Quantum network probing with indefinite routing

Abstract

Quantum network probing is experimental estimation of network parameters by passing probes through the network. In probing with indefinite routing (IR), the probe traces a quantum mechanical superposition of different paths through the network. We consider a quantum network modeled by three identical qudit depolarizing channels, each channel with state preservation probability $\theta$ and dimension d. Using quantum Fisher (QF) information as our measure of merit, we comprehensively assess the advantage associated with maximal IR for estimating the network depolarization rate $1-\theta$. A three-channel network admits three distinctive types of IR, cyclical, directional, and full. Definite routing is the case where the probe is confined to a single path through the network. We find that compared to this baseline case all three IR types yield a gain in QF information, the gain being greatest with full IR. Comparing the information gains with cyclical and directional IR shows that, for $\theta$ above a dimensional threshold, directional IR offers greater advantage, while below this threshold cyclical IR is more advantageous. Our results show further that the joint effect of cyclical IR and directional IR can be synergistic or antagonistic, depending on $\theta$ and d. With our analytical approach, the network output states are quasi-classical, greatly simplifying the derivation of the QF information involved. This approach can be extended to larger networks with different IR schemes.

Appendix

This appendix shows derivations of blocks $\mathbf {G}=\Sigma _{\text {F1},\text {B1}}$, $\mathbf {A}=\Sigma _{\text {F1},\text {F3}}$, and $\mathbf {B}=\Sigma _{\text {F1},\text {B3}}$ in the network outputs in (18), (25), and (32). The Kraus operators for the ith qudit depolarizing channel are $\mathbf {D}_{i0}=\sqrt{\theta _i}\mathbf {I}$ and $\mathbf {D}_{ij}=\frac{1}{d}\sqrt{1-\theta _i}\mathbf {U}_j$ for $j=1,\ldots ,d^2$, where the $\mathbf {U}_j$ are orthogonal unitary operators. Because these $d^2$ unitary operators $\mathbf {U}_i$ are orthogonal, they form an orthonormal basis with respect to the Hilbert–Schmidt inner product, and for any given d-dimensional linear operator $\mathbf {V}$, we have

$$\begin{aligned} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \text {tr}[\mathbf {U}_i^\dagger \mathbf {V}]\mathbf {U}_i = \mathbf {V}\end{aligned}$$

and

$$\begin{aligned} \frac{1}{d} \sum \limits _{i = 1}^{{d^2}} \mathbf {U}_i\mathbf {V}\mathbf {U}_i^\dagger = \text {tr}[\mathbf {V}]\mathbf {I}. \end{aligned}$$

These two identities are used repeatedly in the following calculations. The notation $\bar{\theta }_i = 1-\theta _i$ is used throughout for convenience.

Calculation of $\mathbf {G}=\Sigma _{\text {F1},\text {B1}}$:

$$\begin{aligned} \Sigma _{\text {F1},\text {B1}}= & {} \sum \limits _{i = 0}^{{d^2}} \sum \limits _{j = 0}^{{d^2}} \sum \limits _{k = 0}^{{d^2}} \mathbf {D}_{3k} \mathbf {D}_{2j} \mathbf {D}_{1i} \,\varvec{\sigma }\, \mathbf {D}_{3k}^\dagger \mathbf {D}_{2jk}^\dagger \mathbf {D}_{1i}^\dagger \nonumber \\= & {} \theta _1\theta _2\theta _3 \,\varvec{\sigma }+ \left( \frac{\bar{\theta }_1}{d}\theta _2\theta _3 + \theta _1 \frac{\bar{\theta }_2}{d}\theta _3 +\theta _1\theta _2\frac{\bar{\theta }_3}{d}\right) \mathbf {I}+ \theta _1 \frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}\frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_j^\dagger \nonumber \\&+\, \frac{\bar{\theta }_1}{d} \theta _2 \frac{\bar{\theta }_3}{d} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_i^\dagger + \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\theta _3 \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger \nonumber \\&+\, \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}\frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger . \end{aligned}$$

(38)

The three double sums in (38) are identically

$$\begin{aligned} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_j^\dagger = \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \text {tr}[ \varvec{\sigma }\, \mathbf {U}_k^\dagger ] = \varvec{\sigma }, \end{aligned}$$

(39)

while the triple sum in (38) is

$$\begin{aligned} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger&= \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \text {tr}[ \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_j^\dagger ] \nonumber \\&= \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \,\varvec{\sigma }\, \mathbf {U}_k^\dagger = \mathbf {I}. \end{aligned}$$

(40)

Substituting (39) and (40) back into (38) then yields

$$\begin{aligned} \mathbf {G}= & {} \theta _1\theta _2\theta _3 \,\varvec{\sigma }+ \left( \frac{\bar{\theta }_1}{d}\theta _2\theta _3 + \theta _1 \frac{\bar{\theta }_2}{d}\theta _3 +\theta _1\theta _2 \frac{\bar{\theta }_3}{d}\right) \!\mathbf {I}\nonumber \\&+\, \left( \theta _1 \frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}+ \frac{\bar{\theta }_1}{d}\theta _2 \frac{\bar{\theta }_3}{d}+ \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\theta _3 \right) \varvec{\sigma }+ \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}\mathbf {I}, \end{aligned}$$

(41)

with $g_\sigma (\theta )$ and $g_{\text {\tiny {I}}}(\theta )$ as in (19) for $\theta _1=\theta _2=\theta _3\equiv \theta$.

Calculation of $\mathbf {A}=\Sigma _{\text {F1},\text {F3}}$:

$$\begin{aligned} \Sigma _{\text {F1},\text {F3}}= & {} \sum \limits _{i = 0}^{{d^2}} \sum \limits _{j = 0}^{{d^2}} \sum \limits _{k = 0}^{{d^2}} \mathbf {D}_{3k} \mathbf {D}_{2j} \mathbf {D}_{1i} \,\varvec{\sigma }\, \mathbf {D}_{3k}^\dagger \mathbf {D}_{1i}^\dagger \mathbf {D}_{2j}^\dagger \\= & {} \theta _1\theta _2\theta _3 \,\varvec{\sigma }+ \left( \frac{\bar{\theta }_1}{d}\theta _2\theta _3 + \theta _1 \frac{\bar{\theta }_2}{d} \theta _3 +\theta _1\theta _2\frac{\bar{\theta }_3}{d} \right) \mathbf {I}+ \theta _1 \frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_j^\dagger \\&+\, \frac{\bar{\theta }_1}{d} \theta _2 \frac{\bar{\theta }_3}{d} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_i^\dagger + \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d} \theta _3 \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_i^\dagger \mathbf {U}_j^\dagger \\&+\, \frac{\bar{\theta }_1}{d} \frac{\bar{\theta }_1}{d} \frac{\bar{\theta }_3}{d} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_k^\dagger \mathbf {U}_i^\dagger \mathbf {U}_j^\dagger \\&= \theta _1\theta _2\theta _3 \,\varvec{\sigma }+ \left( \frac{\bar{\theta }_1}{d}\theta _2\theta _3 + \theta _1 \frac{\bar{\theta }_2}{d} \theta _3 +\theta _1\theta _2\frac{\bar{\theta }_3}{d} \right) \!\mathbf {I}\\&+\, \theta _1 \frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d} \varvec{\sigma }+ \frac{\bar{\theta }_1}{d} \theta _2 \frac{\bar{\theta }_3}{d} \varvec{\sigma }+ \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d} \theta _3 d\mathbf {I}+ \frac{\bar{\theta }_1}{d} \frac{\bar{\theta }_1}{d} \frac{\bar{\theta }_3}{d} d\varvec{\sigma }. \end{aligned}$$

This yields $a_\sigma (\theta )$ and $a_{\text {\tiny {I}}}(\theta )$ in (26) for $\theta _1=\theta _2=\theta _3\equiv \theta$.

Calculation of $\mathbf {B}=\Sigma _{\text {F1},\text {B3}}$:

$$\begin{aligned} \Sigma _{\text {F1},\text {B3}}= & {} \sum \limits _{i = 0}^{{d^2}} \sum \limits _{j = 0}^{{d^2}} \sum \limits _{k = 0}^{{d^2}} \mathbf {D}_{3k} \mathbf {D}_{2j} \mathbf {D}_{1i} \,\varvec{\sigma }\, \mathbf {D}_{2j}^\dagger \mathbf {D}_{1i}^\dagger \mathbf {D}_{3k}^\dagger \nonumber \\= & {} \theta _1\theta _2\theta _3 \,\varvec{\sigma }+ \left( \frac{\bar{\theta }_1}{d}\theta _2 \theta _3 + \theta _1 \frac{\bar{\theta }_2}{d}\theta _3 +\theta _1\theta _2 \frac{\bar{\theta }_3}{d}\right) \mathbf {I}+ \theta _1 \frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}\frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_k^\dagger \nonumber \\&+\, \frac{\bar{\theta }_1}{d}\theta _2 \frac{\bar{\theta }_3}{d}\frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_i^\dagger \mathbf {U}_k^\dagger + \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\theta _3 \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger \nonumber \\&+\, \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}\frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger \mathbf {U}_k^\dagger . \end{aligned}$$

(42)

The four sums in (42) are

$$\begin{aligned} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_k^\dagger= & {} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \text {tr}[\varvec{\sigma }]\mathbf {I}\mathbf {U}_k^\dagger = d\mathbf {I}, \end{aligned}$$

(43)

$$\begin{aligned} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_i^\dagger \mathbf {U}_k^\dagger= & {} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \text {tr} [\varvec{\sigma }]\mathbf {I}\mathbf {U}_k^\dagger = d\mathbf {I}, \end{aligned}$$

(44)

$$\begin{aligned} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger= & {} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \mathbf {U}_j \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger \nonumber \\= & {} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \mathbf {U}_j \text {tr}[\varvec{\sigma }\, \mathbf {U}_j^\dagger ] = \varvec{\sigma }, \end{aligned}$$

(45)

and

$$\begin{aligned} \frac{1}{d}\sum \limits _{i = 1}^{{d^2}} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \mathbf {U}_i \,\varvec{\sigma }\, \mathbf {U}_j^\dagger \mathbf {U}_i^\dagger \mathbf {U}_k^\dagger= & {} \frac{1}{d}\sum \limits _{j = 1}^{{d^2}} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}} \mathbf {U}_k \mathbf {U}_j \text {tr}[\varvec{\sigma }\ \mathbf {U}_j^\dagger ] \mathbf {U}_k^\dagger \nonumber \\= & {} \frac{1}{d}\sum \limits _{k = 1}^{{d^2}}\mathbf {U}_k \,\varvec{\sigma }\, \mathbf {U}_k^\dagger = \mathbf {I}. \end{aligned}$$

(46)

Substituting (43), (44), (45), and (46) back into (42) yields

$$\begin{aligned} \mathbf {B}= & {} \theta _1\theta _2\theta _3 \,\varvec{\sigma }+ \left( \frac{\bar{\theta }_1}{d}\theta _2\theta _3 + \theta _1 \frac{\bar{\theta }_2}{d} \theta _3 +\theta _1\theta _2\frac{\bar{\theta }_3}{d} \right) \!\mathbf {I}\\&+\, \theta _1 \frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}d\mathbf {I}+ \frac{\bar{\theta }_1}{d}\theta _2 \frac{\bar{\theta }_3}{d}d\mathbf {I}+ \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\theta _3 \varvec{\sigma }+ \frac{\bar{\theta }_1}{d}\frac{\bar{\theta }_2}{d}\frac{\bar{\theta }_3}{d}\,\mathbf {I}, \end{aligned}$$

with $b_\sigma (\theta )$ and $b_{\text {\tiny {I}}}(\theta )$ as in (33) for $\theta _1=\theta _2=\theta _3\equiv \theta$.