## Brief Discussion about the Teaching of Quantum Mechanics Postulate in Structural Chemistry

Wang Dan,

 基金资助: 兰州大学2019年示范课程建设

 Fund supported: 兰州大学2019年示范课程建设

Abstract

Based on quantum mechanics, structural chemistry reveals the relationship between the structure and the properties of the materials from the microscopic field. This paper analyzes the difficulty in the teaching of quantum mechanics postulates in structural chemistry and puts forward a method to stimulate the students' learning motivation and improve the teaching effect.

Keywords： Structural chemistry ; Quantum mechanics postulate ; Teaching methodology

Wang Dan. Brief Discussion about the Teaching of Quantum Mechanics Postulate in Structural Chemistry. University Chemistry[J], 2020, 35(9): 173-177 doi:10.3866/PKU.DXHX201909018

## 2 量子力学与经典物理学的区别

### 3.1 兴趣的激发

$\begin{array}{l}\int {{\varphi ^*}\varphi {\rm{d}}\tau = A,令} \psi = \frac{1}{{\sqrt A }}\varphi \\\int {{\psi ^*}\psi {\rm{d}}\tau = \int {\frac{1}{{\sqrt A }}{\varphi ^*}\frac{1}{{\sqrt A }}\varphi } } {\rm{d}}\tau = \frac{1}{A}\int {{\varphi ^*}} \varphi {\rm{d}}\tau = \frac{1}{A} \times A = 1\end{array}$

### 3.2 通过提问引导思考

$\mathop v\limits^ \to = \frac{{{\rm{d}}\mathop r\limits^ \to }}{{{\rm{d}}t}}$

$T = \frac{1}{2}m{v^2}$

$< T > = \frac{{{{\int \psi }^*}\mathop T\limits^ \wedge \psi {\rm{d}}\tau }}{{{{\int \psi }^*}\psi {\rm{d}}\tau }}$

“造”动能算符$\mathop T\limits^ \wedge$需要把其经典力学表达式中动量p换成算符形式，注意：平方在算符中为二阶偏导：

$\mathop T\limits^ \wedge = \frac{{{{(\frac{\hbar }{i}\frac{\partial }{{\partial q}})}^2}}}{{2m}} = \frac{1}{{2m}}\frac{{{\hbar ^2}}}{{{i^2}}}\frac{{{\partial ^2}}}{{\partial {q^2}}} = - \frac{{{\hbar ^2}}}{{2m}}\frac{{{\partial ^2}}}{{\partial {q^2}}}$

### 图1

(a)光子通过偏振片；(b)光子被偏振片吸收；(c)光子有几率通过偏振片。

### 3.4 能力提升

${\hat P_{ij}} \psi = C\psi$

${\hat P_{ij}}$为交换算符，其中ij = 1, 2, 3, …, N, ψ为体系的波函数，C为常数，公式(4)表示两个粒子交换坐标后的状态。对于全同粒子，两个粒子交换坐标后，不会改变体系的量子态，ψ${\hat P_{ij}} \psi$描述的是同一个量子态，它们可以相差一个常数因子C

${\hat P_{ij}} ( {\hat P_{ij}} \psi ) = {\hat P_{ij}} (C\psi ) = C( {\hat P_{ij}} \psi ) = C \cdot C\psi = {C^2}\psi$

$\begin{gathered} {\hat P_{ij}} \psi = + \psi \\ {\hat P_{ij}} \psi = - \psi \\ \end{gathered}$

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