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Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 4)

4.1

For the thermal state \displaystyle\rho_\infty=\frac{1}{2}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}:

 \displaystyle <\sigma^x> = {\rm Tr}\left[\sigma^x\rho_\infty\right]
={\rm Tr}\left[\frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}\right]
=0

For the pure state \displaystyle\rho_q=\frac{1}{2}\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}:

 \displaystyle <\sigma^x> = {\rm Tr}\left[\sigma^x\rho_q\right]
={\rm Tr}\left[\frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}\right]
=1

4.2

By mathematical induction, you can prove:

 \hat\phi^k\hat q = \hat q\hat\phi^k +ik\hbar\hat\phi^{k-1}\ (k=1,2,\cdots)

Hence:

 \displaystyle \exp\left(\frac{i}{\varphi_0}\hat\phi\right)\hat q = \sum_{k=0}^\infty\frac{1}{k!}\left(\frac{i}{\varphi_0}\right)^k\hat\phi^k\hat q
= \sum_{k=0}^\infty\frac{1}{k!}\left(\frac{i}{\varphi_0}\right)^k\left(\hat q\hat\phi^k + ik\hbar\hat\phi^{k-1}\right)

  \displaystyle  =\hat q\exp\left(\frac{i}{\varphi_0}\hat\phi\right)+i\hbar\frac{i}{\varphi_0}\sum_{k=1}^\infty\frac{1}{(k-1)!}\left(\frac{i}{\varphi_0}\right)^{k-1}\hat\phi^{k-1}

  \displaystyle  =\hat q\exp\left(\frac{i}{\varphi_0}\hat\phi\right) - \frac{\hbar}{\varphi_0}\exp\left(\frac{i}{\varphi_0}\hat\phi\right)

  \displaystyle  =(\hat q-2e)\exp\left(\frac{i}{\varphi_0}\hat\phi\right)\ \ \left(\varphi_0 = \frac{\hbar}{2e}\right)

4.4

Current conservation:

 I_L = I_C + I_R

Branch currents:

 \displaystyle I_C = C\dot V,\ \dot I_L = -\frac{V}{L},\ I_R = \frac{V}{R}

Hence:

 \displaystyle\dot H = \frac{d}{dt}\left(\frac{1}{2}CV^2 + \frac{1}{2}LI_L^2\right) = CV\dot V+LI_L\dot I_L=V(I_C-I_L)

Without the register:

 \dot H = 0\ \ (\because I_C = I_L)

With the register:

 \displaystyle \dot H = -VI_R = -\frac{V^2}{R} \le 0

4.5


4.6

This is very rough apporximation. I'm not sure if this is really an intended answer.


4.7

 \displaystyle E_{\rm ind}(\phi_-) = -E_J(\Phi)\cos\left(\frac{\phi_-}{\varphi_0}\right) - I\phi_-

 \displaystyle \frac{d}{d\phi_-}E_{\rm ind}(\phi_-) = \frac{E_J(\Phi)}{\varphi_0}\sin\left(\frac{\phi_-}{\varphi_0}\right) - I

\displaystyle\therefore \forall \phi_-;\ \frac{d}{d\phi_-}E_{\rm ind}(\phi_-) < 0 \ \Longleftrightarrow\ I > \frac{E_J(\Phi)}{\varphi_0} = 2I_c\cos\left(\frac{\Phi}{2\varphi_0}\right)

In the textbook (p.57), the critical value is written as 2I_c\cos(\Phi/2\varphi_0) / \varphi_0. I guess the last /\varphi_0 is a typo.

4.8

 \displaystyle \hat Q_+ = -2e \hat N_+ = \sum_n (-2e n) \mid n\rangle_+\langle n\,\mid_+\otimes \mathbb 1_-

 \displaystyle \hat Q^2_+ = \sum_n (-2e n)^2 \mid n\rangle_+\langle n\,\mid_+\otimes \mathbb 1_-

 \displaystyle \hat Q_- = -2e \hat N_- = \sum_n (-2e n)  \mathbb 1_+\otimes\mid n\rangle_-\langle n\,\mid_-

 \displaystyle \hat Q^2_- = \sum_n (-2e n)^2 \mathbb 1_+\otimes\mid n\rangle_-\langle n\,\mid_-

 \displaystyle \cos\left(\frac{\Phi+\hat\phi_+}{\varphi_0}\right) = \frac{1}{2}e^{i\frac{\Phi}{\varphi_0}}e^{i\frac{\hat\phi_+}{\varphi_0}} +\frac{1}{2} e^{-i\frac{\Phi}{\varphi_0}}e^{-i\frac{\hat\phi_+}{\varphi_0}}

  \displaystyle = \frac{1}{2}e^{i\frac{\Phi}{\varphi_0}}\sum_n \mid n\rangle_+\langle n+1\mid_+\otimes\mathbb 1_- + \frac{1}{2}e^{-i\frac{\Phi}{\varphi_0}}\sum_n \mid n+1\rangle_+\langle n\mid_+\otimes\mathbb 1_-

 \displaystyle\cos\left(\frac{\hat\phi_-}{2\varphi_0}\right)\cos\left(\frac{\hat\phi_+}{2\varphi_0}\right)

  \displaystyle =\frac{1}{4}\left[
\left(e^{i\frac{\hat\phi_-}{\varphi_0}}\right)^{\frac{1}{2}}+\left(e^{-i\frac{\hat\phi_-}{\varphi_0}}\right)^{\frac{1}{2}}
\right]\left[
\left(e^{i\frac{\hat\phi_+}{\varphi_0}}\right)^{\frac{1}{2}}+\left(e^{-i\frac{\hat\phi_+}{\varphi_0}}\right)^{\frac{1}{2}}
\right]

  \displaystyle =\frac{1}{4}\left[
\left(\sum_n\mid n\rangle_+\langle n+1\mid_+\right)^{\frac{1}{2}}+ \left(\sum_n\mid n+1\rangle_+\langle n\mid_+\right)^{\frac{1}{2}}
\right]\otimes

      \displaystyle \left[
\left(\sum_n\mid n\rangle_-\langle n+1\mid_-\right)^{\frac{1}{2}}+ \left(\sum_n\mid n+1\rangle_-\langle n\mid_-\right)^{\frac{1}{2}}
\right]

I'm not sure what's the explicit representation of the root squrare of the ladder operators...