## 物理代写|热力学代写thermodynamics代考|Measurements by a Qubit Meter

The foregoing results can be illustrated for a TLS state measured by a qubit meter. For a TLS initially in the general state $\left|\psi_{\mathrm{S}}\right\rangle=c_g|g\rangle+c_g|e\rangle$ and a qubit meter in the state $|0\rangle$, a CNOT gate operation on $\left|\psi_{\mathrm{SM}}(0)\right\rangle=\left|\psi_{\mathrm{S}}\right\rangle|0\rangle$ yields the S-M correlated state
$$\left|\psi_{\mathrm{SM}}\left(\tau_{\mathrm{M}}\right)\right\rangle=c_g|g 0\rangle+c_e|e 1\rangle$$
The state (9.29) implies that the SP basis is ${|0\rangle,|1\rangle}$, whereas the general pointer basis is
$$\left|R_0\right\rangle=a|0\rangle+b|1\rangle, \quad\left|R_1\right\rangle=b^|0\rangle-a^|1\rangle$$
with $|a|^2+|b|^2=1$.
According to (9.17),
$$\left|\tilde{S}_0^{\prime}\right\rangle=a^* c_g|g\rangle+b^* c_e|e\rangle, \quad\left|\tilde{S}_1^{\prime}\right\rangle=b c_g|g\rangle-a c_e|e\rangle .$$
Now
$$\mathbf{E}=\left(\begin{array}{ll} |a|^2 & |b|^2 \ |b|^2 & |a|^2 \end{array}\right)$$
and its determinant
$$D=|a|^4-|b|^4=|a|^2-|b|^2=1-2|b|^2 .$$
When $D \neq 0, c$ can be computed from (9.22), yielding
$$\left|c_g\right|^2=\frac{p_0-|b|^2}{1-2|b|^2}, \quad\left|c_e\right|^2=\frac{p_1-|b|^2}{1-2|b|^2} .$$
When $D=0,|a|^2=|b|^2=1 / 2$, that is, the pointer basis is of the form
$$\left{\left(|0\rangle+e^{i \phi}|1\rangle\right) / \sqrt{2}, \quad\left(|0\rangle-e^{i \phi}|1\rangle\right) / \sqrt{2}\right},$$
where $\phi$ is an arbitrary phase. Such pointer bases, lying in the $x y$-plane of the Bloch sphere, are all the bases that are unbiased with respect to the SP basis. Measurements employing these pointers do not provide information on the system, yielding instead the random results [see (9.28)]
$$p_0=p_1=1 / 2$$

## 物理代写|热力学代写thermodynamics代考|Decoherence of Entangled System–Meter States

In the preceding section we disproved the statement (a) presented at the beginning of Section 9.1. Below we also disprove statement (b) therein.

Let us now account for the meter-bath (M-B) interaction via a term in the total Hamiltonian written in the interaction representation,
$$H^{\mathrm{I}}(t)=H_{\mathrm{SM}}^{\mathrm{I}}(t)+H_{\mathrm{MB}}^{\mathrm{I}}(t)$$
We assume that there is a nondegenerate meter variable $\hat{Q}$ that satisfies the backaction evasion condition for the M-B interaction,
$$\left[\hat{Q}, H_{\mathrm{MB}}^{\mathrm{I}}(t)\right]=0 .$$
This ensures that the eigenstates $\left|Q_n\right\rangle$ of $\hat{Q}$ are invariant under decoherence. The eigenstates $\left|Q_n\right\rangle$ are sometimes called “pointer states,” although the basis $\left{\left|Q_n\right\rangle\right}$ generally differs from the SP basis.

Still, let us consider first the case where the SP basis coincides with $\left{\left|Q_n\right\rangle\right}$. For simplicity, let us assume that the S-M interaction is much stronger (hence faster) than that of M-B. The whole process then consists of three stages. We may identify as stage 1 the entanglement of $\mathrm{S}$ and $\mathrm{M}$ in the Schmidt basis,
$$\sum_n c_n\left|S_n\right\rangle \otimes|M\rangle \otimes|\phi\rangle \rightarrow \sum_n c_n\left|S_n\right\rangle \otimes\left|Q_n\right\rangle \otimes|\phi\rangle$$
where $|\phi\rangle$ is the initial state of the bath B.

# 热力学代考

## 物理代写|热力学代写thermodynamics代考|Measurements by a Qubit Meter

$$\left|\psi_{\mathrm{SM}}\left(\tau_{\mathrm{M}}\right)\right\rangle=c_g|g 0\rangle+c_e|e 1\rangle$$

$$\left.\left|R_0\right\rangle=a|0\rangle+b|1\rangle, \quad\left|R_1\right\rangle=b|0\rangle-a^{\mid} 1\right\rangle$$

$$\left|\tilde{S}_0^{\prime}\right\rangle=a^* c_g|g\rangle+b^* c_e|e\rangle, \quad\left|\tilde{S}_1^{\prime}\right\rangle=b c_g|g\rangle-a c_e|e\rangle .$$

$$\mathbf{E}=\left(\begin{array}{lll} |a|^2 & |b|^2|b|^2 \quad|a|^2 \end{array}\right)$$

$$D=|a|^4-|b|^4=|a|^2-|b|^2=1-2|b|^2 .$$什么时候 $D \neq 0, c$ 可以从 (9.22) 计算得出
$$\left|c_g\right|^2=\frac{p_0-|b|^2}{1-2|b|^2}, \quad\left|c_e\right|^2=\frac{p_1-|b|^2}{1-2|b|^2} .$$

$$p_0=p_1=1 / 2$$

## 物理代写|热力学代写thermodynamics代考|Decoherence of Entangled System–Meter States

(b) 。

$$H^{\mathrm{I}}(t)=H_{\mathrm{SM}}^{\mathrm{I}}(t)+H_{\mathrm{MB}}^{\mathrm{I}}(t)$$

$$\left[\hat{Q}, H_{\mathrm{MB}}^{\mathrm{I}}(t)\right]=0$$

$$\sum_n c_n\left|S_n\right\rangle \otimes|M\rangle \otimes|\phi\rangle \rightarrow \sum_n c_n\left|S_n\right\rangle \otimes\left|Q_n\right\rangle \otimes|\phi\rangle$$

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