4}T^{4}e^{-2r}}
ΔM|fM|\displaystyle\frac{\Delta M}{|\partial_{f}\langle M\rangle|} =1cos2(fgT2)e2f4T4e2r|gT2sin(fgT2)K22fT2ere2f2T2ercos(fgT2)|,\displaystyle=\frac{\sqrt{1-\cos^{2}(fgT^{2})e^{-2f^{4}T^{4}e^{-2r}}}}{|gT^{2}\sin(fgT^{2})K-2\sqrt{2}fT^{2}e^{-r}e^{-2f^{2}T^{2}e^{-r}}\cos(fgT^{2})|},

For a given value TT, there exists a sufficiently large rr such that e2f4T4e2r1e^{-2f^{4}T^{4}e^{-2r}}\approx 1, er0e^{-r}\approx 0, K0K\approx 0 and

Δf\displaystyle\Delta f =ΔM|fM|=1cos2(fgT2)e2f4T4e2r|gT2sin(fgT2)K22fT2ere2f2T2ercos(fgT2)||sin(fgT2)|gT2|sin(fgT2)|=1gT2=1FQ.\displaystyle=\frac{\Delta M}{|\partial_{f}\langle M\rangle|}=\frac{\sqrt{1-\cos^{2}(fgT^{2})e^{-2f^{4}T^{4}e^{-2r}}}}{|gT^{2}\sin(fgT^{2})K-2\sqrt{2}fT^{2}e^{-r}e^{-2f^{2}T^{2}e^{-r}}\cos(fgT^{2})|}\approx\frac{|\sin(fgT^{2})|}{gT^{2}|\sin(fgT^{2})|}=\frac{1}{gT^{2}}=\frac{1}{\sqrt{F_{Q}}}.

Thus, we have achieved the T2T^{2} enhancement in Δf\Delta f.

Proof of the exponential enhancement

Define the following quantities:

P(x,n)\displaystyle P(x,n) j=0nxjj!\displaystyle\equiv\sum_{j=0}^{n}\frac{x^{j}}{j!}
Q(x,n)\displaystyle Q(x,n) j=0n(xjj!)2\displaystyle\equiv\sum_{j=0}^{n}(\frac{x^{j}}{j!})^{2}

We hope to find an appropriate dependence of nn on xx such that Q(x,n)Q(x,n) scales like exe^{x} as xx increases. First of all, P(x,n)P(x,n) is the Taylor expansion of exe^{x}, satisfying P(x,n)exP(x,n)\to e^{x} as n+n\to+\infty, for any xx. Moreover, the larger xx, the larger nn is needed for P(x,n)P(x,n) to approximate exe^{x}. One interesting question is, how large nn should be so that P(x,n)P(x,n) can can approximate exe^{x} reasonably well, i.e., the difference between P(x,n)P(x,n) and exe^{x} can be made arbitrarily small?

By Stirling’s formula and Cauchy–Schwarz inequality, it is easy to show that, for x3x\geq 3, n=x2n=x^{2} is sufficient to make |P(x,n=x2)ex||P(x,n=x^{2})-e^{x}| sufficiently small and Q(x,n=x2)>exQ(x,n=x^{2})>e^{x}. The next question is whether we can further reduce such quadratic dependence of nn on xx to a linear dependence. The answer is yes, as we have the following lemma:

Lemma 1.

For x2x\geq 2, if n=3xn=3x is an integer, we have

(1) exP(x,n)<x2π(e3)3x+1e^{x}-P(x,n)<\frac{x}{\sqrt{2\pi}}\big{(}\frac{e}{3}\big{)}^{3x+1};

(2) Q(x,n)>exQ(x,n)>e^{x}.

Proof.

For x>1x>1, and n>xn>x, we have

exP(x,n)\displaystyle e^{x}-P(x,n) =j=n+1+xjj!\displaystyle=\sum_{j=n+1}^{+\infty}\frac{x^{j}}{j!}
=xn+1(n+1)!(1+xn+2+j=2+(n+1)!xj(n+j+1)!)\displaystyle=\frac{x^{n+1}}{(n+1)!}\left(1+\frac{x}{n+2}+\sum_{j=2}^{+\infty}\frac{(n+1)!x^{j}}{(n+j+1)!}\right)
xn+1(n+1)!(1+xn+2+j=2+x2(n+j)(n+j+1))\displaystyle\leq\frac{x^{n+1}}{(n+1)!}\left(1+\frac{x}{n+2}+\sum_{j=2}^{+\infty}\frac{x^{2}}{(n+j)(n+j+1)}\right)
=xn+1(n+1)!(1+xn+2+x2n+2)\displaystyle=\frac{x^{n+1}}{(n+1)!}\left(1+\frac{x}{n+2}+\frac{x^{2}}{n+2}\right)

Then, if we choose n=3xn=3x, and for x5+12x\geq\frac{\sqrt{5}+1}{2}, we have

exP(x,n)x3x+1(3x+1)!(1+x3x+2+x23x+2)x3x+1(3x+1)!x\displaystyle e^{x}-P(x,n)\leq\frac{x^{3x+1}}{(3x+1)!}\left(1+\frac{x}{3x+2}+\frac{x^{2}}{3x+2}\right)\leq\frac{x^{3x+1}}{(3x+1)!}x

According to Stirling’s Formula

n!2πnn+12en,n!\geq\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n},

we have

(3x+1)!2π(3x+1)(3x+1+12)e(3x+1).(3x+1)!\geq\sqrt{2\pi}(3x+1)^{(3x+1+\frac{1}{2})}e^{-(3x+1)}.

Thus, we obtain

exP(x,n=3x)x3x+1e3x+12π(3x+1)(3x+1+12)x<12π(ex3x+1)3x+1x<x2π(e3)3x+1,e^{x}-P(x,n=3x)\leq\frac{x^{3x+1}e^{3x+1}}{\sqrt{2\pi}(3x+1)^{(3x+1+\frac{1}{2})}}x<\frac{1}{\sqrt{2\pi}}\left(\frac{ex}{3x+1}\right)^{3x+1}x<\frac{x}{\sqrt{2\pi}}\left(\frac{e}{3}\right)^{3x+1},

implying that for sufficiently large xx, the difference between P(x,n=3x)P(x,n=3x) and exe^{x} can be made arbitrarily small. Furthermore, it is easy to show that, for x2x\geq 2,

P(x,n=3x)exx2π(e3)3x+1>1011ex.\displaystyle P(x,n=3x)\geq e^{x}-\frac{x}{\sqrt{2\pi}}\left(\frac{e}{3}\right)^{3x+1}>\frac{10}{11}e^{x}.

From Cauchy–Schwarz inequality, we have:

P(x,n)2=(jnxjj!)2njn(xjj!)2=nQ(x,n).\displaystyle P(x,n)^{2}=\Big{(}\sum_{j}^{n}\frac{x^{j}}{j!}\Big{)}^{2}\leq n\sum_{j}^{n}\Big{(}\frac{x^{j}}{j!}\Big{)}^{2}=nQ(x,n).

which gives

P(x,n=3x)2\displaystyle P(x,n=3x)^{2} 3xQ(x,n=3x)\displaystyle\leq 3xQ(x,n=3x)
Q(x,n=3x)\displaystyle Q(x,n=3x) P(x,n=3x)23x>13x(10ex11)2.\displaystyle\geq\frac{P(x,n=3x)^{2}}{3x}>\frac{1}{3x}\left(\frac{10e^{x}}{11}\right)^{2}.

On the other hand, for x2x\geq 2,

13x(100ex121)\displaystyle\frac{1}{3x}\left(\frac{100e^{x}}{121}\right) >1\displaystyle>1
Q(x,n=3x)\displaystyle Q(x,n=3x) P(x,n=3x)23x>13x(100ex121)ex>ex,\displaystyle\geq\frac{P(x,n=3x)^{2}}{3x}>\frac{1}{3x}\left(\frac{100e^{x}}{121}\right)e^{x}>e^{x},

which completes the proof.

Thus, we have shown that, for n=3xn=3x to be an integer and x2x\geq 2,

Q(x,n=3x)=j=03x(xjj!)2>ex,\displaystyle Q(x,n=3x)=\sum_{j=0}^{3x}(\frac{x^{j}}{j!})^{2}>e^{x},

When n=3xn=3x is not an integer, we have:

Q(x,n=3x)=j=03x(xjj!)2j=03x((3x3)jj!)2>e3x3ex13,\displaystyle Q(x,n=\lfloor 3x\rfloor)=\sum_{j=0}^{\lfloor 3x\rfloor}(\frac{x^{j}}{j!})^{2}\geq\sum_{j=0}^{\lfloor 3x\rfloor}(\frac{(\frac{\lfloor 3x\rfloor}{3})^{j}}{j!})^{2}>e^{\frac{\lfloor 3x\rfloor}{3}}\geq e^{x-\frac{1}{3}},

Hence, no matter 3x3x is an integer or not, we have:

Q(x,n=3x)=j=03x(xjj!)2>ex13.\displaystyle Q(x,n=\lfloor 3x\rfloor)=\sum_{j=0}^{\lfloor 3x\rfloor}(\frac{x^{j}}{j!})^{2}>e^{x-\frac{1}{3}}.
Theorem 1.

For x2x\geq 2,

j=13x(xjj!)2>12ex\displaystyle\sum_{j=1}^{\lfloor 3x\rfloor}(\frac{x^{j}}{j!})^{2}>\frac{1}{2}e^{x}
Proof.

For x2x\geq 2, since ex131>12exe^{x-\frac{1}{3}}-1>\frac{1}{2}e^{x}, we have

j=13x(xjj!)2=Q(x,n=3x)1>ex131>12ex\displaystyle\sum_{j=1}^{\lfloor 3x\rfloor}(\frac{x^{j}}{j!})^{2}=Q(x,n=\lfloor 3x\rfloor)-1>e^{x-\frac{1}{3}}-1>\frac{1}{2}e^{x}

From Eq. (11), we have

FQ\displaystyle{F_{Q}} =4j=0n1(gT)2j+2g2[(j+1)!]2(Δ2Xj+1)=4Δ2Xg2j=1n((gT)jj!)2\displaystyle=4\sum_{j=0}^{n-1}\frac{(gT)^{2j+2}}{g^{2}[(j+1)!]^{2}}(\Delta^{2}X_{j+1})=\frac{4\Delta^{2}X}{g^{2}}\sum_{j=1}^{n}\Big{(}\frac{(gT)^{j}}{j!}\Big{)}^{2}

Substituting x=gTx=gT into the expression of Q(x,n=3x)Q(x,n=\lfloor 3x\rfloor) and for gT2gT\geq 2, we find:

Q(x=gT,n=3x)\displaystyle Q(x=gT,n=\lfloor 3x\rfloor) =j=13gT((gT)jj!)2>12egT\displaystyle=\sum_{j=1}^{\lfloor 3gT\rfloor}\Big{(}\frac{(gT)^{j}}{j!}\Big{)}^{2}>\frac{1}{2}e^{gT}

Hence, for gT2gT\geq 2 and n=3gTn=\lfloor 3gT\rfloor, we have

FQ\displaystyle{F_{Q}} =4Δ2Xg2j=13gT((gT)jj!)2>2Δ2Xg2egT\displaystyle=\frac{4\Delta^{2}X}{g^{2}}\sum\limits_{j=1}^{\lfloor 3gT\rfloor}\Big{(}\frac{(gT)^{j}}{j!}\Big{)}^{2}>\frac{2\Delta^{2}X}{g^{2}}e^{gT}

Moreover, for n=3xn=3x to be an integer and x2x\geq 2, analogous to the above discussion, we can prove:

Q(x,n=3x)n=13xj=0n=3x(xjj!)2>ex3x>ex6,\displaystyle\frac{Q(x,n=3x)}{n}=\frac{1}{3x}\sum_{j=0}^{n=3x}(\frac{x^{j}}{j!})^{2}>\frac{e^{x}}{3x}>\frac{e^{x}}{6},

For the averaged QFI, FQ¯(n,gT)1nFQ\overline{F_{Q}}(n,gT)\equiv\frac{1}{n}F_{Q}, we have:

FQ¯(n=3gT,gT)1nFQ(n=3gT,gT)=4Δ2Xg2j=13gT13gT((gT)jj!)2>(Δ2X)egT3g2,\displaystyle\overline{F_{Q}}(n=\lfloor 3gT\rfloor,gT)\equiv\frac{1}{n}F_{Q}(n=\lfloor 3gT\rfloor,gT)=\frac{4\Delta^{2}X}{g^{2}}\sum\limits_{j=1}^{\lfloor 3gT\rfloor}\frac{1}{\lfloor 3gT\rfloor}\Big{(}\frac{(gT)^{j}}{j!}\Big{)}^{2}>\frac{(\Delta^{2}X)e^{gT}}{3g^{2}},

which also grows exponentially with gTgT.