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

4.7

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

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

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

2022-09-20

Derivation of the first and second Josephson relation

In the following discussion, we assume that the quantum states are quasi-static. We solve the time-independent Schrödinger equation supposing that external potentials are independent of time $v(\mathbf x), \mathbf A(\mathbf x)$ . When we change them slowly enough, the corresponding quantum states (the effective wavefunction) changes so that they keep following the time-independent Schrödinger equation (with the given potentials) at each point of time.

First, we shows that the time-independent Schrödinger equation for the gauge-invariant wavefuntion $\psi_{\rm GI}(\mathbf x)$ has the following form:

　 $\displaystyle E\psi_{\rm GI}(\mathbf x) = \left[\frac{1}{2m_s}(-i\hbar\nabla)^2 + q_sv(\mathbf x)\right]\psi_{\rm GI}(\mathbf x)$ --- (1)

[Proof]

We start from the time-dependent Schrödinger equation of the effective wafefunction:

　 $\displaystyle i\hbar\partial_t\psi(\mathbf x, t) = \left[\frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x, t)\right\}^2 + q_sv(\mathbf x,t)\right]\psi(\mathbf x, t)$ --- (3.5)

When the potentials are static $\displaystyle v(\mathbf x), \mathbf A(\mathbf x)$ , the energy eigenstate is given by:

　 $\displaystyle \psi(\mathbf x,t) = e^{-i\hbar Et}\psi(\mathbf x)$ --- (2)

and (3.5) becomes:

　 $\displaystyle E\psi(\mathbf x) = \left[\frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x)\right\}^2 + q_sv(\mathbf x)\right]\psi(\mathbf x)$ --- (3)

From (2), the gauge-independent wavefunction has the following from:

　 $\displaystyle \psi_{\rm GI}(\mathbf x,t) = e^{-i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi(\mathbf x, t) = e^{-i\hbar Et} e^{-i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi(\mathbf x)$

So we define the time-independent part of $\displaystyle \psi_{\rm GI}(\mathbf x,t)$ as:

　 $\displaystyle \psi_{\rm GI}(\mathbf x) = e^{-i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi(\mathbf x)$

Equivalently, we have:

　 $\displaystyle \psi(\mathbf x) = e^{i\frac{q_s}{\hbar}\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r)\cdot d\mathbf l}\psi_{\rm GI}(\mathbf x)$ --- (4)

By substituting (4) into (3), we get (1). ■

Now we apply (1) to the one-dimensional potential barrier model of Josephson Junctions:

　 $\displaystyle|x| > \frac{d}{2};\ E\psi_{\rm GI}(x) = \frac{1}{2m_s}(-i\hbar\partial_x)^2\psi_{\rm GI}(x)$

　 $\displaystyle |x| < \frac{d}{2};\ E\psi_{\rm GI}(x) = \left\{\frac{1}{2m_s}(-i\hbar\partial_x)^2 + U_0\right\}\psi_{\rm GI}(x)$

where $U_0 > E$ .

For $\displaystyle|x| > \frac{d}{2}$ , the solution can be plane-wave functions:

　 $\displaystyle x>\frac{d}{2};\ \psi_{\rm GI}(x) = \sqrt{n_s}\exp\left[i\frac{\sqrt{2m_sE}}{\hbar}\left(x-\frac{d}{2}\right)+i\varphi_R\right]$

　 $\displaystyle x<-\frac{d}{2};\ \psi_{\rm GI}(x) = \sqrt{n_s}\exp\left[i\frac{\sqrt{2m_sE}}{\hbar}\left(x+\frac{d}{2}\right)+i\varphi_L\right]$

The arbitrariness of the global phase is expressed by the constants $\varphi_R,\,\varphi_L$ . At this point, they can be taken independently, but later, we will show that they are related through the boundary condition of currents $\mathbf J$ .

For $\displaystyle|x| < \frac{d}{2}$ , the solution can be a liner combination of $\cosh$ and $\sinh$ .

　 $\displaystyle \psi_{\rm GI}(x) = \alpha_+\cosh\left(\frac{x}{\xi}\right) + \alpha_-\sinh\left(\frac{x}{\xi}\right)$ --- (5)

　　 $\displaystyle=\frac{\alpha_+}{2}\left(e^{\frac{x}{\xi}}+e^{-\frac{x}{\xi}}\right)+\frac{\alpha_-}{2}\left(e^{\frac{x}{\xi}}-e^{-\frac{x}{\xi}}\right)$

where $\displaystyle \xi = \frac{\hbar}{\sqrt{2m_s(U_0-E)}}$ .

From the boundary condition (continuity of $\psi_{\rm GI}(x)$ ) at $\displaystyle x=\pm\frac{d}{2}$ , we have:

　 $\displaystyle\frac{\alpha_+}{2}\left(e^{\frac{d}{2\xi}}+e^{-\frac{d}{2\xi}}\right)+\frac{\alpha_-}{2}\left(e^{\frac{d}{2\xi}}-e^{-\frac{d}{2\xi}}\right) = \sqrt{n_s}e^{i\varphi_R}$

　 $\displaystyle\frac{\alpha_+}{2}\left(e^{-\frac{d}{2\xi}}+e^{\frac{d}{2\xi}}\right)+\frac{\alpha_-}{2}\left(e^{-\frac{d}{2\xi}}-e^{\frac{d}{2\xi}}\right) = \sqrt{n_s}e^{i\varphi_L}$

By solving these linear equations, we have:

　 $\displaystyle\alpha_{\pm} = \sqrt{n_s}\frac{e^{i\varphi_R}\pm e^{i\varphi_L}}{e^{\frac{d}{2\xi}}\pm e^{-\frac{d}{2\xi}}}$ --- (3.37)

In general, the currents:

　 $\displaystyle \mathbf J(\mathbf x, t) = \frac{\hbar q_sn_s}{m_s}\nabla \varphi(\mathbf x, t)$ --- (3.23)

can be calculated from the gauge-invariant wavefunction $\psi_{\rm GI}(\mathbf x, t)$ as:

　 $\displaystyle \mathbf J(\mathbf x, t) = \frac{q_s}{m_s}{\rm Re}\left[-\psi_{\rm GI}^*(\mathbf x, t)i\hbar\nabla \psi_{\rm GI}(\mathbf x, t)\right]$

(You can confirm it with the direct calculation using the general form $\displaystyle \psi_{\rm GI}(\mathbf x, t) = e^{i\varphi(\mathbf x,t)}\sqrt{n_s(\mathbf x, t)}$ .)

In our particular case (5), we have:

　 $\displaystyle J(x, t) =\frac{q_s}{m_s}{\rm Re}\left[-\left\{e^{-i\hbar Et}\psi_{\rm GI}(x)\right\}^*i\hbar\partial_x\left\{e^{-i\hbar Et}\psi_{\rm GI}(x)\right\}\right]$

　　 $\displaystyle = -\frac{q_s\hbar}{m_s}{\rm Im}\left[\psi_{\rm GI}^*(x)\partial_x\psi_{\rm GI}(x)\right]$

　　 $\displaystyle = -\frac{q_s\hbar}{m_s\xi}{\rm Im}\bigg[ \left\{\alpha_+^*\cosh\left(\frac{x}{\xi}\right)+\alpha_-^*\sinh\left(\frac{x}{\xi}\right)\right\}$
　　　　　　　　　 $\displaystyle\left\{\alpha_+\sinh\left(\frac{x}{\xi}\right)+\alpha_-\cosh\left(\frac{x}{\xi}\right)\right\}\bigg]$

　　 $\displaystyle = -\frac{q_s\hbar}{m_s\xi}{\rm Im}\left[ \alpha_+^*\alpha_-\cosh^2\left(\frac{x}{\xi}\right) + \alpha_-^*\alpha_+\sinh^2\left(\frac{x}{\xi}\right) \right]$

　　 $\displaystyle = -\frac{q_s\hbar}{m_s\xi}\frac{n_s}{e^\frac{d}{\xi}-e^{-\frac{d}{\xi}}}{\rm Im} \bigg[ (e^{-i\varphi_R}+e^{-i\varphi_L})(e^{i\varphi_R}-e^{i\varphi_L})\cosh^2\left(\frac{x}{\xi}\right)$

　　　　　　　　 $\displaystyle +\,(e^{-i\varphi_R}-e^{-i\varphi_L})(e^{i\varphi_R}+e^{i\varphi_L})\sinh^2\left(\frac{x}{\xi}\right) \bigg]$

　　 $\displaystyle = -\frac{q_s\hbar}{m_s\xi}\frac{n_s}{e^\frac{d}{\xi}-e^{-\frac{d}{\xi}}}{\rm Im} \left[ e^{i(\varphi_R-\varphi_L)}-e^{-i(\varphi_R-\varphi_L)}\right]$

　　 $\displaystyle = -\frac{q_s\hbar}{m_s\xi}\frac{2n_s}{e^\frac{d}{\xi}-e^{-\frac{d}{\xi}}}\sin(\varphi_R-\varphi_L)$

Hence, we finally get:

　 $\displaystyle J(x, t) = J_C\sin(\varphi_R-\varphi_L)$ --- (6)

where:

　 $\displaystyle J_C = -\frac{q_s\hbar}{m_s\xi}\frac{n_s}{\sinh\left(\frac{d}{\xi}\right)}$

(6) suggests that the currents are constant and unique inside the insulator, and we enforce the boundary condition that it's the same as the currents outside the insulator:

　 $\displaystyle J = \frac{\sqrt{2n_sE}}{\Lambda}$ --- (3.34)

From this condition, the phase difference $\delta\varphi := \varphi_R-\varphi_L$ can be decided.

In terms of the current intensity:

　 $I = J\times A$ --- (3.32)

we have the similar relationship:

　 $I=I_C\sin(\delta\varphi)$

Now we assume that we slowly apply an external electric field over the circuit (outside the insulator) so that the voltage (the electric potential difference) between the contact points $\displaystyle x=\pm\frac{d}{2}$ to the insulator becomes $V$ . By applying the discussion:

・Derivation of the gauge-independent relation between the phase and the electric field

the phase difference between two points follows the relationship:

　 $\displaystyle\frac{d}{dt}\delta\varphi(t) = \frac{2\pi}{\Phi_0}V$ --- (7)

(We're assuming that $V$ is very small and the change $\delta\varphi(t)$ follows the quasi-static assumption described at the begging of this note.)

So we conclude that we have the oscillating current inside the insulator:

　 $\displaystyle I(t) = I_C\sin\left(\frac{2\pi}{\Phi_0}Vt + \delta\varphi(0)\right)$

Note

As in the discussion in the following link, I need an assumption that $n_s$ is constant and unique to use the relationship (7).

・Derivation of the gauge-independent relation between the phase and the electric field

So I applied the discussion to the wavefunction outside the insulator that (presumably) forms a closed loop. I'm not sure if this is really a common way to justify using (7) in this model.

2022-09-19

Derivation of the gauge-independent relation between the phase and the electric field

The effective wavefunction $\psi(\mathbf x, t)$ and the charge current $\mathbf J$ are given as:

　 $\displaystyle \psi(\mathbf x, t) = e^{i\theta(\mathbf x, t)}\sqrt{n_s(\mathbf x, t)}$ --- (3.4)

　 $\displaystyle \mathbf J = q_sn_s\left(\frac{\hbar}{m_s}\nabla\theta -\frac{q_s}{m_s}\mathbf A\right)$ ---(3.13)

The wavefunction follows the Schrödinger equation:

　 $\displaystyle i\hbar\partial_t\psi(\mathbf x, t) = \left[\frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x, t)\right\}^2 + q_sv(\mathbf x,t)\right]\psi(\mathbf x, t)$ --- (3.5)

Without losing the generality, we can take the Coulomb gauge:

　 $\nabla\cdot\mathbf A=0$ --- (1)

Now, we assume that the charge density $n_s = |\psi(\mathbf x, t)|^2$ is constant and uniform:

　 $\displaystyle \psi(\mathbf x,t) = e^{i\theta(\mathbf x,t)}\sqrt{n_s}$

In this case, the currents should be divergence-free $\nabla\cdot\mathbf J=0$ . Then, by applying $\nabla$ on (3.13), we have:

　 $\nabla^2\theta=0$ --- (2)

Also, we have:

　 $\displaystyle\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}\psi(\mathbf x,t) = \left\{\hbar\nabla\theta(\mathbf x,t)-q_s\mathbf A(\mathbf x,t)\right\}\psi(\mathbf x,t)$

By using (1) and (2), we also have:

　 $\displaystyle\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}^2\psi(\mathbf x,t) = \left\{\hbar\nabla\theta(\mathbf x,t)-q_s\mathbf A(\mathbf x,t)\right\}^2\psi(\mathbf x,t)$

　　 $\displaystyle = \frac{m_s^2}{q_s^2n_s^2}\mathbf J(\mathbf x,t)^2\psi(\mathbf x,t)$

(3.5) becomes:

　 $\displaystyle-\hbar\partial_t\theta(\mathbf x,t) = \frac{m_s}{2q_s^2n_s^2}\mathbf J(\mathbf x,t)^2 + q_sv(\mathbf x,t)$ --- (3)

Since this is valid only on the Coulomb gauge, we will rewrite it as a gauge-independent expression.

The gauge-invariant phase $\varphi(\mathbf x,t)$ is defined as:

　 $\displaystyle \varphi(\mathbf x,t)-\varphi(\mathbf x_0,t) = \left\{\theta(\mathbf x,t)-\theta(\mathbf x_0,t)\right\} -\frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\mathbf A(\mathbf r,t)\cdot d\mathbf l$ --- (3.16)

As the RHS of (3.16) is gauge-invariant, we can take the Coulomb gauge and use (3) without breaking the gauge-invariance of LHS. By applying $\partial_t$ on (3.16), and assuming that the currents are uniform:

　 $\displaystyle \mathbf J(\mathbf x, t) = \mathbf J(\mathbf x_0, t)$

we have:

　 $\displaystyle\partial_t\varphi(\mathbf x,t) - \partial_t\varphi(\mathbf x_0,t) = -\frac{q_s}{\hbar}\left\{v(\mathbf x,t)-v(\mathbf x_0,t)\right\} - \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\partial_t\mathbf A(\mathbf r,t)\cdot d\mathbf l$

　　 $\displaystyle = \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\left\{-\nabla v(\mathbf r, t)-\partial_t\mathbf A(\mathbf r,t)\right\}\cdot d\mathbf l = \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\mathbf E(\mathbf r, t)\cdot d\mathbf l$

where we used the relationship between the electric field $\mathbf E$ and potentials $(\mathbf A,\,v)$ :

　 $\displaystyle \mathbf E(\mathbf x,t) = -\nabla v(\mathbf x,t)-\partial_t\mathbf A(\mathbf x,t)$

Because $\mathbf E(\mathbf x,t)$ is gauge-invariant, the following relationship is gauge-independent:

　 $\displaystyle \partial_t\varphi(\mathbf x,t) - \partial_t\varphi(\mathbf x_0,t) = \frac{q_s}{\hbar}\int_{\mathbf x_0}^\mathbf x\mathbf E(\mathbf r,t)\cdot d\mathbf l = -\frac{q_s}{\hbar}V$

Where $V := V(\mathbf x)-V(\mathbf x_0)$ is a voltage (electric potential difference) between the two points $(\mathbf x,\, \mathbf x_0)$ .

By defining:

　 $\displaystyle\delta\varphi(t) := \varphi(\mathbf x,t)-\varphi(\mathbf x_0,t)$

　 $\displaystyle\Phi_0 := \frac{h}{|q_s|} = -\frac{2\pi\hbar}{q_s}$

We have:

　 $\displaystyle\frac{d}{dt}\delta\varphi(t) = \frac{2\pi}{\Phi_0}V$ --- (4)

Hence, if we apply a constant voltage $V$ between two points, the phase difference grows as:

　 $\displaystyle\delta\varphi(t) = \frac{2\pi}{\Phi_0}Vt + \delta\varphi(0)$

Note:

In the textbook, it says "We have to complete our derivation to regard more general conditions!" on p.27. In the derivation of (4), we used the following two assumptions:

The charge density $n_s = |\psi(\mathbf x,t)|^2$ is constant and uniform (independent of $(\mathbf x,\, t)$ ).

The currents $\mathbf J(\mathbf x,t)$ are uniform (independent of $\mathbf x$ ).

These are the same assumptions used in the next section "3.7 Josephson Junctions". So we can use the relationship (4) in that section.

めもめも

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

4.1

4.2

4.4

4.5

4.6

4.7

Derivation of the first and second Josephson relation

Note

Derivation of the gauge-independent relation between the phase and the electric field

Note: