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: