1. Introduction
Phase estimation is the core topic in precision measurement. Precision measurement of gravity, temperature, weak magnetic strength, and many other parameters can be classified in the category of phase estimation. Since the optical quantum phase estimation revealed the power of quantum resources to beat the standard quantum limit [1], quantum phase estimation has become the most important field in quantum metrology. It has been widely used in many fields, such as magnetic field measurement [2, 3], gravity measurement [4, 5], atomic clocks [6], navigation [7, 8], gravitational wave detection [9, 10], and dark matter [11, 12].
The optimal scheme for quantum phase estimation depends on exploring the optimal probe states. Caves found out that the precision for phase estimation can beat the shot-noise limit (standard quantum limit) by taking a high-intensity coherent state and a low-intensity squeezed vacuum state as the input states of a Mach-Zehnder interferometer [1]. Since then, many states have been proposed as the optimal probe state to fulfill a similar job, such as NOON states [13, 14], entangled coherent states [15-17], two-mode squeezed states [18-21], number squeezed states [22, 23], and so on. Recently, the path-entangled Fock states (PEFSs) taking the form of $(| m,m^{\prime} \rangle \,+| m^{\prime} ,m\rangle )/\sqrt{2}(m\gt m^{\prime} )$ has been proven to be more robust than NOON states in the case of photon losses [24-28]. This has aroused great research interest. However, whether the PEFSs are optimal has not yet been discovered. Thus, in this work, we focus on the general form of the superposition state of the Fock states and optimize it for phase estimation.
To depict the precision of a parameter Φ, quantum Fisher information (QFI) is an available useful concept because it describes the lower bound on the variance of the unbiased estimator $\hat{\phi }$ due to the Cramér-Rao theorem: $\,\rm{Var}\,(\hat{\phi })\geqslant 1/(\nu F)$ [29-32], where Var( · ) is the variance, ν is the number of repeated experiments, and F is the QFI. The method of increasing the mean particle number can achieve an increase in QFI. However, in the current progress of experiments (such as NOON states [33-36]), the mean particle number that can be prepared is very limited. Thus, in this work, we fix the mean particle number and then optimize the probe states taking the form of two entangled superposition Fock states by using QFI as the criterion.
2. Theoretical model
Quantum phase estimation can be applied to optical systems, cold atom systems, and condensed matter systems, such as optical interferometers, atomic interferometers, Bose-Einstein condensates, superconducting circuits, and quantum dots [37-41]. The Mach-Zehnder interferometer is one common theoretical model in optical systems. It consists of beam splitters, mirrors, and phase shifters (see figure 1).
Figure 1. Sketch of the Mach-Zehnder interferometer. The blue rectangles represent the beam splitters, the red frame represents the phase shifter, a and b represent two modes. |
The beam splitter of the Mach-Zehnder interferometer can be described as ${B}_{i}(\theta )=\exp ({\rm{i}}\theta {J}_{i})$ with i = x, y, z. Here, Ji are Schwinger representations of bosons. They take the form of ${J}_{x}=\left({a}^{\dagger }b+a{b}^{\dagger }\right)/2$, ${J}_{y}=\left({a}^{\dagger }b-a{b}^{\dagger }\right)/(2{\rm{i}})$, and ${J}_{z}=\left({a}^{\dagger }a-{b}^{\dagger }b\right)/2$. Ji are the generators of the SU(2) group, which satisfy the commutation relations [Ji, Jj] = iεijkJk with the Levi-Civita symbol εijk.
The phase shifter of the Mach-Zehnder interferometer can be represented as a unitary operator. In this paper, we consider the linear phase shift as
$\begin{eqnarray}U({\phi }_{a})={{\rm{e}}}^{{\rm{i}}{\phi }_{a}{a}^{\dagger }a},\end{eqnarray}$
where Φa is the accumulated phase on the mode a. For two modes (a and b) with such processes, the total phase operator can be described as $U({\phi }_{a})U({\phi }_{b})=\exp ({\rm{i}}{\phi }_{{\rm{tot}}}n/2)\exp ({\rm{i}}\phi {J}_{z})$ with the total phase Φtot = Φa + Φb and the phase difference Φ = Φa - Φb. Here, n = a†a + b†b is the total mean particle number operator.For convenient analysis, we assume that the input state |ψin$\rangle$ directly enters the phase shifter and then passes through the beam splitter Bx(π/2), which can be expressed as
$\begin{eqnarray}U(\phi )={B}_{x}\left(\frac{{\rm{\pi }}}{2}\right){{\rm{e}}}^{\frac{{\rm{i}}}{2}{\phi }_{{\rm{tot}}}n}{{\rm{e}}}^{{\rm{i}}\phi {J}_{z}}.\end{eqnarray}$
We are interested in the phase difference Φ to quantify the sensitivity of the probe state. For a pure state |ψin$\rangle$, the QFI of the final state |ψ$\rangle$ = U(Φ)|ψin$\rangle$ is simply connected with the variance of Jz in the state |ψin$\rangle$, that is [42, 43], $\begin{eqnarray}F=4\left(\langle {{\rm{\partial }}}_{\phi }\psi | {{\rm{\partial }}}_{\phi }\psi \rangle -{\left|\langle {{\rm{\partial }}}_{\phi }\psi | \psi \rangle \right|}^{2}\right)=4{\rm{Var}}({{J}}_{{z}}).\end{eqnarray}$
QFI is an available useful concept to depict the precision of a parameter Φ [44-50]. Therefore, we use QFI as a criterion to optimize the probe states.3. Mathematical form of the optimal probe states
We consider an (N + 1)-dimensional probe state taking the form of 3 ), we can obtain the QFI associated with the probe state |ψin$\rangle$. When k = g and l = h, the corresponding QFI is zero. It means such a state is not an optimal one. When k ≠ g or l ≠ h, the corresponding QFI reduces to
$\begin{eqnarray}| {\psi }_{\,\rm{in}\,}\rangle ={c}_{0}\,| k,l\rangle +{c}_{1}\,| g,h\rangle .\end{eqnarray}$
Here, c0(1) is a complex coefficient, k, l, g, h are integers in [0, N] with the free parameter N. According to equation ( $\begin{eqnarray}F=| {c}_{0}{| }^{2}(1-| {c}_{0}{| }^{2}){(k+h-l-g)}^{2},\end{eqnarray}$
where the condition |c0|2 + |c1|2 = 1 is used.The mean particle number constraint is 5 ) becomes
$\begin{eqnarray}\bar{n}=| {c}_{0}{| }^{2}(k+l-g-h)+g+h.\end{eqnarray}$
Similar to [51], we will discuss $\bar{n}$ in two intervals: $\bar{n}\in (0,N]$ and $\bar{n}\in (N,2N)$. Notice that the phase difference is not encoded into the probe states when $\bar{n}=0$ and $\bar{n}=2N$, thus they are not included in the coming discussions. For convenience, we define a new set of parameters p = k + l and q = g + h. When p = q, the constraint is $\bar{n}=p=q$. Then, the QFI in equation ( $\begin{eqnarray}F=| {c}_{0}{| }^{2}(1-| {c}_{0}{| }^{2}){(2\bar{n}-2l-2g)}^{2}.\end{eqnarray}$
When both l and g reach their minimum values, the QFI can reach its maximum value.Firstly, for the case of $\bar{n}\in (0,N]$, the minimum values of l and g are zero due to $k+l=g+h=\bar{n}$. Under the conditions l = g = 0 ($k=h=\bar{n}$), we optimize the QFI in equation (7 ) to get the maximum value, i.e., $\mathop{{\rm{\max }}}\limits_{p=q}F$, which takes the form of
$\begin{eqnarray}{\max }_{p=q}F={\bar{n}}^{2}.\end{eqnarray}$
We also obtain |c0|2 = 1/2 and the corresponding optimal state of |ψin$\rangle$ is $\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\frac{1}{\sqrt{2}}\left(| 0,\bar{n}\rangle +{{\rm{e}}}^{{\rm{i}}\theta }| \bar{n},0\rangle \right),\end{eqnarray}$
where θ ∈ [0, 2π) is a relative phase. It is an $\bar{n}00\bar{n}$ state.When p ≠ q, |c0|2 can be expressed as $| {c}_{0}{| }^{2}=(\bar{n}-q)/(p-q)$. To make sure |c0|2 is in the region [0, 1], p and q should satisfy $\{p\in [\bar{n},2N],\,q\in [0,\bar{n}]\}$ or $\{p\in [0,\bar{n}],\,q\in [\bar{n},2N]\}$. We first discuss this under the conditions $\{p\in [\bar{n},2N],\,q\in [0,\bar{n}]\}$. Substituting the expression of |c0|2 into the QFI in equation (5 ), we can get
$\begin{eqnarray}F=\frac{(\bar{n}-q)(p-\bar{n})}{{(p-q)}^{2}}{(p+q-2l-2g)}^{2}.\end{eqnarray}$
Thus, when both l and g reach their minimum values (i.e., l = g = 0), the QFI can reach its maximum value. In different regions of p, the minimum value of g is always zero since $g\in [0,\bar{n}]$, while the minimum value of l varies. When $p\in [\bar{n},\,N]$, the minimum value of l is zero; When p ∈ [N, 2N], the minimum value of l is p - N. Hence, by the parameters l and g, we optimize the maximum value of QFI, that is $\mathop{{\rm{\max }}}\limits_{g,l}F$, as $\begin{eqnarray}\mathop{\max }\limits_{g,l}F=\left\{\begin{array}{ll}\frac{(\bar{n}-q)(p-\bar{n})}{{(p-q)}^{2}}{(p+q)}^{2},\quad & p\in [\bar{n},N],\\ \frac{(\bar{n}-q)(p-\bar{n})}{{(p-q)}^{2}}{(2N-p+q)}^{2},\quad & p\in [N,2N].\end{array}\right.\end{eqnarray}$
Based on equation (11 ), we define ${F}_{gl}:= \mathop{{\rm{\max }}}\limits_{g,l}F$ and then we optimize the QFI by parameters p and q. In $p\in [\bar{n},N]$, we obtain the derivatives of Fgl as 12 ), we discuss the solutions of differential equations ∂pFgl = ∂qFgl = 0. (1) If $q=\bar{n}$ or ${p}^{2}-{q}^{2}-4(\bar{n}-p)q=0$, we have ∂pFgl = 0. (2) If $p=\bar{n}$ or ${p}^{2}-{q}^{2}-4(\bar{n}-q)p=0$, we have ∂qFgl = 0. It is worth noting that we can derive ${p}^{2}-{q}^{2}-4(\bar{n}-q)p\ne 0$ from $q=\bar{n}$, and derive ${p}^{2}-{q}^{2}-4(\bar{n}-p)q\ne 0$ from $p=\bar{n}$. Due to p ≠ q, $q=\bar{n}$ and $p=\bar{n}$ cannot hold at the same time, while ${p}^{2}-{q}^{2}-4(\bar{n}-p)q=0$ and ${p}^{2}-{q}^{2}\,-4(\bar{n}-q)p=0$ cannot hold at the same time. The results are that ∂pFgl and ∂qFgl cannot be both 0 at the same time. This means that the sets of the optimal parameters {p, q} should be on the edge of the range $\{p\in [\bar{n},N],\,q\in [0,\bar{n}]\}$, that is, the yellow solid line in figure 2.
$\begin{eqnarray}\displaystyle \begin{array}{rcl}{{\rm{\partial }}}_{p}{F}_{gl} & = & \frac{(p+q)(\bar{n}-q)\left[{p}^{2}-{q}^{2}-4(\bar{n}-p)q\right]}{{(p-q)}^{3}},\\ {{\rm{\partial }}}_{q}{F}_{gl} & = & -\frac{(p+q)(p-\bar{n})\left[{p}^{2}-{q}^{2}-4(\bar{n}-q)p\right]}{{(p-q)}^{3}}.\end{array}\end{eqnarray}$
Based on equation (
Figure 2. In the case of $\bar{n}\in (0,N]$, the sets of optimal parameters {p, q} corresponding to the maximum values of Fgl should be on the yellow solid line for $p\in [\bar{n},N]$ and on the red dash-dotted line for p ∈ [N, 2N]. After some discussion, we obtain that the sets of optimal parameters are {p = N, q = 0} and $\{p=N,\,q=(2-\sqrt{5-4\bar{n}/N})N\}$, which correspond to the blue square and the green dot, respectively. |
We search for the sets of the optimal parameters {p, q} corresponding to the maximum values of Fgl from the yellow solid line in figure 2. (1) For the line $p=\bar{n}$, we obtain Fgl = 0, then the sets of the parameters {p, q} on this line are not optimal. (2) Similarly, for the line $q=\bar{n}$, the sets of the parameters {p, q} are not optimal due to Fgl = 0. (3) For the line q = 0, Fgl is a monotonically increasing function about p, with its maximum value being at p = N. Thus, the set of the optimal parameters is {p = N, q = 0}, corresponding to the blue square in figure 2. (4) For the line p = N, we can get
$\begin{eqnarray}{\left.{{\rm{\partial }}}_{q}{F}_{gl}\right|}_{p=N}=-\frac{(N+q)(N-\bar{n})\left[{N}^{2}-{q}^{2}-4N(\bar{n}-q)\right]}{{(N-q)}^{3}}.\end{eqnarray}$
When $q=(2-\sqrt{5-4\bar{n}/N})N$, we can obtain ∂qFgl|p=N = 0. In the case of $\bar{n}\in (0,N/4)$, we have ∂qFgl|p=N < 0. Then, Fgl is a monotonically decreasing function about q, with its maximum value being at q = 0. Thus, the set of optimal parameters is {p = N, q = 0}. In the case of $\bar{n}\in [N/4,N]$, the solution $q=(2-\sqrt{5-4\bar{n}/N})N$ is in the range of $q\in [0,\bar{n}]$. In this interval, Fgl with respect to q first increases and then decreases, with its maximum value being at $q=(2-\sqrt{5-4\bar{n}/N})N$. Therefore, the set of optimal parameters is $\{p=N,\,q=(2-\sqrt{5-4\bar{n}/N})N\}$, corresponding to the green dot in figure 2.We optimize the maximum value of by Fgl the parameters p and q, i.e., $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$, take the form of 9 ).
$\begin{eqnarray}\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}=\left\{\begin{array}{ll}\bar{n}(N-\bar{n}),\quad & \bar{n}\in \left(0,\frac{N}{4}\right),\\ {N}^{2}\frac{\left(1-\frac{\bar{n}}{N}\right)\left(\frac{\bar{n}}{N}-2+\sqrt{5-\frac{4\bar{n}}{N}}\right){\left(3-\sqrt{5-\frac{4\bar{n}}{N}}\right)}^{2}}{{\left(1-\sqrt{5-\frac{4\bar{n}}{N}}\right)}^{2}},\quad & \bar{n}\in \left[\frac{N}{4},N\right).\end{array}\right.\end{eqnarray}$
Here, when $\bar{n}=N$, it goes back to the case of p = q, and then we get $\mathop{{\rm{\max }}}\limits_{p=q}F={N}^{2}$ and the optimal state takes the form of equation (In p ∈ [N, 2N], we obtain the derivatives of Fgl based on equation (11 ) as 15 ), we discuss the solutions of differential equations ∂pFgl = ∂qFgl = 0. (1) If p = 2N and q = 0, we have ∂pFgl = ∂qFgl = 0. Thus, {p = 2N, q = 0} is one solution. (2) If $q=\bar{n}$, we have ∂pFgl = 0. But, at this time, $p-\bar{n}\ne 0$, 2N - p + q ≠ 0, and $4\bar{n}N-2N(p+q)\,+{(p-q)}^{2}\ne 0$, that is, ∂qFgl ≠ 0. Thus, $q=\bar{n}$ is not the solution. (3) If $p=\bar{n}$, we have ∂qFgl = 0. But, at this time, $\bar{n}-q\ne 0$, 2N - p + q ≠ 0, and $4\bar{n}N-2N(p+q)\,-{(p-q)}^{2}\ne 0$, that is, ∂pFgl ≠ 0. Thus, $p=\bar{n}$ is not the solution. (4) If $4\bar{n}N-2N(p+q)-{(p-q)}^{2}=0$ and $4\bar{n}N-2N(p\,+q)+{(p-q)}^{2}=0$, we have ∂pFgl = ∂qFgl = 0. But, at this time, we obtain p = q, which contradicts the previous condition p ≠ q. So, this is not the solution. Therefore, {p = 2N, q = 0} is the unique solution of ∂pFgl = ∂qFgl = 0.
$\begin{eqnarray}\displaystyle \begin{array}{rcl}{{\rm{\partial }}}_{p}{F}_{gl} & = & \frac{(\bar{n}-q)(2N-p+q)\left[4\bar{n}N-2N(p+q)-{(p-q)}^{2}\right]}{{(p-q)}^{3}},\\ {{\rm{\partial }}}_{q}{F}_{gl} & = & \frac{(p-\bar{n})(2N-p+q)\left[4\bar{n}N-2N(p+q)+{(p-q)}^{2}\right]}{{(p-q)}^{3}}.\end{array}\end{eqnarray}$
Based on equation (We search for the sets of optimal parameters {p, q} corresponding to the maximum values of Fgl by solving the differential equations ∂pFgl = ∂qFgl = 0. The unique solution {p = 2N, q = 0} is not optimal, because it makes the corresponding Fgl in equation (11 ) to be zero. This means that the sets of the optimal parameters {p, q} should be on the edge of the region $\{p\in [N,2N],\,q\in [0,\bar{n}]\}$, that is, the red dash-dotted line in figure 2. (1) For the line $q=\bar{n}$, we have Fgl = 0, then the sets of the parameters {p, q} on this edge are not optimal. (2) For the line p = N, the sets of the parameters {p, q} on this edge are optimal, which are discussed in the range $p\in [\bar{n},N]$. (3) For the line p = 2N, due to $\bar{n}\,\leqslant \,N$, we can get ∂qFgl|p=2N < 0 when $q\ne \bar{n}$. This means that the maximum value of Fgl is taken at q = 0, and the maximum value is Fgl = 0. Thus, the sets of the parameters {p, q} on this edge are not optimal. (4) For the line q = 0 (excluding endpoints {p = 2N, q = 0}), we can get 16 ) are all positive. The term $N(N+4\bar{n})\,-{(p+N)}^{2}$ is monotonic in such a region. For p = 2N, we have $N(N+4\bar{n})-{(p+N)}^{2}\lt 0$. While, for p = N, such a term becomes $4\bar{n}N-3{N}^{2}$. In the case of $\bar{n}\lt 3N/4$, $N(N+4\bar{n})-{(p+N)}^{2}\lt 0$, then, ∂pFgl|q=0 < 0. This indicates that Fgl is a decreasing function with its optimal point at {p = N, q = 0}, i.e., the blue square in figure 2. In the case of $\bar{n}\,\geqslant \,3N/4$, we have $N(N+4\bar{n})\,\,-{(p+N)}^{2}\,\geqslant \,0$. There is one point satisfying ∂pFgl|q=0 = 0, that is, Fgl is a function about p that increases first and then decreases. We substitute the inflection point of Fgl into ∂qFgl and get ∂qFgl|q=0 > 0. This is an increasing function, so q = 0 is not optimal.
$\begin{eqnarray}{\left.{{\rm{\partial }}}_{p}{F}_{gl}\right|}_{q=0}=\frac{\bar{n}(2N-p)\left[N(N+4\bar{n})-{(p+N)}^{2}\right]}{{p}^{3}}.\end{eqnarray}$
The sign of ∂pFgl|q=0 is determined by the sign of the term $N(N+4\bar{n})-{(p+N)}^{2}$, because the other terms in equation (In a word, when p ≠ q, the sets of parameters {p, q} should be on the line p = N, either in $p\in [\bar{n},N]$ or in p ∈ [N, 2N]. The corresponding $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$ is
$\begin{eqnarray}\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}=\left\{\begin{array}{ll}{N}^{2}(1-r)r,\quad & r\in \left(0,\frac{1}{4}\right),\\ {N}^{2}\left(1-r\right)\left(r-2+\sqrt{5-4r}\right)\frac{{\left(3-\sqrt{5-4r}\right)}^{2}}{{\left(\sqrt{5-4r}-1\right)}^{2}},\quad & r\in \left[\frac{1}{4},1\right).\end{array}\right.\end{eqnarray}$
Here, we define $r:= \bar{n}/N$.To finally acquire the optimal states, we compare $\mathop{{\rm{\max }}}\limits_{p=q}F$ in equation (8 ) with $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$ in equation (17 ), where the former is the maximum QFI for p = q and the latter is the one for p ≠ q. In the range of r ∈ (0, 1/4), we have $\mathop{{\rm{\max }}}\limits_{p=q}F\lt \mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$. In the range of r ∈ [1/4, 1), the ratio between them is 19 ) [equation (20 )], the $\mathop{{\rm{\max }}}\limits_{p,q,g,l}F$ (the state $| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle $) for r = 1/4 reduces to the expression in the range of (0, 1/4]. Thus, we change the range r ∈ (0, 1/4) to the range r ∈ (0, 1/4].
$\begin{eqnarray}\kappa =\frac{\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}}{\mathop{{\rm{\max }}}\limits_{p=q}F}=\frac{\left(1-r\right)\left(r-2+\sqrt{5-4r}\right){\left(3-\sqrt{5-4r}\right)}^{2}}{{\left(r-r\sqrt{5-4r}\right)}^{2}}.\end{eqnarray}$
We discuss this ratio in figure 3, which is larger than 1. We combine the case in the range of r ∈ (0, 1) and the case of p = q = N. We define the maximum QFI as $\mathop{{\rm{\max }}}\limits_{p,q,g,l}F$, which is the larger one between $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$ and $\mathop{{\rm{\max }}}\limits_{p=q}F$. Therefore, in the case of $\bar{n}\,\leqslant \,N$, we have $\begin{eqnarray}\mathop{{\rm{\max }}}\limits_{p,q,g,l}F=\left\{\begin{array}{ll}{N}^{2}(1-r)r,\quad & r\in \left(0,\frac{1}{4}\right],\\ {N}^{2}\left(1-r\right)\left(r-2+\sqrt{5-4r}\right)\frac{{\left(3-\sqrt{5-4r}\right)}^{2}}{{\left(\sqrt{5-4r}-1\right)}^{2}},\quad & r\in \left[\frac{1}{4},1\right),\\ {N}^{2},\quad & r=1.\end{array}\right.\end{eqnarray}$
The corresponding optimal state of |ψin$\rangle$ is $\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\left\{\begin{array}{ll}\sqrt{r}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{1-r}| \mathrm{0,\; 0}\rangle ,\quad & r\in \left(0,\frac{1}{4}\right],\\ \sqrt{\frac{r-2+\sqrt{5-4r}}{\sqrt{5-4r}-1}}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{1-r}{\sqrt{5-4r}-1}}| 0,N\left(2-\sqrt{5-4r}\right)\rangle ,\quad & r\in \left[\frac{1}{4},1\right),\\ \frac{1}{\sqrt{2}}\left(| 0,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }| N,0\rangle \right),\quad & r=1.\end{array}\right.\end{eqnarray}$
Here, θ is a relative phase. In equation (
Figure 3. Ratio $\kappa =\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}/\mathop{{\rm{\max }}}\limits_{p=q}F$ in equation ( |
In the case of r ∈ (0, 1/4], the optimal probe state is a separate state. Going through the phase shift $\exp ({\rm{i}}{\phi }_{a}{a}^{\dagger }a+{\rm{i}}{\phi }_{b}{b}^{\dagger }b)$, this state becomes $\left({{\rm{e}}}^{{\rm{i}}{\phi }_{a}N}\sqrt{r}| N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{1-r}| 0\rangle \right)| 0\rangle .$ The phase Φa is encoded in the state, but the phase Φb is not encoded in the state. Therefore, the state for r ∈ (0, 1/4] in equation (20 ) needs to be further optimized with respect to the phase difference Φ = Φa - Φb.
The optimal parameters {p, q} should be close to the previous parameter {p = N, q = 0} with l = g = 0. Thus, we compare the parameters {p = N, q = 0} with l = 1 and g = 0, {p = N - 1, q = 0}, {p = N + 1, q = 0}, and {p = N, q = 1} to search. (1) For the case of {p = N, q = 0} with l = 1 and g = 0, we have k = N - 1 and h = 0. The corresponding state is
$\begin{eqnarray}| {\psi }_{1}\rangle =\sqrt{r}| N-1,1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{1-r}| 0,0\rangle .\end{eqnarray}$
The QFI of this state is F1 = r(1 - r)(N - 2)2. (2) For {p = N - 1, q = 0}, we have k = N - 1 and l = g = h = 0, the phase Φb is still not encoded in the state. (3) For {p = N + 1, q = 0}, we have k = N, l = 1, and g = h = 0, the corresponding state is $\begin{eqnarray}\begin{array}{rcl}| {\psi }_{2}\rangle & = & \sqrt{\frac{Nr}{N+1}}| N,1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{1-\frac{Nr}{N+1}}| \mathrm{0,\; 0}\rangle .\end{array}\end{eqnarray}$
The QFI of the state is F2 = Nr(N + 1 - Nr)(N - 1)2/(N + 1)2. (4) For {p = N, q = 1}, there are two cases. For the first case, we have k = N, l = g = 0, and h = 1, then the phase Φb is still not encoded in the state. For the second case, we have k = N, l = g = 0, and h = 1, then the corresponding state is $\begin{eqnarray}\begin{array}{rcl}| {\psi }_{3}\rangle & = & \sqrt{\frac{Nr-1}{N-1}}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(1-r)}{N-1}}| \mathrm{0,\; 1}\rangle .\end{array}\end{eqnarray}$
Since the coefficients of this state are within the range of [0, 1], the condition that r needs to satisfy is r ≥ 1/N. Then we can obtain N ≥ 4. Based on the above discussion, this state is legitimate when N ≥ 4 and r ∈ [1/N, 1/4]. The QFI of the state is F3 = N(Nr - 1)(1 - r)(N + 1)2/(N - 1)2. It is worth noting that both l and g of the optimal states reach their minimum values for cases (2)-(4).Now we compare the QFIs of the state |ψ1$\rangle$ and the state |ψ2$\rangle$. The ratio between the QFIs of these two states is F1/F2 = (1 - r)(N - 2)2(N + 1)2/[N(N + 1 - Nr)(N - 1)2]. When N ≥ 2, the QFI of the state |ψ2$\rangle$ is larger than the one of the state |ψ1$\rangle$. Next, we compare the QFIs of the state |ψ2$\rangle$ and the state |ψ3$\rangle$. For N = 2 or 3, we can only choose the state |ψ2$\rangle$, because the state |ψ3$\rangle$ is not available. For N ≥ 4 and r ∈ (0, 1/N), we can still only choose the state |ψ2$\rangle$, because the state |ψ3$\rangle$ is not available. For N ≥ 4 and r ∈ [1/N, 1/4], we definite the solution of r for F2 = F3 as
$\begin{eqnarray}{r}_{{{\rm{e}}}_{1}}=\frac{N+1}{2N}\left[1-\frac{N-1}{\sqrt{2({N}^{2}+1)}}\right].\end{eqnarray}$
For N = 4, 5, the state |ψ2$\rangle$ in r ∈ [1/N, 1/4] is better due to ${r}_{{{\rm{e}}}_{1}}\gt 1/4$. For N ≥ 6, in the range of $r\in [1/N,{r}_{{{\rm{e}}}_{1}}]$, the state |ψ2$\rangle$ is better; in the range of $r\in [{r}_{{{\rm{e}}}_{1}},1/4]$, the state |ψ3$\rangle$ is better.In the above discussion, we consider the optimal states for p ≠ q. Next, we compare the above results with the ones for p = q. First, we compare $\mathop{{\rm{\max }}}\limits_{p=q}F$ of the state $| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle $ for p = q in equation (9 ) with F2 of the state |ψ2$\rangle$ for p ≠ q in equation (22 ). The solution of r for ${F}_{2}=\mathop{{\rm{\max }}}\limits_{p=q}F$ is (N + 1)(N - 1)2/[2N(N2 + 1)]. For N = 2, the QFI of the $\bar{n}00\bar{n}$ state in r ∈ [3/20, 1/4] is higher, but the value of $\bar{n}$ cannot be an integer, thus we should choose the state |ψ2$\rangle$. For N ≥ 3, the state |ψ2$\rangle$ is better. Furthermore, if the state |ψ3$\rangle$ is better than the state |ψ2$\rangle$, then the state |ψ3$\rangle$ is naturally better than the $\bar{n}00\bar{n}$ state. In summary, the optimal state in r ∈ (0, 1/4] is
$\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\left\{\begin{array}{lc}\sqrt{\frac{Nr}{N+1}}| N,1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{1-\frac{Nr}{N+1}}| \mathrm{0,\; 0}\rangle ,\quad & N={{\mathbb{Z}}}_{[2,5]}\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left(0,{r}_{{{\rm{e}}}_{1}}\right],\\ \sqrt{\frac{Nr-1}{N-1}}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(1-r)}{N-1}}| \mathrm{0,\; 1}\rangle ,\quad & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{1}},\frac{1}{4}\right].\end{array}\right.\end{eqnarray}$
Here, ${{\mathbb{Z}}}_{[2,5]}$ is the set of integers from 2 to 5.According to the above discussion for the case of $\bar{n}\in (0,N]$, we have
$\begin{eqnarray}\mathop{{\rm{\max }}}\limits_{p,q,g,l}F=\left\{\begin{array}{lc}Nr(N+1-Nr)\frac{{(N-1)}^{2}}{{(N+1)}^{2}},\quad N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left(0,\frac{1}{4}\right]{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left(0,{r}_{{{\rm{e}}}_{1}}\right],\\ N(Nr-1)(1-r)\frac{{(N+1)}^{2}}{{(N-1)}^{2}}, & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{1}},\frac{1}{4}\right],\\ {N}^{2}\left(1-r\right)\left(r-2+\sqrt{5-4r}\right)\frac{{\left(3-\sqrt{5-4r}\right)}^{2}}{{\left(\sqrt{5-4r}-1\right)}^{2}}, & r\in \left(\frac{1}{4},1\right),\\ {N}^{2}, & r=1.\end{array}\right.\end{eqnarray}$
The corresponding optimal state of |ψin$\rangle$ is $\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\left\{\begin{array}{lc}\sqrt{\frac{Nr}{N+1}}| N,1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{1-\frac{Nr}{N+1}}| \mathrm{0,\; 0}\rangle ,\quad N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left(0,\frac{1}{4}\right]\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left(0,{r}_{{{\rm{e}}}_{1}}\right],\\ \sqrt{\frac{Nr-1}{N-1}}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(1-r)}{N-1}}| \mathrm{0,\; 1}\rangle , & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{1}},\frac{1}{4}\right],\\ \sqrt{\frac{r-2+\sqrt{5-4r}}{\sqrt{5-4r}-1}}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{1-r}{\sqrt{5-4r}-1}}| 0,N\left(2-\sqrt{5-4r}\right)\rangle , & r\in \left(\frac{1}{4},1\right),\\ \frac{1}{\sqrt{2}}\left(| 0,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }| N,0\rangle \right), & r=1.\end{array}\right.\end{eqnarray}$
Here, θ is a relative phase. In r ∈ [1/4, 1), the state for r = 1/4 is a separable state. Thus, we change the range r ∈ [1/4, 1) to the range r ∈ (1/4, 1).Next, we discuss the case of $\bar{n}\in (N,2N)$. For $p=q=\bar{n}$, when both l and g reach their minimum values, the QFI in equation (7 ) can reach its maximum value. Under the conditions $l=g=\bar{n}-N$ (k = h = N), we optimize the QFI in equation (7 ) to get the maximum value, i.e., $\mathop{{\rm{\max }}}\limits_{p=q}F$, which takes the form of
$\begin{eqnarray}\mathop{\max }\limits_{p=q}F={(2N-\bar{n})}^{2}.\end{eqnarray}$
The corresponding optimal state of |ψin$\rangle$ is $\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\frac{1}{\sqrt{2}}\left(| N,\bar{n}-N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }| \bar{n}-N,N\rangle \right).\end{eqnarray}$
It is worth noting that such an optimal state is exactly the PEFS $(| m,m^{\prime} \rangle +| m^{\prime} ,m\rangle )/\sqrt{2}$ with m = N and $m^{\prime} =\bar{n}-N$ [24-28].For p ≠ q, we will discuss two regions $\{p\in [\bar{n},2N],\,q\in [0,N]\}$ and $\{p\in [\bar{n},2N],\,q\in [N,\bar{n}]\}$. When $p\in [\bar{n},2N]$ and q ∈ [0, N], we have l ∈ [p - N, N] and g ∈ [0, q]. Then, Fgl, the maximum value of QFI by optimizing the parameters l and g, is equal to $(\bar{n}-q)(p-\bar{n}){(2N-p+q)}^{2}/{(p-q)}^{2}$. It is the same as the one for p ∈ [N, 2N] in equation (11 ). The derivatives of Fgl to p and to q are also equal to the ones in equation (15 ). For {p = 2N, q = 0}, we have ∂pFgl = ∂qFgl = 0. But, this set of parameters {p, q} is not optimal because the corresponding Fgl is zero. Therefore, the sets of optimal parameters must be on the solid yellow lines in figure 4. (1) For the line $p=\bar{n}$, we have Fgl = 0, then the sets of parameters {p, q} on this line are not optimal. (2) For the line q = 0 (excluding endpoints {p = 2N, q = 0}), we can have 30 ) is greater than 0. In this case, Fgl is a monotonically increasing function about q, so q = 0 is not optimal. (3) For the line p = 2N, we can get 31 ) and obtain ∂qFgl|p=2N > 0. Thus, the set of optimal parameters is {p = 2N, q = N} [see blue square in figure 4]. In the case of $\bar{n}\leqslant 7N/4$, when $p\,=\sqrt{N(4\bar{n}-3N)}$, we can obtain ∂pFgl|q=N = 0. The set of parameters $\{p=\sqrt{N(4\bar{n}-3N)},q=N\}$ is an extremum point of ∂pFgl|q=N. Substituting this set of parameters into ∂qFgl in equation (15 ), we have ∂qFgl|q=N > 0. Thus, the set of optimal parameters is $\{p=\sqrt{N(4\bar{n}-3N)},\,q=N\}$ [see green dot in figure 4]. In a word, by the parameters p and q, we optimize the maximum value of Fgl, that is $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$, taking the form as
$\begin{eqnarray}{\left.{{\rm{\partial }}}_{q}{F}_{gl}\right|}_{q=0}=\frac{(p-\bar{n})(2N-p)\left[{(p-N)}^{2}+N(4\bar{n}-N)\right]}{{p}^{3}}.\end{eqnarray}$
When p ≠ 2N, ∂qFgl|q=0 in equation ( $\begin{eqnarray}{\left.{{\rm{\partial }}}_{q}{F}_{gl}\right|}_{p=2N}=\frac{q(2N-\bar{n})\left[{(q-3N)}^{2}+N(4\bar{n}-9N)\right]}{{(2N-q)}^{3}}.\end{eqnarray}$
In the case of $\bar{n}\gt 5N/4$, ${(q-3N)}^{2}+N(4\bar{n}-9N)\gt 0$, then, ∂qFgl|p=2N > 0. Thus, the set of optimal parameters is {p = 2N, q = N} [see blue square in figure 4]. In the case of $\bar{n}\leqslant 5N/4$, when $q=3N-\sqrt{9{N}^{2}-4\bar{n}N}$, we can have ∂qFgl = 0. That is, Fgl is a function about q that increases first and then decreases. Then we substitute the inflection point of Fgl into ∂pFgl and we can get $\begin{eqnarray}{\left.{{\rm{\partial }}}_{p}{F}_{gl}\right|}_{p=2N}=\frac{q(\bar{n}-q)\left[-{(q-N)}^{2}+N(4\bar{n}-7N)\right]}{{(2N-q)}^{3}}.\end{eqnarray}$
It is less than zero, this is a decreasing function, so p = 2N is not optimal. (4) For the line q = N, we can get $\begin{eqnarray}{\left.{{\rm{\partial }}}_{p}{F}_{gl}\right|}_{q=N}=\frac{(\bar{n}-N)(3N-p)(4\bar{n}N-3{N}^{2}-{p}^{2})}{{(p-N)}^{3}}.\end{eqnarray}$
In the case of $\bar{n}\gt 7N/4$, we have ∂pFgl|q=N > 0. Then, Fgl is a monotonically increasing function about p, with its maximum value being at p = 2N. We substitute q = N into equation ( $\begin{eqnarray}\mathop{\max }\limits_{p,q}{F}_{gl}=\left\{\begin{array}{ll}{N}^{2}\frac{\left(r-1\right)\left(\sqrt{4r-3}-r\right){\left(3-\sqrt{4r-3}\right)}^{2}}{{\left(\sqrt{4r-3}-1\right)}^{2}},\quad & r\in \left(1,\frac{7}{4}\right],\\ {N}^{2}(r-1)(2-r),\quad & r\in \left(\frac{7}{4},2\right).\end{array}\right.\end{eqnarray}$
The corresponding sets of optimal parameters are $\{p=\sqrt{N(4\bar{n}-3N)},\,q=N\}$ and {p = 2N, q = N}, respectively.
Figure 4. In the case of $\bar{n}\in (N,\,2N)$, the sets of optimal parameters {p, q} corresponding to the maximum values of Fgl should be on the yellow solid line for q ∈ [0, N] and on the red dash-dotted line for $q\in [N,\,\bar{n}]$. After discussion, we obtain that the sets of optimal parameters are {p = 2N, q = N} and $\{p=\sqrt{N(4\bar{n}-3N)},q=N\}$, which correspond to the blue square and the green dot, respectively. |
In $q\in [N,\bar{n}]$, we have
$\begin{eqnarray}{F}_{gl}=\frac{(p-\bar{n})(\bar{n}-q)}{{(p-q)}^{2}}{\left(4N-p-q\right)}^{2}.\end{eqnarray}$
Here, we use the minimum values of l and g, that is, l = p - N and g = q - N. The derivatives of Fgl to p and q are $\begin{eqnarray}\begin{array}{rcl}{{\rm{\partial }}}_{p}{F}_{gl} & = & \frac{(\bar{n}-q)(4N-p-q)}{{(p-q)}^{3}}\left[(4N-p-q)(2\bar{n}-p-q)\right.\\ & & \left.-2(p-\bar{n})(p-q)\right],\\ {{\rm{\partial }}}_{q}{F}_{gl} & = & \frac{(p-\bar{n})(4N-p-q)}{{(p-q)}^{3}}\left[(4N-p-q)(2\bar{n}-p-q)\right.\\ & & \left.-2(\bar{n}-q)(p-q)\right].\end{array}\end{eqnarray}$
Based on the above equations, we discuss the solutions of differential equations ∂pFgl = ∂qFgl = 0. (1) If $q=\bar{n}$ or $(4N-p-q)(2\bar{n}-p-q)-2(p-\bar{n})(p-q)=0$, we have ∂pFgl = 0. (2) If $p=\bar{n}$ or $(4N-p-q)(2\bar{n}-p-q)\,-2(\bar{n}-q)(p-q)=0$, we have ∂qFgl = 0. We can derive $(4N-p-q)(2\bar{n}-p-q)-2(\bar{n}-q)(p-q)\ne 0$ from $q=\bar{n}$, and derive $(4N-p-q)(2\bar{n}-p-q)\,-2(p-\bar{n})(p-q)\ne 0$ from $p=\bar{n}$. Due to p ≠ q, $q=\bar{n}$ and $p=\bar{n}$ cannot hold at the same time, while $(4N\,-p-q)(2\bar{n}-p-q)-2(\bar{n}-q)(p-q)=0$ and $(4N\,-p-q)(2\bar{n}-p-q)-2(p-\bar{n})(p-q)=0$ cannot hold at the same time. This results in that ∂pFgl and ∂qFgl cannot be both 0 at the same time. This means that the sets of the optimal parameters {p, q} should be on the edge of the region $\{p\in [\bar{n},2N],q\in [N,\bar{n}]\}$, that is, the red dash-dotted lines in figure 4.We search for the sets of the optimal parameters {p, q} corresponding to the maximum values of Fgl from the red dash-dotted lines in figure 4. (1) For the line $p=\bar{n}$, we obtain Fgl = 0, then the sets of parameters {p, q} on this line are not optimal. (2) Similarly, for the line $q=\bar{n}$, the sets of parameters {p, q} are not optimal due to Fgl = 0. (3) For the line p = 2N, we have ∂qFgl|p=2N < 0. Fgl is a monotonically decreasing function about q, with its maximum value being at q = N. Thus, the set of the optimal parameters is {p = 2N, q = N}, corresponding to the blue square in figure 4. (4) For the line of q = N, it has already been discussed in q ∈ [0, N], which on the solid yellow lines in figure 4. We optimize the maximum value of Fgl by the parameters p and q, that is $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$, taking the form of 28 ) with $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$ in equation (37 ), where the former is the maximum QFI for p = q and the latter is the one for p ≠ q. In the range of r ∈ (1, 7/4], the ratio between them is 39 ) [equation (40 )], the $\mathop{{\rm{\max }}}\limits_{p,q,g,l}F$ (the state $| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle $) for r = 7/4 reduces to the expression in the range of r ∈ [7/4, 2). Thus, we change the range r ∈ (7/4, 2) to the range r ∈ [7/4, 2). Therefore, using QFI as the criterion, we find the optimal states which are better than PEFSs.
$\begin{eqnarray}\mathop{\max }\limits_{p,q}{F}_{gl}=\left\{\begin{array}{ll}{N}^{2}\frac{\left(r-1\right)\left(\sqrt{4r-3}-r\right){\left(3-\sqrt{4r-3}\right)}^{2}}{{\left(\sqrt{4r-3}-1\right)}^{2}},\quad & r\in \left(1,\frac{7}{4}\right],\\ {N}^{2}(r-1)(2-r),\quad & r\in \left(\frac{7}{4},2\right).\end{array}\right.\end{eqnarray}$
We compare $\mathop{{\rm{\max }}}\limits_{p=q}F$ in equation ( $\begin{eqnarray}\kappa =\frac{\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}}{\mathop{{\rm{\max }}}\limits_{p=q}F}=\frac{\left(r-1\right)\left(\sqrt{4r-3}-r\right){\left(3-\sqrt{4r-3}\right)}^{2}}{{\left(\sqrt{4r-3}-1\right)}^{2}{\left(2-r\right)}^{2}}.\end{eqnarray}$
We discuss this ratio in figure 5, which is always larger than 1. In the range of r ∈ (7/4, 2), $\mathop{{\rm{\max }}}\limits_{p=q}F\lt \mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$. This means that PEFSs are not optimal. We combine the case in the range of r ∈ (0, 2), and then get the maximum QFI as $\mathop{{\rm{\max }}}\limits_{p,q,g,l}F$, which is the larger one between $\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}$ and $\mathop{{\rm{\max }}}\limits_{p=q}F$. Therefore, in the case of $\bar{n}\in (N,2N)$, we have $\begin{eqnarray}\mathop{{\rm{\max }}}\limits_{p,q,g,l}F=\left\{\begin{array}{ll}{N}^{2}\frac{\left(r-1\right)\left(\sqrt{4r-3}-r\right){\left(3-\sqrt{4r-3}\right)}^{2}}{{\left(\sqrt{4r-3}-1\right)}^{2}},\quad & r\in \left(1,\frac{7}{4}\right],\\ {N}^{2}(r-1)(2-r),\quad & r\in \left[\frac{7}{4},2\right).\end{array}\right.\end{eqnarray}$
The corresponding optimal state of |ψin$\rangle$ is $\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\left\{\begin{array}{ll}\sqrt{\frac{r-1}{\sqrt{4r-3}-1}}| N,N(\sqrt{4r-3}-1)\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{\sqrt{4r-3}-r}{\sqrt{4r-3}-1}}| 0,N\rangle ,\quad & r\in \left(1,\frac{7}{4}\right],\\ \sqrt{r-1}| N,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{2-r}| 0,N\rangle ,\quad & r\in \left[\frac{7}{4},2\right).\end{array}\right.\end{eqnarray}$
Here, θ is a relative phase. In equation (
Figure 5. Ratio $\kappa =\mathop{{\rm{\max }}}\limits_{p,q}{F}_{gl}/\mathop{{\rm{\max }}}\limits_{p=q}F$ in equation ( |
In the case of r ∈ [7/4, 2), the phase Φb is not encoded in the state. Therefore, the state for r ∈ [7/4, 2) in equation (40 ) needs to be further optimized with respect to the phase difference Φ. The possible optimal parameters {p, q} are {p = 2N, q = N} with l = N and g = 1, {p = 2N, q = N + 1}, {p = 2N, q = N - 1}, and {p = 2N - 1, q = N}. (1) For the case of {p = 2N, q = N} with l = N and g = 1, we have k = N and h = N - 1. The corresponding state is
$\begin{eqnarray}| {\psi }_{a}\rangle =\sqrt{r-1}| N,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{2-r}| 1,N-1\rangle .\end{eqnarray}$
The QFI of this state is Fa = (r - 1)(2 - r)(N - 2)2. (2) For {p = 2N, q = N + 1}, we have k = l = h = N and g = 1, the phase Φb is not encoded in the state. (3) For {p =2N, q = N - 1}, we have k = l = N, g = 0, and h = N - 1, the corresponding state is $\begin{eqnarray}\begin{array}{rcl}| {\psi }_{b}\rangle & = & \sqrt{\frac{N(r-1)+1}{N+1}}| N,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)}{N+1}}| 0,N-1\rangle .\end{array}\end{eqnarray}$
The QFI of the state is Fb = N[N(r - 1) +1](2 - r)(N - 1)2/(N + 1)2. (4) For {p = 2N - 1, q = N}, there are two cases. For the first case, we have k = N - 1, l = h = N, and g = 0, the phase Φb is not encoded in the state. For the second case, we have k = h = N, l = N - 1, and g = 0, the corresponding state is $\begin{eqnarray}\begin{array}{rcl}| {\psi }_{c}\rangle & = & \sqrt{\frac{N(r-1)}{N-1}}| N,N-1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)-1}{N-1}}| 0,N\rangle .\end{array}\end{eqnarray}$
As the coefficients of this state are within the range of [0, 1], the condition that r needs to satisfy is r≤2 - 1/N. Then we can obtain N ≥ 4. Based on the above discussion, this state is legitimate when N ≥ 4 and r ∈ [7/4, 2 - 1/N]. The QFI of the state is Fc = N(r - 1)[N(2 - r) - 1](N + 1)2/(N - 1)2. It is worth noting that both l and g of the optimal states reach their minimum values for cases (2)-(4).Now we compare the QFIs of the state |ψa$\rangle$ and the state |ψb$\rangle$. The ratio between the QFIs for these two states is Fa/Fb = (r - 1)(N - 2)2(N + 1)2/[N(Nr - N + 1)(N - 1)2]. When N ≥ 2, the QFI of the state |ψb$\rangle$ is larger than the one of the state |ψa$\rangle$. Next, we compare the QFIs of the state |ψb$\rangle$ and the state |ψc$\rangle$. For N = 2 or N = 3, we can only choose the state |ψb$\rangle$, because the state |ψc$\rangle$ is not available. For N ≥ 4 and r ∈ (2 - 1/N, 2), we can still only choose the state |ψb$\rangle$, because the state |ψc$\rangle$ is not available. For N ≥ 4 and r ∈ [7/4, 2 - 1/N], we definite the solution of r for Fb = Fc as
$\begin{eqnarray}{r}_{{{\rm{e}}}_{2}}=\frac{1}{2N}\left[3N-1+\frac{{N}^{2}-1}{\sqrt{2({N}^{2}+1)}}\right].\end{eqnarray}$
For N = 4, 5, the state |ψb$\rangle$ in r ∈ [7/4, 2 - 1/N] is better due to ${r}_{{{\rm{e}}}_{2}}\lt 7/4$. For N ≥ 6, in the range of $r\in [7/4,{r}_{{{\rm{e}}}_{2}}]$, the state |ψc$\rangle$ is better; in the range of $r\in [{r}_{{{\rm{e}}}_{2}},2-1/N]$, the state |ψb$\rangle$ is better.In the above discussion, we consider the optimal states for p ≠ q. Next, we compare the above results with the ones for p = q. First, we compare $\mathop{{\rm{\max }}}\limits_{p=q}F$ of the state $| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle $ for p = q in equation (29 ) with Fb of the state |ψb$\rangle$ for p ≠ q in equation (42 ). The solution of r for ${F}_{b}=\mathop{{\rm{\max }}}\limits_{p=q}F$ is [2N(N + 1)2 + (N - 1)3]/{N[(N - 1)2 + (N + 1)2]}. For N = 2, the QFI of the state $| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle $ for p = q in r ∈ [7/4, 37/20] is higher, but the value of $\bar{n}$ cannot be an integer, thus we should choose the state |ψb$\rangle$. For N ≥ 3, the state |ψb$\rangle$ is better. Furthermore, if the state |ψc$\rangle$ is better than the state |ψb$\rangle$, then the state |ψc$\rangle$ is naturally better than the state $| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle $ for p = q. In summary, the optimal state in r ∈ [7/4, 2) is
$\begin{eqnarray}\left\{\begin{array}{ll}\sqrt{\frac{N(r-1)}{N-1}}| N,N-1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)-1}{N-1}}| 0,N\rangle , & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[\frac{7}{4},{r}_{{{\rm{e}}}_{2}}\right],\\ \sqrt{\frac{N(r-1)+1}{N+1}}| N,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)}{N+1}}| 0,N-1\rangle ,\quad & N={{\mathbb{Z}}}_{[2,5]}\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{2}},2\right).\end{array}\right.\end{eqnarray}$
According to the above discussion for the case of $\bar{n}\in (N,2N)$, we have
$\begin{eqnarray}\mathop{{\rm{\max }}}\limits_{p,q,g,l}F=\left\{\begin{array}{lc}{N}^{2}\left(r-1\right)\left(\sqrt{4r-3}-r\right)\frac{{\left(3-\sqrt{4r-3}\right)}^{2}}{{\left(\sqrt{4r-3}-1\right)}^{2}}, & r\in \left(1,\frac{7}{4}\right),\\ N(r-1)[N(2-r)-1]\frac{{(N+1)}^{2}}{{(N-1)}^{2}}, & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[\frac{7}{4},{r}_{{{\rm{e}}}_{2}}\right],\\ N[N(r-1)+1](2-r)\frac{{(N-1)}^{2}}{{(N+1)}^{2}},\,N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left[\frac{7}{4},2\right)\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{2}},2\right).\end{array}\right.\end{eqnarray}$
The corresponding optimal state of |ψin$\rangle$ is $\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\left\{\begin{array}{lc}\sqrt{\frac{r-1}{\sqrt{4r-3}-1}}| N,N(\sqrt{4r-3}-1)\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{\sqrt{4r-3}-r}{\sqrt{4r-3}-1}}| 0,N\rangle , & r\in \left(1,\frac{7}{4}\right),\\ \sqrt{\frac{N(r-1)}{N-1}}| N,N-1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)-1}{N-1}}| 0,N\rangle , & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[\frac{7}{4},{r}_{{{\rm{e}}}_{2}}\right],\\ \sqrt{\frac{N(r-1)+1}{N+1}}| N,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)}{N+1}}| 0,N-1\rangle ,N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left[\frac{7}{4},2\right)\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{2}},2\right).\end{array}\right.\end{eqnarray}$
Here, θ is a relative phase. In r ∈ (1, 7/4], the state for r = 7/4 is a separable state. Thus, we change the range r ∈ (1, 7/4] to the range r ∈ (1, 7/4).In summary, the maximum QFI is
$\begin{eqnarray}\mathop{{\rm{\max }}}\limits_{p,q,g,l}F=\left\{\begin{array}{ll}Nr(N+1-Nr)\frac{{(N-1)}^{2}}{{(N+1)}^{2}},\quad \quad N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left(0,\frac{1}{4}\right]\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left(0,{r}_{{{\rm{e}}}_{1}}\right],\\ N(Nr-1)(1-r)\frac{{(N+1)}^{2}}{{(N-1)}^{2}}, & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{1}},\frac{1}{4}\right],\\ {N}^{2}\frac{\left(1-r\right)\left(r-2+\sqrt{5-4r}\right){\left(3-\sqrt{5-4r}\right)}^{2}}{{\left(\sqrt{5-4r}-1\right)}^{2}}, & r\in \left(\frac{1}{4},1\right),\\ {N}^{2}, & r=1,\\ {N}^{2}\frac{\left(r-1\right)\left(\sqrt{4r-3}-r\right){\left(3-\sqrt{4r-3}\right)}^{2}}{{\left(\sqrt{4r-3}-1\right)}^{2}}, & r\in \left(1,\frac{7}{4}\right),\\ N(r-1)[N(2-r)-1]\frac{{(N+1)}^{2}}{{(N-1)}^{2}}, & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[\frac{7}{4},{r}_{{{\rm{e}}}_{2}}\right],\\ N[N(r-1)+1](2-r)\frac{{(N-1)}^{2}}{{(N+1)}^{2}},\,N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left[\frac{7}{4},2\right)\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{2}},2\right).\end{array}\right.\end{eqnarray}$
The corresponding optimal state of |ψin$\rangle$ is $\begin{eqnarray}| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle =\left\{\begin{array}{ll}\sqrt{\frac{Nr}{N+1}}| N,1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{1-\frac{Nr}{N+1}}| \mathrm{0,\; 0}\rangle ,\,N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left(0,\frac{1}{4}\right]\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left(0,{r}_{{{\rm{e}}}_{1}}\right],\\ \sqrt{\frac{Nr-1}{N-1}}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(1-r)}{N-1}}| \mathrm{0,\; 1}\rangle , & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{1}},\frac{1}{4}\right],\\ \sqrt{\frac{r-2+\sqrt{5-4r}}{\sqrt{5-4r}-1}}| N,0\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{1-r}{\sqrt{5-4r}-1}}| 0,N\left(2-\sqrt{5-4r}\right)\rangle , & r\in \left(\frac{1}{4},1\right),\\ \frac{1}{\sqrt{2}}\left(| 0,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }| N,0\rangle \right), & r=1,\\ \sqrt{\frac{r-1}{\sqrt{4r-3}-1}}| N,N(\sqrt{4r-3}-1)\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{\sqrt{4r-3}-r}{\sqrt{4r-3}-1}}| 0,N\rangle , & r\in \left(1,\frac{7}{4}\right),\\ \sqrt{\frac{N(r-1)}{N-1}}| N,N-1\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)-1}{N-1}}| 0,N\rangle , & N\,\geqslant \,6\,{\rm{with}}\,r\in \left[\frac{7}{4},{r}_{{{\rm{e}}}_{2}}\right],\\ \sqrt{\frac{N(r-1)+1}{N+1}}| N,N\rangle +{{\rm{e}}}^{{\rm{i}}\theta }\sqrt{\frac{N(2-r)}{N+1}}| 0,N-1\rangle ,N={{\mathbb{Z}}}_{[2,5]} & {\rm{with}}\,r\in \left[\frac{7}{4},2\right)\,{\rm{or}}\,N\,\geqslant \,6\,{\rm{with}}\,r\in \left[{r}_{{{\rm{e}}}_{2}},2\right).\quad & \end{array}\right.\end{eqnarray}$
The above discussions are based on $\{p\in [0,\bar{n}],\,q\in [\bar{n},2N]\}$. For the other case $\{p\in [\bar{n},2N],\,q\in [0,\bar{n}]\}$, the results are the same as the ones for a=b. Due to replacing k, l with g, h is equivalent to replacing |kl$\rangle$ to |gh$\rangle$ in equation (4 ).
For a fixed mean particle number, we discuss the probe states for phase estimation. The QFI of the $\bar{n}00\bar{n}$ states in equation (8 ) is represented as a blue dashed line in figure 6. The dimension of the $\bar{n}00\bar{n}$ states is $\bar{n}+1$ with $\bar{n}=Nr$. Here, we choose N = 6. The maximum QFI of the optimal probe states in equation (48 ) is depicted as a red solid line in figure 6. Through comparison, it is found that in the interval r ∈ (0, 1), the precision of parameter estimation using the optimal probe states in equation (49 ) is higher. Whereas, in the interval r ∈ [1, 2), the precision of parameter estimations using the $\bar{n}00\bar{n}$ states in equation (9 ) is higher. The reason for this is that, in the interval r ∈ [1, 2), the $\bar{n}00\bar{n}$ states possess a higher dimension, which serves as a metrological resource [51]. We then increase the dimension of the optimal probe states to $\bar{n}+1$, which matches the dimension of the $\bar{n}00\bar{n}$ states. Then we show its maximum QFI as the green dotted line with pentagrams in figure 6. In the interval r ∈ [1, 2), the blue dashed line and the green dotted line with pentagrams coincide. It indicates that our optimal probe states are also optimal after compensating for the dimensional resources.
Figure 6. The QFI of the $\bar{n}00\bar{n}$ states in equation ( |
The impact of particle losses in optical phase estimation needs to be fully considered. We integrate fictitious beam splitters into two arms with the transmission coefficients ${T}_{1}={\cos }^{2}({\varphi }_{1}/2)$ and ${T}_{2}={\cos }^{2}({\varphi }_{2}/2)$ describing particle losses [52, 53]. The fictitious beam splitters can be represented as the operators 2 ) is ρ(Φ) = U(Φ)ρabU(Φ)†. The QFI of the optimal probe states $| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle $ in particle losses is ${F}_{{\rm{opt,loss}}}={\rm{Tr}}\left[\rho (\phi ){L}^{2}\right]$. Here, L is the symmetric logarithmic derivative that satisfies the relation ∂Φρ(Φ) = [ρ(Φ)L + Lρ(Φ)]/2. Similarly, the QFI of the $\bar{n}00\bar{n}$ state in particle losses is ${F}_{\bar{n}00\bar{n},{\rm{loss}}}$. In the case of particle losses, we compare the robustness of the optimal probe states and the one of the $\bar{n}00\bar{n}$ state by the QFI in the cases of $\bar{n}\leqslant N$ and $\bar{n}\gt N$, respectively (see figure 7).
$\begin{eqnarray}{U}_{ac}({\varphi }_{1})={{\rm{e}}}^{{\rm{i}}\frac{{\varphi }_{1}}{2}({a}^{\dagger }c+a{c}^{\dagger })},\quad {U}_{bd}({\varphi }_{2})={{\rm{e}}}^{{\rm{i}}\frac{{\varphi }_{2}}{2}({b}^{\dagger }d+b{d}^{\dagger })}.\end{eqnarray}$
Here, φ1 and φ2 are the corresponding phases. The symbols c and d are two fictitious modes. The transmission coefficients T1,2 = 1 (T1,2 = 0) represent no (all) particles loss in modes a and b, respectively. The total probe state is $| {\psi }_{{\rm{tot}}}\rangle =| {\psi }_{{\rm{in}}}^{{\rm{opt}}}\rangle | 0{\rangle }_{c}| 0{\rangle }_{d}$. Then, the reduced density matrix with respect to modes a and b is $\begin{eqnarray}{\rho }_{ab}={{\rm{Tr}}}_{cd}\left[{U}_{bd}({\varphi }_{2}){U}_{ac}({\varphi }_{1})| {\psi }_{{\rm{tot}}}\rangle \langle {\psi }_{{\rm{tot}}}| {U}_{ac}^{\dagger }({\varphi }_{1}){U}_{bd}^{\dagger }({\varphi }_{2})\right],\end{eqnarray}$
where ${{\rm{Tr}}}_{cd}(\cdot )$ is the partial trace to modes c and d. The final state after the entire operation U(Φ) in equation (
Figure 7. Comparison of QFI between the optimal probe state and the $\bar{n}00\bar{n}$ state with the transmission coefficients T1,2. (a) $\bar{n}=1$ (corresponding to the example of $\bar{n}\leqslant N$), (b) $\bar{n}=11$ (corresponding to the example of $\bar{n}\gt N$). Here, N = 6. |
In figure 7(a), we demonstrate the situation where $\bar{n}=1$ (corresponding to the example of $\bar{n}\leqslant N$). In the blue region, the optimal probe state shows better robustness. Especially, in the black dot region, the QFI of the optimal probe state in particle losses is even greater than the one of the $\bar{n}00\bar{n}$ state without particle losses. This indicates that our optimal probe state is very robust in this region. In figure 7(b), we demonstrate the situation where $\bar{n}=11$ (corresponding to the example of $\bar{n}\gt N$). In the blue region, the optimal probe state with respect to particle losses is more robust than the $\bar{n}00\bar{n}$ state with respect to particle losses. This indicates that even in the case where the particle losses are significant, our optimal probe state is robust.
4. Conclusion
To improve the precision, we discuss the optimal probe states of linear phase estimation. The PEFSs have been proven to be more robust than the NOON states in the case of photon losses, but it has not yet been discussed whether it is optimal. Thus, we focus on the general form of the superposition state of the Fock states, and then use QFI as a criterion to optimize the probe states. Due to the very limited mean particle number in the current experiment, we fix the mean particle number to search for the maximum QFI. The maximum QFI corresponds to the optimal probe states, which are superior than the PEFSs. For $\bar{n}\leqslant N$, our optimal probe states can provide better performance than the $\bar{n}00\bar{n}$ states. For $\bar{n}\gt N$, when the dimension of the probe states is large enough, our optimal probe states can also remain optimal.