The Exercise

A qubit undergoing amplitude damping loses energy to its environment: the excited state $|1\rangle$ decays to the ground state $|0\rangle$ with probability $\gamma$, where $0 \le \gamma \le 1$. The channel is $\mathcal{E}_\gamma(\rho) = E_0 \rho E_0^\dagger + E_1 \rho E_1^\dagger$, with operation elements $E_0 = |0\rangle\langle 0| + \sqrt{1-\gamma}\,|1\rangle\langle 1|$ and $E_1 = \sqrt{\gamma}\,|0\rangle\langle 1|$.

A channel $\mathcal{E}$ on a qubit is fixed by its process matrix $\chi$, the $4 \times 4$ matrix of coefficients in $\mathcal{E}(\rho) = \sum_{m,n} \chi_{mn}\, \tilde{E}_m \rho \tilde{E}_n^\dagger$ over the operators $\tilde{E}_0 = I$, $\tilde{E}_1 = X$, $\tilde{E}_2 = -iY$, $\tilde{E}_3 = Z$, where $I$ is the identity and $X$, $Y$, $Z$ are the Pauli matrices. Process tomography determines $\chi$ by experiment. From the outputs on four prepared input states it obtains, by linearity, the images $\rho'_1 = \mathcal{E}(|0\rangle\langle 0|)$, $\rho'_2 = \mathcal{E}(|0\rangle\langle 1|)$, $\rho'_3 = \mathcal{E}(|1\rangle\langle 0|)$ and $\rho'_4 = \mathcal{E}(|1\rangle\langle 1|)$, and a fixed linear formula turns these four matrices into $\chi$. The exercise runs the reconstruction for amplitude damping and asks for the process matrix it returns.

The Tomography Formula

The tomography matrix $\Lambda$ is the $4 \times 4$ complex matrix whose $2 \times 2$ blocks are built from $I$ and the Pauli matrix $X = \left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right)$: $$\Lambda = \frac{1}{2} \begin{pmatrix} I & X \\ X & -I \end{pmatrix}.$$(1) The process matrix reconstructed from a $4 \times 4$ complex matrix $\rho'$ is the $4 \times 4$ matrix $$\chi(\rho') = \Lambda \, \rho' \, \Lambda .$$(2)^✦ In tomography $\rho'$ is the block matrix of the four outputs, with $\rho'_1$ and $\rho'_2$ in its upper row of blocks and $\rho'_3$ and $\rho'_4$ in its lower row.

The Measured Outputs

The measured block $\rho'_\gamma$ of amplitude damping with parameter $\gamma$ is the $4 \times 4$ complex matrix $$\rho'_\gamma = \left(\begin{array}{cc|cc} 1 & 0 & 0 & \sqrt{1-\gamma} \\ 0 & 0 & 0 & 0 \\ \hline 0 & 0 & \gamma & 0 \\ \sqrt{1-\gamma} & 0 & 0 & 1-\gamma \end{array}\right).$$(3)^✦

The Claim

The answer is the amplitude damping matrix $\chi_\gamma$, the $4 \times 4$ complex matrix $$\chi_\gamma = \frac{1}{4} \left(\begin{array}{cc|cc} (1+\sqrt{1-\gamma})^2 & 0 & 0 & \gamma \\ 0 & \gamma & -\gamma & 0 \\ \hline 0 & -\gamma & \gamma & 0 \\ \gamma & 0 & 0 & (1-\sqrt{1-\gamma})^2 \end{array}\right).$$(4)^✦

What Is Checked

The claim is stated over $4 \times 4$ complex matrices with explicit entries: the tomography matrix $\Lambda$ of (1), with the Pauli matrix $X$ in its blocks, the reconstruction (2), the measured block $\rho'_\gamma$ of (3) and the amplitude damping matrix $\chi_\gamma$ of (4).

What the library proves about them is the identity (5) for every real $\gamma \le 1$, entry by entry. The hypothesis $\gamma \le 1$ enters at the first and last diagonal entries of $\chi_\gamma$, $(1 \pm \sqrt{1-\gamma})^2/4$, which the reconstruction produces as $(2 \pm 2\sqrt{1-\gamma} - \gamma)/4$; the two agree because $(\sqrt{1-\gamma})^2 = 1 - \gamma$.