I need clarification on a few aspects related to Box 8.5 and Exercise 8.34 from the book Quantum Computation and Quantum Information by Nielsen & Chuang . While attempting Exercise 8.34, I encountered several challenges. I start with solving Quantum Process Tomography(QPT) for 1 qubit as given in box 8.5. Using the given expression of $\Lambda$: \begin{equation} \Lambda = \frac{1}{2}\begin{bmatrix} I & X \\ X & -I \end{bmatrix} \end{equation}

So now if I calculate $\beta = \Lambda \otimes \Lambda$ \begin{equation} \Lambda \otimes \Lambda\ = \frac{1}{2}\begin{bmatrix} I & X \\ X & -I \end{bmatrix} \otimes \frac{1}{2}\begin{bmatrix} I & X \\ X & -I \end{bmatrix} \end{equation} or, \begin{equation*} \beta = \frac{1}{4} \left[ \begin{matrix} I & X & & & & & I & X \\ X & -I & & & & & X & -I \\ & & I & X & I & X & & \\ & & X & -I & X & -I & & \\ & & I & X & -I & -X & & \\ & & X & -I & -X & I & & \\ I & X & & & & & -I & -X \\ X & -I & & & & & -X & I \end{matrix} \right]. \end{equation*} This gives a 16 x 16 matrix for $\beta$. (The blank spaces in the above matrix are occupied by null matrices.)

Now $\beta$ can also be calculated using Equation (8.156) from Nielsen Chuang.

\begin{equation} \tilde{E_m} \rho_j \tilde{E_n^\dagger}= \sum_k \beta_{jk}^{mn} \rho_k \end{equation}

Now I try to construct a 16 x 16 matrix using the above relation.

Let me do it to find the element corresponding to $\beta_{00}^{00}$ here m=0, n=0, j=1 and k={1,2,3,4} \begin{equation} \tilde{E_o} \rho_1 \tilde{E_o^\dagger}= \beta_{11}^{00} \rho_1 + \beta_{12}^{00} \rho_2 +\beta_{13}^{00} \rho_3 +\beta_{14}^{00} \rho_4 \end{equation}

for our choice of fixed operators(as given in Box 8.5 of Nielsen Chuang) $\tilde{E_o} =I$ and $\tilde{E_o^\dagger}=I^\dagger=I$. So we get \begin{equation} \rho_1 = \beta_{11}^{00} \rho_1 + \beta_{12}^{00} \rho_2 +\beta_{13}^{00} \rho_3 +\beta_{14}^{00} \rho_4 \end{equation}

Comparing the coefficients we get $\beta_{11}^{00} =1, \beta_{12}^{00}= 0, \beta_{13}^{00}=0, \beta_{14}^{00}= 0$. Likewise rest of the elements for 16 x 16 matrix can be obtained.

These two versions of $\beta$ matrix do not match. So I get stuck there.

In box 8.5 it is said that equations (8.178) & (8.179) are obtained because of our special choice of $\{\tilde{E}_k\}$ and the basis set $\{\rho_k\}.$ I expect that using these I should be able to construct $\beta$ matrix without explicitly constructing it element wise via eq.(8.156), (by expressing the basis in terms of Pauli operators with $\rho_1 = \frac{1}{2}(I+Z)$ and then using the algebra of Pauli operators.) This is because for the two qubit case as in exercise 8.34, $\beta$ matrix is $256\times 256$ whose element wise construction could be tedious. I am guessing there could be some generalised trend which can be followed. Let me know how to proceed for 2 qubit case as asked in Ex 8.34 and how to resolve $\beta$ issue that I am getting as mentioned above.

  • 1
    $\begingroup$ related: quantumcomputing.stackexchange.com/q/17329/55, quantumcomputing.stackexchange.com/questions/11622/55 $\endgroup$
    – glS
    Commented Nov 2, 2023 at 11:15
  • $\begingroup$ so to summarise, the issue here is the computation of the $\beta$ coefficients, yes? You say "the two versions of $\beta$ don't match", but I don't quite understand which two versions you're referring to. $\endgroup$
    – glS
    Commented Nov 2, 2023 at 11:20
  • $\begingroup$ @gIS The two versions of $\beta$ I am talking about are: 1. In the book the authors say $\beta = \Lambda \otimes \Lambda $, so I constructed the $\beta$ matrix using this. 2. The other way of getting elements of $\beta$ is by using $ \tilde{E_m} \rho_j \tilde{E_n^\dagger}= \sum_k \beta_{jk}^{mn} \rho_k$. These 2 versions of $\beta$ are not matching. $\endgroup$ Commented Nov 3, 2023 at 11:54
  • $\begingroup$ I redid the calculations for $\beta$ and, as far as I can tell, the factor of 1/2 should not be there for $\Lambda$. However, note that the factor of 1/2 also appears in the publication. $\endgroup$
    – agq
    Commented Jun 10 at 18:48


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