《Quantum Chemistry》章末总结3
第三章开始迈入量子世界。
Chapter 3 The Schrödinger Equation and a Particle in a Box
The Schrödinger equation is our fundamental equation of quantum mechanics. The solutions to the Schrödinger equation are called wave functions. We will see that a wave function gives a complete quantum-mechanical description of any system.
In this chapter, we present and discuss the version of the Schrodinger equation that does not contain time as a variable. Solutions to the time-independent Schrodinger equation are called stationary-state wave functions because they are independent of time. Many problems of interest to chemists can be treated by using only stationary-state wave functions.
3.1 The Schrödinger Equation Is the Equation for the Wave Function of a Particle
We cannot derive the Schrödinger equation any more than we can derive Newton’s laws, and Newton’s second law, $f=ma$, in particular. We shall regard the Schrödinger equation to be a fundamental postulate, or axiom, of quantum mechanics, just as Newton’s laws are fundamental postulates of classical mechanics.
（ 我们虽然可以通过经典波动方程，配合一些量子化条件，构造出薛定谔方程的形式，但是这并不能算是“推导”，因为我们得到的结果在物理意义上与所谓的薛定谔方程完全不同，通过经典波动方程得到的结果仍然处于经典范畴内。 ）
Let’s start with the classical one-dimensional wave equation for simplicity:
$$
\frac {\partial^2u}{\partial x^2}=\frac 1{v^2}\frac {\partial^2u}{\partial t^2}
$$
We will express the temporal part as $cos\omega t$:
$$
u(x,t)=\psi(x)cos\omega t
$$
Because $\psi(x)$ is the spatial factor of the amplitude $u(x,t)$, we will call $\psi(x)$ the spatial amplitude of the wave. And we can obtain an equation for the spatial amplitude:
$$
\frac {d^2\psi}{dx^2}+\frac {4 \pi^2}{\lambda^2}\psi(x)=0
$$
The total energy of a particle is the sum of its kinetic energy and its potential energy. We now introduce the idea of de Broglie matter waves into the equation, we obtain:
$$
\frac {d^2\psi}{dx^2}+\frac {2m}{\hbar^2}[E-V(x)]\psi(x)=0
$$
Equation is the Schrödinger equation, a differential equation whose solution, $\psi(x)$, describes a particle of mass m moving in a potential field described by $V(x)$.
The equation does not contain time and is called the time-independent Schrödinger equation. The wave functions obtained from equation are called stationary-state wave functions.
Although there is a more general Schrödinger equation that contains a time dependence, we will see that many problems of chemical interest can be described in terms of stationary-state wave functions.
Another form to rewrite the equation:
$$
-\frac {\hbar^2}{2m}\frac {d^2\psi}{dx^2}+V(x)\psi(x)=E\psi(x)
$$
This is a particularly nice way to write the Schrödinger equation when we introduce the idea of an operator in the next section.
（ 这个形式就顺眼多了，因为它用算符表示可以写得非常简洁。 ）
3.2 Classical-Mechanical Quantities Are Represented by Linear Operators in Quantum Mechanics
An operator is a symbol that tells you to do something to whatever follows the symbol. We usually denote an operator by a capital letter with a carat over it (e.g., $\hat A$). Thus, we write:
$$
\hat Af(x)=g(x)
$$
to indicate thatthe operator $\hat A$ operates on $f(x)$ to give a new function $g(x)$.
In quantum mechanics, we deal only with linear operators. An operator is said to be linear if:
$$
\hat A[c_1f_1(x)+c_2f_2(x)]=c_1\hat Af_1(x)+c_2\hat Af_2(x)
$$
（ 关注线性算符的原因是量子力学是基于线性空间（或者，希尔伯特空间$\mathcal{H}$？）讨论问题的。 ）
3.3 The Schrödinger Equation Can Be Formulated as an Eigenvalue Problem
A problem that we frequently encounter in physical chemistry is the following: Given $\hat A$, find a function $\phi(x)$ and a constant $a$ such that:
$$
\hat A\phi(x)=a\phi(x)
$$
Note that the result of operating on the function $\phi(x)$ by $\hat A$ is simply to give $\phi(x)$ back agin, only multiplied by a constant factor. The function $\phi(x)$ is called an eigenfunction of the operator $\hat A$, and $a$ is called an eigenvalue. The problem of determining $\phi(x)$ and $a$ for a given $\hat A$ is called an eigenvalue problem.
Let’s go back to equation mentioned in section 3.1. We can write the left side of equation in the form:
$$
\left[-\frac {\hbar^2}{2m}\frac {d^2}{dx^2}+V(x) \right]\psi(x)=E\psi(x)
$$
If we denote the operator in brackets by $\hat H$:
$$
\hat H\psi(x)=E\psi(x)
$$
We have formulated the Schrödinger equation as an eigenvalue problem. The operator $\hat H$ is called the Hamiltonian operator. The wave function is an eigenfunction, and the energy is an eigenvalue of the Hamiltonian operator.
If $V(x)=0$ in the equation, the energy is all kinetic energy and so we define a kinetic energy operator according to:
$$
\hat T_x=-\frac {\hbar^2}{2m}\frac {d^2}{dx^2}
$$
Furthermore, classically, $T=p^2/2m$, and so we conclude that:
$$
\hat p_x^2=-\hbar^2\frac {d^2}{dx^2}
$$
We can interpret the operator $\hat p_x^2$; by considering the case of two operators acting sequentially.
3.4 Wave Functions Have a Probabilistic Interpretation
In this section, we will study the case of a free particle of mass m constrained to lie along the x axis between $x=0$ and $x=a$. This case is called the problem of a particle in a one-dimensional box.
The terminology free particle means that the particle experiences no potential energy or that $V(x)=0$. We see that the Schrödinger equation for a free particle in a one-dimensional box is:
$$
\frac {d^2\psi}{dx^2}+\frac {2mE}{\hbar^2}\psi(x)=0
$$
The particle is restricted to the region $0\le x\le a$ and so cannot be found outside this region (see Figure 3.1). To implement the condition that the particle is restricted to the region $0\le x\le a$, we must formulate an interpretation of the wave function $\psi(x)$.
Max Born, a German physicist working in scattering theory, pointed $\psi^*(x)\psi(x)dx$ as the probability that the particle is located between $x$ and $x+dx$. Born’s view is now generally accepted.
Because the particle is restricted to the region $0\le x \le a$, the probability that the particle is found outside this region is zero. Consequently, we shall require that $\psi(x)=0$ outside the region $0\le x\le a$. Furthermore, because $\psi(x)$ is a measure of the position of the particle, we shall require $\psi(x)$ to be a continuous function.
$$
\psi(0)=\psi(a)=0
$$
These are boundary conditions that we impose on the problem.
（ 为了获取方程的边界条件，首先需要考虑波函数$\psi(x)$的物理意义。薛定谔最初从电荷入手，他认为电荷量×波函数的平方为空间的电荷密度，据此他设想电荷在整个空间中都有分布。不过几年后波恩发现了这种说法的问题，并提出了概率波修正薛定谔的解释，现在被广泛接受。 ）
3.5 The Energy of a Particle in a Box Is Quantized
The general solution of equation of a one-dimensional box is:
$$
\psi(x)=Acoskx+Bsinkx \quad k=\frac {(2mE)^{1/2}}{\hbar}
$$
Apply the boundary conditions and avoid the trivial or physically uninteresting solution, the other choice is that:
$$
ka=n\pi \quad n=1,2,\cdots
$$
We find that:
$$
E_n=\frac {h^2n^2}{8ma^2} \quad n=1,2,\cdots
$$
Thus, the energy turns out to have only the discrete values and no other values. The energy of the particle is said to be quantized and the integer $n$ is called a quantum number.
The wave function corresponding to $E_n$ is:
$$
\psi(x)=Bsinkx=Bsin\frac {n\pi x}a \quad n=1,2,\cdots
$$
These wave functions are plotted in Figure 3.2. They look just like the standing waves set up in a vibrating string. Note that the energy increases with the number of nodes.
The energy levels, wave functions (a), and probability densities (b) for the particle in a box.
The model of a particle in a one-dimensional box has been applied to the $\pi$ electrons in linear conjugated hydrocarbons. Although butadiene, like all polyenes, is not a linear molecule, we will assume for simplicity that the rr electrons in butadiene move along a straight line whose length can be estimated as equal to two $C=C$ bond lengths ($2\times135pm$) plus one $C-C$ bond ($154pm$) plus the distance of a carbon atom radius at each end ($2 \times 77.0pm=154pm$), giving a total distance of $578pm$. According to the equation, the allowed $\pi$ electronic energies are given by:
$$
E_n=\frac {h^2n^2}{8m_ea^2}\quad n=1,2,\cdots
$$
But the Pauli exclusion principle (which we discuss later but is assumed here to be known from general chemistry) says that each of these states can hold only two electrons (with opposite spins), and so the four $\pi$ electrons fill the first two levels as shown in Figure 3.3.
The first excited state of this system of four $\pi$ electrons is that which has one electron elevated from the $n=2$ state to the $n=3$ state (cf. Figure 3.3), and the
energy to make a transition is:
$$
\Delta E=\frac {h^2}{8m_ea^2}(3^2-2^2)=\frac {(6.626\times10^{-34}J\cdot s)^2\cdot5}{8(9.109\times10^{-31}kg)(578\times10^{-12}m)^2}=9.02\times10^{-19}J
$$
and
$$
\tilde \nu=4.54\times10^4cm^{-1}
$$
Butadiene has an absorption band at $4.64\times10^4cm^{-1}$ , and so we see that this very simple model, called the free-electron model, can be somewhat successful at explaining the absorption spectrum of butadiene.
3.6 Wave Functions Must Be Normalized
Accorrung to the Born interpretation:
$$
\psi_n^*(x)\psi_n(x)dx=B^*Bsin^2\frac {n\pi x}{a}dx
$$
is the probability that the particle is located between $x$ and $x+dx$.
It is certain to be found there and so the probability that the particle lies between 0 and a is unity:
$$
\int_0^a\psi^*_n(x)\psi_n(x)dx=1
$$
So we find:
$$
|B|^2\int_0^asin^2\frac {n\pi x}{a}dx=|B|^2\cdot\frac a2=1
$$
Therefore, $B=(a/2)^{1/2}$, and:
$$
\psi_n(x)= \left(\frac 2a \right)^{1/2}sin\frac {n\pi x}{a} \quad 0\le x\le a,n=1,2,\cdots
$$
A wave function given by the equation is said to be normalized. Because the Hamiltonian operator is a linear operator, if $\psi$ is a solution to $\hat H\psi(x)=E\psi(x)$, then any constant, say, $A$, times $\psi$ is also a solution, and $A$ can always be chosen to produce a normalized solution to the Schrödinger equation.
With some calculation, we can find that the probability density becomes uniform as $n$ increases, which is the expected behavior of a classical particle, which has no preferred position between 0 and a. The results illustrate the correspondence principle, according to which quantum-mechanical results and classical-mechanical results tend to agree in the limit of large quantum numbers. The large quantum number limit is often called the classical limit.
（ 对应原理说明当量子数趋向一个极限时，量子理论的描述与经典理论的描述相符，搭建起微观与宏观世界的桥梁。 ）
3.7 The Average Momentum of a Particle in a Box Is Zero
We can use the probability distribution $\psi^*_n(x)\psi_n(x)dx$ to calculate averages and standard deviations of various physical quantities such as position and momentum.
The average value of $x$, or the mean position of the particle, is given by:
$$
\left\langle x\right\rangle=\frac 2a\int_0^axsin^2\frac {n\pi x}{a}dx=\frac a2
$$
This is the physically expected result because the particle “sees” nothing except the walls at $x = 0$ and $x =a$, and so by symmetry $\left\langle x\right\rangle$ must be $a/2$.
We can calculate the spread about $\left\langle x\right\rangle$ by calculating the variance, $\sigma_x^2$.
$$
\left\langle x^2\right\rangle=\frac 2a\int_0^ax^2sin^2\frac {n\pi x}{a}dx=\frac {a^2}{3}-\frac {a^2}{2n^2\pi^2}
$$
$$
\sigma_x^2=\left\langle x^2\right\rangle-\left\langle x\right\rangle^2= \left(\frac {a}{2\pi n} \right)^2 \left(\frac {\pi^2n^2}3-2 \right)
$$
and so the standard deviation is:
$$
\sigma_x=\frac {a}{2\pi n} \left(\frac {\pi^2n^2}3-2 \right)^{1/2}
$$
We shall see that $\sigma_x$ is directly involved in the Heisenberg uncertainty principle. We can also find that $\left\langle x\right\rangle$, $\left\langle x^2\right\rangle$, and $\sigma_x$ go to the classical limit as $n \rightarrow \infin$.
A problem arises if we wish to calculate the average energy or momentum because these quantities are represented by differential operators.
let’s go back to the Schrödinger equation in operator notation, lf we multiply this equation from the left by $\psi_n^*(x)$ and integrate over all values of $x$, we obtain:
$$
\int\psi_n^(x)\hat H\psi_n(x)dx = \int\psi_n^(x)E_n\psi_n(x)dx = E_n\int\psi_n^*(x)\psi_n(x)dx = E_n
$$
We will set this up as a formal postulate in Chapter 4, but our assumption is that:
$$
\left\langle s\right\rangle=\int\psi_n^*(x)\hat S\psi_n(x)dx
$$
where $\left\langle s\right\rangle$ the quanttun-mechanical operator associated with the physical quantity $s$, and $\left\langle s\right\rangle$ is the average value of sin the state described by the wave function.
For example, the average momentum of a particle in a box in the state described by $\psi_n(x)$ is:
$$
\left\langle p\right\rangle=\int_0^a \left[ \left(\frac 2a \right)^{1/2}sin\frac {n\pi x}{a} \right] \left(-i\hbar\frac {d}{dx} \right) \left[ \left(\frac 2a \right)^{1/2}sin\frac {n\pi x}{a} \right]dx
$$
We simplify the equation and consulting the table of integals, we find the integral is equal to zero, and so:
$$
\left\langle p\right\rangle=0
$$
Thus, a particle in a box is equally likely to be moving in either direction.
3.8 The Uncertainty Principle Says That $\sigma_p\sigma_x\ge\hbar/2$
Now let’s calculate the variance of the momentwn, $\sigma_p^2=\left\langle p^2\right\rangle-\left\langle p\right\rangle^2$, of a particle in a box:
$$
\left\langle p^2\right\rangle=\int \psi_n^*(x)\hat p_x^2\psi_n(x)dx
$$
$$
\left\langle p^2\right\rangle=2m\left\langle E\right\rangle=\frac {n^2h^2}{4a^2}
$$
$$
\sigma_p^2=\frac {n^2\pi^2\hbar^2}{a^2} \quad and \quad \sigma_p=\frac {n\pi\hbar}{a}
$$
If we take the product of $\sigma_x$ and $\sigma_p$, then we have:
$$
\sigma_x\sigma_p=\frac {\hbar}2 \left(\frac {\pi^2n^2}3-2 \right)^{1/2}
$$
The value of the square-root term here is never less than 1, and so we write:
$$
\sigma_x\sigma_p > \frac {\hbar}2
$$
The equation is one version of the Heisenberg uncertainty principle.
Let’s try to summarize what we have teamed concerning the uncertainty principle. A free particle has a definite momentum, but its position is completely indefinite. When we localize a particle by restricting it to a region of length a, it no longer has a definite momentum. If we let the
length a of the region go to zero, so that we have localized the particle precisely and there is no uncertainty in its position, then there is an infinite uncertainty in the momentum. The uncertainty principle says that the minimum product of the two uncertainties is on the order of the Planck constant.
3.9 The Problem of a Particle in a Three-Dimensional Box Is a Simple Extension of the One-Dimensional Case
The simplest three-dimensional quantum-mechanical system is the three-dimensional version of a particle in a box. In this case, the particle is confined to lie within a rectangular parallelepiped with sides of lengths $a,b,c$.
The Schrödinger equation for this system is the three-dimensional extension:
$$
-\frac {\hbar^2}{2m}\nabla^2\psi=E\psi \quad 0\le x\le a,0\le y \le b,0\le z\le c
$$
Where the operator $\nabla$ is called the Laplacian operator. The Laplacian operator appears in many physical problems.
The wave function $\psi(x,y,z)$ satisfies the boundary conditions that it vanishes at all the walls of the box, we will use the method of separation of variables to solve equation.
$$
-\frac {\hbar^2}{2m}\frac {1}{X(x)}\frac {d^2X}{dx^2} =E_x
$$
$$
-\frac {\hbar^2}{2m}\frac {1}{Y(y)}\frac {d^2Y}{dy^2} =E_y
$$
$$
-\frac {\hbar^2}{2m}\frac {1}{Z(z)}\frac {d^2Z}{dz^2} =E_z
$$
where $E_x$,$E_y$, and $E_z$ are constants and where:
$$
E_x+E_y+E_z=E
$$
With the boundary conditions, the solution to the Schrödinger equation in a three-dimensional quantum-mechanical system is:
$$
\psi(x,y,z)=A_xA_yA_zsin\frac {n_x\pi x}{a}sin\frac {n_y\pi y}{b}sin\frac {n_z\pi z}{c}
$$
The normalization constant $A_xA_yA_z$ is found from the equation:
$$
\int_0^adx\int_0^bdy\int_0^cdz\psi^*(x,y,z)\psi(x,y,z)=1
$$
It shows:
$$
A_xA_yA_z=\left(\frac {8}{abc} \right)^{1/2}
$$
Thus, the normalized wave functions of a particle in a three-dimensional box are:
$$
\psi_{n_xn_yn_Z}=\left(\frac 8{abc} \right)^{1/2} sin\frac {n_x\pi x}{a}sin\frac {n_y\pi y}{b}sin\frac {n_z\pi z}{c} \quad n_x,n_y,n_z=1,2,3,\cdots
$$
The energy is:
$$
E_{n_xn_yn_z}=\frac {h^2}{8m} \left(\frac {n_x^2}{a^2}+\frac {n_y^2}{b^2}+\frac {n_z^2}{c^2} \right) \quad n_x,n_y,n_z=1,2,3,\cdots
$$
We should expect by symmetry that the average position of a particle in a three-dimensional box is at the center of the box, but we can show this by direct calculation.
$$
\left\langle \boldsymbol r\right\rangle=\int_0^adx\int_0^bdy\int_0^cdz \boldsymbol{\hat R}\psi^*(x,y,z)\psi(x,y,z)=\frac a2 \boldsymbol i+\frac b2 \boldsymbol j+\frac c2 \boldsymbol k
$$
and:
$$
\left\langle \boldsymbol P\right\rangle=\int_0^adx\int_0^bdy\int_0^cdz \boldsymbol{\hat P}\psi^*(x,y,z)\psi(x,y,z)=0
$$
An interesting feature of a particle in a three-dimensional box occurs when the sides of the box are equal. In this case, $a=b=c$:
$$
E_{n_xn_yn_z}=\frac {h^2}{8ma^2}(n_x^2+n_y^2+n_z^2)
$$
Only one set of values $n_x,n_y,n_z$ corresponds to the lowest energy level. This level, $E_{111}$ is said to be nondegenerate. However, three sets of values of $n_x,n_y,n_z$ correspond to the second energy level, and we say that this level is three-fold degenerate, or:
$$
E_{211}=E_{121}=E_{112}=\frac {6h^2}{8ma^2}
$$
Figure 3.5 shows the distribution of the fust few energy levels of a particle in a cube.
Note that the degeneracy occurs because of the symmetry introduced when the general rectangular box becomes a cube and that the degeneracy is “lifted” when the symmetry is destroyed by making the sides of different lengths. A general principle of quantum mechanics states that degeneracies are the result of underlying symmetry and are lifted when the symmetry is broken.
The wave functions for a particle in a three-dimensional box factor into products of wave functions for a particle in a one-dimensional box. The problem of a particle in a three-dimensional box reduces to three one-dimensional problems.
This is no accident. It is a direct result of the fact that the Hamiltonian operator for a particle in a three-dimensional box is a sum of three independent terms:
$$
\hat H=\hat H_x+\hat H_y+\hat H_z
$$
We say that the Hamiltonian operator is separable.
（ 哈密顿算符可以分离，意味着哈密顿算符的本征值，即能量，可以被分离为多项的加和，这又意味着可以以此来简化问题，比如把三维势箱当成三个一维势箱处理。 ）
总结
本章详细地讨论了势箱问题，可以作为量子力学的一个引子，在下一章，我们将会讨论量子力学公设，从更基础的角度更全面的认识量子力学。