Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 8)

8.1

　 $\displaystyle H\mid 0\rangle = \frac{1}{\sqrt{2}}\left(\mid 0\rangle +\mid 1\rangle\right)$

　 $\displaystyle \therefore\,H^{\otimes N}\mid 00\cdots0\rangle = \frac{1}{2^{N/2}}\left(\mid 0\rangle +\mid 1\rangle\right)\otimes\left(\mid 0\rangle +\mid 1\rangle\right)\otimes\cdots\left(\mid 0\rangle +\mid 1\rangle\right)$

By expanding the tensor product, we have the all possible binary numbers.

As discussed in 6.1.4, for a single qubit, Hadamard gate can be implemented as:

　 $\displaystyle H = e^{i\pi\sigma^z/2}e^{i\pi\sigma^y/2}$

using the coherent drive:

　 $\displaystyle \epsilon(t) = \epsilon_0\cos(\omega_0t+\phi)$

that's on resonance $\omega_0=\Delta$ and $\phi=\pi/2$ .

To apply it to $N$ identical flux qubits, we have to have different gaps $\Delta$ on neighboring qubits so that the drive doesn't affect neighboring qubits.

8.2

Since $H\sigma^z H = \sigma^x$ and $HH=1$ ,

　 $(1\otimes H)U_{\rm CZ}(1\otimes H) = U_{\rm CNOT}$

8.3

　 $\displaystyle H = \sum_{i=1}^2\left\{\frac{\Delta_i}{2}\sigma_i^z + g_i(\sigma_i^+\hat a + \hat a^\dagger\sigma_i^-)\right\}+\omega\hat a^\dagger \hat a$

　　 $\displaystyle = \sum_{i=1}^2 \frac{\omega}{2}\sigma_i^z+\omega\hat a^\dagger \hat a + \sum_{i=1}^2\left\{\frac{\delta_i}{2}\sigma_i^z + g_i(\sigma_i^+\hat a + \hat a^\dagger\sigma_i^-)\right\}$

where $\delta_i := \Delta_i - \omega$ . Hence, we have:

　 $\displaystyle H = H_0 + H_1$

　 $\displaystyle H_0 =\sum_{i=1}^2 \frac{\omega}{2}\sigma_i^z+\omega\hat a^\dagger \hat a$

　 $\displaystyle H_1 = \sum_{i=1}^2\left\{\frac{\delta_i}{2}\sigma_i^z + g_i(\sigma_i^+\hat a + \hat a^\dagger\sigma_i^-)\right\}$

Then,

　 $\displaystyle [H_0,H_1] = \sum_{i,j}\frac{\omega g_j}{2}\left([\sigma_i^z,\sigma_j^+]\hat a + [\sigma_i^z,\sigma_j^-]\hat a^\dagger\right)$

　　　　　　 $\displaystyle {} + \sum_j\omega g_j\left(\sigma_j^+[\hat a^\dagger\hat a,\hat a] + \sigma_j^-[\hat a^\dagger\hat a,\hat a^\dagger]\right)$

　　 $\displaystyle = \sum_{i,j}\frac{\omega g_j}{2}\left(2\sigma_j^+\delta_{ij}\hat a - 2\sigma_j^-\delta_{ij}\hat a^\dagger\right) + \sum_j\omega g_j\left(-\sigma_j^+\hat a + \sigma_j^-\hat a^\dagger\right)$

　　 $\displaystyle = 0$

So in the interaction picture with $H_0$ , the system is described with $H_1$ itself. Note that, in this picture, the photon has no energy. (Photon energy is absorbed in $\delta_i$ ).

We split $H_1$ as:

　 $\displaystyle H_1 = \sum_{i=1}^2\frac{\delta_i}{2}\sigma_i^z + V$

　 $\displaystyle V = \sum_{i=1}^2g_i(\sigma_i^+\hat a + \hat a^\dagger\sigma_i^-)$

and take $V$ as a weak perturbation.

Now we apply the nondegenerate perturbation theory. In general, the transformation generator $S_0$ is given by:

　 $\displaystyle S_0 = \sum_{n\ne m}\frac{\langle n\mid V\mid m\rangle}{E_n-E_m}\mid n\rangle\langle m\mid$

We consider the states with minimum number of photons (or $N=2$ subspace of JC-Ladder), such as:

　 $\mid 00\rangle\otimes\mid 2\rangle,\,\mid 10\rangle\otimes\mid 1\rangle,\,\mid 01\rangle\otimes\mid 1\rangle,\,\mid 11\rangle\otimes\mid 0\rangle$

Since photons have no energy, their respective energy are:

　 $0,\,\delta_1,\,\delta_2,\,\delta_1 + \delta_2$

Then, $S_0$ can be expanded as:

　 $\displaystyle S_0 = \frac{\sqrt{2}g_1}{\delta_1}\mid 10\rangle\otimes\mid 1\rangle\langle 00\mid\otimes\langle 2\mid + \frac{\sqrt{2}g_2}{\delta_2}\mid 01\rangle\otimes\mid 1\rangle\langle 00\mid\otimes\langle 2\mid$

　　　 $\displaystyle {} + \frac{g_1}{\delta_1}\mid 11\rangle\otimes\mid 0\rangle\langle 01\mid\otimes\langle 1\mid + \frac{g_2}{\delta_2}\mid 11\rangle\otimes\mid 0\rangle\langle 10\mid\otimes\langle 1\mid$
　
　　　 $\displaystyle {}- \frac{\sqrt{2}g_1}{\delta_1}\mid 00\rangle\otimes\mid 2\rangle\langle 10\mid\otimes\langle 1\mid - \frac{\sqrt{2}g_2}{\delta_2}\mid 00\rangle\otimes\mid 2\rangle\langle 01\mid\otimes\langle 2\mid$

　　　 $\displaystyle {} - \frac{g_1}{\delta_1}\mid 01\rangle\otimes\mid 1\rangle\langle 11\mid\otimes\langle 0\mid + \frac{g_2}{\delta_2}\mid 10\rangle\otimes\mid 1\rangle\langle 11\mid\otimes\langle 0\mid$

　　 $\displaystyle = \sum_{i=1}^2\frac{g_i}{\delta_i}\left(\sigma_i^+\hat a - \sigma_i^-\hat a^\dagger\right)$

Note that, with similar calculations, you can confirm that the above expression for $S_0$ is valid for any subspace of JC-Ladder though we explicitly calculated in $N=2$ subspace.

So the perturbative correction to the Hamiltonian is given by:

　 $\displaystyle \frac{1}{2}[S_0,V] = \frac{1}{2}\sum_{i,j}\frac{g_ig_j}{\delta_i}\left([\sigma^+_i\hat a,\sigma_j^-\hat a^\dagger]-[\sigma_i^-\hat a^\dagger,\sigma_j^+\hat a]\right)$

　　 $\displaystyle = \frac{1}{2}\sum_{i,j}\left\{\frac{g_ig_j}{\delta_i}[\sigma^+_i\hat a,\sigma_j^-\hat a^\dagger] + (i \leftrightarrow j)\right\}$

Now,

　 $\displaystyle [\sigma^+_i\hat a,\sigma_j^-\hat a^\dagger] = \sigma_i^+\sigma_j^-\hat a\hat a^\dagger - \sigma_j^-\sigma_i^+\hat a^\dagger\hat a$

　　 $\displaystyle = \sigma_i^+\sigma_j^-(\hat a^\dagger\hat a + 1) - \sigma_j^-\sigma_i^+\hat a^\dagger\hat a$

　　 $\displaystyle = [\sigma_i^+,\sigma_j^-]\hat a^\dagger\hat a + \sigma_i^+\sigma_j^-$

　　 $\displaystyle = \delta_{ij}\sigma_i^z\hat a^\dagger\hat a + \sigma_i^+\sigma_j^-$

Hence,

　 $\displaystyle \frac{1}{2}[S_0,V] = \frac{1}{2}\sum_{i,j}\left\{\frac{g_ig_j}{\delta_i}\left(\delta_{ij}\sigma_i^z\hat a^\dagger\hat a + \sigma_i^+\sigma_j^-\right) + (i \leftrightarrow j)\right\}$

　　 $\displaystyle = \sum_i\frac{g_i^2}{\delta_i}\sigma_i^z\hat a^\dagger\hat a + \sum_i\frac{g_i^2}{\delta_i}\sigma_i^+\sigma_i^- + \frac{1}{2}\sum_{i\ne j}\frac{g_ig_j}{\delta_i}(\sigma_i^+\sigma_j^-+\sigma_j^+\sigma_i^-)$

　　 $\displaystyle = \sum_i\frac{g_i^2}{\delta_i}\sigma_i^z\hat a^\dagger\hat a + \frac{1}{2}\sum_i\frac{g_i^2}{\delta_i}(1+\sigma_i^z) + \frac{g_1g_2}{2}\left(\frac{1}{\delta_1}+\frac{1}{\delta_2}\right)(\sigma_1^+\sigma_2^-+\sigma_1^-\sigma_2^+)$

So, the effective Hamiltonian (in the interaction picture) is:

　 $\displaystyle H_{\rm eff} = \sum_{i=1}^2\frac{1}{2}\left(\delta_i + \frac{g_i^2}{\delta_i}\right)\sigma_i^z + \frac{g_1g_2}{2}\left(\frac{1}{\delta_1}+\frac{1}{\delta_2}\right)(\sigma_1^+\sigma_2^-+\sigma_1^-\sigma_2^+)$

　　　 $\displaystyle {} + \sum_{i=1}^2\frac{g_i^2}{\delta_i}\sigma_i^z\hat a^\dagger\hat a + {\rm Const.}$ --- (1-1)

Since $\displaystyle [H_0,S_0] =0$ , we can go back to the original Schroedinger picture without modifying $H_0$ . So getting back to the original Schroedinger picture and dropping the photon energy $\propto \hat a^\dagger\hat a$ , we have:

　 $\displaystyle H_{\rm eff} = \sum_{i=1}^2\frac{1}{2}\left(\Delta_i + \frac{g_i^2}{\delta_i}\right)\sigma_i^z + \frac{g_1g_2}{2}\left(\frac{1}{\delta_1}+\frac{1}{\delta_2}\right)(\sigma_1^+\sigma_2^-+\sigma_1^-\sigma_2^+)$

(2) Defining $\displaystyle\Delta'_i := \Delta_i +\frac{g_i^2}{\delta_i}, J := \frac{g_1g_2}{2}\left(\frac{1}{\delta_1}+\frac{1}{\delta_2}\right)$ , we have:

　 $\displaystyle H_{\rm eff} = \sum_{i=1}^2\frac{\Delta'_i}{2}\sigma_i^z + J(\sigma_1^+\sigma_2^-+\sigma_1^-\sigma_2^+)$

　　 $\displaystyle=\begin{pmatrix}\frac{\Delta'_1+\Delta'_2}{2} & 0 & 0 & 0 \\ 0 & \frac{\Delta'_1-\Delta'_2}{2} & J & 0 \\ 0 & J & -\frac{\Delta'_1-\Delta'_2}{2} & 0 \\ 0 & 0 & 0 & -\frac{\Delta'_1+\Delta'_2}{2}\end{pmatrix}$

Hence states $\mid 00\rangle, \,\mid 11\rangle$ are eigenstates of $H_{\rm eff}$ whereas states $\mid 00\rangle, \,\mid 11\rangle$ are mixed together via time-evolution.

As discussed in 8.3.2, when $\Delta'_1 = \Delta'_2 =: \Delta'$ , the time evolution operator becomes:

　 $\displaystyle U(t) = e^{-iH_{\rm eff}t} = e^{-i\Delta'(\sigma_1^z+\sigma_2^z)}\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & \cos(Jt) & -i\sin(Jt) & 0 \\ 0 & -i\sin(Jt) & \cos(Jt) & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$

So you can implement iSwap with $\displaystyle U(3\pi / 2J)$ up to local transformations:

　 $\displaystyle U_{\rm iSWAP} = e^{-i\pi\Delta'(\sigma_1^z+\sigma_2^z)}U(3\pi/2J)$ --- (2-1)

(3) $\displaystyle \langle \hat a^\dagger\hat a\rangle = \frac{1}{e^{\hbar\omega/kT}-1} \sim 0.67$

From (1-1), the effective energy of qubits $\Delta'$ increase $\displaystyle \frac{2g_i^2}{\delta_i}\langle\hat a^\dagger\hat a\rangle \sim \frac{g_i^2}{\delta_i}$ . This would influence the local transformation in (2-1). But I don't see how it affects the two-qubit exchange???

8.4

We use the unit system with $\hbar = 1$ , and a consistent notation for energy levels $\omega_1,\,\omega_2$ (instead of $\Delta_1,\,\Delta_2$ ).

(1) The eigenvalues of $\mid 01\rangle,\,\mid 10\rangle$ subspace are:

　 $\displaystyle E_1,\,E_2 = \frac{\omega_1+\omega_2 - \sqrt{D_1}}{2},\, \frac{\omega_1+\omega_2 + \sqrt{D_1}}{2}$

where

　 $\displaystyle D_1 = (\omega_2-\omega_1)^2 + 4J^2$

The eigenvalues of $\mid 11\rangle,\,\mid 02\rangle$ subspace are:

　 $\displaystyle E_3,\,E_4 = \frac{-\alpha + \omega_1+3\omega_2 - \sqrt{D_2}}{2},\, \frac{-\alpha + \omega_1+3\omega_2 + \sqrt{D_2}}{2}$

where

　 $\displaystyle D_2 = (\omega_2-\omega_1-\alpha)^2 + 8J^2$

gist.github.com

(2) From the results of (1), energy levels degenerate at $D_1 = 0$ and $D_2=0$ , that is, $\omega_1=\omega_2$ and $\omega_2 = \omega_1 + \alpha$

(3) In $D_1,\, D_2$ , for large $\omega_2$ , we can ignore $J$ . So approximating as $J\sim 0$ , we have:

　 $E_1,\,E_2,\,E_3,\,E_4 = \omega_1,\,\omega_2,\,\omega_1+\omega_2,\,2\omega_2-\alpha$

They respectively correspond to states $\mid 10\rangle,\,\mid 01\rangle,\,\mid 11\rangle,\,\mid 02\rangle$ .

(4) Relabel the eigenvalues as:

　 $E_{10} := E_1,\,E_{01} := E_2,\,E_{11} := E_3,\,E_{02} := E_4$

Then, the phase changes are described as:

　 $\displaystyle\theta(s_1,s_2) = -\int_0^TE_{s_1s_2}(t)\,dt$

(5) Define:

　 $\displaystyle\Delta E(t) := \frac{1}{4}\left[\left\{E_{11}(t) + E_{00}(t)\right\} - \left\{E_{01}(t) + E_{10}(t)\right\}\right]$

Then,

　 $\displaystyle\Delta\theta = \frac{1}{4}\left[\left\{\theta(1,1) + \theta(0,0)\right\} - \left\{\theta(0,1) + \theta(1,0)\right\}\right]$

　　 $\displaystyle = -\int_0^T\Delta E(t)\,dt$

If we modify the adiabatic change $n$ times slower, $\Delta\theta$ is modified as:

　 $\displaystyle\Delta\theta' = -\int_0^{nT}\Delta E\left(\frac{t}{n}\right)\,dt = -n\int_0^T\Delta E(t)\,dt = n \Delta\theta$

So, by adjusting the speed of the adiabatic change, we can achieve arbitrary value of $\Delta\theta$ .

Note that once we achieve the transformation with $\Delta\theta$ , by applying single qubit operations, we can achieve the transformation:

　 $\displaystyle U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i\Delta\theta}\end{pmatrix}$

(6) N-bit Quantum Fourier transformation can be implemented with control-phase gates:

　 $\displaystyle U(k) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{2\pi i/2^k}\end{pmatrix}\ (k=2,\cdots,N)$

So we need $\displaystyle\Delta\theta = \frac{2\pi}{2^k}$ .

8.5

　 $\displaystyle U(t) =\exp\left(-it\sigma_1^z\sigma_2^z\right) = \begin{pmatrix} e^{-it} & 0 & 0 & 0 \\ 0 & e^{it} & 0 & 0 \\ 0 & 0 & e^{it} & 0 \\ 0 & 0 & 0 & e^{-it}\end{pmatrix} = \begin{pmatrix} e^{-i\sigma_z} & \mathbf 0 \\ \mathbf 0 & e^{i\sigma_z} \end{pmatrix}$

　 $\displaystyle U_{\rm C-NOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix} =\begin{pmatrix} {\mathbf 1} & \mathbf 0 \\ \mathbf 0 & \sigma_x \end{pmatrix}$

　 $\displaystyle U_{\rm C-NOT} (\mathbf 1\otimes \sigma_z) U_{\rm C-NOT} = \begin{pmatrix} {\mathbf 1} & \mathbf 0 \\ \mathbf 0 & \sigma_x \end{pmatrix} \begin{pmatrix} \sigma_z & \mathbf 0 \\ \mathbf 0 & \sigma_z \end{pmatrix} \begin{pmatrix} {\mathbf 1} & \mathbf 0 \\ \mathbf 0 & \sigma_x \end{pmatrix}$

　　 $\displaystyle = \begin{pmatrix} \sigma_z & \mathbf 0 \\ \mathbf 0 & \sigma_x\sigma_z\sigma_x \end{pmatrix} = \begin{pmatrix} \sigma_z & \mathbf 0 \\ \mathbf 0 & -\sigma_z \end{pmatrix}$

Hence,

　 $\displaystyle U_{\rm C-NOT} (\mathbf 1\otimes e^{-i\sigma_z}) U_{\rm C-NOT} = \begin{pmatrix} e^{-i\sigma_z} & \mathbf 0 \\ \mathbf 0 & e^{i\sigma_z} \end{pmatrix} = U(t)$

8.6

(1) Asymmetric totally depolarizing channel:

　 $\displaystyle \epsilon(\rho) = \left(1-\sum_{\alpha=x,y,z}p_\alpha\right)\rho + \sum_{\alpha=x,y,z}p_\alpha\sigma^\alpha\rho\sigma^\alpha$

This applies the phase flip (along the $\alpha$ axe) $\sigma^\alpha$ with probability $p_\alpha$ .

Since $\displaystyle \frac{1}{2}\left(\rho + \sum_{\alpha=x,y,z}\sigma^\alpha\rho\sigma^\alpha\right) = \mathbb 1$ , by setting $\displaystyle p_x=p_y=p_z=\frac{p}{4}$ , we have:

　 $\displaystyle\epsilon(\rho) = \left(1-\frac{3p}{4}\right)\rho + \frac{p}{4}\sum_{\alpha=x,y,z}\sigma^\alpha\rho\sigma^\alpha$

　　 $\displaystyle = \left(1-p\right)\rho + \frac{p}{4}\left(\rho+\sum_{\alpha=x,y,z}\sigma^\alpha\rho\sigma^\alpha\right)$

　　 $\displaystyle = (1-p)\rho + p\frac{\mathbb 1}{2}$

In this symmetric case, with probability p, the system is replaced with the completely mixed status $\displaystyle \frac{\mathbb 1}{2}$ .

(2) Random flip channel is implemented as:

　 $\displaystyle \epsilon(\rho) = (1-p)\rho + p\sigma_x\rho\sigma_x$

This is a special case of the asymmetric totally depolarizing channel $p_x = p, p_y = p_z =0$ .

(3) For $\displaystyle \rho = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$ ,

　 $\displaystyle \epsilon(\rho) = \begin{pmatrix} a & (1-p)b \\ (1-p)c & d\end{pmatrix}$

So applying it $N$ times, we have:

　　 $\displaystyle \epsilon^N(\rho) = \begin{pmatrix} a & (1-p)^Nb \\ (1-p)^Nc & d\end{pmatrix}$

(4) Kraus operators for asymmetric totally depolarizing channel are:

　 $\displaystyle A_0 = \sqrt{1-\sum_{\alpha=x,y,z}p_\alpha}\,\mathbb 1,\, A_\alpha = \sqrt{p_\alpha}\sigma^\alpha\ (\alpha=x,y,z)$

Kraus operators for dephasing channel are:

　 $\displaystyle A_0 = \sqrt{\frac{1}{2}(1+p)}\,\mathbb 1,\,A_1 = \sqrt{\frac{1}{2}(1-p)}\sigma^z$

Actually,

　 $\displaystyle A_0\rho A_0^\dagger + A_1\rho A_1^\dagger = \frac{1}{2}(1+p)\begin{pmatrix} a & b \\ c & d\end{pmatrix} + \frac{1}{2}(1-p)\begin{pmatrix} a & -b \\ -c & d\end{pmatrix}$

　　 $\displaystyle = \begin{pmatrix} a & 0 \\ 0 & d\end{pmatrix} + p\begin{pmatrix} 0 & b \\ c & 0\end{pmatrix} =p\begin{pmatrix} a & b \\ c & d\end{pmatrix} + (1-p)\begin{pmatrix} a & 0 \\ 0 & d\end{pmatrix}$

Since ${\rm tr}(\sigma^\alpha\rho\sigma^\alpha) = {\rm tr}\rho$ , for asymmetric totally depolarizing channel,

　 $\displaystyle {\rm tr}\epsilon(\rho) = \left(1-\sum p_\alpha\right){\rm tr}\rho + \sum p_\alpha\, {\rm tr}\rho = {\rm tr}\rho$

For other channels, you can confirm $\displaystyle {\rm tr}\epsilon(\rho) = {\rm tr}\rho$ with similar calculations.

8.7

Define $\displaystyle \sigma_0 = 1$ and you can confirm with a direct calculation that:

　 $\displaystyle {\rm tr}(\sigma^\dagger_{\alpha}\sigma_{\beta})=\delta_{\alpha\beta}\ (\alpha,\beta=x, y, z, 0)$ .

SInce [tex|displaystyle {\rm tr}(A\otimes B)={\rm tr}(A){\rm tr}(B)], it's also true that:

　 $\displaystyle {\rm tr}(S^\dagger_{\alpha}S_{\beta})=\delta_{\alpha\beta}\ (S_\alpha,S_\beta \in \{1,\sigma^x,\sigma^y,\sigma^z\}^{\otimes N})$ .

So, under the inner product $\displaystyle (S_\alpha,S_\beta) := {\rm tr}(S_\alpha^\dagger S_\beta)$ they are linear-independent and become orthobormal basis of $2^{2N}$ dimension linear space.

As the $2^N\times 2^N$ matrix $\rho$ has $2^{2N}$ elements, it can be expressed by a liner combination of $S_\alpha$ .

　 $\rho = \sum s_\alpha S^\alpha$

And the coefficients are determined by $\alpha = (\rho, S_\alpha)$ .

8.8

In the textbook, matrix representation and operator notation are somewhat mixed up. In the operator notation, we have:

　 $\displaystyle \epsilon(\hat\rho) = \frac{1}{d}\sum_{\alpha,\beta}\sum_{i,j}M_{\beta\alpha}{\rm tr}(\hat S^\dagger_\alpha\hat\rho)(\hat S_\beta)_{ij}\mid i\rangle\langle j\mid$ --- (1)

where $\displaystyle (\hat S_\beta)_{ij} := \langle i \mid\hat S_\beta\mid j \rangle$

　 $\displaystyle \epsilon(\hat \rho) = \sum_{\alpha,\beta}\sum_{i,j}\chi_{\alpha\beta}(\hat S_\alpha\hat\rho\hat S^\dagger_\beta)_{ij}\mid i\rangle\langle j\mid$ --- (2)

where $\displaystyle (\hat S_\alpha\hat\rho\hat S^\dagger_\beta)_{ij} := \langle i \mid\hat S_\alpha\hat\rho\hat S^\dagger_\beta\mid j \rangle$

From (1), we have:

　 $\displaystyle \rho_\epsilon := (\mathbb 1\otimes\epsilon)\left(\frac{1}{d}\sum_{i,j}\mid i\rangle\langle j\mid\otimes\mid i\rangle\langle j\mid\right)= \frac{1}{d}\sum_{i,j}\mid i\rangle\langle j\mid\otimes\epsilon\left(\mid i\rangle\langle j\mid\right)$

　　 $\displaystyle = \frac{1}{d^2}\sum_{i,j}\mid i\rangle\langle j\mid\otimes\sum_{\alpha,\beta}\sum_{k,l}M_{\beta\alpha}{\rm tr}(\hat S^\dagger_\alpha\mid i\rangle\langle j\mid)(\hat S_\beta)_{kl}\mid k\rangle\langle l\mid$

　　 $\displaystyle = \frac{1}{d^2}\sum_{\alpha,\beta}\sum_{k,l}\sum_{i,j}\mid i\rangle\langle j\mid\otimes M_{\beta\alpha}(\hat S_\alpha)_{ji}(\hat S_\beta)_{kl}\mid k\rangle\langle l\mid$

　　 $\displaystyle = \frac{1}{d^2}\sum_{\alpha,\beta}\sum_{k,l}\sum_{i,j} M_{\beta\alpha}(\hat S_\alpha)^{\rm T}_{ij}\mid i\rangle\langle j\mid\otimes(\hat S_\beta)_{kl}\mid k\rangle\langle l\mid$

　　 $\displaystyle = \frac{1}{d^2}\sum_{\alpha,\beta}M_{\beta\alpha}\hat S^{\rm T}_\alpha\otimes \hat S_\beta$ .

From (2), we have:

　　 $\displaystyle = \frac{1}{d}\sum_{i,j}\mid i\rangle\langle j\mid\otimes \sum_{\alpha,\beta}\sum_{k,l}\chi_{\alpha\beta}(\hat S_\alpha\mid i\rangle\langle j\mid \hat S^\dagger_\beta)_{kl}\mid k\rangle\langle l\mid$

　　 $\displaystyle = \frac{1}{d}\sum_{\alpha,\beta}\sum_{i,j}\chi_{\alpha\beta}\mid i\rangle\langle j\mid\otimes \hat S_\alpha\mid i\rangle\langle j\mid \hat S^\dagger_\beta$

　　 $\displaystyle = \frac{1}{d}\sum_{\alpha,\beta}\sum_{i,j}\chi_{\alpha\beta}\mid i\rangle\otimes\hat S_\alpha\mid i\rangle\langle j\mid\otimes\langle j\mid\hat S^\dagger_\beta$

　　 $\displaystyle = \sum_{\alpha,\beta}\chi_{\alpha\beta}\mid\alpha\rangle\langle\beta\mid$

Now we prove (8.16). For any $\hat\rho$ ,

　 $\displaystyle\epsilon(\rho) = \frac{1}{d}\sum_{\alpha,\beta}M_{\beta\alpha}{\rm tr}(\hat S_\alpha\hat\rho)\hat S_\beta = \sum_{\alpha,\beta}\chi_{\alpha\beta}\hat S_\alpha\hat\rho\hat S_\beta^\dagger$

On both sides, we apply $\displaystyle {\rm tr}(\hat S_{\beta'}*)$ from left:

　 $\displaystyle \frac{1}{d}\sum_{\alpha,\beta}M_{\beta\alpha}{\rm tr}(\hat S_\alpha\hat\rho){\rm tr}(\hat S_{\beta'}\hat S_\beta) = \sum_{\alpha,\beta}\chi_{\alpha\beta}{\rm tr}(\hat S_{\beta'}\hat S_\alpha\hat\rho\hat S_\beta^\dagger)$

Since $\displaystyle {\rm tr}(\hat S_{\beta'}\hat S_\beta) = d\delta_{\beta\beta'}$ , we have:

　 $\displaystyle \sum_{\alpha}M_{\beta'\alpha}{\rm tr}(\hat S_\alpha\hat\rho) = \sum_{\alpha,\beta}\chi_{\alpha\beta}{\rm tr}(\hat S_{\beta'}\hat S_\alpha\hat\rho\hat S_\beta^\dagger)$ --- (3)

Now we take $\displaystyle \hat\rho = \hat S_{\alpha'}$ , and using the relationship $\displaystyle {\rm tr}(\hat S_\alpha\hat S_{\alpha'}) = d\delta_{\alpha\alpha'}$ , we have:

　 $\displaystyle M_{\beta'\alpha'} = \frac{1}{d} \sum_{\alpha,\beta}\chi_{\alpha\beta}{\rm tr}(\hat S_{\beta'}\hat S_\alpha\hat S_{\alpha'}\hat S_\beta^\dagger)$

8.9

(1) Deduce $M$ and $\chi$ for $\displaystyle \epsilon(\hat\rho) = U\hat\rho U^\dagger$ .

For $M_{\alpha\beta}$ , from Exercise 8.8 (1), we have:

　 $\displaystyle \frac{1}{d}\sum_{\alpha,\beta}M_{\beta\alpha}{\rm tr}(\hat S^\dagger_\alpha\hat \rho)\hat S_\beta = U\hat\rho U^\dagger$

Applying $\displaystyle{\rm tr}(\hat S_{\beta'}\, *)$ and taking $\displaystyle \hat\rho = \hat S_{\alpha'}$ , we have:

　 $\displaystyle d\sum_{\alpha,\beta}M_{\beta\alpha}\delta_{\alpha\alpha'}\delta_{\beta\beta'} = {\rm tr}(\hat S_{\beta'}U\hat S_{\alpha'} U^\dagger)$

　 $\displaystyle\therefore\, M_{\beta\alpha} = \frac{1}{d}{\rm tr}(\hat S_\beta U\hat S_\alpha U^\dagger)$

　 $\displaystyle\therefore\, M_{\alpha\beta} = \frac{1}{d}{\rm tr}(U^\dagger\hat S_\alpha U\hat S_\beta )$

For $\chi_{\alpha\beta}$ ,

　 $\displaystyle \mid\alpha \rangle = \frac{1}{\sqrt{d}}\sum_i\mid i\rangle\otimes \hat S_\alpha\mid i\rangle$

　 $\displaystyle \langle\alpha\mid\beta\rangle = \frac{1}{d}\sum_{i,j}\langle i\mid j\rangle\otimes\langle i\mid \hat S^\dagger_\alpha \hat S_\beta\mid j\rangle= \frac{1}{d}{\rm tr}(\hat S_\alpha^\dagger \hat S_\beta) = \delta_{\alpha\beta}$

Hence, from Exercise 8.8 (3),

　 $\displaystyle\chi_{\alpha\beta} = \langle\alpha\mid(\mathbb 1\otimes\epsilon)(\mid\Psi\rangle\langle\Psi\mid)\mid\beta\rangle$

　　 $\displaystyle = \frac{1}{d}\sum_{i,j}\langle\alpha\mid i\rangle\langle j\mid \otimes\epsilon(\mid i\rangle\langle j\mid)\mid\beta\rangle$

　　 $\displaystyle = \frac{1}{d}\sum_{i,j}\langle\alpha\mid i\rangle\langle j\mid \otimes U\mid i\rangle\langle j\mid U^\dagger \mid\beta\rangle$

　　 $\displaystyle = \frac{1}{d^2}\sum_{i,j,k,l}\langle k\mid i\rangle\langle j\mid l \rangle \otimes\langle k\mid \hat S^\dagger_\alpha U\mid i\rangle\langle j\mid U^\dagger\hat S_\beta\mid l\rangle$

　　 $\displaystyle = \frac{1}{d^2}{\rm tr}(\hat S^\dagger_\alpha U){\rm tr}(U^\dagger\hat S_\beta)$

(2) $M$ and $\chi$ for $\displaystyle U=\exp\left(-i\theta\mathbf n\cdot\mathbf\sigma\right)$ .

We use the Einstein summation convention in the following calculations.

For $M_{\alpha\beta}$ ,

　 $\displaystyle M_{\alpha\beta} = \frac{1}{d}{\rm tr}(U^\dagger \sigma_\alpha U\sigma_\beta )\ \$ where $\sigma_\alpha \in \{\mathbb 1, \sigma_x,\sigma_y,\sigma_z\}$

If either $\sigma_\alpha$ or $\sigma_\beta$ is $\mathbb 1$ , $M_{\alpha\beta} = 0$ .

For other cases, from the relationships:

　 $\displaystyle[\sigma_i,\sigma_j] = 2i\epsilon_{ijk}\sigma_k$

　 $\displaystyle\sigma_i\sigma_j = \delta_{ij}\mathbb 1 + i\epsilon_{ijk}\sigma_k$

　 $\displaystyle \exp(i\theta\mathbf n\cdot\mathbf\sigma) = (\cos\theta)\mathbb 1 + (i\sin\theta)\mathbf n\cdot\mathbf \sigma$

we have:

　 $\displaystyle{\rm tr}(U^\dagger\sigma_i U\sigma_j) = {\rm tr}\left\{\exp(i\theta\mathbf n\cdot\mathbf\sigma)\sigma_i \exp(-i\theta\mathbf n\cdot\mathbf\sigma)\sigma_j\right\}$

　　 $\displaystyle = {\rm tr}\left[\left\{(\cos\theta)\mathbb 1 + (i\sin\theta)n_k\sigma_k\right\}\sigma_i \left\{(\cos\theta)\mathbb 1 - (i\sin\theta)n_l\sigma_l\right\}\sigma_j\right]$

　　 $\displaystyle = \cos^2\theta\, {\rm tr}(\sigma_i\sigma_j)$

　　　 $\displaystyle {} + i\sin\theta\cos\theta\,n_k{\rm tr}(\sigma_k\sigma_i\sigma_j) - i\sin\theta\cos\theta\,n_l{\rm tr}(\sigma_i\sigma_l\sigma_j)$

　　　 $\displaystyle {} + \sin^2\theta\,n_kn_l\,{\rm tr}(\sigma_k\sigma_i\sigma_l\sigma_j)$

　　 $\displaystyle = 2\cos^2\theta\,\delta_{ij} + i\sin\theta\cos\theta\,n_k{\rm tr}(\sigma_k[\sigma_i, \sigma_j]) + \sin^2\theta\,n_kn_l\,{\rm tr}(\sigma_k\sigma_i\sigma_l\sigma_j)$

　(1st term) $\displaystyle = 2\cos^2\theta\,\mathbf e_i\cdot\mathbf e_j$

　(2nd term) $\displaystyle = -2\sin\theta\cos\theta\,n_k\,{\rm tr}(\sigma_k\epsilon_{ijl}\sigma_l) = -4 \sin\theta\cos\theta\,n_k\epsilon_{ijl}\delta_{kl}$

　　 $\displaystyle = -4\sin\theta\cos\theta \, (\mathbf e_i\times\mathbf e_j)\cdot\mathbf n$

　(3rd term) $\displaystyle = \sin^2\theta\,n_kn_l\,{\rm tr}\left\{ (\delta_{ki}\mathbb 1 + i\epsilon_{kim}\sigma_m)(\delta_{lj}\mathbb 1+i\epsilon_{ljn}\sigma_n)\right\}$

　　 $\displaystyle = 2 \sin^2\theta\,n_kn_l\,(\delta_{ki}\delta_{lj}-\epsilon_{kim}\epsilon_{ljm})$

　　 $\displaystyle = 2 \sin^2\theta\,n_kn_l\,\left\{\delta_{ki}\delta_{lj}-(\delta_{kl}\delta_{ij}-\delta_{kj}\delta_{il})\right\}$

　　 $\displaystyle = 2 \sin^2\theta\,(n_in_j - \delta_{ij} + n_jn_i)$

　　 $\displaystyle = 4\sin^2\theta\,(\mathbf e_i\cdot\mathbf n)(\mathbf e_j\cdot\mathbf n)-2\sin^2\theta\,\mathbf e_i\cdot\mathbf e_j$

　 $\displaystyle\therefore M_{ij} = \frac{1}{2}{\rm tr}(U^\dagger \sigma_i U\sigma_j)$

　　 $\displaystyle = (\cos^2\theta-\sin^2\theta)\,(\mathbf e_i\cdot\mathbf e_j) - 2\sin\theta\cos\theta \, (\mathbf e_i\times\mathbf e_j)\cdot\mathbf n + 2\sin^2\theta\,(\mathbf e_i\cdot\mathbf n)(\mathbf e_j\cdot\mathbf n)$

For $\chi_{\alpha\beta}$ ,

　 $\displaystyle \chi_{\alpha\beta}= \frac{1}{4}{\rm tr}(\sigma_\alpha U){\rm tr}(U^\dagger\sigma_\beta)\ \$ where $\sigma_\alpha \in \{\mathbb 1, \sigma_x,\sigma_y,\sigma_z\}$

We use the convention that $n_0 = 0$ , then we have:

　 $\displaystyle \chi_{\alpha\beta}= \frac{1}{4}{\rm tr}\left\{\sigma_\alpha \exp(-i\theta\mathbf n\cdot\mathbf\sigma)\right\}{\rm tr}\left\{\exp(i\theta\mathbf n\cdot\mathbf\sigma)\sigma_\beta\right\}$

　　 $\displaystyle = \frac{1}{4}{\rm tr}\left[ \sigma_\alpha \left\{(\cos\theta)\mathbb 1 - (i\sin\theta)n_k \sigma_k\right\}\right] {\rm tr}\left[ \left\{(\cos\theta)\mathbb 1 + (i\sin\theta)n_l \sigma_l\right\}\sigma_\beta \right]$

　　 $\displaystyle = \left(\cos\theta\,\delta_{\alpha 0}-i\sin\theta\,n_k\delta_{\alpha k}\right) \left(\cos\theta\,\delta_{\beta 0}+i\sin\theta\,n_l\delta_{\beta l}\right)$

　　 $\displaystyle = \cos^2\theta\,\delta_{\alpha 0}\delta_{\beta 0}+i\sin\theta\,n_\beta\delta_{\alpha 0}-i\sin\theta\,n_\alpha\delta_{\beta 0} + \sin^2\theta\, n_\alpha n_\beta$

(3) $M$ and $\chi$ for CNOT $\displaystyle U=U^\dagger = \,\mid 0\rangle\langle 0\mid \otimes \mathbb 1 + \mid 1\rangle\langle 1\mid\otimes\sigma_x$ .

　 $\displaystyle U(\sigma_\alpha\otimes\sigma_\beta) = \mid 0\rangle\langle 0\mid \sigma_\alpha\otimes\sigma_\beta + \mid 1\rangle\langle 1\mid\sigma_\alpha\otimes\sigma_x\sigma_\beta$

Hence for $M_{(\alpha\beta)(\gamma\kappa)}$ ,

　 $\displaystyle M_{(\alpha\beta)(\gamma\kappa)} = \frac{1}{4}{\rm tr}\left\{U(\sigma_\alpha\otimes\sigma_\beta) U(\sigma_\gamma\otimes\sigma_\kappa)\right\}$

　　 $\displaystyle = \frac{1}{4}{\rm tr}\Big\{\left( \mid 0\rangle\langle 0\mid \sigma_\alpha\otimes\sigma_\beta + \mid 1\rangle\langle 1\mid\sigma_\alpha\otimes\sigma_x\sigma_\beta\right)$

　　　　　 $\displaystyle \times\,\left( \mid 0\rangle\langle 0\mid \sigma_\gamma\otimes\sigma_\kappa + \mid 1\rangle\langle 1\mid\sigma_\gamma\otimes\sigma_x\sigma_\kappa\right)\Big\}$

　　 $\displaystyle = \frac{1}{4}\Big\{2(\sigma_\alpha)_{00}\delta_{\beta\kappa}+(\sigma_\alpha)_{01}(\sigma_\gamma)_{10}{\rm tr}(\sigma_\beta\sigma_x\sigma_\kappa)$

　　　　　　 $\displaystyle{}+(\sigma_\alpha)_{10}(\sigma_\gamma)_{01}{\rm tr}(\sigma_\kappa\sigma_x\sigma_\beta)+(\sigma_\alpha)_{11}(\sigma_\gamma)_{11}{\rm tr}(\sigma_x\sigma_\beta\sigma_x\sigma_\kappa)\Big\}$

For $\chi_{(\alpha\beta)(\gamma\kappa)}$ ,

　 $\displaystyle \chi_{(\alpha\beta)(\gamma\kappa)} = \frac{1}{16}{\rm tr}\left\{U(\sigma_\alpha\otimes\sigma_\beta)\right\}{\rm tr}\left\{U(\sigma_\gamma\otimes\sigma_\kappa)\right\}$

　　 $\displaystyle=\frac{1}{16}\left\{(\sigma_\alpha)_{00}\sigma_\beta+(\sigma_\alpha)_{11}\sigma_x\sigma_\beta\right\} \left\{(\sigma_\gamma)_{00}\sigma_\kappa+(\sigma_\gamma)_{11}\sigma_x\sigma_\kappa\right\}$

8.10

The error mapping process is described as:

　 $\displaystyle \epsilon(\rho) = \begin{pmatrix} \rho_{11}e^{-\frac{t}{T_1}} & \rho_{10}e^{-i\Delta t-\frac{t}{T^*_2}} \\ \rho_{01}e^{i\Delta t-\frac{t}{T^*_2}} & \rho_{00} + (1-e^{-\frac{t}{T_1}})\rho_{11} \end{pmatrix}$

In the operator form, this can be described as:

　 $\displaystyle\epsilon\left(\mid 0\rangle\langle 0\mid\right) = \mid 0\rangle\langle 0 \mid$

　 $\displaystyle\epsilon\left(\mid 1\rangle\langle 1\mid\right) = e^{-\frac{t}{T_1}}\mid 1\rangle\langle 1 \mid + \left(1-e^{-\frac{t}{T_1}}\right)\mid 0\rangle\langle 0 \mid$

　 $\displaystyle\epsilon\left(\mid 0\rangle\langle 1\mid\right) = e^{i\Delta t-\frac{t}{T^*_2}}\mid 0\rangle\langle 1 \mid$

　 $\displaystyle\epsilon\left(\mid 1\rangle\langle 0\mid\right) = e^{-i\Delta t-\frac{t}{T^*_2}}\mid 1\rangle\langle 0 \mid$

Now the matrix elements of $\chi$ are calculated through the relationships:

　 $\displaystyle\chi_{\alpha\beta} = \langle\alpha\mid(\mathbb 1\otimes\epsilon)(\mid\Psi\rangle\langle\Psi\mid)\mid\beta\rangle$

　　 $\displaystyle = \frac{1}{d}\sum_{i,j}\langle\alpha\mid i\rangle\langle j\mid \otimes\epsilon(\mid i\rangle\langle j\mid)\mid\beta\rangle$

　　 $\displaystyle = \frac{1}{d^2}\sum_{i,j}\sum_{k,l}\langle k\mid i\rangle\langle j\mid l\rangle\otimes\langle k\mid\hat S^\dagger_\alpha\epsilon(\mid i\rangle\langle j\mid)\hat S_\beta\mid l\rangle$

　　 $\displaystyle = \frac{1}{d^2}\sum_{i,j}\langle i\mid\hat S^\dagger_\alpha\epsilon(\mid i\rangle\langle j\mid)\hat S_\beta\mid j\rangle$

For our particular case, using the notation $\displaystyle \sigma_\alpha,\, \sigma_\beta \in \{\sigma_0 := \mathbb 1,\sigma_x,\sigma_y,\sigma_z\}$ :

　 $\displaystyle\chi_{\alpha\beta} = \frac{1}{4}\Big\{\langle 0\mid \sigma_\alpha\mid 0\rangle\langle 0\mid \sigma_\beta\mid 0\rangle$

　　　　　　 $\displaystyle{}+\langle 1\mid \sigma_\alpha\mid 1\rangle\langle 1\mid \sigma_\beta\mid 1\rangle e^{-\frac{t}{T_1}}+ \langle 1\mid \sigma_\alpha\mid 0\rangle\langle 0\mid \sigma_\beta\mid 1\rangle\left(1- e^{-\frac{t}{T_1}}\right)$

　　　　　　 $\displaystyle{}+\langle 0\mid \sigma_\alpha\mid 0\rangle\langle 1\mid \sigma_\beta\mid 1\rangle e^{i\Delta t-\frac{t}{T^*_2}}$

　　　　　　 $\displaystyle{}+\langle 1\mid \sigma_\alpha\mid 1\rangle\langle 0\mid \sigma_\beta\mid 0\rangle e^{-i\Delta t-\frac{t}{T^*_2}}\Big\}$

Non-zero elements are:

　 $\displaystyle\chi_{00} = \frac{1}{4}\left\{1+e^{-\frac{t}{T_1}}+2e^{-\frac{t}{T^*_2}}\cos(\Delta t)\right\}$

　 $\displaystyle\chi_{0z} = \frac{1}{4}\left\{-1+e^{-\frac{t}{T_1}}+2ie^{-\frac{t}{T^*_2}}\sin(\Delta t)\right\}$

　 $\displaystyle\chi_{z0} = \frac{1}{4}\left\{-1+e^{-\frac{t}{T_1}}-2ie^{-\frac{t}{T^*_2}}\sin(\Delta t)\right\}$

　 $\displaystyle\chi_{zz} = \frac{1}{4}\left\{1+e^{-\frac{t}{T_1}}-2e^{-\frac{t}{T^*_2}}\cos(\Delta t)\right\}$

　 $\displaystyle\chi_{xx} = \frac{1}{4}\left(1-e^{-\frac{t}{T_1}}\right)$

　 $\displaystyle\chi_{xy} = -\chi_{yx} = \frac{i}{4}\left(1-e^{-\frac{t}{T_1}}\right)$

　 $\displaystyle\chi_{yy} = \frac{1}{4}\left(1-e^{-\frac{t}{T_1}}\right)$

　 $\displaystyle\therefore\,\chi_{\epsilon_1} = \frac{1}{4}\begin{pmatrix} 1+e^{-\frac{t}{T_1}}+2e^{-\frac{t}{T^*_2}}\cos(\Delta t) & -1+e^{-\frac{t}{T_1}}+2ie^{-\frac{t}{T^*_2}}\sin(\Delta t) & 0 & 0 \\ -1+e^{-\frac{t}{T_1}}-2ie^{-\frac{t}{T^*_2}}\sin(\Delta t) & 1+e^{-\frac{t}{T_1}}-2e^{-\frac{t}{T^*_2}}\cos(\Delta t) & 0 & 0 \\ 0 & 0 & 1-e^{-\frac{t}{T_1}} & i \left(1-e^{-\frac{t}{T_1}}\right) \\ 0 & 0 & -i \left(1-e^{-\frac{t}{T_1}}\right) & 1-e^{-\frac{t}{T_1}} \end{pmatrix}$

By taking the limit $T_1, T^*_2 \to \infty$ , the error-free version is:

　 $\displaystyle\therefore\,\chi_{\epsilon_0} = \frac{1}{4}\begin{pmatrix} 2+2\cos(\Delta t) & 2i\sin(\Delta t) & 0 & 0 \\ -2i\sin(\Delta t) & 2-2\cos(\Delta t) & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

Hence, the process fidelity is:

　 $\displaystyle F_{\rm prc} = {\rm tr}\left(\chi_{\epsilon_0}^\dagger\chi_{\epsilon_1}\right)$

　　 $\displaystyle = \frac{1}{16}\Big\{2(1+\cos(\Delta t))\left(1+e^{-\frac{t}{T_1}}+2e^{-\frac{t}{T^*_2}}\cos(\Delta t)\right)$

　　　　　 $\displaystyle {} + 2i\sin(\Delta t)\left(-1+e^{-\frac{t}{T_1}}-2ie^{-\frac{t}{T^*_2}}\sin(\Delta t)\right)$

　　　　　 $\displaystyle {} + 2(1-\cos(\Delta t))\left(1+e^{-\frac{t}{T_1}}-2e^{-\frac{t}{T^*_2}}\cos(\Delta t)\right)$

　　　　　 $\displaystyle {} - 2i\sin(\Delta t)\left(-1+e^{-\frac{t}{T_1}}+2ie^{-\frac{t}{T^*_2}}\sin(\Delta t)\right)\Big\}$

　　 $\displaystyle = \frac{1}{4}\left\{1+e^{-\frac{t}{T_1}}+2e^{-\frac{t}{T^*_2}}\cos^2(\Delta t)+2e^{-\frac{t}{T^*_2}}\sin^2(\Delta t)\right\}$

　　 $\displaystyle = \frac{1}{4}\left(1+e^{-\frac{t}{T_1}}+2e^{-\frac{t}{T^*_2}}\right)$

The average fidelity is:

　 $\displaystyle F_{\rm avg} = \frac{2F_{\rm proc}+1}{2+1} = \frac{2}{3}F_{\rm proc}+\frac{1}{3}$

The matrix elements of $M$ can be calculated through the relationship:

　 $\displaystyle M_{\beta'\alpha'} = \frac{1}{d} \sum_{\alpha,\beta}\chi_{\alpha\beta}{\rm tr}(\hat S_{\beta'}\hat S_\alpha\hat S_{\alpha'}\hat S_\beta^\dagger)$