## 物理代写|热力学代写thermodynamics代考|Von Neuman’s Analysis of Quantum Measurements

Here we consider only the coupling of the quantum system to a quantum meter, both isolated from the environment. The standard (von Neumann) treatment of quantum measurements is then as follows.

Suppose that the observable to be measured is represented in its basis of (orthonormal) eigenstates $\left|S_n\right\rangle$, as
$$\hat{S}=\sum_n a_n\left|S_n\right\rangle\left\langle S_n\right|$$
with
$$\left\langle S_m \mid S_n\right\rangle=\delta_{m n}$$
whereas the initial state of the system $S$ is
$$\left|\psi_{\mathrm{s}}\right\rangle=\sum_n c_n\left|S_n\right\rangle$$
The system-meter interaction is turned on in the time interval $\left(0, \tau_{\mathrm{M}}\right)$. This interaction correlates the initial factorized state of $\mathrm{S}$ and $\mathrm{M}$, which then obeys the Schmidt decomposition,
$$\left|\psi_{\mathrm{SM}}(0)\right\rangle=\sum_n c_n\left|S_n\right\rangle \otimes|M\rangle \rightarrow\left|\psi_{\mathrm{SM}}\left(\tau_{\mathrm{M}}\right)\right\rangle=\sum_n c_n\left|S_n\right\rangle \otimes\left|P_n\right\rangle,$$
where the meter states also satisfy orthonormality,
$$\left\langle P_m \mid P_n\right\rangle=\delta_{m n}$$
In the interaction picture, this entangled state is obtained as a result of the action of the system-meter Hamiltonian,
$$H_{\mathrm{SM}}^{\mathrm{I}}(t)=e^{i H_0 t} H_{\mathrm{SM}}(t) e^{-i H_0 t}$$
where $H_0=H_{\mathrm{S}}+H_{\mathrm{M}}, H_{\mathrm{M}}$ being the meter Hamiltonian. For a proper measurement of the observable, we impose the back-action evasion condition,
$$\left[\hat{S}, H_{\mathrm{SM}}^{\mathrm{I}}(t)\right]=0 \text {. }$$
We assume that $H_{\mathrm{SM}}$ vanishes for $t>\tau_{\mathrm{M}}$, so that the state $\left|\psi_{\mathrm{SM}}(t)\right\rangle=\left|\psi_{\mathrm{SM}}\left(\tau_{\mathrm{M}}\right)\right\rangle$ remains constant for $t>\tau_{\mathrm{M}}$

## 物理代写|热力学代写thermodynamics代考|Measurements in Alternative Pointer Bases

Consider, however, another set of basis states of the meter, $\left{\left|R_n\right\rangle\right}$, so that
$$\left|P_n\right\rangle=\sum_{n^{\prime}}\left|R_{n^{\prime}}\right\rangle\left\langle R_{n^{\prime}} \mid P_n\right\rangle$$
In this basis, the entangled system-meter state in (9.4) changes to
$$\left|\psi_{\mathrm{SM}}\left(\tau_{\mathrm{M}}\right)\right\rangle=\sum_n \sqrt{p_n}\left|\tilde{S}n\right\rangle \otimes\left|R_n\right\rangle$$ where $$\left|\tilde{S}_n^{\prime}\right\rangle=\sqrt{p_n}\left|\tilde{S}_n\right\rangle=\sum{n^{\prime}} c_{n^{\prime}}\left\langle R_n \mid P_{n^{\prime}}\right\rangle\left|S_{n^{\prime}}\right\rangle$$
$\left|\tilde{S}n\right\rangle\left(\left|\tilde{S}_n^{\prime}\right\rangle\right)$ being the new normalized (non-normalized) system states, and $$p_n=\left\langle\tilde{S}_m^{\prime} \mid \tilde{S}_m^{\prime}\right\rangle=\sum{n^{\prime}}\left|\left\langle R_n \mid P_{n^{\prime}}\right\rangle\right|^2\left|c_{n^{\prime}}\right|^2$$
The new states are generally not orthogonal,
$$\left\langle\tilde{S}m \mid \tilde{S}_n\right\rangle=\left(p_m p_n\right)^{-1 / 2} \sum{n^{\prime}}\left|c_{n^{\prime}}\right|^2\left\langle R_n \mid P_{n^{\prime}}\right\rangle\left\langle P_{n^{\prime}} \mid R_m\right\rangle$$
except in the case where all $\left|c_n\right|$ are equal.
Thus, we have replaced the meter observable by
$$\hat{R}=\sum_n r_n\left|R_n\right\rangle\left\langle R_n\right|$$ with some arbitrary eigenvalues $r_n$. Applying the projection postulate to a measurement of $\hat{R}$ yields, in view of (9.11), the eigenvalue $r_n$ with the probability $p_n$, and the system is subsequently left in the state $\left|\tilde{S}_n\right\rangle$.

# 热力学代考

## 物理代写|热力学代写thermodynamics代考|Von Neuman’s Analysis of Quantum Measurements

$$\hat{S}=\sum_n a_n\left|S_n\right\rangle\left\langle S_n\right|$$

$$\left\langle S_m \mid S_n\right\rangle=\delta_{m n}$$

$$\left|\psi_{\mathrm{s}}\right\rangle=\sum_n c_n\left|S_n\right\rangle$$

$$\left|\psi_{\mathrm{SM}}(0)\right\rangle=\sum_n c_n\left|S_n\right\rangle \otimes|M\rangle \rightarrow\left|\psi_{\mathrm{SM}}\left(\tau_{\mathrm{M}}\right)\right\rangle=\sum_n c_n\left|S_n\right\rangle \otimes\left|P_n\right\rangle$$

$$\left\langle P_m \mid P_n\right\rangle=\delta_{m n}$$

$$H_{\mathrm{SM}}^{\mathrm{I}}(t)=e^{i H_0 t} H_{\mathrm{SM}}(t) e^{-i H_0 t}$$

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

## 物理代写|热力学代写thermodynamics代考|Measurements in Alternative Pointer Bases

$$\left|P_n\right\rangle=\sum_{n^{\prime}}\left|R_{n^{\prime}}\right\rangle\left\langle R_{n^{\prime}} \mid P_n\right\rangle$$

$$\left|\psi_{\mathrm{SM}}\left(\tau_{\mathrm{M}}\right)\right\rangle=\sum_n \sqrt{p_n}|\tilde{S} n\rangle \otimes\left|R_n\right\rangle$$

$$\left|\tilde{S}n^{\prime}\right\rangle=\sqrt{p_n}\left|\tilde{S}_n\right\rangle=\sum n^{\prime} c{n^{\prime}}\left\langle R_n \mid P_{n^{\prime}}\right\rangle\left|S_{n^{\prime}}\right\rangle$$
$|\tilde{S} n\rangle\left(\left|\tilde{S}n^{\prime}\right\rangle\right)$ 是新的规范化 (非规范化) 系统状态，并且 $$p_n=\left\langle\tilde{S}_m^{\prime} \mid \tilde{S}_m^{\prime}\right\rangle=\sum n^{\prime}\left|\left\langle R_n \mid P{n^{\prime}}\right\rangle\right|^2\left|c_{n^{\prime}}\right|^2$$

$$\left\langle\tilde{S} m \mid \tilde{S}n\right\rangle=\left(p_m p_n\right)^{-1 / 2} \sum n^{\prime}\left|c{n^{\prime}}\right|^2\left\langle R_n \mid P_{n^{\prime}}\right\rangle\left\langle P_{n^{\prime}} \mid R_m\right\rangle$$除了在所有情况下 $\left|c_n\right|$ 是平等的。

$$\hat{R}=\sum_n r_n\left|R_n\right\rangle\left\langle R_n\right|$$

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