2022-10-26

Detailed derivation of the Input-Output theory

Basic definitions

Think of a system consisted of:

A long waveguide with length $d$ accommodating a photon field $\hat b_k$ .
A small resonator at $x=0$ accommodating an electric oscillation $\hat a$ .

that has the RWA Hamiltonian (5.60):

　 $\displaystyle H_{\rm RWA} = \omega\hat a^\dagger\hat a + \sum_k\left(g_k\hat a\hat b_k^\dagger + g_k^*\hat a^\dagger\hat b_k\right) + \sum_k\omega_k\hat b_k^\dagger\hat b_k$ --- (5.60)

where we use the unit system with $\hbar = 1$ . So, the momentum $p$ and the wavenumber $k$ are the same $p=k$ . (In general $p = \hbar k$ .)

The wavelength of the photon field is numbered as $\displaystyle \lambda_n \sim \frac{d}{n}$ . So, the wavenumber is counted as:

　 $\displaystyle k = p = \frac{2\pi}{\lambda_n} = \frac{2\pi}{d}n$

Hence, assuming that $d \gg 1$ , we can replace $\displaystyle \frac{2\pi}{d}\sum_k \sim \int_{-\infty}^\infty dk = 2\int_0^\infty \left|\frac{dk}{d\omega}\right|d\omega$ .

Applying it to the last term of $H_{\rm RWA}$ , we have:

　 $\displaystyle \sum_k\omega_k\hat b_k^\dagger\hat b_k \sim \frac{d}{\pi}\int_0^\infty \omega\hat b(\omega)^\dagger\hat b(\omega)\left|\frac{dk}{d\omega}\right|\,d\omega$

So the operator $\hat b_k$ has a scale dependency as $\displaystyle \hat b \sim \frac{1}{\sqrt{d}}$ . From the second term of $H_{\rm RWA}$ , the coupling constant $g_k$ has the same scale dependency $\displaystyle g_k \sim \frac{1}{\sqrt{d}}$

Now we define $\displaystyle J^{\rm QO}(\omega) := 2d|g(\omega)|^2\left|\frac{dk}{d\omega}\right|$ as a scale independent spectral function. Then, we have the relationship (5.62):

　 $\displaystyle \frac{1}{2\pi}\int_0^{\infty} J^{\rm QO}(\omega)e^{-i\omega t}\,d\omega \sim \sum_k |g_k|^2e^{-i\omega_k t} =: K(t)$ --- (5.62)

Also, in the following discussion, we use the Fourier transformation of the Heaviside step function $\Theta(t)$ :

　 $\displaystyle \int_0^\infty e^{i\omega u}\,du = \int_{-\infty}^\infty\Theta(u)e^{i\omega u}\,du = \pi\delta(\omega) + i\,{\mathcal PV}\left(\frac{1}{\omega}\right)$ --- (0)

Note that (0) is a relationship as a hyperfunction. So it's valid only when they are combined with the operation $\displaystyle \int_{-\infty}^\infty f(\omega)[*]\,d\omega$ .

Formal solution of the photon field $\hat b_k(t)$

From (5.60), we have the following Heisenberg equations (B.22):

　 $\displaystyle \frac{d}{dt}\hat a(t) = -i\omega\hat a(t) - i\sum_k g_k^*\hat b_k(t)$ --- (1)

　 $\displaystyle \frac{d}{dt}\hat b_k(t) = -i\omega_k\hat b_k(t) - ig_k\hat a(t)$ --- (2)

Define $\hat f(t)$ as $\displaystyle \hat b_k(t) = e^{-i\omega_k(t-t_0)}\left\{\hat b_k(t_0)+\hat f(t)\right\}$ with the initial condition $\hat f(t_0)=0$ . Then, from (2), we have:

　 $\displaystyle e^{-i\omega_k(t-t_0)}\frac{d}{dt}\hat f(t) = -ig_k\hat a(t)$

　 $\displaystyle\therefore\, \frac{d}{dt}\hat f(t) = -i e^{i\omega_k(t-t_0)}g_k\hat a(t)$

　 $\displaystyle\therefore\, \hat f(t) = -ig_ke^{-i\omega_kt_0}\int_{t_0}^te^{i\omega_k \tau}\hat a(\tau)\,d\tau$

　 $\displaystyle\therefore\, \hat b_k(t) = e^{-i\omega_k(t-t_0)}\hat b_k(t_0) -ig_k\int_{t_0}^t e^{-i\omega_k(t-\tau)}\hat a(\tau)\,d\tau$ --- (B.23)

A Langevin equation for the resonator

By substituting (B.23) into (1), we have:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -i\omega\hat a(t) - i\sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0) - \int_{t_0}^t\sum_k |g_k|^2e^{-i\omega_k(t-\tau)}\hat a(\tau)\,d\tau$

　 $\displaystyle \therefore\, \frac{d}{dt}\hat a(t) = -i\omega\hat a(t) - i \hat\xi(t) -\int_{t_0}^tK(t-\tau)\hat a(\tau)\,d\tau$ --- (5.61)

where $\displaystyle \hat\xi(t) := \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ .

Markovian approximation

We assume that the resonator resonates with the photon field with some frequency $\omega'$ with a slow modulation $\hat a_{\rm slow}(t)$ .

　 $\hat a(t) = \hat a_{\rm slow}(t)e^{-i\omega' t}$

and the contribution to the integration in (5.61) is dominated by the oscillation $e^{-i\omega' t}$ . So,

　 $\displaystyle \int_{t_0}^tK(t-\tau)\hat a(\tau)\,d\tau \simeq \hat a_{\rm slow}(t)\int_{t_0}^tK(t-\tau)e^{-i\omega'\tau}\,d\tau$

　　 $\displaystyle=\hat a_{\rm slow}(t)e^{-i\omega't}\int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du\ (u:=t-\tau)$

　　 $\displaystyle=\hat a(t)\int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du$

In the limit $t_0 \to -\infty$ , we have:

　 $\displaystyle \int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du = \frac{1}{2\pi}\int_0^\infty J^{\rm QO}(\omega)\left\{\int_0^{\infty}e^{i(\omega'-\omega)u}\,du\right\}d\omega$

　　 $\displaystyle = \frac{1}{2\pi}\int_0^\infty J^{\rm QO}(\omega)\left\{ \pi\delta(\omega'-\omega) + i\,{\mathcal PV}\left(\frac{1}{\omega'-\omega}\right)\right\}\,d\omega$

　　 $\displaystyle \simeq \frac{1}{2\pi}\int_{-\infty}^\infty J^{\rm QO}(\omega)\left\{ \pi\delta(\omega'-\omega) + i\,{\mathcal PV}\left(\frac{1}{\omega'-\omega}\right)\right\}\,d\omega$

　　　( $\because$ Contribution from around $\omega\sim\omega'$ is dominant.)

　　 $\displaystyle = \frac{1}{2}J^{\rm QO}(\omega') +i \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$

　　 $\displaystyle = \frac{\kappa}{2} + i\delta \omega$ --- (B.29)

where,

　 $\displaystyle\kappa := J^{\rm QO}(\omega')$

　 $\displaystyle\delta\omega := \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$ --- (B.30)

By combining these results, we have:

　 $\displaystyle \int_{t_0}^tK(t-\tau)\hat a(\tau)\,d\tau \simeq \left(i\delta\omega + \frac{\kappa}{2}\right)\hat a(t)$ --- (B.26)

And from (5.61), we have:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left\{i(\omega + \delta\omega)+\frac{\kappa}{2}\right\}\hat a(t)-i\hat\xi(t)$

From this result, you can see that $\hat a(t)$ oscillates as $\displaystyle\hat a(t) \sim \hat a(0)e^{-\frac{\kappa}{2}t} e^{-i(\omega+\delta\omega)t}$ . So,

　 $\omega' = \omega + \delta\omega$ .

This means that $\omega'$ is (implicitly) decided from the consistency condition with (B.30):

　 $\displaystyle\delta\omega = \omega' - \omega = \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$

Solution of the resonator $\hat a(t)$

Now we have the Langevin equation for the resonator:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left(i\omega'+\frac{\kappa}{2}\right)\hat a(t)-i\hat\xi(t)$ --- (3)

　 $\displaystyle \hat\xi(t) := \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ --- (4)

Define $\hat f(t)$ as $\displaystyle \hat a(t) = e^{-i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)}\hat f(t)$ . Then from (3), we have:

　 $\displaystyle e^{-i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)}\frac{d}{dt}\hat f(t) = -i\hat\xi(t)$

　 $\displaystyle\therefore\,\frac{d}{dt}\hat f(t) = -i e^{i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)}\hat\xi(t) =-i\sum_k g_k^* e^{i\left\{(\omega'-\omega_k)+\frac{\kappa}{2}\right\}(t-t_0)} \hat b_k(t_0)$

　 $\displaystyle\therefore\,\hat f(t) = -i \sum_k g_k^* \frac{e^{i\left\{(\omega'-\omega_k)+\frac{\kappa}{2}\right\}(t-t_0)}}{i(\omega'-\omega_k)+\frac{\kappa}{2}} \hat b_k(t_0) +C$

Hence, the solution $\hat a(t)$ is:

　 $\displaystyle \hat a(t) = Ce^{-i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)} -i \sum_k g_k^* \frac{e^{i\omega_k(t-t_0)}}{i(\omega'-\omega_k)+\frac{\kappa}{2}} \hat b_k(t_0)$ --- (B.32)

The constant $C$ is determined with the initial condition.

From this result, you can see that the contribution from coupling $g_k$ is dominant around the resonant frequency $\omega_k \sim \omega'$ . So changing the coupling strength $g_k$ outside $\omega_k \sim \omega'$ doesn't change the behavior of the system, and we can safely assume that $g_k$ is almost constant. In terms of $J^{\rm QO}(\omega)$ , this means that $J^{\rm QO}(\omega) \simeq J^{\rm QO}(\omega') = \kappa$

Then, compared to the free plane wave solution $\displaystyle \sum_k e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ , $\displaystyle \hat\xi(t) := \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ is rescaled with $g_k \propto \sqrt{\kappa}$ .

So, we define $\displaystyle\hat b^{\rm in}(t) := \frac{1}{\sqrt{\kappa}}\hat\xi(t)$ to have a renormalized operator, and (3) becomes:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left(i\omega'+\frac{\kappa}{2}\right)\hat a(t) -i\sqrt{\kappa}\hat b^{\rm in}(t)$ --- (5.64)

Note that in the discussions above, we used the initial condition at $t_0 \to -\infty$ . As a result, $\displaystyle\hat\xi(t)$ corresponds to the input photon generated at $t=t_0$ . So we named $\hat b^{\rm in}(t)$ to indicate the input photon.

Input-Output relations

It's also possible to consider the boundary condition at $t_0 \to \infty$ in the Markovian approximation. In this case, the calculation to get (B.29) becomes:

　 $\displaystyle \int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du = \frac{1}{2\pi}\int_{0}^\infty J^{\rm QO}(\omega)\left\{\int_0^{-\infty}e^{i(\omega'-\omega)u}\,du\right\}d\omega$

　　 $\displaystyle = -\frac{1}{2\pi}\int_{0}^\infty J^{\rm QO}(\omega)\left\{\int_0^{\infty}e^{i(\omega-\omega')u}\,du\right\}d\omega$

　　 $\displaystyle \simeq - \frac{1}{2\pi}\int_{-\infty}^\infty J^{\rm QO}(\omega)\left\{ \pi\delta(\omega-\omega') + i\,{\mathcal PV}\left(\frac{1}{\omega-\omega'}\right)\right\}\,d\omega$

　　 $\displaystyle = -\frac{1}{2}J^{\rm QO}(\omega') +i \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$

　　 $\displaystyle = -\frac{\kappa}{2} + i\delta \omega$ --- (B.29)'

Note that we used the fact that $\delta(x)$ is an even function and $\displaystyle{\mathcal PV}\left(\frac{1}{x-x'}\right)$ is an odd function.

Compared to (B.29), (B.29)' has an opposite sign for $\kappa$ , and (5.64) becomes:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left(i\omega'-\frac{\kappa}{2}\right)\hat a(t) -i\sqrt{\kappa}\hat b^{\rm out}(t)$ --- (5.64)'

where $\displaystyle\sqrt{\kappa}\hat b^{\rm out}(t) = \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ corresponds to the photon field pulling back to the current time $t$ from the final status at $t=t_0 \sim \infty$ .

By equating (5.64) and (5.64)', we have:

　 $\hat b^{\rm out}(t) = \hat b^{\rm in}(t) - i\sqrt{\kappa}\hat a(t)$ --- (B.38)

2022-10-24

Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 7)

7.1

Using the unit system $\hbar = 1$ and adding constant $\displaystyle \frac{\Delta}{2}$ to the total energy, the Hamiltonian is:

　 $\displaystyle \hat H = \sum_k \omega_k\hat a_k^\dagger \hat a_k + \frac{\Delta}{2}\sigma^z + \sum_k\left(g_k\sigma^+\hat a_k+g_k^*\hat a_k^\dagger\sigma^-\right) + \frac{\Delta}{2}$

The anzatz is:

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

where $\displaystyle \mid 0, 1_k\rangle = \hat a_k^\dagger \mid 0, {\rm vac}\rangle$

Then,

　 $\displaystyle \omega_k\hat a_k^\dagger \hat a_k \mid\psi(t)\rangle = \omega_k\psi_k(t)\mid 0, 1_k\rangle$

　 $\displaystyle \frac{\Delta}{2}\sigma^z \mid\psi(t)\rangle = \frac{\Delta}{2}c_1(t)\mid 1, {\rm vac}\rangle - \frac{\Delta}{2} \psi_k(t)\mid 0,1_k\rangle$

　 $\displaystyle g_k\sigma^+\hat a_k \mid\psi(t)\rangle = g_k\psi_k(t)\mid 1,{\rm vac}\rangle$

　 $\displaystyle g_k^*\hat a_k^\dagger\sigma^- \mid\psi(t)\rangle = g_k^*c_1(t)\mid 0,1_k\rangle$

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

　 $\displaystyle i\dot c_1(t) = \Delta c_1(t) + \sum_k g_k\psi_k(t)$

　 $\displaystyle i\dot \psi_k(t) = \omega_k\psi_k(t) + g_k^*c_1(t)$

7.2

From (5.61), we have a Langevin equation for the qubit's amplitude.

　 $\displaystyle \frac{d}{dt}c_1(t) = -i\Delta c_1(t) - i\hat\xi(t)-\frac{1}{2\pi}\int_0^t\int J(\omega)e^{-i\omega(t-\tau)}c_1(\tau)\,d\omega\,d\tau$

We ignore the input photons $\hat\xi(t)$ , and assume $J(\omega)=2\pi\alpha\,(\omega\in \mathbb{R})$ , then,

　 $\displaystyle \frac{1}{2\pi}\int_0^t\int J(\omega)e^{-i\omega(t-\tau)}c_1(\tau)\,d\omega\,d\tau = \alpha \int_0^t\int_{-\infty}^\infty e^{-i\omega(t-\tau)}c_1(\tau)\,d\omega\,d\tau$

　　 $\displaystyle = 2\pi\alpha\int_0^t\delta(t-\tau)c_1(\tau)\,d\tau = 2\pi\alpha c_1(t)$

So, the decay rate is $2\pi\alpha$ and no Lamb shift.

7.3

　 $\displaystyle \rho_{\rm qb} = \begin{pmatrix}\rho_{11} & \rho_{10} \\ \rho_{01} & \rho_{00}\end{pmatrix} = \begin{pmatrix} |\beta|^2e^{-\gamma t} & \alpha\beta^*e^{-i\Delta t-\frac{\gamma}{2}t} \\ \alpha^*\beta e^{i\Delta t-\frac{\gamma}{2}t} & 1 + (|\alpha|^2 -1)e^{-\gamma t}\end{pmatrix}$

　　 $\displaystyle = \begin{pmatrix} |\beta|^2e^{-\gamma t} & \alpha\beta^*e^{-i\Delta t-\frac{\gamma}{2}t} \\ \alpha^*\beta e^{i\Delta t-\frac{\gamma}{2}t} & 1 -|\beta|^2 e^{-\gamma t}\end{pmatrix}\ \left(\because |\alpha|^2+|\beta|^2 = 1\right)$

Eigenvalues are $\displaystyle \lambda_{\pm} = \frac{1\pm\sqrt{1-4|\beta|^2e^{-\gamma t}(1-e^{-\gamma t})}}{2}$

At $t=0$ and $t \to \infty$ , we have $\lambda_{\pm} = 0, 1$ . So $S=-(\lambda_-\log\lambda_- + \lambda_+\log\lambda_+)=0$ .

Defining $\displaystyle f(t) = e^{-\gamma t}(1-e^{-\gamma t})$ --- (1)

we have:

　 $\displaystyle \lambda_{\pm} = \frac{1\pm\sqrt{1-4|\beta|^2f(t)}}{2}$

And,

　 $\displaystyle\frac{dS}{dt} = \frac{\partial S}{\partial \lambda_-}\frac{d\lambda_-}{df(t)}\frac{df(t)}{dt} + \frac{\partial S}{\partial \lambda_+}\frac{d\lambda_+}{df(t)}\frac{df(t)}{dt}$

　　 $\displaystyle = \log\left(\frac{\lambda_+}{\lambda_-}\right)\frac{d\lambda_-}{df(t)}\frac{df(t)}{dt}\ \left(\because\,\frac{d\lambda_+}{df(t)} = -\frac{d\lambda_-}{df(t)}\right)$

Hence, $S$ becomes maximum at $\displaystyle \frac{df(t)}{dt} = 0$ .

From (1), $f(t)$ takes maximum $\displaystyle f(t)=\frac{1}{4}$ when $\displaystyle e^{-\gamma t} = \frac{1}{2}$ .

So, $S$ takes maximum at $\displaystyle \lambda_\pm = \frac{1\pm\sqrt{1-|\beta|^2}}{2}$ .

　 $\displaystyle \therefore\, \max S = -\frac{1-\sqrt{1-|\beta|^2}}{2}\log \frac{1-\sqrt{1-|\beta|^2}}{2}$

　　　　　　　　　　 $\displaystyle {}- \frac{1+\sqrt{1-|\beta|^2}}{2}\log \frac{1+\sqrt{1-|\beta|^2}}{2}$

gist.github.com

7.4

???

7.5

I assume that $g_k\sim gdk^{1/2}$ is a typo of $g_k\sim gd^{-1/2}$ .

(1) When photons propagate both directions, we can use the relationship:

　 $\displaystyle \frac{2\pi}{d}\sum_k \sim \int_{-\infty}^\infty dk = 2\int_0^\infty \left|\frac{dk}{d\omega}\right|d\omega$ .

(cf. Detailed derivation of the Input-Output theory)

On the other hand, if photons propagate only one direction, the relationship above becomes:

　 $\displaystyle \frac{2\pi}{d}\sum_k \sim \int_{0}^\infty dk = \int_0^\infty \left|\frac{dk}{d\omega}\right|d\omega$ .

So,

　 $\displaystyle J^{\rm QO}(\omega) = 2\pi\sum_k|g_k|^2\delta(\omega-\omega_k) = \frac{|g|^2}{v}\int_0^\infty\delta(\omega-\omega_k)\,d\omega_k$

　 $\displaystyle \therefore\,\gamma = \frac{|g|^2}{v}$

(2) From (1), $\displaystyle g_k = \frac{g}{\sqrt{d}} = \sqrt{\frac{\gamma v}{d}}$ . As the pre-factor of $c_1(t)$ in (7.20) came from $g_k$ , we have:

　 $\displaystyle\Psi_+^{\rm out}(0, t) = \Psi_+^{\rm in}(0,t)-i\sqrt{\frac{\gamma}{v}}c_1(t)$

(3) Since $\xi(x,t,t_0)$ contains only right-moving waves $k>0$ ,

　 $\displaystyle\xi (x,t,t_0) = \sum_{k>0} g_k e^{ikx-i\omega_k(t-t_0)}\psi_k(t_0)$

So from the definition of $\displaystyle\Psi_+(x,t,t_0)$ ,

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

with the assumption $\displaystyle g_k = \frac{g}{\sqrt{d}}$ , we have:

　 $\displaystyle\xi (x,t,t_0) = g\Psi_+(x,t,t_0) = \sqrt{\gamma v}\Psi_+(x,t,t_0)$

Hence, from (7.16):

　 $\displaystyle i\dot c_1(t) = \left(\Delta'-i\frac{1}{2}\gamma\right)c_1(t) + \xi(0,t,t_0)$ --- (7.16)

we have:

　 $\displaystyle i\dot c_1(t) = \left(\Delta'-i\frac{1}{2}\gamma\right)c_1(t) + \sqrt{\gamma v}\Psi_+(0,t,t_0)$

Defining $f(t)$ as $\displaystyle c_1(t) = f(t)e^{-i\left(\Delta'-i\frac{1}{2}\gamma\right)t}$ , we have:

　 $\displaystyle i\,\dot f(t)e^{-i\left(\Delta'-i\frac{1}{2}\gamma\right)t} = \sqrt{\gamma v}\Psi_+(0,t,t_0)$

　 $\displaystyle\therefore f(t) = -i\sqrt{\gamma v}\int_{t_0}^t\Psi^{\rm in}_+(0,\tau)e^{i\left(\Delta'-i\frac{1}{2}\gamma\right)\tau}\,d\tau$

　 $\displaystyle\therefore c_1(t) = -i\sqrt{\gamma v}\int_{t_0}^t\Psi^{\rm in}_+(0,\tau)e^{i\left(\Delta'-i\frac{1}{2}\gamma\right)(\tau-t)}\,d\tau$

For $\displaystyle \Psi_+^{\rm in}(0, t) = a_+e^{(-i\omega + 0^+)t}$ , we have:

　 $\displaystyle c_1(t) = -i\sqrt{\gamma v}a_+\int_{t_0}^te^{(-i\omega + 0^+)\tau}e^{i\left(\Delta'-i\frac{1}{2}\gamma\right)(\tau-t)}\,d\tau$

　　 $\displaystyle = -i\sqrt{\gamma v}a_+e^{(-i\omega+0^+)t}\frac{1}{i(\Delta'-\omega)+\frac{\gamma}{2}}$

Hence,

　 $\displaystyle\Psi_+^{\rm out}(0, t) = \Psi_+^{\rm in}(0,t)-i\sqrt{\frac{\gamma}{v}}c_1(t)$

　　 $\displaystyle = a_+e^{(-i\omega + 0^+)t} -\gamma a_+e^{(-i\omega + 0^+)t}\frac{1}{i(\Delta'-\omega)+\frac{\gamma}{2}}$

　　 $\displaystyle = a_+e^{(-i\omega + 0^+)t}\left\{1-\gamma \frac{1}{i(\Delta'-\omega)+\frac{\gamma}{2}}\right\}$

　 $\displaystyle\therefore\, T_\omega = 1-\gamma \frac{1}{i(\Delta'-\omega)+\frac{\gamma}{2}},\,R_\omega = 0$

(4) When a qubit is placed at the end of a semi-infinite transmission line, we have only reflected photons. If we virtually (or mathematically) invert the direction of the reflected photons, it's the same as the chiral model discussed here. Note that we suppose that input photons are reflected not only by a qubit but also the end of the transmission line.

7.6

(1) $\displaystyle H =\sum_{i=1}^N\Delta_i\mid i\rangle\langle i \mid + \sum_k\omega_k\hat a_k^\dagger\hat a_k + \sum_{i=1}^N\sum_k\left(\beta_ig_k\sigma^+_i\hat a_k + \beta^*_ig_k^*\sigma^-_i\hat a^\dagger\right)$

　 $\displaystyle \mid\psi(t)\rangle = \sum_ic_i(t)\mid i,{\rm vac}\rangle + \sum_k\psi_k(t)\mid0,1_k\rangle$

　 $\displaystyle H\mid\psi(t)\rangle = \sum_i\Delta_ic_i(t)\mid i,{\rm vac}\rangle + \sum_k\omega_k\psi_k(t)\mid 0,1_k\rangle$
　　　　　　　　 $\displaystyle {}+\sum_{i,k}\left\{\beta_ig_k\psi_k(t)\mid i,{\rm vac}\rangle + \beta^*_ig_k^*c_i(t)\mid 0,1_k\rangle\right\}$

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

　 $\displaystyle i\dot c_i(t) = \Delta_ic_i(t) + \sum_k\beta_ig_k\psi_k(t)$ --- (1-1)

　 $\displaystyle i\dot \psi_k(t) = \omega_k\psi_k(t) + \sum_n\beta_i^*g_k^*c_i(t)$ --- (1-2)

(2) Defining $f(t)$ as $\displaystyle\psi_k(t) = e^{-i\omega_k(t-t_0)}\left\{\psi_k(t_0)+f(t)\right\}$ with a initial condition $f(t_0)=0$ , from(1-2), we have:

　 $\displaystyle\dot f(t) = -ie^{i\omega_k(t-t_0)}\sum_i\beta_i^*g_k^*c_i(t)$

　 $\displaystyle\therefore\, f(t) = -i\sum_i\beta_i^*g_k^*e^{-i\omega_kt_0}\int_{t_0}^te^{i\omega_k\tau}c_i(\tau)\,d\tau$

　 $\displaystyle\therefore\, \psi_k(t) = e^{-i\omega_k(t-t_0)}\psi_k(t_0)-i\sum_i\beta_i^*g_k^*\int_{t_0}^te^{-i\omega_k(t-\tau)}c_i(\tau)\,d\tau$ --- (2-1)

Substituting this result into (1-1), we have:

　 $\displaystyle\dot c_i(t) = -i\Delta_ic_i(t)-i\beta_i\xi(t) - |\beta_i|^2\int_{t_0}^tK(t-\tau)c_i(\tau)\,d\tau$

where

　 $\displaystyle \xi(t) = \sum_kg_ke^{-i\omega_k(t-t_0)}\psi_k(t_0)$

　 $\displaystyle K(t) = \sum_k|g_k|^2e^{-i\omega_kt}$

By applying the Markovian approximation, we have:

　 $\displaystyle\int_{t_0}^tK(t-\tau)c_i(\tau)\,d\tau = \frac{\gamma_i}{2} + i\delta\omega_i$

where $\displaystyle \gamma_i = J^{\rm CO}(\omega'_i),\,\delta\omega_i = \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega_i-\omega'_i|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'_i-\omega}\,d\omega$

　 $\displaystyle\therefore\,\dot c_i(t) = -\left\{i\left(\Delta_i + |\beta_i|^2\delta\omega_i\right)+ |\beta_i|^2\frac{\gamma_i}{2}\right\}c_i(t) - i\beta_i\xi(t)$

　　　　 $\displaystyle = -\left(i\Delta'_i+\frac{|\beta_i|^2\gamma_i}{2}\right)c_i(t)-i\beta_i\xi(t)$ --- (2-2)

(3) From (2-1), we have:

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

　　 $\displaystyle = \Psi_\pm(x,t,t_0) - i \sum_i\beta_i^*\sum_{k>0}\int_{t_0}^tg_k^*\frac{e^{\pm ikx-i\omega_k(t-\tau)}}{\sqrt{d}}c_i(\tau)\,d\tau$

From the assumption $\displaystyle g_k = \frac{g}{\sqrt{d}}$ , based on the same argument as (7.18), we have $\displaystyle \gamma_i = \frac{2|g|^2}{v} := \gamma$ . Then, by following the same argument (7.19) and (7.20), we have:

　 $\displaystyle\Psi_\pm^{\rm out}(0, t) = \Psi_\pm^{\rm in}(0,t) - i\sqrt{\frac{\gamma}{2v}}\sum_i\beta_i^*c_i(t)$ --- (3-1)

(4) From the assumption $\displaystyle g_k=\frac{g}{\sqrt{d}} = \sqrt{\frac{\gamma v}{2d}}$ , we have:

　 $\displaystyle\xi(t) = \sqrt{\frac{\gamma v}{2}}\left\{\Psi_+(t,t_0)+\Psi_-(t,t_0)\right\}$

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

　 $\displaystyle\dot c_i(t) = -\left(i\Delta'_i+\frac{|\beta_i|^2\gamma_i}{2}\right)c_i(t)-i\beta_i\sqrt{\frac{\gamma v}{2}}\left\{\Psi_+(t,t_0)+\Psi_-(t,t_0)\right\}$ --- (4-1)

Defining $f(t)$ as $\displaystyle c_i(t) = e^{-\left(i\Delta'_i+\frac{|\beta_i|^2\gamma}{2}\right)(t-t_0)}f(t)$ with an initial condition $f(t_0)=0$ , from (4-1), we have a solution:

　 $\displaystyle c_i(t) = -i\beta_i\sqrt{\frac{\gamma v}{2}}\int_{t_0}^t\left\{\Psi_+(0,\tau)+\Psi_-(0,\tau)\right\}e^{-\left(\Delta'_i+\frac{|\beta_i|^2\gamma}{2}\right)(t-\tau)}\,d\tau$

With input photons $\displaystyle\Psi_\pm^{\rm in}(0,t) = a_\pm e^{(-i\omega+0^+)t}$ , by taking $t_0\to-\infty$ , we have:

　 $\displaystyle c_i(t) = -i\beta_i\sqrt{\frac{\gamma v}{2}}\frac{a_+ + a_-}{i(\Delta'_i-\omega)+\frac{|\beta_i|^2\gamma}{2}}e^{(-i\omega+0^+)t}$

Hence, from(3-1), we have:

　 $\displaystyle \Psi_\pm^{\rm out}(0, t) = b_\pm e^{(-i\omega+0^+)t} = e^{(-i\omega+0^+)t}\left[a_\pm-\frac{\gamma}{2}\sum_i|\beta_i|^2\frac{a_+ + a_-}{i(\Delta'_i-\omega)+\frac{|\beta_i|^2\gamma}{2}}\right]$

　 $\displaystyle \therefore\,T_\omega = 1 + R_\omega,\,R_\omega = -\frac{\gamma}{2}\sum_i\frac{|\beta_i|^2}{i(\Delta'_i-\omega)+\frac{|\beta_i|^2\gamma}{2}}$

Hence the spectrum has multiple peaks at each $\omega\sim \Delta'_i$ with height and width $\propto |\beta_i|^2$ .

7.7

　 $\displaystyle H=\Delta\mid 1\rangle\langle 1\mid + \sum_k\omega_k\hat a_k^\dagger\hat a_k+\sum_k\left(g_{a,k}\mid 1\rangle\langle a\mid \hat a_k + g_{b,k}\mid 1\rangle\langle b\mid \hat a_k +\,{\rm h.c.}\right)$

　　 $\displaystyle = \Delta\mid 1\rangle\langle 1\mid + \sum_k\omega_k\hat a_k^\dagger\hat a_k+\sum_k\left(\sqrt{|g_{a,k}|^2+|g_{b,k}|^2}\mid 1\rangle\langle 0_k\mid \hat a_k +\,{\rm h.c.}\right)$

where $\displaystyle \mid 0_k\rangle := \frac{1}{\sqrt{|g_{a,k}|^2+|g_{b,k}|^2}}\left(g_{a,k}\mid a\rangle + g_{b,k}\mid b\rangle\right)$

Around the resonance frequency $\omega_k \sim \Delta'$ , suppose that the couplings $g_{a,k},\,g_{b,k}$ are almost constant $g_{a,k}=g_a,\, g_{b,K}=g_b$ , the system is effectively described with:

　 $\displaystyle H_{\rm eff} = \Delta\mid 1\rangle\langle 1\mid + \sum_k\omega_k\hat a_k^\dagger\hat a_k+\sum_k\left(\sqrt{|g_{a}|^2+|g_{b}|^2}\mid 1\rangle\langle 0\mid \hat a_k +\,{\rm h.c.}\right)$

where $\displaystyle \mid 0\rangle := \frac{1}{\sqrt{|g_{a}|^2+|g_{b}|^2}}\left(g_{a}\mid a\rangle + g_{b}\mid b\rangle\right)$

So the system has effectively a single ground state $\mid 0\rangle$ , and the state orthogonal to $\mid 0\rangle$ becomes a dark state.

7.8

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7.9

???

7.10

　 $\displaystyle U \simeq \mid g\rangle\langle g\mid\otimes D(\alpha_-) + \mid e\rangle\langle e\mid \otimes D(\alpha_+)$

　 $\displaystyle U\rho_{\rm qubit}\otimes\mid{\rm vac}\rangle\langle{\rm vac}\mid U^\dagger$

　　 $\displaystyle =\, \mid g\rangle\langle g\mid \rho_{\rm qubit}\mid g\rangle\langle g\mid \otimes \mid\alpha_-\rangle\langle \alpha_-\mid + \mid e\rangle\langle e\mid\rho_{\rm qubit}\mid e\rangle\langle e\mid \otimes \mid\alpha_+\rangle\langle \alpha_+\mid$

　　　 $\displaystyle +\, \mid g\rangle\langle g\mid \rho_{\rm qubit}\mid e\rangle\langle e\mid \otimes \mid\alpha_-\rangle\langle \alpha_+\mid + \mid e\rangle\langle e\mid\rho_{\rm qubit}\mid g\rangle\langle g\mid \otimes \mid\alpha_+\rangle\langle \alpha_-\mid$

　 $\displaystyle \therefore\,\epsilon_{+1}(\rho_{\rm qubit}) = {\rm Tr_{cavity}}\left(\prod_{+1}\mid d\rangle\langle g\otimes D(\alpha_-) + \mid e\rangle\langle e\mid \otimes D(\alpha_+)\right)$

　　　 $\displaystyle = \,\mid g\rangle\langle g\mid \rho_{\rm qubit}\mid g\rangle\langle g\mid {\rm Tr_{cavity}}\left(\prod_{+1}\mid\alpha_-\rangle\langle \alpha_-\mid\right)$

　　　　 $\displaystyle + \,\mid e\rangle\langle e\mid \rho_{\rm qubit}\mid e\rangle\langle e\mid {\rm Tr_{cavity}}\left(\prod_{+1}\mid\alpha_+\rangle\langle \alpha_+\mid\right)$

　　　　 $\displaystyle + \,\mid g\rangle\langle g\mid \rho_{\rm qubit}\mid e\rangle\langle e\mid {\rm Tr_{cavity}}\left(\prod_{+1}\mid\alpha_-\rangle\langle \alpha_+\mid\right)$

　　　　 $\displaystyle + \,\mid e\rangle\langle e \mid \rho_{\rm qubit}\mid g\rangle\langle g\mid {\rm Tr_{cavity}}\left(\prod_{+1}\mid\alpha_+\rangle\langle \alpha_-\mid\right)$

　　　 $\displaystyle = \,\mid g\rangle\langle g\mid \rho_{\rm qubit}\mid g\rangle\langle g\mid M_n + \mid e\rangle\langle e\mid \rho_{\rm qubit}\mid e\rangle\langle e\mid (1-M_n)$

　　　　 $\displaystyle + \,\mid g\rangle\langle g\mid \rho_{\rm qubit}\mid e\rangle\langle e\mid {\rm Tr_{cavity}}\left(\prod_{+1}\mid\alpha_-\rangle\langle \alpha_+\mid\right)$

　　　　 $\displaystyle + \,\mid e\rangle\langle e \mid \rho_{\rm qubit}\mid g\rangle\langle g\mid {\rm Tr_{cavity}}\left(\prod_{+1}\mid\alpha_+\rangle\langle \alpha_-\mid\right)$

　 $\displaystyle \therefore\, p_{+1} = {\rm Tr}(\epsilon_{+1}(\rho_{\rm qubit}))$

　　 $\displaystyle = M_n \langle g\mid\rho_{\rm qubit}\mid g\rangle + (1-M_n) \langle e\mid\rho_{\rm qubit}\mid e\rangle$

　　 $\displaystyle = M_n {\rm Tr}(\rho_{\rm qubit}\mid g\rangle\langle g\mid) + (1-M_n) {\rm Tr}(\rho_{\rm qubit}\mid e\rangle\langle e\mid)$

Likewise,

　 $\displaystyle p_{-1} = {\rm Tr}(\epsilon_{-1}(\rho_{\rm qubit}))$

　　 $\displaystyle = (1-M_n) {\rm Tr}(\rho_{\rm qubit}\mid g\rangle\langle g\mid) + M_n {\rm Tr}(\rho_{\rm qubit}\mid e\rangle\langle e\mid)$

2022-10-21

Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 6) : Part2

6.10

(1) $\displaystyle H = (2\Delta + \alpha)\mid 2\rangle\langle 2\mid + \Delta \mid 1 \rangle\langle 1 \mid$

　　 $\displaystyle {} + \epsilon_0\cos\omega t\left(\mid 2\rangle\langle 1\mid + \mid 1\rangle\langle 0\mid + \mid 1\rangle\langle 2\mid + \mid 0\rangle\langle 1\mid\right)$

Define $H_0$ as:

　 $\displaystyle H_0 = 2\omega \mid 2 \rangle\langle 2 \mid + \omega \mid 1 \rangle\langle 1 \mid$

Then,

　 $\displaystyle H = H_0 + V_1 + V_2$

　 $\displaystyle V_1 = (2\delta + \alpha) \mid 2 \rangle\langle 2 \mid+\delta \mid 1 \rangle\langle 1 \mid$

　 $\displaystyle V_2 = \epsilon_0\cos\omega t\left(\mid 2\rangle\langle 1\mid + \mid 1\rangle\langle 0\mid + \mid 1\rangle\langle 2\mid + \mid 0\rangle\langle 1\mid\right)$

So, the interaction Hamiltonian is:

　 $\displaystyle H_I(t) = e^{iH_0t}(V_1 + V_2)e^{-iH_0t} = V_1 + e^{iH_0t}V_2e^{-iH_0t}$

Since:

　 $\displaystyle [iH_0t, \mid 2\rangle\langle 1\mid ] = i\omega t \mid 2\rangle\langle 1\mid$

　 $\displaystyle [iH_0t, \mid 1\rangle\langle 0\mid ] = i\omega t \mid 1\rangle\langle 0\mid$

We have:

　 $\displaystyle e^{iH_0t}V_2e^{-iH_0t} = \epsilon_0\cos\omega t\left(e^{i\omega t}\mid 2\rangle\langle 1\mid+e^{i\omega t}\mid 1\rangle\langle 0\mid+ {\rm h.c.}\right)$

By dropping the non-RWA terms such as $\displaystyle e^{\pm 2i\omega t}$ , we have:

　 $\displaystyle H_I(t) = (2\delta + \alpha) \mid 2 \rangle\langle 2 \mid+\delta \mid 1 \rangle\langle 1 \mid+ \frac{\epsilon_0}{2}\left(\mid 2\rangle\langle 1\mid + \mid 1 \rangle\langle 0\mid + {\rm h.c.}\right)$

(2) By dropping the third state, we have:

　 $\displaystyle H_I = \delta\mid 1\rangle\langle 1\mid + \frac{\epsilon_0}{2}\left(\mid 1 \rangle\langle 0\mid + \mid0\rangle\langle 1\mid\right)$

For $\displaystyle \mid\Psi(t)\rangle = c_0(t)\mid 0\rangle + c_1(t)\mid 1\rangle$

　 $\displaystyle H_I\mid\Psi(t)\rangle = c_0(t)\frac{\epsilon_0}{2} +c_1(t)\delta\mid 1 \rangle + c_1(t)\frac{\epsilon_0}{2}\mid 0\rangle$

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

　 $\displaystyle i\dot c_0(t) = \frac{\epsilon_0}{2}c_1(t)$ --- (1)

　 $\displaystyle i\dot c_1(t) = \frac{\epsilon_0}{2}c_0(t) + \delta c_1(t)$ --- (2)

By combining them, we have:

　 $\displaystyle\ddot c_1(t) + i\delta\dot c_1(t) + \left(\frac{\epsilon_0}{2}\right)^2 c_1(t) = 0$

　 $\therefore c_1(t) = e^{-i\frac{\delta}{2}t}\left(A e^{i\frac{D}{2}t} + B e^{-i\frac{D}{2}t}\right)$ , where $\displaystyle D=\sqrt{\delta^2+\epsilon^2}$

From the initial condition $c_1(0) = 0$ , $A+B=0$ , so,

　 $\displaystyle c_1(t) = Ae^{-i\frac{\delta}{2}t}\left(e^{i\frac{D}{2}t} - e^{-i\frac{D}{2}t}\right)= 2Aie^{-i\frac{\delta}{2}t}\sin\frac{D}{2}t$

From (2), $\displaystyle c_0(t) = \frac{2i}{\epsilon_0}\dot c_1(t) -\frac{2}{\epsilon_0}\delta c_1(t)$

The initial condition for $c_0(t)$ is:

　 $\displaystyle c_0(0) = \frac{2i}{\epsilon_0}2Ai\frac{D}{2} = 1$

　 $\displaystyle \therefore A = -\frac{\epsilon_0}{2D}$

　 $\displaystyle\therefore c_1(t) = -\frac{\epsilon_0i}{\sqrt{\delta^2+\epsilon_0^2}}e^{-\frac{\delta}{2}t}\sin\left(\frac{1}{2}\sqrt{\delta^2 + \epsilon_0^2}t\right)$

So, if $\delta = 0$ , $|c_1(t)|^2 = 1$ at $\displaystyle \frac{\epsilon_0}{2}T_{\rm flip} = \frac{\pi}{2}$ .

　 $\displaystyle \therefore T_{\rm flip} = \frac{\pi}{\epsilon_0}$

(3) The interaction Hamiltonian is:

　 $\displaystyle H_I = \alpha \mid 2 \rangle\langle 2 \mid + \frac{\epsilon_0}{2}\left(\mid 2\rangle\langle 1\mid + \mid 1 \rangle\langle 0\mid + {\rm h.c.}\right)$

In order to diagonalize the qubit space, introduce the states $\mid \pm\rangle$ as:

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

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

Since $\displaystyle \mid 1\rangle \langle 0\mid +\, {\rm h.c.} = \mid +\rangle\langle +\mid - \mid -\rangle\langle -\mid$ ,

　 $\displaystyle H_I = \alpha \mid 2 \rangle\langle 2 \mid + \frac{\epsilon_0}{2}\mid +\rangle\langle +\mid - \frac{\epsilon_0}{2}\mid -\rangle\langle -\mid$

　　　　　　　　 $\displaystyle {} + \frac{\epsilon_0}{2\sqrt{2}}\left(\mid 2\rangle\langle + \mid - \mid 2\rangle\langle - \mid +\, {\rm h.c.}\right)$

　　 $\displaystyle = H_0 + V$

where,

　 $\displaystyle H_0 = \alpha \mid 2 \rangle\langle 2 \mid + \frac{\epsilon_0}{2}\mid +\rangle\langle +\mid - \frac{\epsilon_0}{2}\mid -\rangle\langle -\mid$

　 $\displaystyle V = \frac{\epsilon_0}{2\sqrt{2}}\left(\mid 2\rangle\langle + \mid - \mid 2\rangle\langle - \mid +\, {\rm h.c.}\right)$

We will apply an additional transformation with $H_0$ to a new interaction picture, then, the new interaction Hamiltonian is:

　 $\displaystyle V_I(t) = e^{iH_0t}Ve^{-iH_0t}$

　　 $\displaystyle = \frac{\epsilon_0}{2\sqrt{2}}\left\{ e^{i\left(\alpha-\frac{\epsilon_0}{2}\right)t}\mid 2\rangle\langle + \mid - e^{i\left(\alpha+\frac{\epsilon_0}{2}t\right)}\mid 2\rangle\langle - \mid +\, {\rm h.c.}\right\}$

　 $\displaystyle\because\, [iH_0t,\mid 2\rangle\langle + \mid] = it\left(\alpha-\frac{\epsilon_0}{2}\right)\mid 2\rangle\langle +\mid$

　　 $\displaystyle[iH_0t,\mid 2\rangle\langle - \mid] = it\left(\alpha+\frac{\epsilon_0}{2}\right)\mid 2\rangle\langle -\mid$

Since $\epsilon_0 << \alpha$ ,

　 $\displaystyle V_I(t) \simeq \frac{\epsilon_0}{2\sqrt{2}}\left\{e^{i\alpha t}\left(\mid 2\rangle\langle + \mid - \mid 2\rangle\langle - \mid\right) + e^{-i\alpha t}\left(\mid +\rangle\langle 2 \mid - \mid -\rangle\langle 2 \mid\right) \right\}$

For $\displaystyle \mid\Psi(t)\rangle = c_2(t)\mid 2\rangle + c_+(t)\mid+\rangle + c_-(t)\mid -\rangle$

　 $\displaystyle V_I(t)\mid\Psi(t)\rangle = \frac{\epsilon_0}{2\sqrt{2}}\left\{ c_2(t)e^{-i\alpha t}\mid +\rangle - c_2(t)e^{-i\alpha t}\mid -\rangle + c_+(t)e^{i\alpha t}\mid 2\rangle -c_-(t)e^{i\alpha t}\mid 2\rangle\right\}$

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

　 $\displaystyle i\dot c_+(t) = \frac{\epsilon_0}{2\sqrt{2}}e^{-i\alpha t}c_2(t)$ --- (1)

　 $\displaystyle i\dot c_-(t) = -\frac{\epsilon_0}{2\sqrt{2}}e^{-i\alpha t}c_2(t)$ --- (2)

　 $\displaystyle i\dot c_2(t) =\frac{\epsilon_0}{2\sqrt{2}}e^{i\alpha t}\left\{c_+(t)-c_-(t)\right\}$ --- (3)

From (3),

　 $\displaystyle \frac{\epsilon_0}{2\sqrt{2}}\left\{c_+(t)-c_-(t)\right\} = ie^{-i\alpha t}\dot c_2(t)$

　 $\displaystyle \therefore\,\frac{\epsilon_0}{2\sqrt{2}}\left\{\dot c_+(t)-\dot c_-(t)\right\} = \alpha e^{-i\alpha t}\dot c_2(t) + i e^{-i\alpha t}\ddot c_2(t)$

By substituting (1), (2), we have:

　 $\displaystyle \ddot c_2(t) -i\alpha \dot c_2(t) + \frac{\epsilon^2_0}{4}c_2(t) = 0$

　 $\displaystyle c_2(t) = e^{i\frac{\alpha}{2}t}\left(Ae^{i\frac{D}{2}t}+B e^{-i\frac{D}{2}t}\right)$ , where $\displaystyle D = \sqrt{\alpha^2+\epsilon^2}\simeq \alpha$

From the initial condition $c_2(0) = 0$ , $A+B=0$ , so,

　 $\displaystyle c_2(t) = 2iAe^{i\frac{\alpha}{2}t}\sin\frac{\alpha}{2}t$

Then $\displaystyle \dot c_2(0) = iA\alpha$

From the initial condition $c_+(0) - c_-(0) = \pm 1$ (where we suppose that $\displaystyle \mid\Psi(0)\rangle =\, \mid \pm \rangle$ ) and (3),

　 $\displaystyle -A\alpha = \pm \frac{\epsilon_0}{2\sqrt{2}}$

　 $\displaystyle \therefore A = \pm\frac{\epsilon_0}{2\sqrt{2}\alpha}$

　 $\displaystyle \therefore c_2(t) = \pm i \frac{\epsilon_0}{\sqrt{2}\alpha}e^{i\frac{\alpha}{2}t}\sin\frac{\alpha}{2}t$

Hence, the excitation probability is:

　 $\displaystyle \overline{|c_2(t)|^2} = \frac{\epsilon_0^2}{4\alpha^2}$

When $\alpha = 0.05\Delta$ , to suppress the leakage below 1%,

　 $\displaystyle\frac{\epsilon_0^2}{4(0.05\Delta)^2} \le 0.01$

　 $\therefore \epsilon_0 \le 0.01\Delta$

(4) $\displaystyle\alpha = 2\pi\times 400\times 10^9 = 8\pi\times 10^{11}$

Suppose that $\displaystyle T_{\rm flip} = \frac{\pi}{\epsilon_0} = 10 \times 10^{-9} = 10^{-8}$

The error rate is $\displaystyle \frac{\epsilon_0^2}{4\alpha^2} = \frac{\pi^2}{4\alpha^2 T_{\rm flip}^2} \sim 10^{-8}$

To suppress the error rate $\displaystyle \frac{\epsilon_0^2}{4\alpha^2} \sim 0.01$ ,

　 $\displaystyle \epsilon_0 \sim 2\alpha\times 10^{-1} = 16\pi\times 10^{10}$

　 $\displaystyle T_{\rm flip} = \frac{\pi}{\epsilon_0} \sim 10^{-11}$

6.11

www.wolframcloud.com

6.12

gist.github.com

6.13

???

6.14

Since the current rotates in the opposite directions in two loops, the external flux $\Phi$ in the potential is replaced by $\Phi_{\rm up}-\Phi_{\rm down}$ .

6.15

$\varphi_{L,R}$ is (classically) determined by (6.52):

　 $\displaystyle \frac{L_J}{L}\varphi = \sin(\varphi)$

When $L_J \ll L$

　 $\varphi_{R} = \pi - \epsilon$ , and $\sin(\varphi)=\sin(\pi-\epsilon) \simeq \epsilon$

　 $\displaystyle \therefore \frac{L_J}{L}(\pi-\epsilon) = \epsilon$

　 $\displaystyle \therefore \epsilon = \frac{L_J}{L+L_J}\pi \simeq \frac{L_J}{L}\pi$

　 $\displaystyle\therefore \varphi_R = \pi - \frac{L_J}{L}\pi$

Likewise,

　 $\displaystyle\therefore \varphi_L = -\pi + \frac{L_J}{L}\pi$

For the inductor $L$ , $\displaystyle \hat I = \frac{\hat\phi}{L} = \frac{\varphi_0}{L}\hat\varphi$

Suppose that the wavefunction of $\mid L\rangle$ and $\mid R\rangle$ is approximately Gaussian:

　 $\displaystyle\phi_{L,R}(\varphi) = \frac{1}{\sqrt{Z}}\exp\left\{-\frac{1}{2\sigma^2_\varphi}\left(\varphi - \varphi_{L,R}\right)^2\right\}$

　 $\displaystyle\langle L\mid \hat I\mid R\rangle \simeq \frac{\varphi_0}{Z L} \int_{-\infty}^{\infty} \exp\left\{-\frac{1}{2\sigma^2_\varphi}\left(\varphi - \varphi_{L}\right)^2\right\}\varphi\exp\left\{-\frac{1}{2\sigma^2_\varphi}\left(\varphi - \varphi_{R}\right)^2\right\}\,d\varphi$

Since $\varphi_{L} = -\varphi_{R}$ , the integrand is an odd function.

　 $\displaystyle \therefore\langle L\mid \hat I\mid R\rangle \simeq 0$

The diagonal elements are:

　 $\displaystyle\langle L\mid \hat I\mid L\rangle \simeq \frac{\varphi_0}{Z L} \int_{-\infty}^{\infty} \exp\left\{-\frac{1}{\sigma^2_\varphi}\left(\varphi - \varphi_{L}\right)^2\right\}\varphi\,d\varphi$

　　 $\displaystyle = \frac{\varphi_0\varphi_L}{Z L} \int_{-\infty}^{\infty} \exp\left(-\frac{\varphi'^2}{\sigma^2_\varphi}\right)\,d\varphi' = \frac{\varphi_0\varphi_L}{L} = -\frac{\Phi_0}{2L}\left(1-\frac{L_J}{L}\right) = \tilde I_C$

Likewise,

　 $\displaystyle\langle R\mid \hat I\mid R\rangle \simeq \tilde I_C$

Note that $\displaystyle\langle R\mid \hat I\mid R\rangle = - \langle L\mid \hat I\mid L\rangle$

Using the new basis:

　 $\displaystyle \mid 0\rangle = \frac{1}{\sqrt{2}}\left(\mid L\rangle - \mid R\rangle\right)$

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

we have:

　 $\displaystyle\langle 0 \mid \hat I \mid 0 \rangle = \langle 1 \mid \hat I \mid 1 \rangle = 0$

　 $\displaystyle\langle 0 \mid \hat I \mid 1 \rangle = \langle L \mid \hat I \mid L \rangle = \tilde I_C$

　 $\displaystyle \therefore \hat I = \tilde I_C\sigma^x$

6.16

(1) $\displaystyle H = \frac{\Delta}{2}\sigma^z_1 + \frac{\Delta}{2}\sigma^z_2 + J\sigma_1^x\sigma_2^x = H_0 + V$

where,

　 $\displaystyle H_0 = \frac{\Delta}{2}\sigma^z_1 + \frac{\Delta}{2}\sigma^z_2$

　 $\displaystyle V = J\sigma_1^x\sigma_2^x = J(\sigma^+_1 + \sigma^-_1)(\sigma^+_2 + \sigma^-_2) =\sigma_1^+\sigma_2^+ + \sigma_1^-\sigma_2^- + \sigma_1^+\sigma_2^- + \sigma_1^-\sigma_2^+$

Since $\displaystyle [\sigma_z,\sigma^{\pm}] = \pm 2\sigma^{\pm}$ ,

　 $[H_0, \sigma_1^+\sigma_2^+] = 2\Delta \sigma_1^+\sigma_2^+,\,[H_0, \sigma_1^-\sigma_2^-] = -2\Delta \sigma_1^-\sigma_2^-$

　 $[H_0, \sigma_1^+\sigma_2^-] = 0,\,[H_0, \sigma_1^-\sigma_2^+] = 0$

Hence, the interaction Hamiltonian is:

　 $\displaystyle H_I = e^{iH_0t}Ve^{-iH_0t} = J\left(e^{2i\Delta t}\sigma_1^+\sigma_2^+ + e^{-2i\Delta t}\sigma_1^-\sigma_2^- + \sigma^+_1\sigma^-_2 + \sigma^-_1\sigma^+_2\right)$

The first two terms oscillates two times faster than the energy gap $\Delta$ , so as an approximation (of a weak interaction), we can drop these terms. Hence, getting back to the original Schroedinger picture, we have:

　 $\displaystyle H_{\rm RWA} = \frac{\Delta}{2}\sigma^z_1 + \frac{\Delta}{2}\sigma^z_2 + J( \sigma^+_1\sigma^-_2 + \sigma^-_1\sigma^+_2)$

　Using $\{\mid 11\rangle, \mid 10\rangle, \mid 01\rangle, \mid 00\rangle\}$ as basis, the matrix representation becomes:

　 $\displaystyle H_{\rm RWA} = \frac{\Delta}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} + \frac{\Delta}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}\otimes \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$

　　　　　　　 $\displaystyle {} + J \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\otimes \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix} + J \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}\otimes \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$

　 $\displaystyle = \frac{\Delta}{2}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{pmatrix} + \frac{\Delta}{2}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{pmatrix}$

　　　　 $\displaystyle {} + J\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}+J\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}$

　 $\displaystyle =\begin{pmatrix} \Delta & 0 & 0 & 0 \\ 0 & 0 & J & 0 \\ 0 & J & 0 & 0 \\ 0 & 0 & 0 & -\Delta\end{pmatrix}$

By rearranging the order of basis as $\{\mid 11\rangle, \mid 00\rangle, \mid 10\rangle, \mid 01\rangle\}$ , we have:

　 $\displaystyle H_{\rm RWA} = \begin{pmatrix} \Delta & 0 & 0 & 0 \\ 0 & -\Delta & 0 & 0 \\ 0 & 0 & 0 & J \\ 0 & 0 & J & 0\end{pmatrix}$

So in the even space $\{\mid 11\rangle, \mid 00\rangle\}$ , $\mid 11\rangle$ and $\mid 00\rangle$ are the eigenvectors with eigenvalues $\Delta$ and $-\Delta$ . In the odd space $\{\mid 10\rangle, \mid 01\rangle\}$ , $\displaystyle\frac{1}{\sqrt{2}}\left(\mid 10\rangle \pm \mid 01\rangle\right)$ are the eigenvectors with eigenvalues $\pm J$ .

(2)(3) By applying the same matrix representation for the full Hamiltonian, we have:

　 $\displaystyle H = \begin{pmatrix} \Delta & J & 0 & 0 \\ J & -\Delta & 0 & 0 \\ 0 & 0 & 0 & J \\ 0 & 0 & J & 0\end{pmatrix}$

In the odd space, we have the same eigenvectors (with the same eigenvalues) as (1). In the even space, the eigenvalues are $\displaystyle\pm\sqrt{\Delta^2+J^2}$ .

The ground state with the eigenvalue $\displaystyle - \sqrt{\Delta^2+J^2}$ is:

　 $\displaystyle \mid \tilde{00}\rangle \propto \left(\Delta - \sqrt{\Delta^2 + J^2}\right)\mid 11\rangle + J\mid 00\rangle$

　　 $\displaystyle \propto \left\{ 1-\sqrt{1-\left(\frac{J}{\Delta}\right)^2}\right\} \mid 11\rangle + \frac{J}{\Delta}\mid 00\rangle$

　　 $\displaystyle\simeq \frac{1}{2}\left(\frac{J}{\Delta}\right)^2\mid 11\rangle + \frac{J}{\Delta}\mid 00\rangle \propto \frac{J}{2\Delta}\mid 11\rangle + \mid 00\rangle$

　 $\displaystyle \therefore\, \mid \tilde{00}\rangle \simeq \sqrt{\frac{1}{1+\frac{J^2}{4\Delta^2}}}\left(\frac{J}{2\Delta} \mid 11\rangle + \mid 00\rangle\right) = \frac{J}{2\Delta} \mid 11\rangle + \mid 00\rangle + O(\left(\frac{J}{\Delta}\right)^2)$

Likewise, the eigenstate with the eigenvalue $\displaystyle \sqrt{\Delta^2+J^2}$ is:

　 $\displaystyle \mid \tilde{11}\rangle \propto J\mid 11\rangle - \left(\Delta - \sqrt{\Delta^2 + J^2}\right)\mid 00\rangle$

　 $\displaystyle \therefore\, \mid \tilde{11}\rangle \simeq \mid 11\rangle - \frac{J}{2\Delta} \mid 00\rangle + O(\left(\frac{J}{\Delta}\right)^2)$

Hence, the probability of finding the exited state $\displaystyle \mid 11\rangle$ in $\displaystyle \mid \tilde{00}\rangle$ is:

　 $\displaystyle P = |\langle 11\mid\tilde{00}\rangle|^2 = \frac{J^2}{4\Delta^2}$

6.17

(1) Using $\{\mid 11\rangle, \mid 10\rangle, \mid 01\rangle, \mid 00\rangle\}$ as basis, the matrix representation of non-interactive term is:

　 $\displaystyle \frac{\Delta_1}{2}\sigma^z_1 + \frac{\Delta_2}{2}\sigma^z_2 = \frac{\Delta_1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} + \frac{\Delta_2}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}\otimes \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$

　　 $\displaystyle = \frac{\Delta_1}{2}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{pmatrix} + \frac{\Delta_2}{2}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{pmatrix}$

　　 $\displaystyle =\begin{pmatrix} \frac{1}{2}(\Delta_1+\Delta_2)& 0 & 0 & 0 \\ 0 & \frac{\delta}{2} & 0 & 0 \\ 0 & 0 & -\frac{\delta}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}(\Delta_1+\Delta_2)\end{pmatrix}$

The interaction term is the same as in Problem 6.16. So, by rearranging the order of basis as $\{\mid 11\rangle, \mid 00\rangle, \mid 10\rangle, \mid 01\rangle\}$ , the full Hamiltonian becomes:

　　 $\displaystyle H=\begin{pmatrix} \frac{1}{2}(\Delta_1+\Delta_2)& J & 0 & 0 \\ J & -\frac{1}{2}(\Delta_1+\Delta_2)& 0 & 0 \\ 0 & 0 & \frac{\delta}{2} & J \\ 0 & 0 & J & -\frac{\delta}{2} \end{pmatrix}$

(2) The exact eigenvalues of full Hamiltonian are:

　In the even space: $\displaystyle \pm\sqrt{\frac{1}{4}(\Delta_1+\Delta_2)^2+J^2}$

　In the odd space: $\displaystyle \pm\sqrt{\frac{1}{4}\delta^2+J^2} = \pm\frac{\delta}{2}\left\{1+2\left(\frac{J}{\delta}\right)^2\right\} + O\left(\left(\frac{J}{\delta}\right)^4\right)$

On the other hand, if we apply the second-order non-degenerate perturbation in the odd space, the eigenvalues without interaction is $\displaystyle E_{\pm} = \pm\frac{\delta}{2}$ , and the correction to the eigenvalue $E_\pm$ is:

　 $\displaystyle\frac{J^2}{2}\frac{2}{E_\pm - E_\mp} = \pm\frac{J^2}{\delta}$

Hence the approximate eigenvalues are:

　 $\displaystyle E'_\pm = \pm\frac{\delta}{2} \pm\frac{J^2}{\delta} = \pm\frac{\delta}{2}\left\{1+2\left(\frac{J}{\delta}\right)^2\right\}$

This coincides with the exact result except $O\left(\left(\frac{J}{\delta}\right)^4\right)$ . This approximation fails when $\delta \sim 0$ (due to the degeneracy.)

(3) Without interactions, the state $\mid 01\rangle$ stays in the same one:

　 $\displaystyle \mid\psi^{\rm eff}_{01}(t)\rangle = e^{-i\frac{\Delta_2}{2}t}\mid 01 \rangle$

For $\displaystyle \mid\psi^{\rm exact}_{01}(t)\rangle = c_{10}(t)\mid 10\rangle + c_{01}(t)\mid 01\rangle$ ,

　 $\displaystyle H \mid\psi^{\rm exact}_{01}(t)\rangle = \left(\frac{\delta}{2}c_{10}(t)+Jc_{01}(t)\right)\mid 10\rangle + \left(-\frac{\delta}{2}c_{01}(t)+Jc_{10}(t)\right)\mid 01\rangle$