Network realization of triplet-type quantum stochastic systems

Abstract

This paper focuses on a problem of network synthesis for a class of quantum stochastic systems. The systems under consideration are of triplet-type form and stem from linear quantum optics and linear quantum circuits. A new quantum network realization approach is proposed by generalizing the scattering operator from the scalar form to a unitary matrix in network components. It shows that the triplet-type quantum stochastic system can be approximated by a quantum network which consists of some one-degree-of-freedom generalized open-quantum harmonic oscillators (1DGQHOs) via series, concatenation and feedback connections.

Appendix

The proof of Lemma 3

For ${\hat{R}}=(a_{jk})_{2 \times 2}$, the rank condition (17) implies that there exist vectors ${\hat{A}}$ and ${\hat{B}}$ defined in (20) such that (21) holds. Recalling (16), ${\hat{A}}=(a_1, a_2,\ldots , a_c)^T$ and ${\hat{B}}=(b_1, b_2,\ldots , b_c)^T$, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} {\hat{A}}^T\Delta P=(a_1\alpha _1+a_2\alpha _2+\cdots +a_c\alpha _c)+i(a_1\beta _1+a_2\beta _2+\cdots +a_c\beta _c),\\ {\hat{B}}^T\Delta P=(b_1\alpha _1+b_2\alpha _2+\cdots +b_c\alpha _c)+i(b_1\beta _1+b_2\beta _2+\cdots +b_c\beta _c). \end{array}\right. } \end{aligned}$$

(37)

Substituting (21) into (37) leads to

$$\begin{aligned} {\left\{ \begin{array}{ll} {\hat{A}}^T\Delta P=-a_{12}+ia_{11},\\ {\hat{B}}^T\Delta P=a_{21}+ia_{22}. \end{array}\right. } \end{aligned}$$

(38)

It can be checked by $(A)^{\sharp }=(a_{ij})^*$ and $\mathfrak {I}(A)=(A - A^\sharp )/2i$ that

$$\begin{aligned} \mathfrak {I}\left\{ K_{12}^\dag S_{12}(I-S_{21}S_{12})^{-1}K_{21} \right\}= & {} \mathfrak {I}\left\{ -\left[ K_{12}^\dag S_{12}(I-S_{21}S_{12})^{-1}K_{21}\right] ^\sharp \right\} \nonumber \\= & {} \mathfrak {I}\left\{ -K_{12}^T S_{12}^\sharp (I-S_{21}^\sharp S_{12}^\sharp )^{-1}K_{21}^\sharp \right\} . \end{aligned}$$

(39)

Then in terms of (15), we have

$$\begin{aligned}&\mathfrak {I}\left\{ K_{12}^\dag S_{12}(I-S_{21}S_{12})^{-1}K_{21}+K_{12}^T(I-S_{21}^T S_{12}^T)^{-1}S_{21}^TK_{21}^\sharp \right\} \nonumber \\&\quad =\mathfrak {I}\left\{ -K_{12}^T S_{12}^\sharp (I-S_{21}^\sharp S_{12}^\sharp )^{-1}K_{21}^\sharp +K_{12}^T(I-S_{21}^T S_{12}^T)^{-1}S_{21}^TK_{21}^\sharp \right\} \nonumber \\&\quad =\mathfrak {I}\left\{ K_{12}^T[(I-S_{21}^TS_{12}^T)^{-1}S_{21}^T-S_{12}^\sharp (I-S_{21}^\sharp S_{12}^\sharp )^{-1}]K_{21}^\sharp \right\} =\mathfrak {I}\left\{ K_{12}^T\Delta K_{21}^\sharp \right\} . \qquad \quad \end{aligned}$$

(40)

On the other hand, one has by (19) and (38) that

$$\begin{aligned} \mathfrak {I}\left\{ K_{12}^T\Delta K_{21}^\sharp \right\}= & {} \mathfrak {I}\left\{ \begin{bmatrix} {\hat{A}}^T\Delta P&-i{\hat{A}}^T\Delta P\\ i {\hat{B}}^T\Delta P&{\hat{B}}^T\Delta P \end{bmatrix}\right\} =\mathfrak {I}\left\{ \begin{bmatrix} -a_{12}+ia_{11}&ia_{12}+a_{11}\\ ia_{21}-a_{22}&a_{21}+ia_{22} \end{bmatrix}\right\} \\= & {} \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} ={\hat{R}}, \end{aligned}$$

which together with (40) proves the lemma. $\square$

The proof of Theorem 1

In order to compute the triplet of quantum feedback network $G_{\text {final-net}}$ constructed in the theorem, we determine the triplet of the reduced Markov model system $G_\mathrm{red}=(S_\mathrm{red}, L_\mathrm{red}, H_\mathrm{red})$ obtained in the theorem. With $S, R=(R_{jk}), K=\begin{bmatrix} K_{1}&K_{2}&\ldots&K_{n}\end{bmatrix}, R_j, S_{jk}, K_{jk}, j,k \in \mathbb {N}_n$, as defined above, from (11) in Lemma 1 one can obtain each element of triplet $G_\mathrm{red}$ known as $S_\mathrm{red}, L_\mathrm{red}$ and $H_\mathrm{red}$. As $S_\mathrm{red}$ and $L_\mathrm{red}$ are available directly in the form of (11), only $H_\mathrm{red}$ need to be calculated further

$$\begin{aligned} H_\mathrm{red}&=\sum ^n_{j=1}H_j+\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}\mathfrak {I}\left\{ \begin{bmatrix} L_{jk}^\dag&L_{kj}^\dag \end{bmatrix}\begin{bmatrix} I&-S_{jk}\\ -S_{kj}&I \end{bmatrix}^{-1}\begin{bmatrix} L_{jk} \\ L_{kj}\end{bmatrix} \right\} . \end{aligned}$$

From (10), one has that

$$\begin{aligned} \mathbb {S}_{jk}^{-1}=\begin{bmatrix} (I-S_{jk}S_{kj})^{-1}&S_{jk}(I-S_{kj}S_{jk})^{-1}\\ S_{kj}(I-S_{jk}S_{kj})^{-1}&(I-S_{kj}S_{jk})^{-1} \end{bmatrix}, \end{aligned}$$

which leads to

$$\begin{aligned} H_\mathrm{red}&=\frac{1}{2}\sum ^n_{j=1}x_j^TR_jx_j\\&\quad +\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}\mathfrak {I}\left\{ \begin{bmatrix} L_{jk}^\dag&L_{kj}^\dag \end{bmatrix} \begin{bmatrix} (I-S_{jk}S_{kj})^{-1}&S_{jk}(I-S_{kj}S_{jk})^{-1}\\ S_{kj}(I-S_{jk}S_{kj})^{-1}&(I-S_{kj}S_{jk})^{-1} \end{bmatrix}\begin{bmatrix} L_{jk} \\ L_{kj}\end{bmatrix}\right\} . \end{aligned}$$

Expanding, recombining, and changing the order of summation gives

$$\begin{aligned} H_\mathrm{red}&=\frac{1}{2}\sum ^n_{j=1}x_j^TR_jx_j+\sum ^{n-1}_{j=1}\sum ^n_{k=1, k\ne j}\mathfrak {I}\left\{ L_{jk}^\dag (I-S_{jk}S_{kj})^{-1}L_{jk}\right\} \\&\quad +\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}\mathfrak {I}\left\{ L_{jk}^\dag S_{jk}(I-S_{kj}S_{jk})^{-1}L_{kj}+ L_{kj}^\dag S_{kj}(I-S_{jk}S_{kj})^{-1}L_{kj}\right\} . \end{aligned}$$

With $L_{jk}=K_{jk}x_j$, and the definition of sym(A) in the theorem, one has that

$$\begin{aligned} H_\mathrm{red}&=\frac{1}{2}\sum ^n_{j=1}x_j^T\Biggl [R_j+2\mathrm{sym}\Biggl (\sum ^n_{k=1, k\ne j}\mathfrak {I}\left\{ K_{jk}^\dag (I-S_{jk}S_{kj})^{-1}K_{jk} \right\} \Biggr )\Biggr ]x_j\\&\quad +\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jk}^\dag S_{jk}(I-S_{kj}S_{jk})^{-1}K_{kj}+ K_{kj}^T (I-S_{kj}^TS_{jk}^T)^{-1}S_{kj}^TK_{kj}^\sharp \right\} x_k . \end{aligned}$$

Then it follows from (32) and (34) that

$$\begin{aligned} H_\mathrm{red}&=\frac{1}{2}\sum ^n_{j=1}x_j^TR_{jj}x_j+\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\left( R_{jk} -\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} \right) x_k \nonumber \\&=\frac{1}{2}x^TRx-\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} x_k. \end{aligned}$$

(41)

With the triplet $S_\mathrm{red}=\mathrm{diag}(S_{11}, S_{22},\ldots , S_{nn}), L_\mathrm{red}=(L_{11}^T, L_{22}^T,\ldots , L_{nn}^T)^T$ and $H_\mathrm{red}$ of $G_\mathrm{red}$ as determined above, one can let $G^0_\mathrm{red}=(0, 0, H_\mathrm{red})$ and $G^j_\mathrm{red}=(S_{jj}, L_{jj}, 0)$ for $j\in \mathbb {N}_n$. It is obvious that according to the concatenation product rule (5) $G_\mathrm{red}=(S_\mathrm{red}, L_\mathrm{red}, H_\mathrm{red})$ can be decomposed as

$$\begin{aligned} G_\mathrm{red}=G^0_\mathrm{red}\boxplus (G^1_\mathrm{red}\boxplus G^2_\mathrm{red}\boxplus \cdots \boxplus G^n_\mathrm{red}). \end{aligned}$$

Now, using the series product rules (8) and (25), one obtains that

$$\begin{aligned}&G^n_\mathrm{red}\vartriangleleft \cdots \vartriangleleft G^1_\mathrm{red}\\&\quad =\bigg (S_{nn\twoheadleftarrow 11}, \sum ^n_{j=1}S_{nn\twoheadleftarrow (j+1)(j+1)}L_{jj}, \ \sum ^n_{k=1}H_k\\&\qquad +\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} x_k\bigg )\\&\quad =\bigg (S_{nn}\ldots S_{22}S_{11},\ \sum ^n_{j=1}S_{nn\twoheadleftarrow (j+1)(j+1)} S^\dag _{nn\twoheadleftarrow {(j+1)(j+1)}}K_{j}x_j,\\&\qquad \sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} x_k\bigg )\\&\quad =\bigg (S_{nn}\ldots S_{22}S_{11},\ \sum ^n_{j=1}K_{j}x_j, \ \sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} x_k\bigg )\\&\quad =\bigg (S,\ \begin{bmatrix} K_{1}&K_{2}&\ldots&K_{2} \end{bmatrix}x,\ \sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} x_k\bigg ), \end{aligned}$$

where $H_k=0$ for $k \in \mathbb {N}_n$ .

Then by (36) and (41) one can compute the quantum feedback network $G_{\text {final-net}}$ constructed in the theorem as

$$\begin{aligned} G_{\text {final-net}}&=(S_{\text {final-net}}, L_{\text {final-net}}, H_{\text {final-net}})\nonumber \\&=G^0_\mathrm{red}\boxplus (G^n_\mathrm{red}\vartriangleleft \cdots \vartriangleleft G^2_\mathrm{red}\vartriangleleft G^1_\mathrm{red})\nonumber \\&=\bigg (0,\ 0,\ \frac{1}{2}x^TRx-\sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} x_k\bigg )\boxplus \nonumber \\&\ \quad \bigg (S, \ \begin{bmatrix} K_{1}&K_{2}&\ldots&K_{2} \end{bmatrix}x, \ \sum ^{n-1}_{j=1}\sum ^n_{k=j+1}x_j^T\mathfrak {I}\left\{ K_{jj}^TS^T_{kk\twoheadleftarrow (j+1)(j+1)}K_{kk}^\sharp \right\} x_k\bigg )\nonumber \\&=\left( S,\ Kx,\ \frac{1}{2}x^TRx\right) =\left( S,\ L,\ H\right) . \end{aligned}$$

(42)

It is easy to see from (42) that the triplet of $G_{\text {final-net}}$ has

$$\begin{aligned} S_{\text {final-net}}=S,\ \ \ \ L_{\text {final-net}}=L,\ \ \ \ H_{\text {final-net}}=H, \end{aligned}$$

that is, the triplet of $G_{\text {final-net}}$ is equal to the triplet of $G_\mathrm{sys}$. Therefore, according to Definition 2, $G_{\text {final-net}}$ has realized the triplet-type quantum stochastic system $G_\mathrm{sys}$. $\square$