# Quantum Process Tomography for 2 qubits

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 $$\Lambda$$: $$$$\Lambda = \frac{1}{2}\begin{bmatrix} I & X \\ X & -I \end{bmatrix}$$$$

So now if I calculate $$\beta = \Lambda \otimes \Lambda$$ $$$$\Lambda \otimes \Lambda\ = \frac{1}{2}\begin{bmatrix} I & X \\ X & -I \end{bmatrix} \otimes \frac{1}{2}\begin{bmatrix} I & X \\ X & -I \end{bmatrix}$$$$ or, $$\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.

$$$$\tilde{E_m} \rho_j \tilde{E_n^\dagger}= \sum_k \beta_{jk}^{mn} \rho_k$$$$

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} $$$$\tilde{E_o} \rho_1 \tilde{E_o^\dagger}= \beta_{11}^{00} \rho_1 + \beta_{12}^{00} \rho_2 +\beta_{13}^{00} \rho_3 +\beta_{14}^{00} \rho_4$$$$

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 $$$$\rho_1 = \beta_{11}^{00} \rho_1 + \beta_{12}^{00} \rho_2 +\beta_{13}^{00} \rho_3 +\beta_{14}^{00} \rho_4$$$$

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.

• – glS
Commented Nov 2, 2023 at 11:15
• 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.
– glS
Commented Nov 2, 2023 at 11:20
• @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. Commented Nov 3, 2023 at 11:54
• 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.
– agq
Commented Jun 10 at 18:48