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PHY5524课程简介
Preamble: This is a graduate level course in Statistical Mechanics. In this course we will discuss both the formal aspects and applications of statistical mechanics. I urge that you look through several other books listed below and initiate discussion among yourselves on various aspects of this beautiful subject. Pertinent discussions about the homework sets are acceptable as long as these discussions are good learning experiences for you, and not merely to come up with a homework solution. You are always welcome at my office and don’t need to make an appointment.
Prerequisites
Other useful references:
- Statistical Mechanics, S. K. Ma (World-Scientific, Singapore)
- Statistical Mechanics, Landau and Lifshitz (Pergamon press)
- Statistical Mechanics an intermediate course, by Morandi, Napoli, and Ercolessi (World Scientific)
- Statistical Physics: An Introductory Course, Daniel J. Amit and Yosef Verbin (World Scientific)
- Phase transition and critical phenomena, Nigel Goldenfeld (Addition Wesley)
- Statistical Mechanics Donald A. McQuarrie.
- Equilibrium Statistical Physics, M. Plischke and B. Bergersen, World Scientific
Review of basic notions in classical thermodynamics: Zero-th, first law, definition of heat and internal energy, second law, Carnot cycle, absolute thermodynamic temperature, entropy and irreversibility, entropy of systems on a $(V, p)$ diagram, liquid-vapor equilibria and Clapeyron’s equation, Helmholtz and Gibbs free energy, response functions a thin but extremely well written book is Fermi’s book on Thermodynamics, Dover publication
PHY5524 Statistical Mechanics HELP（EXAM HELP， ONLINE TUTOR）
Consider the problem of an isolated particle of mass $m$ moving freely in a one-dimensional box of size $L$. Such a particle satisfies the following one-dimensional Schrödinger equation:
$$
-\frac{\hbar^2}{2 m} \frac{d^2 \varphi(x)}{d x^2}=\epsilon \varphi(x) \text {, with } \varphi(0)=\varphi(L)=0 .
$$
1.a) Obtain the eigenvalues and the normalized eigenvectors of the Schrödinger equation.
Now consider an isolated system of 4 non-interactingparticles of mass $m$ placed in such one-dimensional box. The energy of the system equals $E=63 \epsilon_0$, where
$$
\epsilon_0=\frac{\hbar^2 \pi^2}{2 m L^2}
$$
is the lowest eigenvalue of the Schrödinger equation. Find the entropy of the system for the following cases:
a) Eigenvalues and normalized eigenvectors:
The solution to the Schrödinger equation can be written as a linear combination of sine functions:
$$
\varphi_n(x)=\sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right), \quad n=1,2, \ldots
$$
Substituting this expression into the Schrödinger equation yields:
$$
\frac{\hbar^2 \pi^2}{2 m L^2} n^2 \varphi_n(x)=\epsilon_n \varphi_n(x)
$$
Therefore, the eigenvalues are given by:
$$
\epsilon_n=\frac{\hbar^2 \pi^2}{2 m L^2} n^2
$$
b) Entropy of the system:
Since the particles are non-interacting, we can treat them as independent systems. Each particle can occupy any of the eigenstates is given by $\$ E=4 \backslash$ epsilon_{n_0} $\$$, where $\$ n_{-} 0 \$$ is the index of the eigenstate with energy $\$ \backslash$ epsilon_{n_0}\$ such that $\$ 4 \backslash$ epsilon_{n_0-1} Veq $E<4 \backslash$ epsilon_{n_0}\$. Using the expression for the eigenvalues derived in part (a), we find:
$$
n_0=\sqrt{\frac{2 m E L^2}{\pi^2 \hbar^2}}=\sqrt{\frac{504}{\pi^2}} \approx 10.14
$$
Therefore, the total number of accessible energy levels for the system is $\$ n _{t e x t{\max }}=10 \$$. Since each particle can occupy any of these energy levels independently, the total number of microstates of the system is given by:
$$
\Omega=\left(\sum_{n=1}^{10} 1\right)^4=10^4=10000
$$
where we have used the fact that there are 10 possible energy levels for each particle. The entropy of the system is then given by:
$$
S=k_B \ln \Omega=k_B \ln 10000 \approx 9.21 k_B
$$
where $\$ \mathrm{k} B \$$ is the Boltzmann constant.
a system of 4 distinguishable spinless particles;
For a system of 4 distinguishable spinless particles, the total number of microstates is given by the product of the number of microstates for each particle:
$$
\Omega=\Omega_1 \cdot \Omega_2 \cdot \Omega_3 \cdot \Omega_4
$$
Each particle can occupy any of the energy levels independently. Therefore, the number of microstates for each particle is equal to the total number of energy levels in the box, which is given by the number of eigenfunctions we derived in part $(a): \$ n_{-}{\backslash \operatorname{text}{\max }}=10 \$$.
Thus, the total number of microstates is:
$$
\Omega=10^4=10000
$$
The entropy of the system is then given by:
$$
S=k_B \ln \Omega=k_B \ln 10000 \approx 9.21 k_B
$$
where $\$ \mathrm{k} B \mathrm{~B}$ is the Boltzmann constant.
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