2022-11-03

Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 9)

9.1

(1) $\displaystyle U = e^{-i\frac{\theta(\epsilon)}{2}\sigma^y} = \cos\frac{\theta}{2}\,\mathbb 1 - i\sin\frac{\theta}{2}\,\sigma^y$

　 $\displaystyle U\sigma_zU^\dagger = \left(\cos\frac{\theta}{2}\,\mathbb 1 - i\sin\frac{\theta}{2}\,\sigma^y\right)\sigma_z\left(\cos\frac{\theta}{2}\,\mathbb 1 + i\sin\frac{\theta}{2}\,\sigma^y\right)$

　　 $\displaystyle = \cos\theta\,\sigma_z + \sin\theta\,\sigma_x$

So, by setting $\displaystyle \tan\theta = \frac{\Delta_{\rm min}}{\epsilon}$ , we have:

　 $\displaystyle H = \frac{\epsilon}{2}\sigma^z + \frac{\Delta_{\rm min}}{2}\sigma^x = \frac{1}{2}\sqrt{\epsilon^2+\Delta_{\rm min}^2}(\cos\theta\,\sigma_z + \sin\theta\,\sigma_x) = \frac{1}{2}\Delta(\epsilon)U\sigma^zU^\dagger$

where $\displaystyle \Delta(\epsilon) := \sqrt{\epsilon^2+\Delta_{\rm min}^2}$

(2) We apply an adiabatic change to $\epsilon(t)$ .

　 $\displaystyle \mid\Psi(t)\rangle = U(t)\mid\xi(t)\rangle$

　 $\displaystyle H\mid\Psi(t)\rangle = \frac{1}{2}\Delta(\epsilon)U(t)\sigma^zU(t)^\dagger U(t)\mid\xi(t)\rangle = \frac{1}{2}\Delta(\epsilon)U(t)\sigma^z\mid\xi(t)\rangle$

On the other hand,

　 $\displaystyle i\frac{d}{dt}\mid\Psi(t)\rangle = i\frac{d}{dt}\left\{U(t)\mid\xi(t)\rangle\right\} = i\dot U(t)\mid\xi(t)\rangle + iU(t)\frac{d}{dt}\mid\xi(t)\rangle$

Hence, from the Schroedinger equation $\displaystyle H\mid\Psi(t)\rangle = i\frac{d}{dt}\mid\Psi(t)\rangle$ , we have:

　 $\displaystyle i\frac{d}{dt}\mid\xi(t)\rangle = -iU^\dagger(t)\dot U(t)\mid\xi(t)\rangle + \frac{1}{2}\Delta(\epsilon)\sigma^z\mid\xi(t)\rangle$

From the definition of $U$ , we have:

　 $\displaystyle\dot U(t) = -i\frac{\dot\theta(t)}{2}\sigma^y U(t)$

So, $\displaystyle -iU^\dagger(t)\dot U(t) = -\frac{\dot\theta(t)}{2}\sigma^y$

　 $\displaystyle\therefore\, i\frac{d}{dt}\mid\xi(t)\rangle = -\frac{\dot\theta(t)}{2}\sigma^y\mid\xi(t)\rangle + \frac{1}{2}\Delta(\epsilon)\sigma^z\mid\xi(t)\rangle$ --- (2-1)

By differentiating the both sides of $\displaystyle \epsilon(t)\tan\theta(t) = \Delta_{\rm min}$ ,

　 $\displaystyle \dot\epsilon(t)\tan\theta(t) + \epsilon(t)\frac{\dot\theta(t)}{\cos^2\theta(t)} = 0$

　 $\displaystyle\therefore\, \dot\theta(t) = -\frac{\dot\epsilon(t)}{\epsilon(t)}\sin\theta\cos\theta = -\frac{\Delta_{\rm min}}{\Delta^2(\epsilon)}\dot\epsilon(t)$

So, from (2-1), we have:

　 $\displaystyle\therefore\, i\frac{d}{dt}\mid\xi(t)\rangle = \left\{\frac{1}{2}\Delta(\epsilon)\sigma^z+\frac{\Delta_{\rm min}\dot\epsilon(t)}{\Delta^2(\epsilon)}\sigma^y\right\}\mid\xi(t)\rangle$

(3) Since $\displaystyle \frac{d}{dt} = \dot\epsilon(t)\frac{d}{d\epsilon}$ , we have:

　 $\displaystyle\therefore\, i\frac{d}{d\epsilon}\mid\xi(\epsilon)\rangle = \left\{\frac{1}{2\dot\epsilon}\Delta(\epsilon)\sigma^z+\frac{\Delta_{\rm min}}{\Delta^2(\epsilon)}\sigma^y\right\}\mid\xi(\epsilon)\rangle$

(4) $\displaystyle H_{\rm eff} = \frac{1}{\Delta}\sigma^z + \frac{\Delta_{\min}\dot\epsilon}{2\Delta^2}\sigma^y =\frac{\Delta}{2}\begin{pmatrix}1 & -ig \\ ig & -1\end{pmatrix}$

where $\displaystyle g = \frac{\Delta_{\min}\dot\epsilon}{\Delta^3}$

The eigenvalues of $H_{\rm eff}$ are $\displaystyle E_{\pm} = \pm\frac{\Delta}{2}\sqrt{1+g^2}$

The eigenvector with $E_-$ is given by:

　 $\displaystyle \mid 0 \rangle \propto \begin{pmatrix}\frac{i}{g}\left(-1\sqrt{1+g^2}\right) \\ 1 \end{pmatrix} = \begin{pmatrix}i\frac{g}{2} \\ 1\end{pmatrix}$

Hence, the correction is $\displaystyle \frac{g}{2} = \frac{\Delta_{\rm min}\dot\epsilon}{2\Delta^3}$

9.2

The change should be slower than the spontaneous photon emission time of the qubit.

9.3

　 $\displaystyle H=\frac{1}{2}\sum_{i=1}^{N-1}J_i\sigma_i^z\sigma_{i+1}^z$

Fix the direction of the 1st spin as $\mid\psi_1\rangle = \mid 0\rangle$ , and decide the remaining spins with the following induction:

　 $\displaystyle \mid \psi_{i+1}\rangle = -{\rm sign}(J_i)\mid\psi_i\rangle\ (i=2, 3,\cdots)$

9.4

(1) We usually choose $H_0$ as a non-interacting system such as $\displaystyle H_0 = \sum_i\Delta_i\sigma_i^x$ in (9.18) or (9.12). So the time evolution under $H_0$ is local.

(2) We need at least one gate for one interaction. So the number of gates is $\sim dN^d\ (d=1,2,3)$ .

9.5

The variance of sample average $\overline x$ scales as $\displaystyle \frac{1}{M}$ .

　 $\displaystyle V(\overline x) = V\left(\sum_{i=1}^M\frac{x_i}{M}\right) = \frac{1}{M^2}\sum_{i=1}^MV(x_i) = \frac{V(x)}{M}$

So, the deviation scales as $\displaystyle \sqrt{V(\overline x)} \sim \frac{1}{\sqrt{M}}$

Since the Hamiltonian has $N^2$ terms, the total variance of the estimated Energy (as a sample average) scales as $\displaystyle \frac{N^4}{M}$ . So, to keep the total variance below $\epsilon$ , we have:

　 $\displaystyle \frac{N^4}{M} < \epsilon$

　 $\displaystyle \therefore M > \frac{N^4}{\epsilon}$

9.6

In the limit of $\displaystyle |J_{\rm f}| \gg 1$ , we ignore other terms. Then the ground state is apparently the one where all spins have the same direction. Flipping one spin, we have the minimum gap $\Delta_{\rm min} = 2|J_{\rm f}|$ .

As the initial state (the ground state of $H_0$ is a superposition of all states, it has an overlap with the ground state of $\displaystyle |J_{\rm f}| \gg 1$ . So, with the adiabatic process (in the limit of $\dot\epsilon\sim 0$ ), it achieves the state where all spins have the same direction, that is, the state of qubit 1 is copied to the N intermediate qubits.

9.7

In this case, "Success in $M$ experiments" means "Success at least one of $M$ experiments". So the success probability is calculated as "1 - probability of fails in all $M$ experiments." Hence, the success probability is:

　 $p = 1 - (1-p_{\rm ramp})^M$

If $p_{\rm ramp} = 0.99$ , we need $M \ge 2$ to achieve $p=0.9999$ .

Practically, $t_{\rm TTS}$ is defined as $M t_{\rm ramp}$ where $M$ is the number of experiments to achieve the required success probability.

2022-10-30

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

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

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(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:

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

8.11

gist.github.com

8.12

Define $p_d$ as a probability of depolarization, and the totally depolarization channel is:

　 $\displaystyle\epsilon(\rho) = (1-p_d)\rho + p_d\frac{\mathbb 1}{d}$

Hence, for a pure state $\displaystyle\mid\psi\rangle\langle\psi\mid$

　 $\displaystyle \epsilon(\mid\psi\rangle\langle\psi\mid) = (1-p_d)\mid\psi\rangle\langle\psi\mid + p_d\frac{\mathbb 1}{d}$

So the fidelity for this particular state is:

　 $\displaystyle\langle\psi\mid \epsilon(\mid\psi\rangle\langle\psi\mid)\mid\psi\rangle = (1-p_d)+\frac{p_d}{d}$

Since this is independent of the state, the average over all pure states is given by:

　 $\displaystyle F_{\rm avg} = (1-p_d)+\frac{p_d}{d}$

Note that the depolarization parameter $p$ in the textbook (p.220) is a probability of not decaying: $p = 1 - p_d$ . So (8.24) should be:

　 $\displaystyle F_{\rm avg} = (1-p_d)+\frac{p_d}{d} = p + \frac{1-p}{d}$

2022-10-27

Derivation of Input-Output relations for waveguide-QED

Target system

The system is similar to the following one.

enakai00.hatenablog.com

Here we restrict the states with only one excitation (either qubit or a propagating photon).

　 $\displaystyle \mid\psi(t)\rangle = c_1(t)\mid 1,{\rm vac}\rangle + \sum_k\psi_k(t)\mid 0,1_k\rangle$ --- (7.13)

The Heisenberg equations are:

　 $\displaystyle i\dot c_1(t) = \Delta c_1(t) + \sum_kg_k\psi_k(t),\ i\dot\psi_k(t) = \omega_k\psi_k(t) + g_k^* c_1(t)$ --- (7.14)

Using the same argument to derive (B.23) in the above article, we have the formal solution for $\psi_k(t)$ as:

　 $\displaystyle\psi(t) = e^{-i\omega_k(t-t_0)}-ig_k^*\int_{t_0}^te^{-i\omega_k(t-\tau)}c(\tau)\,d\tau$ --- (1)

Position space

Wavefunction of the photon field is defined as:

　 $\displaystyle \Psi(x, t) := \frac{1}{\sqrt{d}}\langle x\mid\psi(t)\rangle = \frac{1}{\sqrt{d}}\sum_k\psi_k(t)\langle x \mid 1_k\rangle = \frac{1}{\sqrt{d}}\sum_k\psi_k(t)e^{ikx}$

We included $\displaystyle\frac{1}{\sqrt{d}}$ so that the wavefunction is scale-independent.

We split it into right-moving part $\Psi_+(x, t)$ and left-moving part $\Psi_-(x, t)$ as:

　 $\displaystyle \Psi_+(x, t) = \frac{1}{\sqrt{d}}\sum_{k > 0}\psi_k(t)e^{\pm ikx}$

By substituting (1), we have:

　 $\displaystyle \Psi_\pm(x, t) = \Psi_\pm(x, t, t_0) -i \frac{1}{\sqrt{d}}\sum_{k > 0}g_k^*\int_{t_0}^t e^{\pm ikx-i\omega_k(t-\tau)}c_1(\tau)\,d\tau$ --- (7.17)

　 $\displaystyle \Psi_\pm(x,t,t_0) := \frac{1}{\sqrt{d}}\sum_{k>0}e^{\pm ikx-i\omega_k(t-t_0)}$

where $\displaystyle \Psi_\pm(x,t,t_0)$ represents a group of plane waves corresponds to input ( $t_0=-\infty$ ) or output ( $t_0=\infty$ ) photon beam. Note that in some locations, they are just formal expressions. For example, $\Psi_+(x<0, t, \infty)$ corresponds to a right-moving output beam (before reaching the resonator) that doesn't exist in reality. It's a formal extension of the real output beam in $x>0$ to $x<0$ .

Approximate calculation

We will apply approximation to the last term in (7.17) with the following assumptions:

　 $\displaystyle g_k = \frac{g}{\sqrt{d}},\ \omega = vk,\,\frac{2\pi}{d}\sum_{k>0} \sim \int_0^\infty \frac{dk}{d\omega}\,d\omega = \frac{1}{v}\int_0^\infty \frac{dk}{d\omega}\,d\omega$

The last one is the same as in this. Now,

　 $\displaystyle -i \frac{1}{\sqrt{d}}\sum_{k > 0}g_k^*\int_{t_0}^t e^{\pm ikx-i\omega_k(t-\tau)}c_1(\tau)\,d\tau$

　　 $\displaystyle = -i\frac{g}{2\pi v} \int_{t_0}^t\left\{\int_0^\infty e^{\pm i\omega\frac{x}{v}-i\omega(t-\tau)}d\omega\right\}c_1(\tau)\,d\tau$

　　 $\displaystyle = -i\frac{g}{2\pi v} \int_{t_0}^t\left\{\int_0^\infty e^{i\omega(\tau-t\pm \frac{x}{v})}d\omega\right\}c_1(\tau)\,d\tau$ --- (2)

As the integration with $\omega$ will be dominant around $\displaystyle \tau \sim t\mp\frac{x}{v}$ ,

If the integral interval of $\tau$ contains $\displaystyle \tau \sim t\mp\frac{x}{v}$ , we replace $c_1(\tau)$ with $c_1\left(t\mp\frac{x}{v}\right)$ , and extend the integral interval of $\tau$ to $-\infty$ to $\infty$ .

If not, we approximate as $(2) \sim 0$ .

The followings are the cases that result in $(2) \sim 0$ . They corresponds to the following trivial cases.

$\displaystyle \Psi_+(x<0, t)$ and $t_0 = -\infty$ : In this case, $\displaystyle \Psi_+(x<0, t)$ is an input beam before reaching to the resonator. So it's naturally the same as $\displaystyle \Psi_+(x<0, t, -\infty)$ .

$\displaystyle \Psi_+(x>0, t)$ and $t_0 = \infty$ : In this case, $\displaystyle \Psi_+(x>0, t)$ is an output beam leaving the resonator. So it's naturally the same as $\displaystyle \Psi_+(x>0, t, \infty)$ .

$\displaystyle \Psi_-(x>0, t)$ and $t_0 = -\infty$ : In this case, $\displaystyle \Psi_+(x>0, t)$ is an input beam before reaching to the resonator. So it's naturally the same as $\displaystyle \Psi_-(x>0, t, -\infty)$ .

$\displaystyle \Psi_-(x<0, t)$ and $t_0 = \infty$ : In this case, $\displaystyle \Psi_+(x<0, t)$ is an output beam leaving the resonator. So it's naturally the same as $\displaystyle \Psi_-(x<0, t, \infty)$ .

Note: Strictly speaking, $\displaystyle \Psi_+(x>0, t<\infty)$ is different from the asymptotic plane waves $\displaystyle\Psi_+(x>0, t, t_0=\infty)$ . In this rough approximation, we simply drop this information. Instead, we extract that information from $\displaystyle \Psi_+(x>0, t<\infty) - \Psi_+(x>0, t, t_0=-\infty)$ . In other words, we shouldn't use the trivial cases above to extract any information about the photon-resonator interaction.

In other cases, (2) represents the difference of the beam before and after interacting with the resonator. In the following calculation, we implicitly exclude the trivial cases above.

For the case $t_0=-\infty$ ,

　 $\displaystyle (2) = -i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty \int_0^\infty e^{i\omega(\tau-t\pm \frac{x}{v})}\,d\omega\,d\tau$

　　 $\displaystyle = -i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty\pi\delta\left(\tau-t\pm\frac{x}{v}\right) + i\,{\mathcal PV}\left(\frac{1}{\tau-t\pm\frac{x}{v}}\right)\,d\tau$

　　 $\displaystyle =-i\frac{g}{2 v}c_1\left(t\mp\frac{x}{v}\right)$

For the case $t_0=\infty$ ,

　 $\displaystyle (2) = i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty \int_0^\infty e^{i\omega(\tau-t\pm \frac{x}{v})}\,d\omega\,d\tau$

　　 $\displaystyle = i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty\pi\delta\left(\tau-t\pm\frac{x}{v}\right) + i\,{\mathcal PV}\left(\frac{1}{\tau-t\pm\frac{x}{v}}\right)\,d\tau$