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 $\displaystyle \lambda_\pm = \frac{1\pm\sqrt{1-|\beta|^2}}{2}$ .

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

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

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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

???

7.9

???

7.10

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

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

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

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

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

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

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

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

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

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

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

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

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