# Prove $\beta=\Lambda\otimes\Lambda$, where $\Lambda=\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}$ for single qubit tomography

In the Section on single qubit quantum process tomography, Box 8.5, Page 393, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, and in Prescription for experimental determination of the dynamics of a quantum black box, the procedure is formulated as

Choosing, \begin{align} \tilde{E}_0=I,\tilde{E}_1=X,\tilde{E}_2=-iY,\tilde{E}_3=Z\\ \end{align}

Suppose the input states are $$|0\rangle,|1\rangle,|+\rangle=\dfrac{|0\rangle+|1\rangle}{\sqrt{2}},|-\rangle=\dfrac{|0\rangle+i|1\rangle}{\sqrt{2}}$$ are prepared, and the four states

\begin{align} \mathcal{E}(\rho_1)=\rho_1'=&\mathcal{E}(|0\rangle\langle 0|)\\ \mathcal{E}(\rho_4)=\rho_4'=&\mathcal{E}(|1\rangle\langle 1|)\\ \mathcal{E}(\rho_2)=\rho_2'=&\mathcal{E}(|+\rangle\langle +|)-i\mathcal{E}(|-\rangle\langle -|)-\dfrac{(1-i)}{2}(\rho_1'+\rho_4')\\ \mathcal{E}(\rho_3)=\rho_3'=&\mathcal{E}(|+\rangle\langle +|)+i\mathcal{E}(|-\rangle\langle -|)-\dfrac{(1+i)}{2}(\rho_1'+\rho_4')\\ \end{align} are determined using quantum state tomography, where \begin{align} \rho_1=|0\rangle\langle 0|,\rho_2=|1\rangle\langle 0|=X\rho_1,\rho_3=|0\rangle\langle 1|=\rho_1X,\rho_4=|1\rangle\langle 1|=X\rho_1X \end{align}

with $$X=|0\rangle\langle 1|+|1\rangle\langle 0|=\begin{bmatrix}0&1\\1&0\end{bmatrix}$$

From equation $$\beta=\Lambda\otimes\Lambda$$ we can determine $$\beta$$, and $$\rho_j'=\mathcal{E}(\rho_j)$$ determines $$\vec{\lambda}$$ using the equation $$\mathcal{E}(\rho_j)=\sum_k\lambda_{jk}\rho_k$$.

Then it went on to say that,

Due to the particular choice of basis, and the Pauli matrix representation of $$\tilde{E}_i$$ , we may express the $$\beta$$ matrix as the Kronecker product $$\beta=\Lambda\otimes\Lambda$$, where $$\Lambda=\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}$$, so that $$\chi$$ may be expressed conveniently as $$\chi=\Lambda\begin{bmatrix}\rho_1'&\rho_2'\\\rho_3'&\rho_4'\end{bmatrix}\Lambda$$ in terms of block matrices.

We have the equation $$\tilde{E}_m\rho_j\tilde{E}_n^\dagger=\sum_k\beta_{jk}^{mn}\rho_k$$

Is there any way to obtain the expression $$\beta=\Lambda\otimes\Lambda$$ without explicitly calculating each individual term?

Where does the tensor product comes into picture ?

Note: A similar question has been asked before Quantum process tomography, for one and two qubit, but did not attract any response.

Quantum Process tomography

Let a fixed set of operators $$\tilde{E}_i$$, which form a basis for the set of operators on the state space so that $$E_i=\sum_m e_{im}\tilde{E}_m$$, then the quantum operation can be represented as, \begin{align} \mathcal{E}(\rho)&=\sum_i E_i\rho E_i^\dagger \tag{8.150}\label{eq0}\\ &=\sum_{m,n}\tilde{E}_m\rho\tilde{E}^\dagger_n \chi_{mn} \tag{8.152}\label{eq2}\\ \end{align} where $$\chi_{mn}=\sum_i e_{im}e^*_{in}$$ are the entries of a matrix that is positive hermitian since $$\chi_{mn}^*=(\sum_i e_{im}e^*_{in})^*=\sum_i e_{in}e^*_{im}=\chi_{nm}$$ and $$\chi_{mm}=\sum_i e_{im}e^*_{im}=\sum_i |e_{im}|^2\geq 0$$.

Let $$\rho_j,1\leq j\leq d^2$$ be a fixed, linearly independent basis for the space of $$d\times d$$ matrices,i.e., any $$d\times d$$ matrix can be written as a unique linear combination of the $$\rho_j$$ and it is possible to determine $$E(\rho_j)$$ by state tomography, for each $$\rho_j$$.

Each $$E(\rho_j)$$ may be expressed as a linear combination of the basis states, $$\mathcal{E}(\rho_j)=\sum_k \lambda_{jk}\rho_k \tag{8.155}\label{eq5}$$

And, since $$\mathcal{E}(\rho_j)$$ is known from the state tomography, $$\lambda_{jk}$$ can be determined by standard linear algebraic algorithms. we can also write $$\tilde{E}_m\rho_j\tilde{E}_n^\dagger=\sum_k\beta_{jk}^{mn}\rho_k \tag{8.156}\label{eq6}$$

where $$\beta_{jk}^{mn}$$ are complex numbers that can be determined by standard algorithms from linear algebra given the $$\tilde{E}_m$$ operators and the $$\rho_j$$ operators.

Combining both obtain, \begin{align} \sum_k\sum_{mn}\chi_{mn}\beta_{jk}^{mn}\rho_k&=\sum_{k}\lambda_{jk}\rho_k\tag{8.157}\label{eq7}\\ \implies \sum_{mn}\beta_{jk}^{mn}\chi_{mn}&=\lambda_{jk} \tag{8.158}\label{eq8}\\ \implies \beta\vec{\chi}&=\vec{\lambda} \tag{8.161}\label{eq61} \end{align} Let $$\kappa$$ be the generalized inverse such that $$\beta\kappa\beta=\beta\Leftrightarrow \beta_{jk}^{mn}=\sum_{st,xy}\beta_{jk}^{st}\kappa_{st}^{xy}\beta_{xy}^{mn}$$

We can prove that $$\chi$$ defined by $$\chi_{mn}\equiv\sum_{jk}\kappa_{jk}^{mn}\lambda_{mn}\Leftrightarrow\vec{\chi}=\kappa\vec{\lambda}\tag{8.160}\label{eq60}$$ satisfies the equation $$\sum_{mn}\beta_{jk}^{mn}\chi_{mn}=\lambda_{jk}\Leftrightarrow\beta\vec{\chi}=\vec{\lambda}$$

Direct Approach

Choosing, \begin{align} \tilde{E}_1=I,\tilde{E}_2=X,\tilde{E}_3=-iY,\tilde{E}_4=Z\\ \end{align} \begin{align} \rho_1=|0\rangle\langle 0|,\rho_2=|1\rangle\langle 0|=X\rho_1,\rho_3=|0\rangle\langle 1|=\rho_1X,\rho_4=|1\rangle\langle 1|=X\rho_1X \end{align}

$$\tilde{E}_m\rho_j\tilde{E}_n^\dagger=\sum_k\beta_{jk}^{mn}\rho_k$$ $$\tilde{E}_1\rho_1\tilde{E}_1^\dagger=\rho_1=|0\rangle\langle 0|=1\rho_1+0\rho_2+0\rho_3+0\rho_4\\ \implies \boxed{\beta_{11}^{11}=1,\beta_{12}^{11}=0,\beta_{13}^{11}=0,\beta_{14}^{11}=0}\\ \tilde{E}_1\rho_2\tilde{E}_1^\dagger=X\rho_1=|1\rangle\langle 0|=0\rho_1+1\rho_2+0\rho_3+0\rho_4\\ \boxed{\beta_{21}^{11}=0,\beta_{22}^{11}=1,\beta_{23}^{11}=0,\beta_{24}^{11}=0}\\ \tilde{E}_1\rho_3\tilde{E}_1^\dagger=\rho_1X=|0\rangle\langle 1|=0\rho_1+0\rho_2+1\rho_3+0\rho_4\\ \boxed{\beta_{31}^{11}=0,\beta_{32}^{11}=0,\beta_{33}^{11}=1,\beta_{34}^{11}=0}\\ \tilde{E}_1\rho_4\tilde{E}_1^\dagger=X\rho_1X=|1\rangle\langle 0|=0\rho_1+0\rho_2+0\rho_3+1\rho_4\\ \boxed{\beta_{41}^{11}=0,\beta_{42}^{11}=1,\beta_{43}^{11}=0,\beta_{44}^{11}=1}\\$$ But this is quite cumbersome, and

$$\Lambda\otimes\Lambda=\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}\otimes\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}\\ =\dfrac{1}{2}\begin{bmatrix}1&0&0&1\\0&1&1&0\\0&1&-1&0\\1&0&0&-1\end{bmatrix}\otimes\dfrac{1}{2}\begin{bmatrix}1&0&0&1\\0&1&1&0\\0&1&-1&0\\1&0&0&-1\end{bmatrix}$$

Looks like the calculation in my attempt do not match with the expression $$\Lambda\otimes\Lambda$$ !

Cross-posted on math.SE

• "We can determine $\beta, \lambda$". Sorry but what are $\beta$ and $\lambda$?
– JSdJ
Nov 25, 2022 at 19:58
• @JSdJ $\beta$ is a matrix with $(jk,mn)^{th}$ element is $\beta_{jk}^{mn}$, please have a look at "quantumcomputing.stackexchange.com/questions/29111/…". Nov 25, 2022 at 20:11
• @JSdJ Or please have a look into the original publication, arxiv.org/pdf/quant-ph/9610001.pdf Nov 27, 2022 at 10:35
• so if I understand: you have the channel in terms of its action on a set of states $\rho_i$ (spanning the relevant space of Hermitian ops), define the ops $\tilde E_i$ as some new operatorial basis (here the Pauli matrices as you write), define the coefficients $\beta$ as those in the decomposition of $\tilde E_m \rho_j \tilde E_n^\dagger$ in terms of the $\rho_i$, and define $\lambda$ as the coefficients in the expansion of each $\mathcal E(\rho_i)$ in terms of the $\rho_i$. You ask the relation between $\beta$ and $\lambda$. Is that correct?
– glS
Nov 30, 2022 at 20:06
• @glS Thanks for responding. The relation between $\beta$ and $\vec{\lambda}$ are given by the relation $\beta\vec{\chi}=\vec{\lambda}$, so finding $\chi$ is the final objective. But, before that how do we even reach $\beta=\Lambda\otimes\Lambda$ where $\Lambda=\dfrac{1}{2}\begin{bmatrix}I&X\\X&-I\end{bmatrix}$? Nov 30, 2022 at 23:49