Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 6) : Part1

6.1

　 $\displaystyle \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\, \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\, \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$

Unitary operation for a rotation around $\alpha$ axis is:

　 $\displaystyle e^{-i\theta\frac{\sigma_\alpha}{2}} = I\cos\frac{\theta}{2}-i\sigma_\alpha\sin\frac{\theta}{2}$

An operator to swap x and y is:

　 $\displaystyle U = e^{-i\pi\frac{\sigma_y}{2}}e^{-i\frac{\pi}{2}\frac{\sigma_z}{2}}= \begin{pmatrix} 0 & \frac{-1-i}{\sqrt{2}} \\ \frac{1-i}{\sqrt{2}} & 0\end{pmatrix}$

　 $U\sigma_zU^\dagger=-\sigma_z,\, U\sigma_xU^\dagger=\sigma_y,\, U\sigma_yU^\dagger=\sigma_x$

An operator to swap x and z is:

　 $\displaystyle U = e^{-i\pi\frac{\sigma_x}{2}}e^{-i\frac{\pi}{2}\frac{\sigma_y}{2}}= \frac{i}{\sqrt{2}}\begin{pmatrix} -1 & -1 \\ -1 & 1\end{pmatrix}$

　 $U\sigma_zU^\dagger=\sigma_x,\, U\sigma_xU^\dagger=\sigma_z,\, U\sigma_yU^\dagger=-\sigma_y$

6.2

The general solution to the linear interaction case is given by (5.10) as:

　 $\displaystyle\hat a(t) = e^{-i\omega t}\hat a(0) + \alpha(t)$

　 $\displaystyle\hat a^\dagger(t) = e^{i\omega t}\hat a^\dagger(0) + \alpha^*(t)$

Note that this is valid not only for $t\ge 0$ , but also for $t<0$ , given the boundary condition at $t=0$ as $\hat a(0) = \hat a$ (annihilation operator in the Schroedinger picture).

Then, if you start from $\mid 1\rangle$ , the expectation value of the number operator is:

　 $\displaystyle \langle \hat n(t)\rangle = \langle 1\mid \hat a^\dagger(t)\hat a(t)\mid 1\rangle= \langle 1\mid\left\{ \hat a^\dagger(0)\hat a(0)+\alpha^*(t)\alpha(t)\right\}\mid 1\rangle = 1 + |\alpha(t)|^2 \ge 1$

Hence if you start from $\mid 1\rangle$ , you can never reach $\mid 0\rangle$ (even approximately) that satisfies $\displaystyle \langle \hat n(t)\rangle=0$ in the future ( $t\ge 0$ ) or in the past ( $t<0$ ). In other words, $\mid 1\rangle$ is not reachable from $\mid 0\rangle$ .

To consider the effect of nonlinearity, write down a Schroedinger equation in the interaction picture. First, without the nonlinear term:

　 $\displaystyle H = H_0 + f\hat a+f^*\hat a^\dagger,\ H_0 = \omega\hat a^\dagger\hat a$

The interaction Hamiltonian is:

　 $\displaystyle H_I(t) := e^{-i H_0}(f\hat a + f^*\hat a^\dagger)e^{iH_0} = e^{i\omega t}f\hat a + e^{-i\omega t}f^*\hat a^\dagger$

where I used the relationship:

　 $\displaystyle e^{A}Be^{-A} = \sum_{n=0}^\infty\frac{1}{n!}(C_A)^n B$

　 $C_AB = [A, B],\,(C_A)^2B = [A, [A, B]], \cdots$

By redefining the coupling factor as $g = e^{i\omega t}f$ , we have:

　 $\displaystyle H_I(t) = g\hat a + g^*\hat a^\dagger$

For $\displaystyle \mid\Psi_I(t)\rangle = \sum_{n=0}^\infty c_n\mid n\rangle$ ,

　 $\displaystyle H_I(t) \mid\Psi_I(t)\rangle = \sum_{n=0}^\infty c_n\left(g\sqrt{n}\mid n-1\rangle+g^*\sqrt{n+1}\mid n+1\rangle\right)$

　　 $\displaystyle = \sum_{n=0}^\infty\left(c_{n+1}g\sqrt{n+1}+c_{n-1}g^*\sqrt{n}\right)\mid n\rangle$

　 $\displaystyle i\frac{d}{dt}\mid\Psi_I(t)\rangle = i\sum_{n=0}^\infty\dot c_n\mid n\rangle$

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

　 $\displaystyle i\dot c_0 = gc_1$

　 $\displaystyle i\dot c_1 = g\sqrt{2}c_2 + g^*c_0$

　 $\displaystyle i\dot c_2 = g\sqrt{3}c_3 + g^*\sqrt{2}c_1$

　　 $\vdots$

You can see that each state $c_n$ couples with $c_{n-1}$ and $c_{n+1}$ . The coupling to $c_{n+1}$ is slightly stronger than the coupling to $c_{n-1}$ . Hence if you start from $\mid 0\rangle$ , the state diffuses into higher energy states, and cannot converge into a single energy state.

Now let's add the nonlinear interaction $\displaystyle U\hat a^\dagger\hat a^\dagger\hat a\hat a = U\left\{(\hat a^\dagger\hat a)^2 - \hat a^\dagger\hat a\right\}$ .

The last term $\displaystyle - U\hat a^\dagger \hat a$ can be included in the base Hamiltonian $H_0$ by shifting the frequency from $\omega$ to $\omega - U$ . Then the interaction Hamiltonian becomes:

　 $\displaystyle H_I(t) = g\hat a + g^*\hat a^\dagger + U(\hat a^\dagger\hat a)^2$

And the corresponding equations of motion are:

　 $i\dot c_0 = gc_1$

　 $\displaystyle i\dot c_1 = g\sqrt{2}c_2 + g^*c_0 + Uc_1$

　 $\displaystyle i\dot c_2 = g\sqrt{3}c_3 + g^*\sqrt{2}c_1 + 4Uc_2$

　　 $\vdots$

You see that each state $c_n$ has a self-coupling term with the strength $n^2U$ . So, if U is sufficiently large, the higher energy states decouples from other states, such as:

　 $i\dot c_2 \sim 4Uc_2$

So, as a first order approximation, we have:

　 $i\dot c_0 = gc_1$

　 $\displaystyle i\dot c_1 = g^*c_0 + Uc_1$

Assuming that g is a constant, this can be explicitly solved with the boundary condition $c_0(0) = 1,\,c_1(0)=0$ .

　 $\displaystyle c_1(t) = \frac{-2ig^*}{\sqrt{U^2+4g^*g}}e^{-i\frac{U}{2}t}\sin\left(\frac{1}{2}\sqrt{U^2+4g^*g}t\right)$

When $g >> U$ , it is achievable to get $|c_1(t)|^2\sim 1$ .

Note: The condition to ignore the higher energy states $U >> g$ conflicts with the last assumption $g >> U$ . To avoid it, we may need higher order nonlinearity...?

6.3

The convention for the matrix representation:

　 $\displaystyle \mid 1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix},\,\mid 0\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix},\,\rho_{ij} = \langle i\mid\rho\mid j\rangle$

(1) In the interaction picture,

　 $\displaystyle \mid\Psi(t)\rangle = e^{-i\delta\omega\frac{\sigma_z}{2}t}\left(c_0\mid 0\rangle + c_1\mid 1\rangle\right) =e^{i\frac{\delta\omega}{2}t}c_0\mid 0\rangle + e^{-i\frac{\delta\omega}{2}t}c_1\mid 1\rangle$

　 $\displaystyle\rho(t) = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty d\delta\omega\,e^{-\frac{(\delta\omega)^2}{2\sigma^2}}\mid\Psi(t)\rangle\langle\Psi(t)\mid$

　　 $\displaystyle = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty d\delta\omega\,e^{-\frac{(\delta\omega)^2}{2\sigma^2}}\left(|c_0|^2\mid 0\rangle\langle 0 \mid + |c_1|^2 \mid 1\rangle\langle 1\mid + c_1c_0^*e^{-i\delta\omega t}\mid 1\rangle\langle 0\mid + c_0c_1^*e^{i\delta\omega t}\mid 0\rangle\langle 1\mid\right)$

　　 $= \begin{pmatrix}\rho_{11} & \rho_{10} \\ \rho_{01} & \rho_{00}\end{pmatrix}$

Hence,

　 $\displaystyle\rho_{00} = |c_0|^2,\, \rho_{11}=|c_1|^2$

　 $\displaystyle\rho_{10} = \frac{c_1c_0^*}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty d\delta\omega\, \exp\left\{-\frac{(\delta\omega)^2}{2\sigma^2}-i\delta\omega t\right\}$

　　 $\displaystyle = \frac{c_1c_0^*}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty d\delta\omega\, \exp\left[-\frac{1}{2\sigma^2}\left\{(\delta\omega+i\sigma^2 t)^2\right\}\right]\exp\left(-\frac{1}{2\sigma^2}\sigma^4t^2\right)$

　　 $\displaystyle = c_1c_0^*e^{-\frac{\sigma^2}{2}t^2}$

Likewise,

　 $\displaystyle \rho_{01} = c_0c_1^*e^{-\frac{\sigma^2}{2}t^2}$

(2) Again, in the interaction picture,

　 $\displaystyle\frac{d\rho}{dt} = \frac{\sigma^2t}{2}(\sigma_z\rho\sigma_z-\rho)$

　　 $\displaystyle \Leftrightarrow \frac{d\rho_{11}}{dt}=0,\, \frac{d\rho_{00}}{dt}=0,\,\frac{d\rho_{01}}{dt}=-\sigma^2t\rho_{01},\,\frac{d\rho_{10}}{dt}=-\sigma^2t\rho_{10}$

(3) In the Schroedinger picture,

　 $\displaystyle\sigma_x\rho\sigma_x = \begin{pmatrix}\rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11}\end{pmatrix}$

Hence,

　 $\displaystyle \sigma_x e^{-iHt} \sigma_x e^{-iHt} \rho e^{iHt} \sigma_x e^{iHt} \sigma_x = \sigma_x e^{-iHt} \sigma_x e^{-iHt} \begin{pmatrix} \rho_{11} & \rho_{10} \\ \rho_{01} & \rho_{00}\end{pmatrix} e^{iHt} \sigma_x e^{iHt} \sigma_x$

　　 $\displaystyle = \sigma_x e^{-iHt} \sigma_x\begin{pmatrix} \rho_{11} & e^{-2i(\omega+\delta\omega)t} \rho_{10} \\ e^{2i(\omega+\delta\omega)t} \rho_{01} & \rho_{00}\end{pmatrix} \sigma_x e^{iHt} \sigma_x$

　　 $\displaystyle = \sigma_x e^{-iHt} \begin{pmatrix} \rho_{00} & e^{2i(\omega+\delta\omega)t} \rho_{01} \\ e^{-2i(\omega+\delta\omega)t} \rho_{10} & \rho_{11}\end{pmatrix} e^{iHt} \sigma_x$

　　 $\displaystyle = \sigma_x\begin{pmatrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11}\end{pmatrix} \sigma_x = \begin{pmatrix} \rho_{11} & \rho_{10} \\ \rho_{01} & \rho_{00}\end{pmatrix} =\rho$

6.4

(1) In general,

　 $\displaystyle\langle O\rangle = {\rm Tr}\left[O\rho\right]$

Hence,

　 $\displaystyle\frac{d}{dt}\langle O\rangle = {\rm Tr}\left[O\dot\rho\right]$

　　 $\displaystyle = -i{\rm Tr}\left[OH\rho-O\rho H\right]+(n+1)\frac{\gamma}{2}{\rm Tr}\left[2O\sigma^-\rho\sigma^+-O\sigma^+\sigma^-\rho-O\rho\sigma^+\sigma^-\right]$

　　　　 $\displaystyle +\,n\frac{\gamma}{2}{\rm Tr}\left[2O\sigma^+\rho\sigma^--O\sigma^-\sigma^+\rho-O\rho\sigma^-\sigma^+\right]$

　　 $\displaystyle = -i{\rm Tr}\left[[O, H]\rho\right]+(n+1)\frac{\gamma}{2}{\rm Tr}\left[\left(2\sigma^+O\sigma^--O\sigma^+\sigma^--\sigma^+\sigma^-O\right)\rho\right]$

　　　　 $\displaystyle +\,n\frac{\gamma}{2}{\rm Tr}\left[\left(2\sigma^-O\sigma^+-O\sigma^-\sigma^+-\sigma^-\sigma^+O\right)\rho\right]$

　　 $\displaystyle = -i{\rm Tr}\left[[O, H]\rho\right]+(n+1)\frac{\gamma}{2}{\rm Tr}\left[\left( \sigma^+[O, \sigma^-] + [\sigma^+,O]\sigma^-\right)\rho\right]$

　　　　 $\displaystyle +\,n\frac{\gamma}{2}{\rm Tr}\left[\left(\sigma^-[O,\sigma^+] + [\sigma^-,O]\sigma^+\right)\rho\right]$

　　 $\displaystyle = -i\langle[O,H]\rangle + (n+1)\frac{\gamma}{2}\langle \sigma^+[O,\sigma^-] + [\sigma^+,O]\sigma^-\rangle$

　　　　 $\displaystyle +\,n\frac{\gamma}{2}\langle \sigma^-[O,\sigma^+] + [\sigma^-,O]\sigma^+\rangle$ --- (1)

In the zero temperature limit $n=0$ , you get (6.73).

(2) Basic definitions and relationships:

　 $\displaystyle \sigma^{\pm} = \frac{1}{2}(\sigma_x\pm i\sigma_y),\ \{\sigma_i,\sigma_j\} = 2\delta_{ij}I\,(i,j=x,y,z)$

　 $\displaystyle [\sigma_x,\sigma_y] = 2i\sigma_z,\ [\sigma_y,\sigma_z] = 2i\sigma_x,\ [\sigma_z,\sigma_x] = 2i\sigma_y$

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

　 $\displaystyle \sigma^{\pm}\sigma_z = {\mp}\sigma^{\pm},\ \sigma_z\sigma^{\pm} = \pm\sigma^{\pm}$

　 $\displaystyle\sigma^+\sigma^- = \frac{1}{2}(1+\sigma_z),\ \sigma^-\sigma^+ = \frac{1}{2}(1-\sigma_z)$

Note that these relationships hold at any time $t$ in the Heisenberg picture since:

　 $\displaystyle A(0)B(0) = C(0)\ \Rightarrow$

　　 $\displaystyle A(t)B(t) = U^\dagger(t)A(0)U(t)U^\dagger(t)B(0)U(t) = U^\dagger(t)A(0)B(0)U(t)$

　　　　 $\displaystyle = U^\dagger(t)C(0)U(t) = C(t)$

Now, from (1),

　 $\displaystyle \frac{d}{dt}\langle \sigma_z\rangle = (n+1)\frac{\gamma}{2}\langle\sigma^+[\sigma_z,\sigma^-]+[\sigma^+,\sigma_z]\sigma^-\rangle+n\frac{\gamma}{2}\langle\sigma^-[\sigma_z,\sigma^+]+[\sigma^-,\sigma_z]\sigma^+\rangle$

　　 $\displaystyle = -2(n+1)\gamma\langle\sigma^+\sigma^-\rangle + 2n\gamma\langle\sigma^-\sigma^+\rangle =-(n+1)\gamma(1+\langle\sigma_z\rangle)+n\gamma(1-\langle\sigma_z\rangle)$

　　 $\displaystyle = -(2n+1)\gamma\left(\langle\sigma_z\rangle + \frac{1}{2n+1}\right)$

　 $\displaystyle\therefore\, \langle\sigma_z(t)\rangle = \left\{\langle\sigma_z(0)\rangle+\frac{1}{2n+1}\right\}e^{-(2n+1)\gamma t}-\frac{1}{2n+1}$ --- (2)

　 $\displaystyle\frac{d}{dt}\langle\sigma^+\rangle = -i\frac{\Delta}{2}\langle[\sigma^+,\sigma_z]\rangle+(n+1)\frac{\gamma}{2}\langle\sigma^+[\sigma^+,\sigma^-]\rangle+n\frac{\gamma}{2}\langle[\sigma^-,\sigma^+]\sigma^+\rangle$

　　 $\displaystyle = i\Delta\langle\sigma^+\rangle + (n+1)\frac{\gamma}{2}\langle\sigma^+\sigma_z\rangle - n\frac{\gamma}{2}\langle\sigma_z\sigma^+\rangle$

　　 $\displaystyle = i\Delta\langle\sigma^+\rangle-(n+1)\frac{\gamma}{2}\langle\sigma^+\rangle -n\frac{\gamma}{2}\langle\sigma^+\rangle$

　　 $\displaystyle = \left(i\Delta-\frac{2n+1}{2}\gamma\right)\langle\sigma^+\rangle$

　 $\displaystyle\therefore\,\langle\sigma^+(t)\rangle = \langle\sigma^+(0)\rangle e^{i\Delta t-\frac{2n+1}{2}\gamma t}$ --- (3)

Similarly,

　 $\displaystyle\frac{d}{dt}\langle\sigma^-\rangle = \left(-i\Delta-\frac{2n+1}{2}\gamma\right)\langle\sigma^-\rangle$

　 $\displaystyle\therefore\,\langle\sigma^-(t)\rangle = \langle\sigma^-(0)\rangle e^{-i\Delta t-\frac{2n+1}{2}\gamma t}$ --- (4)

In the zero temperature limit $n=0$ , from (2)(3)(4),

　 $\displaystyle \langle\sigma_z(t)\rangle = \left\{\langle\sigma_z(0)\rangle+1\right\}e^{-\gamma t}-1$ --- (2)'

　 $\displaystyle \langle\sigma^+(t)\rangle = \langle\sigma^+(0)\rangle e^{i\Delta t-\frac{\gamma}{2}t}$ --- (3)'

　 $\displaystyle \langle\sigma^-(t)\rangle = \langle\sigma^-(0)\rangle e^{-i\Delta t-\frac{\gamma}{2}t}$ --- (4)'
　
(3) From the expression:

　 $\displaystyle\rho = \frac{1}{2}I + \frac{1}{2}\sum_{i=x,y,z}S_i\sigma_i$

The elements of Bloch vector $S_i$ are given by:

　 $\displaystyle S_i = {\rm Tr}[\sigma_i\rho] = \langle\sigma_i\rangle$

Hence,

　 $\displaystyle S_x^2 = \langle\sigma_x\rangle^2 = \left(\langle\sigma^+\rangle+\langle\sigma^-\rangle\right)^2$

　 $\displaystyle S_y^2 = \langle\sigma_y\rangle^2 = \left\{-i\left(\langle\sigma^+\rangle-\langle\sigma^-\rangle\right)\right\}^2 = - \left(\langle\sigma^+\rangle-\langle\sigma^-\rangle\right)^2$

　 $\displaystyle S_x^2+S_y^2 = 4\langle\sigma^+\rangle\langle\sigma^-\rangle = 4\langle\sigma^+(0)\rangle\langle\sigma^-(0)\rangle e^{-\gamma t}\ \left(\because (3')(4')\right)$

　　 $\displaystyle = \left\{S_x^2(0) + S_y^2(0)\right\}e^{-\gamma t} = \left\{1-S_z^2(0)\right\}e^{-\gamma t}\ \left(\because\, S_x^2(0)+S_y^2(0)+S_z^2(0) = 1\right)$

Likewise,

　 $\displaystyle S_z^2 = \langle\sigma_z\rangle^2 = \left[\left\{\langle\sigma_z(0)\rangle+1\right\}e^{-\gamma t}-1\right]^2\ \left(\because (2')\right)$

　　 $\displaystyle = \left[\left\{\langle S_z(0)\rangle+1\right\}e^{-\gamma t}-1\right]^2$

So, by setting $A = \langle S_z(0)\rangle$ , we have:

　 $\displaystyle 0\le S_x^2+S_y^2+S_z^2 = \left\{(A+1)e^{-\gamma t}-1\right\}^2 + (1-A^2)e^{-\gamma t}$

　　 $\displaystyle = (A+1)^2e^{-\gamma t}(e^{-\gamma t}-1) + 1 \le 1$

(3) Since $\displaystyle \rho =\begin{pmatrix}\rho_{11} & \rho_{10} \\ \rho_{01} & \rho_{00}\end{pmatrix},\,\sigma^+ = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix},\, \frac{1}{2}(1+\sigma_z) = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$

　 $\displaystyle \langle\sigma^+(t)\rangle = {\rm Tr}[\sigma^+\rho(t)] = \rho_{01}(t)$

　 $\displaystyle\frac{1}{2}\left\{1 + \langle\sigma_z(t)\rangle\right\}= {\rm Tr}\left[\frac{1}{2}\left(1+\sigma_z\right)\rho(t)\right]=\rho_{11}(t)$

Hence, from (2) and (3),

　 $\displaystyle \rho_{11}(t) = \frac{n}{2n+1} + \left(\rho_{11}(0)-\frac{n}{2n+1}\right)e^{-(2n+1)\gamma t}$

　 $\displaystyle \rho_{01}(t) = \rho_{01}(0)e^{i\Delta t -\frac{2n+1}{2}\gamma t}$

By defining $\displaystyle P_1 = \frac{n}{2n+1},\,T_1 = \frac{1-2P_1}{\gamma} = \frac{1}{(2n+1)\gamma}$ , we have:

　 $\displaystyle\rho_{11}(t) = P_1 + \left\{\rho_{11}(0)-P_1\right\}e^{-\frac{t}{T_1}}$

　 $\displaystyle\rho_{01}(t) = \rho_{01}(0)e^{i\Delta t-\frac{t}{2T_1}}$

　 $\displaystyle\rho_{10}(t) = \rho_{01}^*(t) = \rho_{10}(0)e^{-i\Delta t-\frac{t}{2T_1}}$

　 $\displaystyle\rho_{00}(t) = 1 - \rho_{11}(t) = (1-P_1)+\left\{\rho_{00}-(1-P_1)\right\}e^{-\frac{t}{T_1}}$

　 $\displaystyle \therefore \rho(t) = \begin{pmatrix} P_1 + \left\{\rho_{11}(0)-P_1\right\}e^{-\frac{t}{T_1}} & \rho_{10}(0)e^{-i\Delta t-\frac{t}{2T_1}} \\ \rho_{01}(0)e^{i\Delta t-\frac{t}{2T_1}} & (1-P_1)+\left\{\rho_{00}(0)-(1-P_1)\right\}e^{-\frac{t}{T_1}} \end{pmatrix}$

　　 $\displaystyle = \begin{pmatrix} P_1 + \left\{\rho_{11}(0)-P_1\right\}e^{-\frac{t}{T_1}} & \rho_{10}(0)e^{-i\Delta t-\frac{t}{2T_1}} \\ \rho_{01}(0)e^{i\Delta t-\frac{t}{2T_1}} & (1-P_1)+\left\{P_1-\rho_{11}(0)\right\}e^{-\frac{t}{T_1}} \end{pmatrix}$

　 $\displaystyle \left(\because\,\rho_{00}(0) = 1 - \rho_{11}(0)\right)$

6.5

Using the unit system $\hbar = 1$ ,

　 $\displaystyle\frac{d}{dt}\rho = -i[H,\rho] +(n+1)\frac{\gamma}{2}\left(2\sigma^-\rho\sigma^+-\sigma^+\sigma^-\rho-\rho\sigma^+\sigma^-\right)$

　　 $\displaystyle {}+n\frac{\gamma}{2}\left(2\sigma^+\rho\sigma^--\sigma^-\sigma^+\rho-\rho\sigma^-\sigma^+\right) + \frac{\gamma_\phi}{2}\left(\sigma_z\rho\sigma_z-\rho\right)$

The contribution from the dephasing term to $\displaystyle \frac{d}{dt}\langle\sigma_z\rangle$ is:

　 $\displaystyle\frac{\gamma_\phi}{2}{\rm Tr}\left[\sigma_z^2\rho\sigma_z - \sigma_z\rho\right] = 0$

Hence, the solution of $\langle\sigma_z\rangle$ is the same as 6.4.

The contribution from the dephasing term to $\displaystyle \frac{d}{dt}\langle\sigma_+\rangle$ is:

　 $\displaystyle\frac{\gamma_\phi}{2}{\rm Tr}\left[\sigma^+\sigma_z\rho\sigma_z-\sigma^+\rho\right]=-\gamma_\phi\langle\sigma^+\rangle$

Hence, the equation for $\displaystyle\frac{d}{dt}\langle\sigma^+\rangle$ in 6.4 is modified as:

　 $\displaystyle \frac{d}{dt}\langle\sigma^+\rangle = \left(i\Delta-\frac{2n+1}{2}\gamma-\gamma_\phi\right)\langle\sigma^+\rangle = \left(i\Delta - \frac{1}{2T_1}-\frac{1}{T_2}\right)\langle\sigma^+\rangle$

　 $\displaystyle\therefore\,\langle\sigma^+\rangle = \langle\sigma^+(0)\rangle e^{i\Delta t-\left(\frac{1}{2T_1}+\frac{1}{T_2}\right)t}$

　 $\displaystyle\rho_{01}(t) = \langle\sigma(t)\rangle = \rho_{01}(0) e^{i\Delta t-\left(\frac{1}{2T_1}+\frac{1}{T_2}\right)t} = \rho_{01}(0) e^{i\Delta t-\frac{1}{T_2^*}t}$

where $\displaystyle \frac{1}{T_2^*} = \frac{1}{2T_1}+\frac{1}{T_2}$

6.6

????

6.7

For simplicity, we use the truncated model. At the symmetry point $n_g = 1/2$ , the Hamiltonian becomes:

　 $\displaystyle H = E_C\left( \mid 0 \rangle\langle 0 \mid + \mid 1 \rangle\langle 1 \mid\right) + \frac{E_J}{2}\left(\mid 0\rangle \langle 1 \mid + \mid 1\rangle\langle 0\mid\right)$

For $\displaystyle \mid\psi(t)\rangle = c_0(t)\mid 0\rangle + c_1(t)\mid 1\rangle$ , we have:

　 $\displaystyle H\mid\psi(t)\rangle = \left\{E_Cc_0(t) + \frac{E_J}{2}c_1(t)\right\}\mid 0\rangle + \left\{E_Cc_1(t) + \frac{E_J}{2}c_0(t)\right\}\mid 1\rangle$

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

　 $\displaystyle i \dot c_0(t) = E_Cc_0(t) +\frac{E_J}{2}c_1(t)$

　 $\displaystyle i \dot c_1(t) = E_Cc_1(t) +\frac{E_J}{2}c_0(t)$

$\displaystyle \therefore\, \frac{d}{dt}\left\{c_0(t) + c_1(t)\right\} = -i\left(E_C + \frac{E_J}{2}\right)\left\{c_0(t) + c_1(t)\right\}$

　 $\displaystyle \frac{d}{dt}\left\{c_0(t) - c_1(t)\right\} = -i\left(E_C - \frac{E_J}{2}\right)\left\{c_0(t) - c_1(t)\right\}$

With the initial condition $c_0(0) = 1,\, c_1(0) = 0$ , we have:

　 $\displaystyle c_0(t) = e^{-iE_Ct}\cos\left(\frac{E_J}{2}t\right),\ c_1(t) = ie^{-iE_Ct}\sin\left(\frac{E_J}{2}t\right)$

So, the probability of finding the state $\mid 1\rangle$ is:

　 $\displaystyle P(t) = |c_1(t)|^2 = \sin^2\left(\frac{E_J}{2}t\right)$

6.8

Since $E_n = 4E_Cn^2$ , the energy gap is $4E_C = \Delta \sim 80[\mu{\rm eV}]$ .

The additional energy for a Cooper pair $2e$ is:

　 $\displaystyle 2eV = \frac{2e^2}{4\pi\epsilon_0 d} = e \times 2 \times e \times \frac{1}{4\pi\epsilon_0} \times \frac{1}{d}$

　　 $\displaystyle \sim e \times 2 \times 1.6\times 10^{-19} \times 9\times 10^9 \times \frac{10^6}{4} = 720[\mu{\rm eV}]$

From $80(1+n_g)^2 = 720$ , we have $n_g = 2$ .

In general, the Hamiltonian becomes:

　 $H = 4E_C\hat n^2 + 2\hat n eV\,$ where $\hat n=\hat a^\dagger \hat a$

6.9

(4) The standard Hamiltonian for the LC resonator is:

　 $\displaystyle H = \frac{1}{2C}\left(\hat q-q_g\right)^2 + \frac{1}{2L}\hat \phi^2$

Impedance $\displaystyle Z = \sqrt{\frac{L}{C}}$

Natural frequency $\displaystyle \omega = \frac{1}{\sqrt{LC}}$

Relation to the ladder operators:

　 $\displaystyle\hat\phi = \sqrt{\frac{Z}{2}}\left(\hat a^\dagger + \hat a\right)$

　 $\displaystyle\hat q = \sqrt{\frac{1}{2Z}}i\left(\hat a^\dagger - \hat a\right)$

The transmon Hamiltonian is:

　 $H = \frac{1}{2C_\Sigma}\left(\hat q-q_g\right)^2 + \frac{E_J}{2\varphi_0^2}\hat \phi^2 + O(\hat\phi^4)$

Hence, the corresponding inductance is:

　 $\displaystyle L = \frac{\varphi_0^2}{E_J}$

Using the relationships $\displaystyle \varphi_0 = \frac{1}{2e},\, C_\Sigma = \frac{e^2}{2E_C}$ , the corresponding impedance is:

　 $\displaystyle Z = \sqrt{\frac{L}{C}} = \sqrt{\frac{1}{(2e)^2E_J}\frac{2E_C}{e^2}} = \frac{1}{(2e)^2}\sqrt{\frac{8E_C}{E_J}}$

(1)

　 $\displaystyle \omega_{01} = \sqrt{\frac{1}{LC}} = \sqrt{(2e)^2 E_J\frac{2E_C}{e^2}} = \sqrt{8E_CE_J}$

(2)

　 $\displaystyle \hat q = \sqrt{\frac{1}{Z}}\frac{i}{\sqrt{2}}\left(\hat a^\dagger - \hat a\right) = 2e\left(\frac{E_J}{8E_C}\right)^{\frac{1}{4}}\frac{i}{\sqrt{2}}\left(\hat a^\dagger - \hat a\right)$

　 $\displaystyle \hat\phi = \sqrt{Z}\frac{1}{\sqrt{2}}\left(\hat a^\dagger+\hat a\right) = \frac{1}{2e}\left(\frac{8E_C}{E_J}\right)^{\frac{1}{4}}\frac{1}{\sqrt{2}}\left(\hat a^\dagger + \hat a\right)$

(3) The quartic term is:

　 $\displaystyle V = -\frac{1}{24}\frac{E_J}{\varphi_0^4}\hat\phi^4 = -\frac{2e^4}{3}E_J\hat\phi^4$

　　 $\displaystyle = -\frac{2e^4}{3}E_J\frac{1}{(2e)^4}\frac{8E_C}{E_J}\frac{1}{4}\left(\hat a^\dagger + \hat a\right)^4 = -\frac{E_C}{12}\left(\hat a^\dagger + \hat a\right)^4$

Hence the first order contribution to the diagonal elements is:

　 $\displaystyle \langle n\mid V\mid n \rangle = -\frac{E_C}{12}\langle n\mid \left(\hat a^\dagger + \hat a\right)^4\mid n \rangle$

めもめも

このブログに記載の内容は個人の見解であり、必ずしも所属組織の立場、戦略、意見を代表するものではありません。

Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 6) : Part1

6.1

6.2

6.3

6.4

6.5

6.6

6.7

6.8

6.9