### 物理代写|热力学代写thermodynamics代考|AMME2262

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|热力学代写thermodynamics代考|The Problem of Equilibration for Physical Observables

The statistical ensemble $\rho(t)$ is not stationary at short $t$ if $\rho(0)$ is out of equilibrium. Yet, if the right-hand side of (1.14) initially depends on $t$, it cannot approach at large $t$ any time-independent “equilibrium ensemble.” Furthermore, any mixed state $\rho(t)$ returns arbitrarily “near” its initial state $\rho(0)$ at certain times $t$ (analogously, but not identically, to pure-state Poincaré recurrences). In what follows, we examine the apparent contradiction of such recurrences with equilibration.
According to (1.14), there exists at least one $\rho_{\operatorname{mn}}(0) \neq 0$ with
$$\omega:=\left(E_{n}-E_{m}\right) / \hbar \neq 0 .$$
We consider observables represented by Hermitian operators
$$X=\sum_{m, n} X_{m n}|m\rangle\langle n|, \quad X_{m n}:=\langle m|X| n\rangle,$$
with expectation values
$$\langle X\rangle(t):=\operatorname{Tr}[\rho(t) X] .$$
For the observable that represents an interlevel transition,
$$X=\hat{X}+\hat{X}^{\dagger}, \quad \hat{X}:=|m\rangle\langle n| / \rho_{m n}(0),$$
we find from (1.14) that
$$\operatorname{Tr}[\rho(t) X]=2 \cos (\omega t) .$$
Thus, the mean value of $X$ in the ensemble $\rho(t)$ exhibits permanent oscillations, allowing us to conclude that quantum mechanics and equilibration are, in general,incompatible. Nevertheless, as shown below, equilibration can approximately hold true for a restricted class of observables $X$ and initial conditions $\rho(0)$.

## 物理代写|热力学代写thermodynamics代考|Equilibration Conditions

The key requirement on the initial condition $\rho(0)$ is that the ensemble-averaged level populations $p_{n}$ can be split into a locally averaged level population density $h(E)$ and “unbiased fluctuations” $\delta p_{n}$, whose average within the interval around $E$ is vanishingly small compared to $h(E)$,
$$p_{n}=h\left(E_{n}\right)+\delta p_{n} .$$
This requirement should hold within any energy interval, which contains many levels $E_{n}$, but is still exceedingly small on any experimentally resolvable scale.

This initial condition is the result of a preparation process, during which the system was still entangled with the outside world. The reduced initial state (at $t=0$ ) of the system (obtained by tracing out the outside world) must therefore be a mixed state. Any time-dependent system Hamiltonian will cause the spreading of occupation probabilities over neighboring energy levels. Since the levels are so dense, the spreading randomizes the $p_{n}$ ‘s in accordance with (1.25). This preparation process stands in contrast to a “sudden” (discontinuous) parametric change of the Hamiltonian, dubbed quantum quench.
1.2.2 Energy Density
Let us define the ensemble-averaged energy density
$$\rho(E):=\langle\delta(E-H)\rangle,$$
$\rho(E) d E$ being the probability to find an energy value between $E$ and $E+d E$. From (1.16) it follows that
$$\rho(E)=\sum_{n} p_{n} \delta\left(E-E_{n}\right)$$

## 物理代写|热力学代写thermodynamics代考|The Problem of Equilibration for Physical Observables

$$\omega:=\left(E_{n}-E_{m}\right) / \hbar \neq 0 .$$

$$X=\sum_{m, n} X_{m n}|m\rangle\langle n|, \quad X_{m n}:=\langle m|X| n\rangle,$$

$$\langle X\rangle(t):=\operatorname{Tr}[\rho(t) X] .$$

$$X=\hat{X}+\hat{X}^{\dagger}, \quad \hat{X}:=|m\rangle\langle n| / \rho_{m n}(0)$$

$$\operatorname{Tr}[\rho(t) X]=2 \cos (\omega t) .$$

## 物理代写|热力学代写thermodynamics代考|Equilibration Conditions

$$p_{n}=h\left(E_{n}\right)+\delta p_{n} .$$

$1.2 .2$ 能量密度

$$\rho(E):=\langle\delta(E-H)\rangle$$
$\rho(E) d E$ 是找到介于两者之间的能量值的概率 $E$ 和 $E+d E$. 从 (1.16) 可以得出
$$\rho(E)=\sum_{n} p_{n} \delta\left(E-E_{n}\right)$$

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