物理代写|费曼图代写Feynman diagram代考|PHY-811

物理代写|费曼图代写Feynman diagram代考|The fundamental postulate of statistical mechanics

Consider an isolated system of $N$ noninteracting, identical particles confined to a region of volume $V$. The Hamiltonian is $H=\sum_{i=1}^N h(i)$, where $h(i)$ is the operator that represents the energy of particle $i$. The single-particle states are obtained by solving the Schrödinger equation $h\left|\phi_1\right\rangle=\epsilon_1\left|\phi_1\right\rangle$, where $v$ stands for all the quantum numbers that characterize the state. The energy $\epsilon_1$ depends on $V$. For example. for a system of noninteracting particles confined to a cube of side length $L$. $\epsilon_{k \sigma}=\hbar^2 k^2 / 2 m$, and if periodic boundary conditions are adopred, then $k_x \cdot k_y, k_z=0, \pm 2 \pi / L \cdot \pm 4 \pi / L \ldots$. The total energy of the system is $E=$ $\sum n_1, \epsilon_1$, where $n_1$ is the number of particles in state $\left|\phi_v\right\rangle$, and the total number of particles is $N=\sum_v n_1$. A macrostate of the system is defined by specifying the values of $N, V$, and $E$.

At the microscopic level, there are many different ways to distribute the energy $E$ among the $N$ particles that comprise the system. Each of these difPerent ways defines a particular microstate that is consistent with the given macrostate. A microstate is a quantum state of the system described by a wave function $\psi(1.2 \ldots N)$. The number of microstates that are consistent with a given macrostate is a function of $N, V$, and $E$, and it is denoted by $\Omega(N, V, E)$. For a macroscopic system consisting of a large number of particles, of the order of Avogadro’s number $\left(6.22 \times 10^{23}\right), \Omega(N, V, E)$ will be, in general, a fantastically large number. The fundamental postulate of statistical mechanics asserts the following: an isolated system in equilibrium, in a given macrostate, is equally likely to be in any of the microstates that are consistent with the given macrostate.

物理代写|费曼图代写Feynman diagram代考|The canonical ensemble

A canonical ensemble is representative of a system at a fixed temperature $T$. We consider a small system $A$ in contact with a heat reservoir $R$ at temperature $T$ (see Figure 5.2). The combined system, $A+R$, is isolated. and its total cnergy is $E_0$, which is a constant. System $A$ is small in the sense that its degrees of freedom are far fewer than those of $R$. What is the probability $p_n$ of finding $A$ in state $\left|\psi_n\right\rangle$ with encrgy $E_n$, once equilibrium has been attained? If $A$ is in stare $\left|\psi_n\right\rangle$. then the number of states of the combined system is simply $\Omega_k\left(E_0-E_n\right)$, which is the number of states accessible to $R$. Therefore, $p_n$ is proportional to $\Omega_R\left(E_0-E_n\right)$.
$$p_n=C \Omega_R\left(E_0-E_n\right)=C e^{\ln \Omega_R\left(E_0-E_n\right)} .$$
We can expand $\ln \Omega_R\left(E_0-E_n\right)$,
\begin{aligned} \operatorname{In} \Omega_R\left(E_0-E_n\right) &=\ln \Omega_R\left(E_0\right)-\left.\frac{\partial \ln \Omega_R}{\partial E}\right|{E_0} E_n+O\left(E_n^2\right) \ &=\ln \Omega_k\left(E_0\right)-\left.\frac{\partial \ln \Omega_R}{\partial E}\right|{E_R+E_n} E_n+O\left(E_n^2\right) \ &=\ln \Omega_R\left(E_0\right)-\left.\frac{\partial \ln \Omega_R}{\partial E}\right|_{E_R} E_n+O\left(E_n^2\right) . \end{aligned}
Since $E_n \ll E_0$. we neglect terms of order higher than $E_n$. Using Eq. (5.2), the above equation may be written as
$$\ln \Omega_R\left(E_0-E_n\right)=\ln \Omega_K\left(E_0\right)-\beta E_n$$
where $\beta=1 / k T$ and $T$ is the temperature of the reservoir. Therefore,
$$p_n=C e^{\ln \Omega_R\left(E_0\right)-\beta E_n}=C \Omega_R\left(E_0\right) e^{-\beta E_n}=C^{\prime} e^{-\beta E_n} .$$

物理代写|费曼图代写Feynman diagram代考|The canonical ensemble

$$p_n=C \Omega_R\left(E_0-E_n\right)=C e^{\ln \Omega_R\left(E_0-E_n\right)} .$$

$$\operatorname{In} \Omega_R\left(E_0-E_n\right)=\ln \Omega_R\left(E_0\right)-\frac{\partial \ln \Omega_R}{\partial E}\left|E_0 E_n+O\left(E_n^2\right) \quad=\ln \Omega_k\left(E_0\right)-\frac{\partial \ln \Omega_R}{\partial E}\right| E_R+E_n E_n$$

$$\ln \Omega_R\left(E_0-E_n\right)=\ln \Omega_K\left(E_0\right)-\beta E_n$$

$$p_n=C e^{\ln \Omega_R\left(E_0\right)-\beta E_n}=C \Omega_R\left(E_0\right) e^{-\beta E_n}=C^{\prime} e^{-\beta E_n} .$$

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