Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 5)

5.1

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

You can redefine $\hat q' := \hat q - q_g$ with the same commutation relation $[\hat \phi,\,\hat q'] = i\hbar$ .

So you can set $q_g = 0$ without losing the generality of the discussion.

Now,

　 $\displaystyle \begin{cases} \hat\phi = \sqrt{\frac{\hbar Z}{2}}(\hat a+\hat a^\dagger) \\ \hat q = \sqrt{\frac{\hbar}{2Z}}i(\hat a^\dagger - \hat a)\end{cases} \ \ \Longleftrightarrow \ \ \begin{cases} \hat a = \sqrt{\frac{1}{2\hbar Z}}\hat\phi + i\sqrt{\frac{Z}{2\hbar}}\hat q \\ \hat a^\dagger = \sqrt{\frac{1}{2\hbar Z}}\hat\phi - i\sqrt{\frac{Z}{2\hbar}}\hat q\end{cases}$

and,

　 $[a,\hat a^\dagger] = 1$

Hence,

　 $\langle 0\mid\hat\phi\mid 0\rangle = \langle 0\mid\hat q\mid 0\rangle = 0\ \ (\because\, \hat a\mid 0\rangle = 0,\,\langle 0\mid \hat a^\dagger =0)$

　 $\displaystyle \langle 0\mid\hat\phi^2\mid 0\rangle = \frac{\hbar Z}{2}\langle 0\mid\hat a\hat a^\dagger\mid 0\rangle = \frac{\hbar Z}{2}\ \ (\because\, \hat a\hat a^\dagger = \hat a^\dagger \hat a + 1)$

　 $\displaystyle \langle 0\mid\hat q^2\mid 0\rangle = \frac{\hbar}{2Z}\langle 0\mid\hat a\hat a^\dagger\mid 0\rangle = \frac{\hbar}{2Z}$

By scaling back with $q_g$ ,

　 $\displaystyle \langle 0\mid\hat q\mid 0\rangle = q_g$

　 $\displaystyle \langle 0\mid(\hat q - q_g)^2\mid 0 \rangle = \frac{\hbar}{2Z}$

　 $\displaystyle {\rm (LHS)} = \langle 0\mid \hat q^2\mid 0\rangle -2 q_g\langle 0\mid\hat q\mid 0\rangle + q_g^2 = \langle 0\mid \hat q^2\mid 0\rangle - q_g^2$

　 $\displaystyle \therefore \langle 0\mid \hat q^2\mid 0\rangle = \frac{\hbar}{2Z}+q_g^2$

For $\hat V$ and $\hat I$ , use the relationship $\displaystyle \hat V = \frac{\hat q}{C},\ \hat I = -\frac{\hat\phi}{L}$

5.2

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5.3

　 $D(\alpha)^\dagger \hat a D(\alpha) = \hat a + \alpha\ \Leftrightarrow\ [\hat a,\, D(\alpha)] = \alpha D(\alpha)$

Hence, the coherent state is an eigen state of $\hat a$ .

　 $\hat a\mid\alpha\rangle = \alpha\mid\alpha\rangle$

So,

　 $\langle\alpha\mid (\hat a +\hat a^\dagger) \mid\alpha\rangle = \alpha + \alpha^*$

　 $\langle\alpha\mid (\hat a +\hat a^\dagger)^2 \mid\alpha\rangle = \langle\alpha\mid (\hat a^2 +\hat a^{\dagger 2} +2\hat a^\dagger \hat a +1) \mid\alpha\rangle = \alpha^2+\alpha^{*2}+2\alpha^*\alpha+1$

Now,

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

　 $\displaystyle (\Delta\hat\phi)^2 = \langle\alpha\mid \hat\phi^2 \mid\alpha\rangle - \langle\alpha\mid \hat\phi \mid\alpha\rangle^2$

　　 $\displaystyle =\frac{\hbar Z}{2}\left\{ \alpha^2+\alpha^{*2}+2\alpha^*\alpha+1 - (\alpha+\alpha^*)^2 \right\}=\frac{\hbar Z}{2}$

Likewise,

　 $\displaystyle (\Delta\hat q)^2=\frac{\hbar}{2Z}$

Hence,

　 $\displaystyle \Delta\hat\phi\Delta\hat q = \frac{\hbar}{2}$

5.4

Assuming that the lowest energy can be realized with a coherent state $\mid\alpha\rangle$ , find the corresponding $\alpha$ .

　 $\displaystyle \hat H = \hbar\omega\hat a^\dagger \hat a + i\hbar \Omega_0(\hat a^\dagger - \hat a)$

　 $\displaystyle \langle\alpha\mid\hat H\mid\alpha\rangle = \hbar\omega \alpha^*\alpha + i\hbar\Omega_0 \alpha^* - i\hbar\Omega_0\alpha$

　 $\displaystyle \frac{\partial}{\partial \alpha^*}\langle\alpha\mid\hat H\mid\alpha\rangle = 0\ \Leftrightarrow\ \alpha = -i\frac{\Omega_0}{\omega}$

With this $\alpha$ , the coherent state $\mid\alpha\rangle$ becomes an eigen state as below:

　 $\displaystyle \hat H\mid\alpha\rangle = \left\{(\hbar\omega\hat a^\dagger \alpha + i\hbar\Omega_0\hat a^\dagger) - i\hbar\Omega_0\alpha\right\}\mid\alpha\rangle =-\hbar\frac{\Omega_0^2}{\omega}\mid\alpha\rangle$

Note: this is not a rigorous proof because it's based on the assumption mentioned above. I'm not sure how I can justify the assumption.

5.5

　 $\displaystyle \hat a(t) = U^\dagger(t)\hat a(0)U(t)$

　 $\displaystyle\therefore\ \hat a(t)\mid\Psi (0)\rangle = U^\dagger(t)\hat a(0)U(t)\mid\Psi (0)\rangle = U^\dagger(t)\hat a(0)\mid\Psi (t)\rangle$

On the other hand,

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

　 $\displaystyle\therefore\ \hat a(t)\mid\Psi (0)\rangle = \left\{e^{-i\omega t}\hat a(0)+\alpha(t)\right\}\mid\Psi (0)\rangle = \alpha(t)\mid\Psi (0)\rangle\ \ (\because \,\mid\Psi (0)\rangle = \mid 0\rangle)$

Hence,

　 $\displaystyle U^\dagger(t)\hat a(0)\mid\Psi (t)\rangle = \alpha(t)\mid\Psi (0)\rangle$

　 $\displaystyle \therefore\ \hat a(0)\mid\Psi (t)\rangle = \alpha(t)U(t)\mid\Psi (0)\rangle = \alpha(t)\mid\Psi (t)\rangle$

By the way, if $\Omega(t)$ is a constant $\Omega(t)=\Omega_0$ ,

　 $\displaystyle \alpha(t) = \int_0^te^{-i\omega(t-\tau)}\Omega_0d\tau = -i\frac{\Omega_0}{\omega}(1-e^{-i\omega t})$

So, the state $\mid\Psi(t)\rangle$ oscillates between $\mid 0 \rangle$ and the ground state found in Exercise 5.4.

5.6

(i) Coherent state

Definition

　Coherent state: $\displaystyle \rho = \mid\alpha\rangle\langle\alpha\mid,\ \alpha = \frac{1}{\sqrt{2}}(x_0 + ip_0)$

　Integration variable of the Fourier transformation: $\displaystyle \mathbf Y = \begin{pmatrix}y \\ p_y\end{pmatrix},\ \eta = \frac{1}{\sqrt{2}}(y+ip_y)$

　Displacement operator: $\displaystyle D(\mathbf Y) = e^{i(p_y\hat x-y\hat p)}=e^{\eta\hat a^\dagger-\eta^*\hat a} = e^{\eta\hat a^\dagger}e^{-\eta^*\hat a}e^{-\frac{1}{2}\eta^*\eta}$

　　( $\displaystyle\because \,e^{A+B} = e^Ae^Be^{-\frac{1}{2}[A, B]}$ if $[A, [A, B]] = [B, [A, B]] = 0$ )

Now,

　 $\displaystyle {\rm tr}[\rho D(\mathbf Y)] = \langle\alpha\mid e^{\eta\hat a^\dagger}e^{-\eta^*\hat a}e^{-\frac{1}{2}\eta^*\eta}\mid\alpha\rangle =e^{\eta\alpha^*}e^{-\eta^*\alpha}e^{-\frac{1}{2}\eta^*\eta}$

　　 $\displaystyle = \exp\left\{\frac{1}{2}(y+ip_y)(x_0-ip_0)-\frac{1}{2}(y-ip_y)(x_0+ip_0)-\frac{1}{4}(y^2+p_y^2)\right\}$

So,

　 $\displaystyle\mathbf R=\frac{1}{2}(x+ip_x)$

　 $\displaystyle W_\rho(\mathbf R) = \frac{1}{(2\pi)^2}\int dy\,dp_y\,e^{iyp_x-ixp_y}{\rm tr}[\rho D(\mathbf Y)]$

　　 $\displaystyle = \frac{1}{(2\pi)^2}\int dy\,dp_y\, \exp\left\{iyp_x-ixp_y + \frac{1}{2}(y+ip_y)(x_0-ip_0)-\frac{1}{2}(y-ip_y)(x_0+ip_0)-\frac{1}{4}(y^2+p_y^2)\right\}$

　　 $\displaystyle = \frac{1}{(2\pi)^2}\int dy\,dp_y\, \exp\left[-\frac{1}{4}\left\{y-2ip_x-(x_0-ip_0)+(x_0+ip_0)\right\}^2\right]\exp\left\{\frac{1}{4}(-2ip_x+2ip_0)^2\right\}$

　　　　 $\displaystyle\times\exp\left[-\frac{1}{4}\left\{p_y+2ix-i(x_0-ip_0)-i(x_0+ip_0)\right\}^2\right]\exp\left\{\frac{1}{4}(-2ix+2ix_0)^2\right\}$

　　 $\displaystyle =\frac{1}{(2\pi)^2}(\sqrt{4\pi})^2e^{-(p_x-p_0)^2}e^{-(x-x_0)^2}=\frac{1}{\pi}e^{-\left\{(x-x_0)^2+(p_x-p_0)^2\right\}}$

(ii) Number state $\mid 1\rangle$

　 $\rho = \mid 1\rangle\langle 1\mid$

　 $\displaystyle {\rm tr}[\rho D(\mathbf Y)] = \langle 1 \mid e^{\eta\hat a^\dagger} e^{-\eta^*\hat a}e^{-\frac{1}{2}\eta^*\eta}\mid 1\rangle =\langle 1\mid(1+\eta\hat a^\dagger)(1-\eta^*\hat a)\mid 1\rangle e^{-\frac{1}{2}\eta^*\eta}$

　　 $\displaystyle =\left(\langle 1\mid + \eta\langle 0\mid\right)\left(\mid1\rangle -\eta^*\mid 0\rangle\right) e^{-\frac{1}{2}\eta^*\eta} =(1-\eta\eta^*)e^{-\frac{1}{2}\eta^*\eta}$

　　 $\displaystyle =\left(1-\frac{1}{2}y^2-\frac{1}{2}p_y^2\right)e^{-\frac{1}{4}(y^2+p_y^2)}$

　 $\displaystyle W_\rho(\mathbf R) = \frac{1}{(2\pi)^2}\int dy\,dp_y\,e^{iyp_x-ixp_y}{\rm tr}[\rho D(\mathbf Y)]$

　　 $\displaystyle = \frac{1}{(2\pi)^2}\int dy\,dp_y\,\exp\left\{iyp_x-ixp_y-\frac{1}{4}y^2-\frac{1}{4}p_y^2\right\}\left(1-\frac{1}{2}y^2-\frac{1}{2}p_y^2\right)$

　　 $\displaystyle = \frac{1}{(2\pi)^2}\int dy\,dp_y\,\exp\left\{-\frac{1}{4}(y+2ip_x)^2-\frac{1}{4}(p_y-2ix)^2\right\}$

　　　　 $\displaystyle\times\exp\left\{\frac{1}{4}(2ip_x)^2+\frac{1}{4}(2ix)^2\right\}\left(1-\frac{1}{2}y^2-\frac{1}{2}p_y^2\right)$

　　 $\displaystyle = \frac{1}{(2\pi)^2}e^{-(x^2+p_x^2)} \int dy'\,dp'_y\,\exp\left(-\frac{1}{4}y'^2\right)\exp\left(-\frac{1}{4}p_y'^2\right)$

　　　　 $\displaystyle\times\left\{1-\frac{1}{2}(y'-2ip_x)^2-\frac{1}{2}(p'_y+2ix)^2\right\}\ (y'=y+2ip_x,\,p_y'=p_y-2ix)$

　　 $\displaystyle = \frac{1}{(2\pi)^2}e^{-(x^2+p_x^2)} \int dy'\,dp'_y\,\exp\left(-\frac{1}{4}y'^2\right)\exp\left(-\frac{1}{4}p_y'^2\right)$

　　　　 $\displaystyle \times \left\{ 1+2(x^2+p_x^2)- \frac{1}{2} (y'^2+p_y'^2) \right\}$

　　 $\displaystyle\left(\because\,\int dy'\exp\left(-\frac{1}{4}y'^2\right)y' = 0,\, \int dp_y'\exp\left(-\frac{1}{4}p_y'^2\right)p_y' = 0\right)$

　　 $\displaystyle = \frac{1}{(2\pi)^2}e^{-(x^2+p_x^2)} \left[ (2\sqrt{\pi})^2\left\{1+2(x^2+p_x^2)\right\}-\frac{1}{2}\times 2\times 2\sqrt{\pi}\times \frac{4}{2} \sqrt{4\pi}\right]$

　　 $\displaystyle = \frac{1}{\pi}\left\{2(x^2+p_x^2)-1\right\}e^{-(x^2+p_x^2)}$

(iii) Schrödinger cat state

　 $\displaystyle\mid\Psi\rangle = \frac{1}{\sqrt{2}}\left(\mid\alpha\rangle+e^{i\theta}\mid-\alpha\rangle\right)$

Note: strictly speaking, since $\displaystyle \langle -\alpha\mid\alpha\rangle = e^{-2\alpha^*\alpha}$ , the normalization factor should be $\displaystyle \sqrt{2+2\cos\theta e^{-2\alpha^*\alpha}}$ . This is approximately $\sqrt{2}$ if $\alpha$ is sufficiently large.

　 $\displaystyle\rho = \mid\Psi\rangle\langle\Psi\mid = \frac{1}{2}\left( \mid\alpha\rangle\langle\alpha\mid + \mid -\alpha\rangle\langle -\alpha\mid + e^{i\theta} \mid -\alpha\rangle\langle \alpha\mid + e^{-i\theta} \mid\alpha\rangle\langle -\alpha\mid\right)$

Thanks to the linearity of trace, the Wigner function is the sum of the ones for above four terms.

　 $\displaystyle W_\rho(\mathbf R) = W_1(\mathbf R) + W_2(\mathbf R) + W_3(\mathbf R) + W_4(\mathbf R)$

$W_1(\mathbf R)$ and $W_2(\mathbf R)$ are the same as ones for the coherent states:

　 $\displaystyle W_1(\mathbf R) = \frac{1}{2\pi}e^{-\left\{(x-x_0)^2+(p_x-p_0)^2\right\}}$

　 $\displaystyle W_2(\mathbf R) = \frac{1}{2\pi}e^{-\left\{(x+x_0)^2+(p_x+p_0)^2\right\}}$

For the third term,

　 $\displaystyle{\rm tr}\left[\frac{1}{2}e^{i\theta}\mid -\alpha\rangle\langle\alpha\mid D(\mathbf Y)\right] = \frac{1}{2}e^{i\theta}\langle\alpha\mid e^{\eta\hat a^\dagger} e^{-\eta^*\hat a}\mid -\alpha\rangle e^{-\frac{1}{2}\eta^*\eta}= \frac{1}{2}e^{i\theta}\langle\alpha\mid -\alpha\rangle e^{\eta\alpha^*}e^{\eta^*\alpha}e^{-\frac{1}{2}\eta^*\eta}$

　　 $\displaystyle = \frac{1}{2}e^{i\theta}e^{-2\alpha^*\alpha}\exp\left\{ \frac{1}{2}(y+ip_y)(x_0-ip_0)+\frac{1}{2}(y-ip_y)(x_0+ip_0)-\frac{1}{4}(y^2+p_y^2)\right\}$

　 $\displaystyle W_3(\mathbf R) = \frac{1}{(2\pi)^2}\int dy\,dp_y\,e^{iyp_x-ixp_y}{\rm tr}\left[\frac{1}{2}e^{i\theta}\mid -\alpha\rangle\langle\alpha\mid D(\mathbf Y)\right]$

　　 $\displaystyle = \frac{1}{8\pi^2}e^{i\theta}e^{-2\alpha^*\alpha}\int dy\,dp_y\,\exp\left\{ iyp_x-ixp_y+\frac{1}{2}(y+ip_y)(x_0-ip_0)+\frac{1}{2}(y-ip_y)(x_0+ip_0)-\frac{1}{4}(y^2+p_y^2)\right\}$

　　 $\displaystyle = \frac{1}{8\pi^2}e^{i\theta}e^{-2\alpha^*\alpha}\int dy\,dp_y\,\exp\left[-\frac{1}{4}\left\{y-2ip_y-(x_0-ip_0)-(x_0+ip_0)\right\}^2\right] \exp\left\{ \frac{1}{4}(-2ip_x-2x_0)^2\right\}$

　　 $\displaystyle\times\exp\left[-\frac{1}{4}\left\{p_y+2ix-i(x_0-ip_0)+i(x_0+ip_0)\right\}^2\right] \exp\left\{\frac{1}{4}(2ix-2p_0)^2\right\}$
　
　 $\displaystyle = \frac{1}{8\pi^2}e^{i\theta}e^{-2\alpha^*\alpha} 4\pi e^{(x_0+ip_x)^2 - (x+ip_0)^2} = \frac{1}{2\pi}e^{i\theta} e^{-(x_0^2+p_0^2)}e^{(x_0+ip_x)^2 - (x+ip_0)^2}$

Likewise,

　 $\displaystyle W_4(\mathbf R) = \frac{1}{2\pi}e^{-i\theta}e^{-(x_0^2+p_0^2)}e^{(-x_0+ip_x)^2 - (x-ip_0)^2}$

Finally, we have:

　 $\displaystyle W_\rho(\mathbf R) = \frac{1}{2\pi}e^{-\left\{(x-x_0)^2+(p_x-p_0)^2\right\}} + \frac{1}{2\pi}e^{-\left\{(x+x_0)^2+(p_x+p_0)^2\right\}}$

　　 $\displaystyle + \frac{1}{2\pi}e^{i\theta} e^{-(x_0^2+p_0^2)}e^{(x_0+ip_x)^2 - (x+ip_0)^2} + \frac{1}{2\pi}e^{-i\theta}e^{-(x_0^2+p_0^2)}e^{(-x_0+ip_x)^2 - (x-ip_0)^2}$

Coherent state is Gaussian and saturates the uncertainty relation. Others are not.

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5.7

From the relationship $\phi = LI$ , choose locations where the average flax $\phi$ becomes maximum.

(1) The both end for $\lambda/2$ , and one end with the capacitor for $\lambda/4$ .

(2) The both end and the middle point for $\lambda/2$ , and $d/3$ and $d$ for $\lambda/4$ .

(3) The points where the average flax is the same for $\omega_o$ and $\omega_1$ .

5.8

(a) The systems's Hamiltonian is:

　 $\displaystyle H = \omega\hat a^\dagger\hat a + \Omega_0e^{i\omega_dt}\hat a + \Omega^*_0e^{-i\omega_dt}\hat a^\dagger$

　　 $\displaystyle = \omega_d\hat a^\dagger\hat a + \delta\hat a^\dagger\hat a + \Omega_0e^{i\omega_dt}\hat a + \Omega^*_0e^{-i\omega_dt}\hat a^\dagger\ (\delta = \omega-\omega_d)$

Take the external force $\omega_d$ as frame of reference, that is, set $\displaystyle H_0 = \omega_d\hat a^\dagger\hat a$ . Then, the interaction Hamiltonian is:

　 $\displaystyle H_I = e^{iH_0t}(\delta\hat a^\dagger\hat a + \Omega_0e^{i\omega_dt}\hat a + \Omega^*_0e^{-i\omega_dt}\hat a^\dagger)e^{-iH_0t}$

　　 $\displaystyle = \delta\hat a^\dagger\hat a + \Omega_0e^{i\omega_dt} \cdot e^{i\omega_dt\hat a^\dagger\hat a}\hat ae^{-i\omega_dt\hat a^\dagger\hat a} + \Omega^*_0e^{-i\omega_dt}\cdot e^{i\omega_dt\hat a^\dagger\hat a}\hat a^\dagger e^{-i\omega_dt\hat a^\dagger\hat a}$

Using the relationship:

　 $\displaystyle e^ABe^{-A} = \sum_{n=0}^\infty (C_A)^nB$ , where $\displaystyle C_AX := [A, X]$

we have,

　 $\displaystyle e^{i\omega_dt\hat a^\dagger\hat a}\hat ae^{-i\omega_dt\hat a^\dagger\hat a} = e^{-i\omega_dt}\hat a, \, e^{i\omega_dt\hat a^\dagger\hat a}\hat a^\dagger e^{-i\omega_dt\hat a^\dagger\hat a} = e^{i\omega_dt}\hat a^\dagger$

Hence, we have a time-independent interaction Hamiltonian:

　 $H_I = \delta\hat a^\dagger\hat a + \Omega_0\hat a + \Omega_0^*\hat a^\dagger$

Suppose that the system is in a thermal bath with the zero-temperature $n=0$ , from (B.15), the interaction with the thermal bath is described as (in the interaction picture):

　 $\displaystyle \frac{d}{dt}\rho = \frac{\kappa}{2}(2\hat a\rho\hat a^\dagger - \hat a^\dagger\hat a\rho -\rho\hat a^\dagger\hat a)$ --- (1)

Also, the interaction with the external force is described as:

　 $\displaystyle \frac{d}{dt}\rho = -i[H_I,\rho]$ --- (2)

By combining (1) and (2), we have:

　 $\displaystyle \frac{d}{dt}\rho = -i[\delta\hat a^\dagger\hat a + \Omega_0\hat a + \Omega_0^*\hat a^\dagger,\rho] + \frac{\kappa}{2}(2\hat a\rho\hat a^\dagger - \hat a^\dagger\hat a\rho -\rho\hat a^\dagger\hat a)$

(b) For $\displaystyle\langle \hat a\rangle$ ,

　 $\displaystyle \frac{d}{dt}\langle \hat a\rangle = {\rm Tr}[\hat a\dot\rho]$

　　 $\displaystyle = -i{\rm Tr}\left[\delta \hat a\hat a^\dagger\hat a\rho +\Omega_0\hat a^2\rho + \Omega_0^*\hat a\hat a^\dagger\rho-\delta\hat a\rho\hat a^\dagger\hat a-\Omega_0\hat a\rho\hat a -\Omega_0^*\hat a \rho\hat a^\dagger\right]$

　　　　 $\displaystyle {}+\frac{\kappa}{2}{\rm Tr}\left[2\hat a^2\rho\hat a^\dagger - \hat a\hat a^\dagger\hat a\rho-\hat a\rho\hat a^\dagger\hat a\right]$

Using the relationships $\displaystyle {\rm Tr}[ABC] = {\rm Tr}[CAB],\, {\rm Tr}[\rho] = 1,\, [\hat a,\hat a^\dagger] = 1$ , we have:

　 $\displaystyle \frac{d}{dt}\langle \hat a\rangle = -\left(\frac{\kappa}{2}+i\delta\right)\langle \hat a \rangle-i\Omega_0^* = -\left(\frac{\kappa}{2}+i\delta\right)\left(\langle a\rangle+\frac{i\Omega_0^*}{\frac{\kappa}{2}+i\delta}\right)$

With the initial condition $\displaystyle \mid\psi(0)\rangle = \mid 0\rangle$ (i.e. $\displaystyle \langle a(0)\rangle = 0$ ), we have:

　 $\displaystyle \langle a\rangle = \frac{i\Omega_0^*}{\frac{\kappa}{2}+i\delta}\left\{e^{-\left(\frac{\kappa}{2}+i\delta\right)t}-1\right\}$ --- (3)

For $\displaystyle\langle \hat n\rangle = \langle \hat a^\dagger\hat a\rangle$ ,

　 $\displaystyle \frac{d}{dt}\langle \hat n\rangle = {\rm Tr}[\hat a^\dagger\hat a\rho]$

　　 $\displaystyle = -i{\rm Tr}[\delta(\hat a^\dagger\hat a)^2\rho + \Omega_0a^\dagger\hat a^2\rho+\Omega_0^*\hat a^\dagger \hat a\hat a^\dagger\rho-\delta \hat a^\dagger\hat a\rho\hat a^\dagger\hat a -\Omega_0\hat a^\dagger\hat a\rho\hat a-\Omega_0^*\hat a^\dagger\hat a\rho\hat a^\dagger]$

　　　　 $\displaystyle {}+\frac{\kappa}{2}{\rm Tr}[2\hat a^\dagger\hat a^2\rho\hat a^\dagger-(\hat a^\dagger\hat a)^2\rho - \hat a^\dagger\hat a\rho\hat a^\dagger\hat a ]$

　　 $\displaystyle = -i\Omega^*_0\langle a^\dagger\rangle + i\Omega_0\langle a\rangle - \kappa\langle\hat n\rangle$

Using (3), we have:

　 $\displaystyle\frac{d}{dt}\langle \hat n\rangle = - \frac{|\Omega_0|^2}{\frac{\kappa}{2}+i\delta}\left\{e^{-\left(\frac{\kappa}{2}+i\delta\right)t}-1\right\}- \frac{|\Omega_0|^2}{\frac{\kappa}{2}-i\delta}\left\{e^{-\left(\frac{\kappa}{2}-i\delta\right)t}-1\right\}-\kappa\langle n\rangle$

　　 $\displaystyle = -|\Omega_0|^2\left\{\frac{1}{\frac{\kappa}{2}+i\delta}e^{-\left(\frac{\kappa}{2}+i\delta\right)t}+\frac{1}{\frac{\kappa}{2}-i\delta}e^{-\left(\frac{\kappa}{2}-i\delta\right)t}\right\} -\kappa\left(\langle n\rangle -N\right)$

where $\displaystyle N = \frac{|\Omega_0|^2}{\kappa}\left(\frac{1}{\frac{\kappa}{2}+i\delta} +\frac{1}{\frac{\kappa}{2}-i\delta}\right) = \frac{|\Omega_0|^2}{\left(\frac{\kappa}{2}\right)^2+\delta^2}$

By setting $\displaystyle \langle \hat n\rangle - N = f(t)e^{-\kappa t}$ , we have:

　 $\displaystyle \frac{df(t)}{dt} = -|\Omega_0|^2\left\{\frac{1}{\frac{\kappa}{2}+i\delta}e^{\left(\frac{\kappa}{2}-i\delta\right)t}+\frac{1}{\frac{\kappa}{2}-i\delta}e^{\left(\frac{\kappa}{2}+i\delta\right)t}\right\}$

　 $\displaystyle \therefore\, f(t) = -|\Omega_0|^2\left\{\frac{1}{\left(\frac{\kappa}{2}+i\delta\right)\left(\frac{\kappa}{2}-i\delta\right)}e^{\left(\frac{\kappa}{2}-i\delta\right)t} + \frac{1}{\left(\frac{\kappa}{2}-i\delta\right)\left(\frac{\kappa}{2}+i\delta\right)}e^{\left(\frac{\kappa}{2}+i\delta\right)t}\right\}+C$

　　 $\displaystyle =-N\left\{e^{\left(\frac{\kappa}{2}-i\delta\right)t}+e^{\left(\frac{\kappa}{2}+i\delta\right)t}\right\}+C$

　 $\displaystyle \therefore\, \langle \hat n\rangle = -N\left\{e^{-\left(\frac{\kappa}{2}+i\delta\right)t}+e^{-\left(\frac{\kappa}{2}-i\delta\right)t}\right\}+Ce^{-\kappa t} + N$

　　　 $\displaystyle = -2Ne^{-\frac{\kappa}{2}t}\cos(\delta t) + Ce^{-\kappa t} + N$

Using the initial condition $\langle\hat n(0)\rangle = 0$ ,

　 $\displaystyle \langle \hat n\rangle = N\left\{1+e^{-\kappa t}-2\cos(\delta t)e^{\frac{\kappa}{2}t}\right\}$

　 $\displaystyle \langle \hat n\rangle = N = \frac{|\Omega_0|^2}{\left(\frac{\kappa}{2}\right)^2+\delta^2}$

Hence, at $\displaystyle\delta = \frac{\kappa}{2}$ , $\langle \hat n\rangle$ becomes half. So, FHW = $\displaystyle 2 \times \frac{\kappa}{2} = \kappa$

5.9

Without the external driving:

　 $\displaystyle \partial_t\rho = \frac{\kappa}{2}(\overline n+1)(2\hat a\rho\hat a^\dagger - \hat a^\dagger\rho - \rho\hat a^\dagger \hat a) + \frac{\kappa}{2}\overline n(2\hat a^\dagger\rho\hat a-\hat a\hat a^\dagger\rho-\rho\hat a\hat a^\dagger)$ --- (1)

　 $\displaystyle\frac{d}{dt}\langle \hat a^\dagger \hat a\rangle = {\rm tr}[\hat a^\dagger\hat a(\partial_t\rho)]$ --- (2)

Substituting (1) into (2), and using the relationships:

${\rm tr}[ABC] = {\rm tr}[BCA]$

$\hat a\hat a^\dagger = \hat a^\dagger \hat a + 1$

You can reorder each term in the normal order, that is, $\displaystyle {\rm tr}[\hat a^{\dagger n}\hat a^m\rho]$ , and you get:

　 $\displaystyle \frac{d}{dt}\langle \hat a^\dagger \hat a\rangle = \kappa\left(\overline n{\rm tr}[\rho]- {\rm tr}[\hat a^\dagger\hat a\rho]\right)$

Since ${\rm tr}[\rho] = 1$ ,

　 $\displaystyle \frac{d}{dt}\langle \hat a^\dagger \hat a\rangle = \kappa\left(\overline n -\langle \hat a^\dagger \hat a\rangle\right)$

　 $\displaystyle \frac{d}{dt}n(t) = \kappa\left\{\overline n -n(t)\right\}$

　 $\displaystyle\therefore\, n(t) = \overline n + \{n(0)-\overline n\}e^{-\kappa t}$

For the decoherence, substituting (1) into $\displaystyle \partial_t\langle 0\mid\rho(t)\mid 1\rangle$ , and using the relationships:

$\hat a\mid n\rangle = \sqrt{n}\mid n-1\rangle$

$\hat a^\dagger\mid n\rangle = \sqrt{n+1}\mid n+1\rangle$

You get:

　 $\displaystyle \partial_t\langle 0\mid\rho(t)\mid 1\rangle = \kappa(\overline n+1)\langle 1\mid\rho(t)\mid 2\rangle - 2\kappa\left(\overline n+\frac{1}{2}\right)\langle 0\mid\rho(t)\mid 1\rangle$

Assuming that all the off-diagonal elements decays at the similar pace $\displaystyle\langle 1\mid\rho(t)\mid 2\rangle\sim \langle 0\mid\rho(t)\mid 1\rangle$ , we have