Projection methods for quantum channel construction

Abstract

We consider the problem of constructing quantum channels, if they exist, that transform a given set of quantum states $\{\rho _1, \ldots , \rho _k\}$ to another such set $\{\hat{\rho }_1, \ldots , \hat{\rho }_k\}$. In other words, we must find a completely positive linear map, if it exists, that maps a given set of density matrices to another given set of density matrices, possibly of different dimension. Using the theory of completely positive linear maps, one can formulate the problem as an instance of a positive semidefinite feasibility problem with highly structured constraints. The nature of the constraints makes projection-based algorithms very appealing when the number of variables is huge and standard interior-point methods for semidefinite programming are not applicable. We provide empirical evidence to this effect. We moreover present heuristics for finding both high-rank and low-rank solutions. Our experiments are based on the method of alternating projections and the Douglas–Rachford reflection method.

Appendix: Background

1.1 Matrix representation of $\mathcal {L}$ and $\mathcal {L}^\dagger$

In this section, we show that the matrices $L$ in (9) and $L^\dagger$ in (10) are indeed matrix representations of the linear map $\mathcal {L}$ [defined in (5)] and its Moore–Penrose generalized inverse, respectively, under a specific choice of basis of the vector space ${\mathcal H} ^{nm}$.

1.1.1 Choice of orthonormal basis of ${\mathcal H} ^s$

We choose the standard orthonormal basis for the vector space ${\mathcal H} ^\ell$ of $\ell \times \ell$ Hermitian matrices (over the reals) as follows. Let $e_j\in \mathbb {R}^\ell$ be the $j$th standard unit vector for $j=1,\ldots ,\ell$. Then $e_ie_j^T\in \mathbb {R}^{\ell \times \ell }$ is zero everywhere except the $(i,j)$th entry, which is 1. For $i,j=1,\ldots ,\ell$, define the $(i,j)$th basis matrix as follows:

$$\begin{aligned} E_{ij} = {\left\{ \begin{array}{ll} \frac{1}{\sqrt{2}} \left( e_i e_j^T + e_j e_i^T\right) &{}\quad \text { if }\; i< j, \\ \frac{\mathsf {i}\,}{\sqrt{2}} \left( e_j e_i^T - e_i e_j^T\right) &{}\quad \text { if }\; i > j, \\ e_j e_j^T &{} \quad \text { if } i=j. \end{array}\right. } \end{aligned}$$

(11)

Then ${\mathcal E} _{{\mathrm {real,offdiag}}}\cup {\mathcal E} _{{\mathrm {imag,offdiag}}}\cup {\mathcal E} _{{{\mathrm{{diag}}}}}$ forms an orthonormal basis of ${\mathcal H} ^\ell$, where

• ${\mathcal E} _{\mathrm {real,offdiag}}:=\{E_{ij} : 1\le i < j \le \ell \}$ collects the real zero-diagonal basis matrices,

• ${\mathcal E} _{\mathrm {imag,offdiag}}:=\{E_{ij} : 1\le j < i \le \ell \}$ collects the imaginary zero-diagonal basis matrices, and

• ${\mathcal E} _{{{\mathrm{{diag}}}}}:=\{E_{jj} : 1\le j \le \ell \}$ collects the real diagonal basis matrices.

We define a total ordering $<$ on the tuples $(i,j)$ for $i,j=1,\ldots ,\ell$, so that the matrices are ordered with ${\mathcal E} _{\mathrm {real,offdiag}}< {\mathcal E} _{\mathrm {imag,offdiag}}< {\mathcal E} _{{{\mathrm{{diag}}}}}$ in the element-wise sense [as stated in (6)]. For any $(i,j), (\tilde{i}, \tilde{j})\in \{1,\ldots ,\ell \}^2$, we say that $(i,j) \prec (\tilde{i}, \tilde{j})$ if one of the following holds.

• Case 1: $i<j$ (so that $E_{ij}$ is a real matrix with zero diagonal).

  • $i<j$ and $\tilde{i} \ge \tilde{j}$.

  • $i<j$ and $\tilde{i}<\tilde{j}$, but $\tilde{j}> j$.

  • $i<j$ and $\tilde{i}<\tilde{j} = j$, but $\tilde{i}>i$.

• Case 2: $i>j$ (so that $E_{ij}$ is a imaginary matrix with zero diagonal).

  In this case we must have $\tilde{i}\ge \tilde{j}$.

  • $j<i$ and $\tilde{j} = \tilde{i}$.

  • $j<i$ and $\tilde{j}<\tilde{i}$, but $\tilde{i}> i$.

  • $j<i$ and $\tilde{j}<\tilde{i} = i$, but $\tilde{j}>j$.

• Case 3: $i=j$ (so that $E_{jj}$ is a real diagonal matrix).

  In this case we must have $\tilde{i}=\tilde{j}$.

  • $j<\tilde{j}$.

For instance, when $\ell =3$, our orthonormal basis of choice is given in the following order:

$$\begin{aligned} \begin{array}{lll} E_{12} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0&{}\quad 1&{}\quad 0\\ 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{bmatrix},\ &{} E_{13} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0 \end{bmatrix},\ &{} E_{23} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 1&{}\quad 0 \end{bmatrix},\ \\ E_{21} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0&{}\quad \mathsf {i}\,&{}\quad 0\\ ii&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{bmatrix},\ &{} E_{31} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0&{}\quad 0&{}\quad \mathsf {i}\,\\ 0&{}\quad 0&{}\quad 0\\ ii&{}\quad 0&{}\quad 0 \end{bmatrix},\ &{} E_{32} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad \mathsf {i}\,\\ 0&{}\quad -\mathsf {i}\,&{}\quad 0 \end{bmatrix},\\ E_{11} = \begin{bmatrix} 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{bmatrix},\ &{} E_{22} = \begin{bmatrix} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{bmatrix},\ &{} E_{33} = \begin{bmatrix} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1 \end{bmatrix}. \end{array} \end{aligned}$$

We also work with $nm\times nm$ block matrices, with each block of size $m\times m$. On the space ${\mathcal H} ^{nm}$, for any $1\le i , j \le nm$, let

$$\begin{aligned} s := \left\lceil \frac{i}{m} \right\rceil , \quad t := \left\lceil \frac{j}{m} \right\rceil , \quad p := i-m(s-1), \quad \text {and}\quad q := j-m(t-1). \end{aligned}$$

(12)

Note that $1\le s,t\le n$ are the block indices and $1\le p,q\le m$ are the intra-block indices. For instance, consider the matrix $e_ie_j^T\in {\mathcal H} ^{nm}$ (having only 1 nonzero entry at position $(i,j)$). The nonzero entry is at the $(p,q)$th entry in the $(s,t)$th block (which is of size $m\times m$). The orthonormal matrices $E_{ij}$ defined in (11) are related to the block indices $(s,t)$ as described in Table 7.

Table 7 The nonzero blocks of $E_{ij}$ for any fixed $i,j=1,\ldots ,nm$

Unlike ${\mathcal H} ^s$, we order the blocked orthonormal matrices $E_{ij}$ in ${\mathcal H} ^{nm}$ via the following total ordering $<$ on the set $\mathcal {I}:=\{(i,j): 1\le i,j \le nm\}$. Let $1\le i,j,\tilde{i}, \tilde{j}\le nm$ with $(i,j)\ne (\tilde{i}, \tilde{j})$. Then letting

$$\begin{aligned} \begin{array}{llll} s:=\left\lceil \frac{i}{m}\right\rceil , &{}t:=\left\lceil \frac{j}{m}\right\rceil , &{}p=i-m(s-1), &{}q=j-m(t-1), \\ \tilde{s}\,:=\left\lceil \frac{\tilde{i}}{m}\right\rceil , &{}\tilde{t}\,:=\left\lceil \frac{\tilde{j}}{m}\right\rceil , &{}\tilde{p}=\tilde{i}-m(\tilde{s}-1), &{}\tilde{q}=\tilde{j}-m(\tilde{t}-1), \end{array} \end{aligned}$$

the relation $(i,j)<(\tilde{i}, \tilde{j})$ holds if and only if one of the following holds.

• $\{p,q\}\ne \{\tilde{p},\tilde{q}\}$ and $\left( \min \{p,q\},\ \max \{p,q\}\right) \prec \left( \min \{\tilde{p},\tilde{q}\},\ \max \{\tilde{p},\tilde{q}\}\right)$.

• $\{p,q\}=\{\tilde{p},\tilde{q}\}$ and $(s,t)\prec (\tilde{s}, \tilde{t})$.

• $\{p,q\}=\{\tilde{p},\tilde{q}\}$, $(s,t)=(\tilde{s}, \tilde{t})$ and $p<q$. (Then $\tilde{q} = p < q = \tilde{p}$.)

In other words, we order the 2-tuples $(i,j)$ in $\mathcal {I}$ by grouping all those with the same intra-block index $(p,q)$ and block indices $s<t$ together, for some $p<q$, followed by those tuples with intra-block index $(q,p)$ and block indices $s<t$. As an example, when $m=2$ and $n=3$, the following list gives the first few entries of $\mathcal {I}$:

$$\begin{aligned} (1,4),\ (1,6),\ (3,6),\ (2,3),\ (2,5),\ (4,5),\ (1,2),\ (3,4),\ \ldots \end{aligned}$$

These first 2-tuples $(i,j)$ in $\mathcal {I}$ have the corresponding 4-tuples $(s,t,p,q)$ defined as in (12), given as follows:

$$\begin{aligned}&(1,2,1,2),\ (1,3,1,2),\ (2,3,1,2),\ (1,2,2,1),\ (1,3,2,1),\ (2,3,2,1),\\&\quad (1,1,1,2),\ (2,2,1,2),\ \ldots \end{aligned}$$

Note that the first three 2-tuples have the same intra-block index $(p,q)=(1,2)$. The immediately following three 2-tuples have the intra-block index $(q,p)=(2,1)$, and so on.

1.1.2 Symmetric vectorization of Hermitian matrices

Using the ordered orthonormal basis of ${\mathcal H} ^s$ described in (11) in Sect. 1, we can define the corresponding symmetric vectorization of Hermitian matrices. Since any Hermitian matrix in ${\mathcal H} ^s$ can be expressed as a unique linear combination of the orthonormal basis matrices $E_{ij}$, the map

$$\begin{aligned} {{\mathrm{{sHvec}}}}:{\mathcal H} ^s \rightarrow \mathbb {R}^{s^2}: H\mapsto v, \end{aligned}$$

where $v\in \mathbb {R}^{s^2}$ is the unique vector such that $H = \sum _{i,j=1}^{s} v_{ij} E_{ij}$, is well-defined. The map ${{\mathrm{{sHvec}}}}$ is a linear isometry (i.e., ${{\mathrm{{sHvec}}}}$ is a linear map and $\Vert {{\mathrm{{sHvec}}}}(H)\Vert ^2={{\mathrm{{trace}}}}(H^2)$ for all $H\in {\mathcal H} ^s$), and its adjoint is given by

$$\begin{aligned} {{\mathrm{{sHMat}}}}: \mathbb {R}^{s^2}\rightarrow {\mathcal H} ^s: v\mapsto \sum _{i,j=1}^{s} v_{ij} E_{ij}, \end{aligned}$$

(13)

which is also the inverse map of ${{\mathrm{{sHvec}}}}$. For instance,

$$\begin{aligned} {{\mathrm{{sHvec}}}}\left( \begin{bmatrix}1&\quad \sqrt{2}-\mathsf {i}\,\\ \sqrt{2}+\mathsf {i}\,&\quad 3\end{bmatrix}\right) = \begin{bmatrix} 2&\quad -\sqrt{2}&\quad 1&\quad 3\end{bmatrix}^T. \end{aligned}$$

1.1.3 Ordering the rows and columns in the matrix representation of $\mathcal {L}$

In the following, we compute matrix representations $L_A$ and $L_T$ of the linear maps $\mathcal {L}_A:{\mathcal H} ^{nm}\rightarrow \otimes _{j=1}^m{\mathcal H} ^m$ and $\mathcal {L}_T:{\mathcal H} ^{nm}\rightarrow {\mathcal H} ^n$, respective. The matrix representation $\mathcal {L}=(\mathcal {L}_A(\cdot ), \mathcal {L}_T(\cdot ))$ is then chosen to be $L = \begin{bmatrix} L_A \\ L_T \end{bmatrix}$, with $L_A\in \mathbb {R}^{km^2\times n^2m^2}$ and $L_T\in \mathbb {R}^{m^2\times n^2m^2}$.

Any matrix representation for the linear map $\mathcal {L}_A$ (resp. $\mathcal {L}_T$) depends on the choice of the ordered orthonormal bases for ${\mathcal H} ^{nm}$ and for ${\mathcal H} ^m\times \cdots \times {\mathcal H} ^m$ (resp. for ${\mathcal H} ^{nm}$ and for ${\mathcal H} ^m$). For ${\mathcal H} ^{nm}$, we use the ordered orthonormal basis defined in Sect. 1. For ${\mathcal H} ^m\times \cdots \times {\mathcal H} ^m$, we use the orthonormal basis

$$\begin{aligned} \begin{array}{llll} (E_{12}, 0, \ldots ,0),\ &{} (0,E_{12},\ldots ,0), \ &{} \ldots , \ &{} (0,0,\ldots ,E_{12}), \\ (E_{21}, 0, \ldots ,0), \ &{} (0,E_{21},\ldots ,0), \ &{} \ldots , \ &{} (0,0,\ldots ,E_{21}),\ \\ (E_{13}, 0, \ldots ,0),\ &{} (0,E_{13},\ldots ,0), \ &{} \ldots , \ &{} (0,0,\ldots ,E_{13}),\ \\ (E_{31}, 0, \ldots ,0),\ &{} (0,E_{31},\ldots ,0), \ &{} \ldots , \ &{} (0,0,\ldots ,E_{31}),\ \ldots ,\ (0,0,\ldots ,E_{mm}). \end{array} \end{aligned}$$

(14)

We first construct a matrix representation $L_A$ of $\mathcal {L}_A$ by rows. Recall that

$$\begin{aligned} \mathcal {L}_A(P) = \left( \sum _{s,t=1}^n (A_1)_{st} P_{st}, \sum _{s,t=1}^n (A_2)_{st} P_{st}, \ldots , \sum _{s,t=1}^n (A_k)_{st} P_{st}\right) \end{aligned}$$

for any $P\in {\mathcal H} ^{nm}$, so the rows of $\mathcal {L}_A$ are determined by the linear functionals

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathbf {L}_{{\ell ,p,q,{\mathrm{Re}\,}}}(P):=\sqrt{2} {\mathrm{Re}\,}\left( \sum _{s,t=1}^n (A_\ell )_{st} (P_{st} )\right) _{pq} \\ \mathbf {L}_{{\ell ,p,q,{\mathrm{Im}\,}}}(P):=\sqrt{2} {\mathrm{Im}\,}\left( \sum _{s,t=1}^n (A_\ell )_{st} (P_{st} )\right) _{pq} \\ \mathbf {D}_{{\ell ,q}}(P):= \left( \sum _{s,t=1}^n (A_\ell )_{st} (P_{st} )\right) _{qq} \end{array}\right. } \end{aligned}$$

for some $\ell \in \{1,\ldots ,k\}$ and $p<q\in \{1,\ldots ,m\}$. Defining vectors $\alpha _{{\ell ,p,q,{\mathrm{Re}\,}}},\alpha _{{\ell ,p,q,{\mathrm{Im}\,}}},\beta _{{\ell ,q}}\in \mathbb {R}^{(nm)^2}$ by

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathbf {L}_{{\ell ,p,q,{\mathrm{Re}\,}}}(P)=(\alpha _{{\ell ,p,q,{\mathrm{Re}\,}}})^T{{\mathrm{{sHvec}}}}(P), \\ \mathbf {L}_{{\ell ,p,q,{\mathrm{Im}\,}}}(P)=(\alpha _{{\ell ,p,q,{\mathrm{Im}\,}}})^T{{\mathrm{{sHvec}}}}(P), \\ \mathbf {D}_{{\ell ,q}}(P)=(\beta _{{\ell ,q}})^T{{\mathrm{{sHvec}}}}(P), \end{array}\right. } \end{aligned}$$

for all $\ell \in \{1,\ldots ,k\}$, $p<q\in \{1,\ldots ,m\}$, we get that

$$\begin{aligned} L_A = \begin{bmatrix} \alpha _{{1,1,2,{\mathrm{Re}\,}}} \cdots&\alpha _{{k,1,2,{\mathrm{Re}\,}}}&\alpha _{{1,1,2,{\mathrm{Im}\,}}} \cdots&\alpha _{{k,m,m,{\mathrm{Im}\,}}}&\beta _{{1,1}}&\cdots&\beta _{{k,1}}&\beta _{{1,2}}&\cdots&\beta _{{k,m}} \end{bmatrix}^T. \end{aligned}$$

Now we proceed to find the vectors $\alpha _{{\ell ,p,q,{\mathrm{Re}\,}}},\alpha _{{\ell ,p,q,{\mathrm{Im}\,}}},\beta _{{\ell ,q}}$.

1.1.4 Computing the rows of $L_A$

Fix any $\ell \in \{1,\ldots ,k\}$. For any $i,j\in \{1,\ldots ,nm\}$, let $(s,t,p,q)$ be defined as in (12).

If $s<t$, then using Table 7 we get

$$\begin{aligned} \sum _{\tilde{s},\tilde{t}=1}^n(A_\ell )_{\tilde{s} \tilde{t}} (E_{ij})_{\tilde{s} \tilde{t}}&=\ \frac{1}{\sqrt{2}} \left( (A_{\ell })_{st} e_p e_q^T + (A_{\ell })_{ts} e_q e_p^T\right) \\&=\ \frac{1}{\sqrt{2}} \left( {\mathrm{Re}\,}(A_{\ell })_{st} \left( e_p e_q^T + e_q e_p^T\right) + \mathsf {i}\,{\mathrm{Im}\,}(A_{\ell })_{st} \left( e_p e_q^T - e_q e_p^T\right) \right) \end{aligned}$$

If $s>t$, then

$$\begin{aligned} \sum _{\tilde{s},\tilde{t}=1}^n(A_\ell )_{\tilde{s} \tilde{t}} (E_{ij})_{\tilde{s} \tilde{t}}&=\ \frac{\mathsf {i}\,}{\sqrt{2}} \left( -(A_{\ell })_{st} e_p e_q^T + (A_{\ell })_{ts} e_q e_p^T \right) \\&=\ \frac{1}{\sqrt{2}} \left( -{\mathrm{Im}\,}(A_{\ell })_{ts} (e_p e_q^T + e_q e_p^T) - \mathsf {i}\,{\mathrm{Re}\,}(A_{\ell })_{ts} (e_p e_q^T - e_q e_p^T) \right) \end{aligned}$$

If $s=t$, then

$$\begin{aligned} \sum _{\tilde{s},\tilde{t}=1}^n(A_\ell )_{\tilde{s} \tilde{t}} (E_{ij})_{\tilde{s} \tilde{t}} =&\ (A_{\ell })_{ss} E_{pq}. \end{aligned}$$

Fix any $\ell =1,\ldots ,k$ and $\hat{p}<\hat{q}$ from $\{1,\ldots ,m\}$. Then for all $i,j\in \{1,\ldots ,nm\}$, defining $(s,t,p,q)$ as in (12), we have

$$\begin{aligned} \mathbf {L}_{{\ell ,\hat{p},\hat{q},{\mathrm{Re}\,}}}(E_{ij}) = {\left\{ \begin{array}{ll} {\mathrm{Re}\,}(A_{\ell })_{st} &{}\text { if}\;\; s<t\, \text { and}\, \{p,q\}=\{\hat{p},\hat{q}\}\\ -{\mathrm{Im}\,}(A_{\ell })_{ts} &{}\text { if}\;\; s>t\, \text { and}\, \{p,q\}=\{\hat{p},\hat{q}\}\\ (A_{\ell })_{ss} &{}\text { if}\;\; s=t, p<q\, \text { and}\, (p,q)=(\hat{p},\hat{q})\\ 0 &{}\text { otherwise}, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \mathbf {L}_{{\ell ,\hat{p},\hat{q},{\mathrm{Im}\,}}}(E_{ij}) = {\left\{ \begin{array}{ll} {\mathrm{Im}\,}(A_{\ell })_{st} &{}\text { if}\;\; s<t\,\text { and}\,(p,q)=(\hat{p},\hat{q}) \\ -{\mathrm{Im}\,}(A_{\ell })_{st} &{} \text { if}\;\; s<t \,\text { and}\, (p,q)=(\hat{q},\hat{p}) \\ -{\mathrm{Re}\,}(A_{\ell })_{ts} &{}\text { if}\;\; s>t\,\text { and} (p,q)=(\hat{p},\hat{q}) \\ {\mathrm{Re}\,}(A_{\ell })_{ts} &{}\text { if}\;\; s>t\,\text { and}\, (p,q)=(\hat{q},\hat{p}) \\ (A_{\ell })_{ss} &{}\text { if}\;\; s=t, p>q\, \text { and}\, (p,q)=(\hat{q},\hat{p}) \\ 0 &{}\text { otherwise}, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \mathbf {D}_{{\ell ,\hat{q}}}(E_{ij}) = {\left\{ \begin{array}{ll} \sqrt{2} {\mathrm{Re}\,}(A_{\ell })_{st} &{} \text { if}\;\; s<t\, \text { and}\, p=q=\hat{q} \\ -\sqrt{2} {\mathrm{Im}\,}(A_{\ell })_{st} &{} \text { if}\;\; s>t\, \text { and}\, p=q=\hat{q} \\ (A_{\ell })_{ss} &{}\text { if}\, s=t\, \text { and}\;\; p=q=\hat{q}. \end{array}\right. } \end{aligned}$$

Therefore for any $\hat{p}<\hat{q}=1,\ldots ,m$ and any $i,j=1,\ldots , nm$, using $(s,t,p,q)$ defined in (12) and the definitions of $M_{{\mathrm{Re}\,}}, M_{{\mathrm{Im}\,}}, M_D$ on Page 6,

$$\begin{aligned} \left( \alpha _{{\ell ,\hat{p},\hat{q},{\mathrm{Re}\,}}}\right) _{i,j} =&\ \alpha _{{\ell ,\hat{p},\hat{q},{\mathrm{Re}\,}}}^T {{\mathrm{{sHvec}}}}(E_{ij}) = \mathbf {L}_{{\ell ,\hat{p},\hat{q},{\mathrm{Re}\,}}}(E_{ij}) \\ =&\ {\left\{ \begin{array}{ll} \frac{1}{\sqrt{2}} (M_{{\mathrm{Re}\,}})_{\ell ,st} &{}\text { if}\;\; s<t\, \text { and}\, \{p,q\}=\{\hat{p},\hat{q}\} \\ -\frac{1}{\sqrt{2}} (M_{{\mathrm{Im}\,}})_{\ell ,ts} &{}\text { if}\;\; s>t\, \text { and}\, \{p,q\}=\{\hat{p},\hat{q}\} \\ (M_D)_{\ell ,s} &{}\text { if}\;\; s=t, p<q\, \text { and}\, (p,q)=(\hat{p},\hat{q}) \\ 0 &{}\text { otherwise}, \end{array}\right. } \end{aligned}$$

implying that $\alpha _{{\ell ,\hat{p},\hat{q},{\mathrm{Re}\,}}}^T$ is one of the first $k$ rows of the matrix $\begin{bmatrix} I_{t(m-1)}\otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\ 0 \end{bmatrix}$. (Here note that the number of pairs $(\hat{p},\hat{q})$ with $1\le \hat{p}< \hat{q} \le m$ is $t(m-1)=\frac{1}{2}m(m-1)$. The zero block corresponds to the index pairs $(i,j)$ with $p=q$, where $(s,t,p,q)$ are the block indices defined in (12).) Similarly,

$$\begin{aligned} \left( \alpha _{{\ell ,\hat{p},\hat{q},{\mathrm{Im}\,}}}\right) _{i,j} =&\ \alpha _{{\ell ,\hat{p},\hat{q},{\mathrm{Im}\,}}}^T {{\mathrm{{sHvec}}}}(E_{ij}) = \mathbf {L}_{{\ell ,p,q,{\mathrm{Im}\,}}}(E_{ij}) \\ =&\ {\left\{ \begin{array}{ll} \frac{1}{\sqrt{2}} (M_{{\mathrm{Im}\,}})_{\ell ,st} &{}\text { if}\;\; s<t\, \text { and}\, (p,q)=(\hat{p},\hat{q}) \\ -\frac{1}{\sqrt{2}} (M_{{\mathrm{Im}\,}})_{\ell ,st} &{} \text { if}\;\; s<t\, \text { and}\, (p,q)=(\hat{q},\hat{p}) \\ -\frac{1}{\sqrt{2}} (M_{{\mathrm{Re}\,}})_{\ell ,ts} &{}\text { if}\;\; s>t\, \text { and}\, (p,q)=(\hat{p},\hat{q}) \\ \frac{1}{\sqrt{2}} (M_{{\mathrm{Re}\,}})_{\ell ,ts} &{}\text { if}\;\; s>t\, \text { and}\, (p,q)=(\hat{q},\hat{p}) \\ (M_D)_{\ell ,s} &{}\text { if}\;\; s=t, p>q\, \text { and}\, (p,q)=(\hat{q},\hat{p}) \\ 0 &{}\text { otherwise}, \end{array}\right. } \end{aligned}$$

so $\alpha _{{\ell ,\hat{p},\hat{q},{\mathrm{Im}\,}}}^T$ is one of the last $k$ rows of the matrix $\begin{bmatrix} I_{t(m-1)}\otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\ 0 \end{bmatrix}$. Finally,

$$\begin{aligned} (\beta _{{\ell ,\hat{q}}})_{ij} = \mathbf {D}_{{\ell ,\hat{q}}}(E_{ij}) = {\left\{ \begin{array}{ll} (M_{{\mathrm{Re}\,}})_{\ell ,st} &{} \text { if}\;\; s<t\, \text { and}\, p=q=\hat{q} \\ -(M_{{\mathrm{Im}\,}})_{\ell ,st} &{} \text { if}\;\; s>t\, \text { and}\, p=q=\hat{q} \\ (M_D)_{\ell ,s} &{}\text { if}\;\; s=t\, \text { and}\, p=q=\hat{q}, \end{array}\right. } \end{aligned}$$

so $\beta _{{\ell ,\hat{q}}}^T$ is one of the rows of the matrix $\begin{bmatrix} 0&\ I_{m}\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}\end{bmatrix}$.

Hence, a matrix representation of $\mathcal {L}_A$ is given by

$$\begin{aligned} \begin{bmatrix} I_{t(m-1)}\otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \\ 0&\quad I_m \otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}\end{bmatrix}. \end{aligned}$$

1.1.5 Computing $L_T$

Recall the linear map $\mathcal {L}_T:{\mathcal H} ^{nm}\rightarrow {\mathcal H} ^n : P\mapsto [{{\mathrm{{trace}}}}(P_{st})]_{s,t=1,\ldots ,n}$, which defines the second component of $\mathcal {L}$. We compute a matrix representation $L_T$ of $\mathcal {L}_T$ by columns, i.e., by considering $\mathcal {L}_T(E_{ij})\in {\mathcal H} ^n$ for $i,j=1,\ldots ,nm$. Defining $(s,t,p,q)$ as in (12), we have

$$\begin{aligned} \mathcal {L}_T(E_{ij}) = {\left\{ \begin{array}{ll} E_{st}&{}\quad \text{ if } p=q, \\ 0 &{}\quad \text { otherwise.} \end{array}\right. } \end{aligned}$$

Hence the $(i,j)$th column of $L_T$ is given by

$$\begin{aligned} {{\mathrm{{sHvec}}}}(\mathcal {L}_T(E_{ij})) = {\left\{ \begin{array}{ll} {{\mathrm{{sHvec}}}}(E_{st})&{}\quad \text{ if } p=q, \\ 0 &{}\quad \text { otherwise.} \end{array}\right. } \end{aligned}$$

This implies that $L_T=\begin{bmatrix} 0&\ e_m^T\times I_{n^2} \end{bmatrix}$, where the zero block corresponds to the $(i,j)$ pairs with $p\ne q$, and each row $e_m^T$ in the Kronecker product corresponds to those $(i,j)$ pairs with the same block indices $(s,t)$ (and there are $m$ pairs of $(i,j)$ with the same block indices that have nonzero intra-block traces).

1.1.6 Alternative column orderings, eliminating redundant rows

Combining the results from the previous two sections, we arrive at a matrix representation of $\mathcal {L}$:

$$\begin{aligned} L= \begin{bmatrix} L_A\\L_T \end{bmatrix} = \begin{bmatrix} I_{t(m-1)} \otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \\ 0&\quad I_m\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}\\ 0&\quad e_m^T\otimes I_{n^2} \end{bmatrix}. \end{aligned}$$

(15)

In the final matrix representation that we use, some of the rows of the second block rows are linearly dependent of the other rows, if the linear map $\mathcal {L}_A$ contains the unital constraints. Hence we remove those rows and replace the original matrix representation $L$ in (15) by the following matrix:

$$\begin{aligned} L=&\ \begin{bmatrix} I_{t(m-1)} \otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \\ 0&\begin{bmatrix} I_{m-1}\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \end{bmatrix} \\ 0&\quad e_m^T\otimes I_{n^2} \end{bmatrix} \\ =&\ \begin{bmatrix} I_{t(m-1)} \otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0&\quad 0 \\ 0&\quad I_{m-1}\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \\ 0&\quad e_{m-1}^T\otimes I_{n^2}&\quad I_{n^2} \end{bmatrix}. \end{aligned}$$

Note that we can use alternative ordering for the off-diagonal entries inside the blocks. While this does not change the ordering of the columns in the second block column in (15) (which correspond to the diagonal entries inside the blocks), it can affect the column ordering of $N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}$ (resulting in, e.g., $N_\mathrm{final}$).

1.1.7 Pseudoinverse of $L$

Using the block diagonal structure of $L$ and the fact that

$$\begin{aligned} \begin{bmatrix} I_{m-1}\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \\ e_{m-1}^T\otimes I_{n^2}&\quad I_{n^2} \end{bmatrix}^\dagger = \begin{bmatrix} I_{m-1}\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}^\dagger&\quad e_{m-1}\otimes (M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D})_\mathrm{null} \\ -e_{m-1}^T\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}^\dagger&\quad I_{n^2} - (m-1) (M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D})_\mathrm{null} \end{bmatrix} \end{aligned}$$

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(which can be easily verified to be the pseudoinverse), it is immediate that

$$\begin{aligned} L^\dagger =&\ \begin{bmatrix} I_{t(m-1)} \otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \\ 0&\begin{bmatrix} I_{m-1}\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0 \\ e_{m-1}^T\otimes I_{n^2}&\quad I_{n^2} \end{bmatrix}^\dagger \end{bmatrix} \\ =&\ \begin{bmatrix} I_{t(m-1)} \otimes N_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}&\quad 0&\quad 0 \\ 0&\quad I_{m-1}\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}^\dagger&\quad e_{m-1}\otimes (M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D})_\mathrm{null} \\ 0&\quad -e_{m-1}^T\otimes M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D}^\dagger&\quad I_n - (m-1) (M_{{\mathrm{Re}\,}{\mathrm{Im}\,}D})_\mathrm{null} \end{bmatrix}. \end{aligned}$$