### 物理代写|热力学代写thermodynamics代考|NEM2201

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• 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|Thermodynamic Variables

The number of energy levels below any given $E$ is
$$N(E):=\sum_{n} \theta\left(E-E_{n}\right),$$
the Heaviside (step) function $\theta(x)$ being equal to 1 for $x>0$ and 0 for $x \leq 0$. The entropy associated with this number of levels is defined as
$$\mathcal{S}(E):=k_{\mathrm{B}} \ln N(E),$$
where $k_{\mathrm{B}}$ is Boltzmann’s constant. Commonly, this entropy is an extensive quantity, since it scales for a system with $f \mathrm{DOF}$ as
$$\mathcal{S}(E) / k_{\mathrm{B}}=O(f) .$$
Equation (1.5) implies that for macroscopic $f=O\left(10^{23}\right)$, the level density is staggering even on extremely small energy scales. Hence, the step function $\theta(x)$ in (1.5) may be assumed to be washed out. The level number $N(E)$ then becomes a smooth function of $E$, whose well-defined derivative represents the density of states
$$\Omega(E)=\sum_{n} \delta\left(E-E_{n}\right),$$ the delta-function $\delta(x)=\theta^{\prime}(x)$ being also assumed to be washed out over many energy levels.

The coarse-grained entropy defined by (1.6) leads to the definition of temperature, which applies whether the system is at equilibrium or not:
$$T(E):=1 / S^{\prime}(E) .$$
In accordance with Nernst’s third law of thermodynamics, the entropy and temperature converge to zero as the energy approaches the ground-state value, $E \rightarrow E_{0}$. For macroscopic values of $E-E_{0}$, the dependence of $S$ on $E$ is logarithmic. It then follows from (1.7) and (1.9) that
$$k_{\mathrm{B}} T(E)=O\left(\frac{E-E_{0}}{f}\right),$$
so that, for any macroscopic energy change $\Delta E$,
$$T(E+\Delta E)=T(E)\left[1+O\left(\frac{\Delta E}{E-E_{0}}\right)\right] .$$
All these relations may fail at extremely low temperatures, which are beyond our consideration here (but cf. references in this chapter).

## 物理代写|热力学代写thermodynamics代考|States and Dynamics

A quantum mechanical state of the system is given by a density operator $\rho(t)$, whose evolution,
$$\rho(t)=U_{t} \rho(0) U_{t}^{t},$$
is governed by the unitary propagator
$$U_{t}:=\exp (-i H t / \hbar)=\sum_{n} \exp \left(-i E_{n} t / \hbar\right)|n\rangle\langle n| .$$
Equations (1.12) and (1.13) yield, for an arbitrary initial state $\rho(0)$,
$$\rho(t)=\sum_{m, n} \rho_{m n}(0) e^{-i\left(E_{m}-E_{n}\right) t / \hbar}|m\rangle\langle n|,$$
where $\sum_{m, n}$ is a summation over all $m, n=0,1,2, \ldots, \rho_{m n}(t):=\langle m|\rho(t)| n\rangle$ being the matrix elements of $\rho(t)$.

The ensemble-averaged occupation probability $p_{E_{n}}$ of an eigenvalue $E_{n}$ is given by the expectation value of the projector (1.3) onto the corresponding eigenspace,
$$p_{E_{n}}:=\operatorname{Tr}\left[P_{E_{n}} \rho(t)\right]=\sum_{E_{m}=E_{n}} \rho_{m m}(t)=\sum_{E_{m}=E_{n}} p_{m},$$

where the level population $p_{n}$ is the time-independent expectation value of the observable $|n\rangle\langle n|$,
$$p_{n}:=\operatorname{Tr}[|n\rangle\langle n| \rho(t)]=\rho_{n n}(t)=\rho_{n n}(0),$$
normalized by
$$1=\operatorname{Tr} \rho(t)=\sum_{n} \rho_{n n}(t)=\sum_{n} p_{n}=\sum_{E_{n}} p_{E_{n}} .$$
In what follows, we shall employ the energy basis in which all the non-diagonal elements of $P_{E_{n}} \rho(0) P_{E_{n}}$ vanish,
$$\rho_{m n}(0)=0 \quad \text { if } \quad m \neq n \quad \text { and } \quad E_{m}=E_{n} .$$

## 物理代写|热力学代写thermodynamics代考|Thermodynamic Variables

$$N(E):=\sum_{n} \theta\left(E-E_{n}\right),$$
Heaviside (step) 函数 $\theta(x)$ 等于 $1 x>0$ 和 0 为 $x \leq 0$. 与该级别数相关的樀定义为
$$\mathcal{S}(E):=k_{\mathrm{B}} \ln N(E),$$

$$\mathcal{S}(E) / k_{\mathrm{B}}=O(f) .$$

$$\Omega(E)=\sum_{n} \delta\left(E-E_{n}\right),$$
delta函数 $\delta(x)=\theta^{\prime}(x)$ 也被假定在许多能级上被淘汰。

$$T(E):=1 / S^{\prime}(E) .$$

$$k_{\mathrm{B}} T(E)=O\left(\frac{E-E_{0}}{f}\right),$$

$$T(E+\Delta E)=T(E)\left[1+O\left(\frac{\Delta E}{E-E_{0}}\right)\right] .$$

## 物理代写|热力学代写thermodynamics代考|States and Dynamics

$$\rho(t)=U_{t} \rho(0) U_{t}^{t},$$

$$U_{t}:=\exp (-i H t / \hbar)=\sum_{n} \exp \left(-i E_{n} t / \hbar\right)|n\rangle\langle n| .$$

$$\rho(t)=\sum_{m, n} \rho_{m n}(0) e^{-i\left(E_{m}-E_{n}\right) t / \hbar}|m\rangle\langle n|,$$

$$p_{E_{n}}:=\operatorname{Tr}\left[P_{E_{n}} \rho(t)\right]=\sum_{E_{m}=E_{n}} \rho_{m m}(t)=\sum_{E_{m}=E_{n}} p_{m},$$

$$p_{n}:=\operatorname{Tr}[|n\rangle\langle n| \rho(t)]=\rho_{n n}(t)=\rho_{n n}(0),$$

$$1=\operatorname{Tr} \rho(t)=\sum_{n} \rho_{n n}(t)=\sum_{n} p_{n}=\sum_{E_{n}} p_{E_{n}} .$$

$$\rho_{m n}(0)=0 \quad \text { if } \quad m \neq n \quad \text { and } \quad E_{m}=E_{n} .$$

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