# 物理代写|热力学代写thermodynamics代考|MECH3024

## 物理代写|热力学代写thermodynamics代考|Dynamically Modified Dephasing due to a Quantum Bath

The foregoing results, obtained for dephasing due to classical noise (frequency fluctuations), remain valid, under proper substitutions, for $\sigma_z$-coupling to a quantum bath. Namely, (11.26) is then
$$H_{\mathrm{I}}=\sigma_z B .$$
Upon performing calculations similar to those made in Section 11.3.3, we obtain that in the absence of modulations, (11.166) and (11.168) remain valid if we substitute
$$\Phi_{\mathrm{r}}(t) \rightarrow \Phi_{\mathrm{s}}(t) / 2$$
where
$$\Phi_{\mathrm{s}}(t)=2 \operatorname{Re} \Phi_T(t)=\langle\tilde{B}(t) \tilde{B}(0)+\tilde{B}(0) \tilde{B}(t)\rangle_{\mathrm{B}}$$
is the symmetrized bath correlation function. Then, in (11.166) and (11.168) [cf. (11.167) and (11.180)]
$$\gamma_{\mathrm{d} 0}(t)=2 \int_0^t \Phi_{\mathrm{s}}\left(t^{\prime}\right) d t^{\prime} .$$
The spectrum of $\Phi_{\mathrm{s}}(t)$ [cf. (11.38)] is the symmetrized bath-response spectrum,
$$G_{\mathrm{s}}(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d t \Phi_{\mathrm{s}}(t) e^{i \omega t}=G_T(\omega)+G_T(-\omega) .$$

## 物理代写|热力学代写thermodynamics代考|Dynamical Decoupling

The dynamical control of dephasing considered above requires $\Omega(t)$ never to become too small [see (11.177)]. An alternative control method, called dynamical decoupling (DD), employs short $\pi$-pulses. Originally, DD was suggested for dephasing due to bosonic baths or a classical Gaussian noise, where an exact solution can be obtained. Here, we use our general approach to show that the same theory holds for arbitrary baths, if only the system-bath coupling is sufficiently weak.

We assume the interaction-picture Hamiltonian (11.134) to have $\bar{\epsilon}(t)=1$ and
$$\Omega(t)=\pi \sum_{i=1,2, \ldots} \delta\left(t-t_i\right) \quad\left(0<t_1<t_2<\ldots\right),$$
corresponding to a sequence of impulsive (infinitely short) $\pi$-pulses that are not necessarily equidistant. We next transform this Hamiltonian to a doubly rotating frame by the unitary transformation,
$$U_{\mathrm{dr}}(t)=\exp \left[-\frac{i}{2} \int_0^t d t^{\prime} \Omega\left(t^{\prime}\right) \sigma_x\right]= \begin{cases}(-1)^k, & n(t)=2 k, \ (-1)^k i \sigma_x, & n(t)=2 k-1,\end{cases}$$
where $k$ is an integer and $n(t)=\sum_i \theta\left(t-t_i\right)$ is the number of the pulses in the interval $(0, t)$. Then, the Hamiltonian becomes
$$\tilde{H}(t)=\epsilon(t) \sigma_z \tilde{B}(t)$$
where
$$\epsilon(t)=e^{i \phi(t)}=(-1)^{n(t)},$$
$\phi(t)$ being given by the second line in (11.142).
We can now employ the universal ME (11.45), with $H_{\mathrm{S}}=0$ and $\tilde{S}(\tau, t)=$ $S(t)=\epsilon(t) \sigma_z .$ It has then the form
$$\dot{\rho}=\frac{1}{2} \gamma_{\mathrm{d}}^{\prime}(t)\left(\sigma_z \rho \sigma_z-\rho\right)$$

## 物理代写|热力学代写thermodynamics代考|Dynamically Modified Dephasing due to a Quantum Bath

$$H_{\mathrm{I}}=\sigma_z B$$

$$\Phi_{\mathrm{r}}(t) \rightarrow \Phi_{\mathrm{s}}(t) / 2$$

$$\Phi_{\mathrm{s}}(t)=2 \operatorname{Re} \Phi_T(t)=\langle\tilde{B}(t) \tilde{B}(0)+\tilde{B}(0) \tilde{B}(t)\rangle_{\mathrm{B}}$$

$$\gamma_{\mathrm{d} 0}(t)=2 \int_0^t \Phi_{\mathrm{s}}\left(t^{\prime}\right) d t^{\prime} .$$

$$G_{\mathrm{s}}(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d t \Phi_s(t) e^{i \omega t}=G_T(\omega)+G_T(-\omega)$$

## 物理代写|热力学代写thermodynamics代考|Dynamical Decoupling

$$\Omega(t)=\pi \sum_{i=1,2, \ldots} \delta\left(t-t_i\right) \quad\left(0<t_1<t_2<\ldots\right),$$

$$U_{\mathrm{dr}}(t)=\exp \left[-\frac{i}{2} \int_0^t d t^{\prime} \Omega\left(t^{\prime}\right) \sigma_x\right]=\left{(-1)^k, \quad n(t)=2 k,(-1)^k i \sigma_x, \quad n(t)=2 k-1,\right.$$

$$\tilde{H}(t)=\epsilon(t) \sigma_z \tilde{B}(t)$$

$$\epsilon(t)=e^{i \phi(t)}=(-1)^{n(t)},$$
$\phi(t)$ 由 (11.142) 中的第二行给出。

$$\dot{\rho}=\frac{1}{2} \gamma_{\mathrm{d}}^{\prime}(t)\left(\sigma_z \rho \sigma_z-\rho\right)$$

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