Feedback Control of Linear Dynamical Quantum Systems

Abstract

This chapter introduces and treats the topic of quantum feedback control on linear quantum systems. Two distinct approaches to feedback control of quantum systems are considered: measurement-based feedback control and coherent feedback control. For measurement-based feedback control, the notions of controlled Hudson–Parthasarathy QSDEs and controlled quantum filtering equations are introduced. Three control problems are formulated and solved: measurement-based linear quadratic Gaussian control, coherent linear quadratic Gaussian control, and coherent feedback $H^{\infty }$ control. Examples are provided to illustrate each control method.

Section 5.2 contains materials reprinted from Automatica [12], with permission from Elsevier.

Section 5.3 and the associated appendices contain materials reprinted, with permission, from [34] $\copyright$ 2008 IEEE.

Appendices

Appendices

Appendix A: Proof of Theorem 5.2

The proof of Theorem 5.2 will use the following lemma.

Lemma 5.4

Consider a real symmetric matrix X and corresponding operator-valued quadratic form $x^{\top } Xx$ for the system (5.54). Then the following statements are equivalent:

• (i) There exists a constant $\lambda \ge 0$ such that $\langle \rho ,x^{\top }Xx\rangle \le \lambda$ for all Gaussian states $\rho$.

• (ii) The matrix X is negative semidefinite.

Proof

$(i) \Rightarrow (ii)$. To establish this part of the lemma, consider a Gaussian state $\rho$ which has mean $\bar{x}$ and covariance matrix $Y=[\mathrm{Tr}(\rho x_ix_j)]$, with symmetrized covariance $\nicefrac {1}{2}(Y+Y^{\top }) \ge 0$ and $Y-Y^{\top } =2 \imath \mathbb {J}_n$. Then, we can write

$$\begin{aligned} \langle \rho ,x^{\top } Xx\rangle= & {} \sum _{i=1}^{n}\sum _{j=1}^{n}X_{ij}\langle \rho ,x_i x_j\rangle \nonumber \\= & {} \sum _{i=1}^{n}\sum _{j=1}^{n}X_{ij}[Y_{ij}+\bar{x}_i \bar{x}_j]\nonumber \\= & {} \bar{x}^T X \bar{x} + \mathrm{Tr}[XY]. \end{aligned}$$

(5.88)

Now for any constant $\alpha > 0$, consider the inequality of part (i) where $\rho$ is a Gaussian state with mean $\alpha \bar{x}$ and covariance matrix Y. Then, it follows from the bound $\langle \rho ,x^{\top }Xx\rangle \le \lambda$ and (5.88) that $\alpha ^2 \bar{x}^{\top } X \bar{x} + \mathrm{Tr}[XY] \le \lambda \$ for all $\alpha > 0$. From this, it immediately follows that $\bar{x}^{\top } X \bar{x} \le 0$. However, $\bar{x}$ can be chosen arbitrarily. Hence, we can conclude that condition (ii) of the lemma is satisfied.

$(ii) \Rightarrow (i)$. Suppose that the matrix X is negative semidefinite and let $\rho$ be any Gaussian state and suppose that $\rho$ has mean $\bar{x}$ and covariance matrix $Y \ge 0$. Then, it follows from (5.88) that $\langle \rho ,x^{\top } Xx\rangle = \bar{x}^{\top } X \bar{x} + \mathrm{Tr}[XY]$. However, $X \le 0$, $Y + Y^{\top } \ge 0$, and $Y-Y^{\top } =2 \imath \mathbb {J}_n$ implies $\bar{x}^{\top } X \bar{x} \le 0$, and $\mathrm{Tr}[XY] = \nicefrac {1}{2}\mathrm{Tr}[X(Y+Y^{\top })] \le 0$. Hence, $\langle \rho ,x^{\top } Xx\rangle \le 0$ and condition (ii) is satisfied with $\lambda = 0$. $\Box$

Proof of Theorem 5.2. Let the system be dissipative with $V(x )=x^{\top } X x$. By Itō’s rule, the product table (5.60), and the quantum stochastic differential equation (5.54), we have

$$\begin{aligned}&d\langle V(x(t)) \rangle \nonumber \\&\quad = \langle dx^{\top }(t) X x(t) + x^{\top }(t) X dx(t) + dx^{\top }(t) X dx(t) \rangle \nonumber \\&\quad =\langle x^{\top }(t)({ A}^{\top } X+X{ A}) x(t) + \beta _w^{\top }(t) { B}^{\top } X x(t) + \nonumber \\&\quad \quad x^{\top }(t) X { B} \beta _w(t)+ \lambda _0 \rangle dt, \end{aligned}$$

(5.89)

where $\lambda _0$ is given by (5.59). We now note that $\langle V(x(t)) \rangle = \langle \rho , E_0 [ V(x(t)) ] \rangle ,$ where $E_t$ ($t \ge 0$) denotes the conditional expectation map² with respect to the vacuum state $|\Omega \rangle$ of the field (e.g., see [51, p. 215]), and $\rho$ is an initial Gaussian state of the system. Combining this with the integral of (5.89) and (5.56) we find that

$$\begin{aligned}&\left\langle \rho , \int _0^t E_0 [ x^{\top }(s)({ A}^{\top } X+X{ A}) x(s) + \beta _w^{\top }(s) { B}^{\top } X x(s) +\right. \\&\qquad \left. x^{\top }(s) X { B} \beta _w(s) + \lambda _0 + r(x(s),\beta _w(s) ] ds \right\rangle \le \lambda t. \end{aligned}$$

Let $t \rightarrow 0$ to obtain

$$\begin{aligned}&\left\langle \rho , x^{\top }({ A}^{\top }X+X{ A}) x + \beta _w^{\top }{ B}^{\top }X + x^{\top } X{ B}\beta _w + \lambda _0 + \right. \\&\quad \left. [x^{\top } \beta _w^{\top } ] R \left[ \begin{array}{c} x \\ \beta _w \end{array} \right] \right\rangle \le \lambda . \end{aligned}$$

Here, x and $\beta _w$ denote the initial conditions. An application of Lemma 5.4 implies (5.57). Also, (5.58) is a straightforward consequence of this inequality when $\mathsf {R}$ is replaced by $\mathsf {R}+\epsilon I$ where $\epsilon > 0$.

To establish the converse part of the theorem, we first assume that (5.57) is satisfied. Then with $V(x) = x^{\top }Xx$, it follows from (5.89) that

$$\begin{aligned} \langle V(x(t)) \rangle - \langle V(x(0)) \rangle + \int _o^t\langle r(x(s),\beta _w(s))\rangle ds \le \lambda _0 t \end{aligned}$$

for all $t > 0$ and all $\beta _w(t)$. Hence, inequality (5.56) is satisfied with $\lambda$ given by (5.59).

If matrix inequality (5.58) is satisfied, then it follows by similar reasoning that there exists an $\epsilon > 0$ such that

$$\begin{aligned}&\langle V(x(t)) \rangle - \langle V(x(0)) \rangle + \int _0^t\langle r(x(s),\beta _w(s))+ \\&\quad \epsilon (x(s)^{\top }x(s) + \beta _w(s)^{\top }\beta _w(s))\rangle ds \le \lambda _0 t. \end{aligned}$$

Hence, inequality (5.56) is satisfied with $\lambda =\lambda _0$ given by (5.59) and with $\mathsf {R}$ replaced by $\mathsf {R}+\epsilon I$. $\Box$

Appendix B: Proof of Lemma 5.2

The proof of Lemma 5.2 will use the following lemma.

Lemma 5.5

If S is a Hermitian matrix, then there is a real constant $\alpha _0$ such that $\alpha I + S \ge 0$ for all $\alpha \ge \alpha _0$.

Proof

Since S is Hermitian, it has real eigenvalues and is diagonalizable. Hence, $S=V^*EV$ for some real diagonal matrix E and orthogonal matrix V. Now let $\alpha _0=-\lambda$, where $\lambda$ is the smallest eigenvalue of S. The result follows since $\alpha I + S=V^*(\alpha I + E)V$ while $\alpha I +E \ge 0$ for all $\alpha \ge \alpha _0$. $\Box$

Proof of Lemma 5.2. The main idea is to explicitly construct matrices $R \in \mathbb {R}^{2n_K \times 2n_K}$, $\mathsf {K} \in \mathbb {C}^{(n_{v_K}+n_y) \times 2n_K}$, $B_{K1} \in \mathbb {R}^{2n_K \times 2n_{v_K}}$, and $B_{K0} \in \mathbb {R}^{2n_u \times 2n_{v_K}}$, with $n_{v_K} \ge n_u$, such that (see Chap. 2)

$$\begin{aligned} A_K= & {} 2 \mathbb {J}_{n_K} (R+\mathfrak {I}\{\mathsf {K}^*\mathsf {K}\}), \end{aligned}$$

(5.90)

$$\begin{aligned} \left[ \begin{array}{cc} B_{K1}&B_K \end{array} \right]= & {} 2\imath \mathbb {J}_{n_K} [\begin{array}{cc} -\mathsf {K}^*&\mathsf {K}^{\top } \end{array}]\Gamma _{n_{v_K}+n_y}, \end{aligned}$$

(5.91)

$$\begin{aligned} C_K= & {} [\begin{array}{cc} I_{2n_u \times 2n_u}&0_{2n_u \times 2(n_{v_K}+n_y-n_u)} \end{array}]P_{n_{v_K}+n_y}^{\top } \nonumber \\&\times \left[ \begin{array}{c} \mathsf {K}+\mathsf {K}^{\#} \\ -\imath \mathsf {K} + \imath \mathsf {K}^{\#} \end{array} \right] , \end{aligned}$$

(5.92)

$$\begin{aligned} \left[ \begin{array}{cc} B_{K0}&0_{2n_u \times 2(n_{v_K}+n_y)}\end{array}\right]= & {} [\begin{array}{cc} I_{2n_u \times 2n_u}&0_{2n_u \times 2(n_{v_K}+n_y-n_u)}\end{array}], \end{aligned}$$

(5.93)

are satisfied. To this end, let $Z=\frac{1}{2}\mathbb {J}_{n_K}^{-1}A=-\frac{1}{2}\mathbb {J}_{n_K} A$. We first construct matrices $\mathsf {K}_{b2}$, $\mathsf {K}_{b1}$, $B_{K1,1}$, and $B_{K1,2}$ according to the following procedure:

• 1. Construct the matrix $\mathsf {K}_{b2}$ according to (5.75).

• 2. Construct a real symmetric $2n_K \times 2n_K$ matrix $W_1$ such that the matrix

  $$\begin{aligned} W_2= & {} W_1 \\&\;+\imath \left( \frac{Z-Z^{\top }}{2}-\frac{1}{4}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{cc} 0 &{} I \\ -I &{} 0 \end{array} \right] P_{n_u}C_K -\mathfrak {I}(\mathsf {K}_{b2}^*\mathsf {K}_{b2}) \right) , \end{aligned}$$

  is non-negative definite. It follows from Lemma 5.5 that such a matrix $W_1$ always exists.

• 3. Construct a matrix $\mathsf {K}_{b1}$ such that $\mathsf {K}_{b1}^* \mathsf {K}_{b1}=W_2$, where $\mathsf {K}_{b1}$ has at least 1 row. This can be done, for example, using the singular value decomposition of $W_2$ (in this case, $\mathsf {K}_{b1}$ will have $2n_K$ rows).

• 4. Construct the matrices $B_{K1,1}$ and $B_{K1,2}$ according to Eqs. (5.74) and (5.76), respectively.

Let $R=\frac{1}{2}(Z+Z^{\top })$. We now show that there exists an integer $n_{v_K} \ge n_u$ such conditions (5.90)–(5.93) are satisfied with the matrix R as defined and with $B_{K1}=[\begin{array}{cc} B_{K1,1}&B_{K1,2} \end{array}]$ and

$$\begin{aligned} \mathsf {K}=\left[ \begin{array}{c} \nicefrac {1}{2}\left[ \begin{array}{cc} I &{} \imath I \end{array}\right] P_{n_u}C_k \\ \mathsf {K}_{b1} \\ \mathsf {K}_{b2} \end{array} \right] . \end{aligned}$$

(5.94)

First note that necessarily $n_{v_K} \ge n_u+1 > n_u$ since $B_{K1}$ has at least $2n_{u}+2$ columns. Also, by virtue of our choice of $\mathsf {K}_{b1}$, we have

$$\begin{aligned} \mathfrak {I}(\mathsf {K}_{b1}^* \mathsf {K}_{b1})= & {} \mathfrak {I}\{W_2\}\\= & {} \nicefrac {1}{2}(Z-Z^{\top })-\nicefrac {1}{4}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{cc} 0 &{} I \\ -I &{} 0 \end{array} \right] P_{n_u}C_K-\mathfrak {I}\{\mathsf {K}_{b2}^* \mathsf {K}_{b2}\}, \end{aligned}$$

and hence

$$\begin{aligned} \mathfrak {I}(\mathsf {K}^*\mathsf {K})= & {} \mathfrak {I}(\mathsf {K}_{b1}^* \mathsf {K}_{b1}) + \mathfrak {I}(\mathsf {K}_{b2}^* \mathsf {K}_{b2})+\nicefrac {1}{4}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{cc} 0 &{} I \\ -I &{} 0 \end{array} \right] P_{n_u} C_K\\= & {} \nicefrac {1}{2}(Z-Z^{\top }). \end{aligned}$$

Since $R=\frac{Z+Z^{\top }}{2}$, we have $R+\mathfrak {I}(\mathsf {K}^* \mathsf {K})=Z$. Therefore, (5.90) is satisfied.

Now, we observe that $\imath \mathbb {J}_{n_K} B_K\mathrm{diag}_{n_y}(M^*)P_{n_y}^{\top }=[\begin{array}{cc} T&-T^\# \end{array}] \$ for some $2n_K \times n_y$ complex matrix T. But by taking the conjugate transpose of both sides of (5.75) which defined $\mathsf {K}_{b2}$, we conclude that $T=-\mathsf {K}_{b2}^*$. Hence,

$$\begin{aligned} B_K=2\imath \mathbb {J}_{n_K} [\begin{array}{cc} -\mathsf {K}_{b2}^*&\mathsf {K}_{b2}^{\top }\end{array} ]P_{n_y}\mathrm{diag}_{n_y}(M). \end{aligned}$$

(5.95)

From (5.74) which defined $B_{K1,1}$, we obtain

$$\begin{aligned} B_{K1,1}= & {} \mathbb {J}_{n_K} C_K ^{\top }\mathrm{diag}_{n_u}(\mathbb {J}) \nonumber \\= & {} \mathbb {J}_{n_K} C_K ^{\top }\mathrm{diag}_{n_u}(\mathbb {J})(2\mathrm{diag}_{n_u}(M^*))\mathrm{diag}_{n_u}(M) \nonumber \\= & {} \imath \mathbb {J}_{n_K} C_K ^{\top }\mathrm{diag}_{n_u}\left( \left[ \begin{array}{cc} -1 &{} 1 \\ \imath &{} \imath \end{array} \right] \right) \mathrm{diag}_{n_u}(M) \nonumber \\= & {} \imath \mathbb {J}_{n_K} C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{cc}-I &{} I \\ \imath I &{} \imath I\end{array} \right] P_{n_u}\mathrm{diag}_{n_u}(M). \end{aligned}$$

(5.96)

Combining (5.76), (5.95) and (5.96) gives us

$$\begin{aligned}&[\begin{array}{ccc}B_{K1,1}&B_{K1,2}&B_{K}\end{array}] \\&\quad =2\imath \mathbb {J}_{n_K} \left[ \nicefrac {1}{2}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{cc} -I &{} I \\ \imath I &{} \imath I \end{array} \right] P_{n_u} \left[ \begin{array}{cc}-\mathsf {K}_{b1}^*&\mathsf {K}_{b1}^{\top }\end{array} \right] P_{n_{v_K}-n_u} \right. \\&\quad \quad \left. \begin{array}{cc}&\left[ \begin{array}{cc} -\mathsf {K}_{b2}^*&\mathsf {K}_{b2}^{\top }\end{array} \right] P_{n_y} \end{array}\right] P_{n_{w_K}}^{\top }P_{n_{w_K}}\mathrm{diag}_{n_{w_K}}(M)\\&\quad =2\imath \mathbb {J}_{n_K} \left[ \begin{array}{ccc} -\nicefrac {1}{2}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{c} I \\ -\imath I \end{array} \right]&-\mathsf {K}_{b1}^*&-\mathsf {K}_{b2}^* \end{array}\right. \\&\quad \left. \begin{array}{ccc}\nicefrac {1}{2}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{c} I \\ \imath I \end{array} \right]&\mathsf {K}_{b1}^{\top }&\mathsf {K}_{b2}^{\top }\end{array} \right] P_{n_{w_K}} \mathrm{diag}_{n_{w_K}}(M) \\&\quad =2\imath \mathbb {J}_{n_K} \left[ \begin{array}{ccc} -\nicefrac {1}{2}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{c} I \\ -\imath I \end{array} \right]&-\mathsf {K}_{b1}^*&-\mathsf {K}_{b2}^* \end{array} \right. \\&\quad \left. \begin{array}{ccc}\nicefrac {1}{2}C_K^{\top }P_{n_u}^{\top }\left[ \begin{array}{c} I \\ \imath I \end{array} \right]&\mathsf {K}_{b1}^{\top }&\mathsf {K}_{b2}^{\top }\end{array} \right] \Gamma \\&\quad =2\imath \mathbb {J}_{n_K} \left[ \begin{array}{cc} -\mathsf {K}^*&\mathsf {K}^{\top }\end{array} \right] \Gamma . \end{aligned}$$

Therefore, (5.91) is also satisfied. Moreover, it is straightforward to verify (5.92) by substituting $\mathsf {K}$ as defined by (5.94) into the right-hand side of (5.92). Finally, since $n_{v_K} \ge n_u$, it follows that $[\begin{array}{cc} B_{K0}&0_{2n_u \times 2(n_{v_K}+n_y-n_u)} \end{array}]$ is precisely the right-hand side of (5.93). This completes the proof of Theorem 5.4. $\Box$