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Exercise 8.30 of Nielson & Chuang's QCQI says qcqi8.30 Equation 7.144, which is mentioned in the text, is $$\begin{bmatrix} a & b\\ b^* & 1-a \end{bmatrix}\rightarrow\begin{bmatrix} (a-a_0)e^{-t/T_1}+a_0 & be^{-t/T_2}\\ b^*e^{-t/T_2} & (a-a_0)e^{-t/T_1}+1-a_0 \end{bmatrix}\tag{7.144}$$ My attempt is as follows.

(1) The first part.

Assume $$\rho=\begin{bmatrix} a & b\\ b^* & 1-a \end{bmatrix}$$ After applying amplitude damping $$\mathcal{E}_{AD}(\rho)=E_0\rho E_0^\dagger+E_1\rho E_1^\dagger$$ $$=\begin{bmatrix} 1 & 0\\ 0 & \sqrt{1-\gamma} \end{bmatrix}\begin{bmatrix} a & b\\ b^* & 1-a \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & \sqrt{1-\gamma} \end{bmatrix}+\begin{bmatrix} 0 & \sqrt{\gamma}\\ 0 & 0 \end{bmatrix}\begin{bmatrix} a & b\\ b^* & 1-a \end{bmatrix}\begin{bmatrix} 0 & 0\\ \sqrt{\gamma} & 0 \end{bmatrix}$$ $$=\begin{bmatrix} a & \sqrt{1-\gamma}b\\ \sqrt{1-\gamma}b^* & (1-\gamma)(1-a) \end{bmatrix}+\begin{bmatrix} \gamma(1-a) & 0\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} \gamma-\gamma a+a & \sqrt{1-\gamma}b\\ \sqrt{1-\gamma}b^* & (1-\gamma)(1-a) \end{bmatrix}$$ Compare this result with (7.144), $$\because\sqrt{1-\gamma}b=be^{-t/T_2}$$ $$\therefore\frac{\textrm{ln}(1-\gamma)}{2}=-\frac{t}{T_2}$$ $$T_2=-\frac{2t}{\textrm{ln}(1-\gamma)}$$ In (7.144), when $a=a_0$, the coefficient of $|0\rangle\langle0|$ is also $a_0$. So, in the case of $\mathcal{E}_{AD}(\rho)$ $$\because a_0=\gamma-\gamma a_0+a_0$$ $$\therefore a_0=1$$ $$\because\gamma-\gamma a+a=(a-a_0)e^{-t/T_1}+a_0$$ $$\therefore\frac{\gamma-\gamma a+a-1}{a-1}=\frac{(1-\gamma)(a-1)}{a-1}=1-\gamma=e^{-t/T_1}$$ $$\textrm{ln}(1-\gamma)=-\frac{t}{T_1}$$ $$\therefore T_1=-\frac{t}{\textrm{ln}(1-\gamma)}=\frac{T_2}{2}$$ Which is quite different from the target. Did I do wrong somewhere?

(2) The second part. First, apply the phase damping, $$\mathcal{E}(\rho)=E_0\rho E_0^\dagger+E_1\rho E_1^\dagger$$ $$=\begin{bmatrix} 1 & 0\\ 0 & \sqrt{1-\gamma'} \end{bmatrix}\begin{bmatrix} a & b\\ b^* & 1-a \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & \sqrt{1-\gamma'} \end{bmatrix}+\begin{bmatrix} 0 & 0\\ 0 & \sqrt{\gamma'} \end{bmatrix}\begin{bmatrix} a & b\\ b^* & 1-a \end{bmatrix}\begin{bmatrix} 0 & 0\\ 0 & \sqrt{\gamma'} \end{bmatrix}$$ $$=\begin{bmatrix} a & \sqrt{1-\gamma'}b\\ \sqrt{1-\gamma'}b^* & (1-\gamma')(1-a) \end{bmatrix}+\begin{bmatrix} 0 & 0\\ 0 & \gamma'(1-a) \end{bmatrix}=\begin{bmatrix} a & \sqrt{1-\gamma'}b\\ \sqrt{1-\gamma'}b^* & 1-a \end{bmatrix}$$ Then, amplitude damping $$\mathcal{E}_{AD}(\mathcal{E}(\rho))=\begin{bmatrix} \gamma-\gamma a+a & \sqrt{(1-\gamma)(1-\gamma')}b\\ \sqrt{(1-\gamma)(1-\gamma')}b^* & (1-\gamma)(1-a) \end{bmatrix}$$ $$T_2'=-\frac{2t}{\textrm{ln}(1-\gamma)+\textrm{ln}(1-\gamma')}\leq2T_1$$ Which is also far from the target.

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  • $\begingroup$ Why do you take $a=a_0$? $\endgroup$
    – DaftWullie
    Aug 22, 2023 at 3:35
  • $\begingroup$ @DaftWullie I try to calculate the value of $a_0$ $\endgroup$
    – Jintao Yu
    Aug 23, 2023 at 1:50
  • $\begingroup$ My point is that's an assumption you're making. If you set $a_0$ to different values, you'll get different answers. By the way, in your reported version of (7.144), you must have a sign error on one of your diagonal terms, because the trace of the output is not 1. $\endgroup$
    – DaftWullie
    Aug 23, 2023 at 4:26
  • $\begingroup$ I suggest that you take $a_0=1$. This makes physical sense because we know that if you let $t\rightarrow\infty$, then under amplitude damping, $\langle 1|\rho|1\rangle\rightarrow 0=1-a_0$. $\endgroup$
    – DaftWullie
    Aug 23, 2023 at 4:31
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    $\begingroup$ Looks like it's just a typo. For amplitude damping T2=2T1, see, e.g., theory.caltech.edu/~preskill/ph219/chap3_15.pdf (page 32). In general, you should get T2<2T1 $\endgroup$
    – EvgeniyZh
    Aug 31, 2023 at 7:42

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