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

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