# The Methods of Quantum Mechanics

As er have seen, if we want to describe a free particle, which is partially localized, we could use a wave packet. De Broglie's hypothesis does not tell us what type of wave one can associate with a particle that is not free and that is acted on by a force. If a particle is acted on by a force, its momentum and its energy will not be constant. Therefore, it is meaningless to talk about a λ\lambda and ν\nu, because these quantities are changing.

The Schrödinger theory tells us how to obtain the wavefunction ψ(x,t)\psi(x,t) associated with a particle, when we specify the forces acting on the particle, by giving the potential energy associated with the forces.

Potential energy

In quantum mechanics the potential energy is often refered to simply as the potential.

TIP

真的很難想像量子力學的一個重要轉機發生在一個原本不是做物理的人身上。De Broglies.

# Schrödinger Equation

由于波函数是一个概率波,是一个数学上的形式,没有必要一定要实数,为了数学上的简便性,他使用了复数的形式。如何在平面坐标上表示一个复数。The wave function itself has no physical meaning. From Euler's identity

e±iθ=cosθ±isinθe^{\pm i \theta} = \cos \theta \pm i \sin \theta

then any 'point' in the conplex plane can be described as

Ae±iθ=A(cosθ±isinθ)A e^{\pm i \theta} = A (\cos \theta \pm i \sin \theta)

A free particle wave function can be described as

ψ(x,t)=Aei(kxωt)=A[cos(kxωt)+isin(kxωt)]\psi (x, t) = A e^{i(kx-\omega t)} = A [\cos (kx - \omega t) + i \sin(kx - \omega t)]

This complex form of wave function has real and imaginary parts, both of which have the same wavelength and frequency:

λ=2πk;ν=ω2π\lambda = \frac{2\pi}{k}; \nu = \frac{\omega}{2\pi}

Hence, the corresponding momentum and energy are:

p=h/λ;E=hνp = h/\lambda; E = h\nu

It is ψ2|\psi|^2 that has physical significance.

ψ2=ψψ=Aei(kxωt)Aei(kxωt)=AA|\psi|^2= \psi^{*} \psi = A^{*} e^{-i(kx-\omega t)} A e^{i(kx-\omega t)} = A^{*} A

where ψ\psi^{*} is the complex conjugate of ψ\psi.

Operators

In mathematics an operator is represented by a symbol or group of symbols and indecates an operation to b performed.

Thus, for example, when the operator ix-i\hbar \frac{\partial}{\partial x} is placed in front of a function, it indicates that the function is to be differentiated with respect to xx and then multiplied by i-i\hbar, where ii is the imaginary number 1\sqrt {-1}. Another operator is iti \hbar \frac{\partial}{\partial t}.

Applying the operator ix-i\hbar \frac{\partial}{\partial x} to the complex form of wave function ψ(x,t)=Aei(kxωt)\psi (x, t) = A e^{i(kx-\omega t)}.

iψx=iiAkei(kxωt)=h2π2πλAei(kxωt)=pψ-i\hbar \frac{\partial \psi }{\partial x} = -i \hbar iAke^{i(kx - \omega t)} = \frac{h}{2\pi} \frac{2\pi}{\lambda} A e^{i(kx - \omega t)} = p \psi

Silimarly, applying iti \hbar \frac{\partial}{\partial t} to ψ\psi, one can obtain

iψt=i(iω)Aei(kxωt)=h2π2πνAei(kxωt)=Eψi \hbar \frac{\partial \psi}{\partial t} = i \hbar (-i\omega) A e^{i(kx - \omega t)} = \frac{h}{2\pi} 2\pi \nu A e^{i(kx - \omega t)} = E \psi

这里的operator就是在数学里面强行地去找物理意义,和Plank去拟合黑体辐射有异曲同工之妙,属于是去凑答案。

The total energy of a particle is equal to the kinetic energy plus the potential energy,

E=12mv2+Ep=p22m+EpE = \frac{1}{2} mv^2 + E_p = \frac{p^2}{2m} + E_p

Multiplying both sides of this equation by the wavefunction ψ\psi we obtain

Eψ=p22mψ+EpψE\psi = \frac{p^2}{2m} \psi + E_p \psi

WARNING

When the particle is not free, the relationship cannot be proved, but we postulate that it still holds.

Then we get

iψt=12mppψ+Epψ=12m(iψx)(iψx)+Epψ=22m2ψx2+Epψi \hbar \frac{\partial \psi}{\partial t} = \frac{1}{2m} p p \psi + E_p \psi = \frac{1}{2m} \left(-i\hbar \frac{\partial \psi }{\partial x}\right) \left(-i\hbar \frac{\partial \psi }{\partial x}\right) + E_p\psi = - \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} + E_p \psi

which is conventionally written as

22m2ψx2+Epψ=iψt- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} + E_p \psi = i \hbar \frac{\partial \psi}{\partial t}

This equation is known as the one-dimensional time-dependent Schrödinger equation. If the potential energy EpE_p is known, this equation can be solved in principle, and the solution will yield the possible wavefunctions that we can associate with the particle.

First Principle

The Schrödinger equation cannot be derived from first principles; it is the first principle, which cannot be mathematically derived, just as Newton's laws of motion are not derivable. The justification lies in the fact that its predictions agree with the experiment.

我总算是知道什么是薛定谔方程了

The Schrödinger equation is to quantum machanics what Newton's second law is to classical physics.

# Expectation Values

One of the consequences of the uncertainty principle is that the position, momentum, energy, and so forth of a particle cannot, in general, be precisely determined. We can, however, talk about the average value of these quantities, i.e., expectation values. Can we determine these expectation values without doing the experiment if we know the wavefunction ψ(x,t)\psi (x,t) associated with the system? The answer is yes.

PP, the probalility density of a particle's position (at a certain time, or assume PP independent of time). The probalility of finding the particle located within dxdx, or Δx\Delta x, centered around xx is

P(x)dxP(x) dx

P(x)Δx=1,Δx0\sum P(x) \Delta x = 1, \Delta x \to 0

When Δx0\Delta x \to 0,

P(x)Δx=1P(x)dx=1\sum P(x) \Delta x = 1 \Rightarrow \int_{-\infty}^\infty P(x) dx = 1

x¯=[P()Δx]++xn[P(xn)Δx]++[P()Δx]\bar{x} = - \infty[P(-\infty) \Delta x] + \cdots + x_n[P(x_n) \Delta x] + \cdots + \infty[P(\infty) \Delta x]

x¯=xi[P(xi)Δx]\Rightarrow \bar{x} = \sum{x_i} [P(x_i) \Delta x]

When Δx0\Delta x \to 0,

x¯=xP(x)dx=xψ2dx=xψψdx\bar{x} = \int_{-\infty}^{\infty} x P(x) dx = \int_{-\infty}^{\infty} x |\psi|^2 dx = \int_{-\infty}^{\infty} x \psi^{*}\psi dx

For this relation to be valid, ψ\psi must be normalized so that the sum of their probabilities for all xx is unity, that is,

ψψdx=1\int_{-\infty} ^{\infty} \psi^{*} \psi dx = 1

It is customary to write the numerator with the xx between the two wavefunctions as

x¯=ψxψdx\bar{x} = \int_{-\infty}^{\infty} \psi^{*}x\psi dx

Similarly, the potential energy expectation can be obtained.

Ep¯=ψEp(x)ψdx\bar{E_p} = \int_{-\infty}^{\infty} \psi^{*} E_p(x) \psi dx

To find p¯\bar{p}, the operator ix-i\hbar \frac{\partial}{\partial x} is used.

p¯=ψpψdx=ψ(iψx)dx\bar{p} = \int_{-\infty}^{\infty} \psi^{*} p \psi dx = \int_{-\infty}^{\infty} \psi^{*} \left(-i\hbar \frac{\partial \psi}{\partial x}\right)dx

To find E¯\bar{E}, the operator iti \hbar \frac{\partial}{\partial t} is used.

E¯=ψEψdx=ψ(iψt)dx\bar{E} = \int_{-\infty}^{\infty} \psi^{*} E \psi dx = \int_{-\infty}^{\infty} \psi^{*} \left( i \hbar \frac{\partial \psi}{\partial t} \right) dx

# Separation of variables

How to solve Schrödinger equantion. One standard technique for solving partial differential eqautions is the method of separation of variables. The method consists in trying a solution that is a product of two functions: one a function of xx alone and the oher a function of tt alone; that is, let

ψ(x,t)=χ(x)Γ(t)\psi(x,t) = \chi(x) \Gamma(t)

WARNING

Why we can separate xx and tt into two functions?

Let us substitute it into Schrödinger equantion,

22m2ψx2+Ep(x)ψ=iψt-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}+E_{p}(x) \psi=i \hbar \frac{\partial \psi}{\partial t}

22mΓ(t)d2χ(x)dx2+Ep(x)χ(x)Γ(t)=iχ(x)dΓdt-\frac{\hbar^{2}}{2 m} \Gamma(t) \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x) \chi(x) \Gamma(t)=i \hbar \chi(x) \frac{d \Gamma}{d t}

If we devide through by χ(x)Γ(t)\chi(x) \Gamma(t), we get

22m1χ(x)d2χ(x)dx2+Ep(x)=i1Γ(t)dΓ(t)dt-\frac{\hbar^{2}}{2 m} \frac{1}{\chi(x)} \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x)=i \hbar \frac{1}{\Gamma(t)} \frac{d \Gamma(t)}{d t}

The right side of the equation is a function of tt alone, and the left side is a function of xx. Because xx and tt are independent variables, the equation will be correct if both sides are equal to the same constant GG which is called the separation constant, that is,

22m1χ(x)d2χ(x)dx2+Ep(x)=G-\frac{\hbar^{2}}{2 m} \frac{1}{\chi(x)} \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x) = G

i1Γ(t)dΓ(t)dt=G i \hbar \frac{1}{\Gamma(t)} \frac{d \Gamma(t)}{d t} = G

We now have two ordinary differential equations, one involving xx alone and the other involving tt alone.

How to ask questions?

方式1: 什么是薛定谔方程? 方式2: 为什么我们需要一个东西叫做薛定谔方程? 这里的第一个提问方式是非常被动的。首先容易被名字说吓住,此时知道是什么之后,马上就停止了。而第二个问题才是积极的方式,可以让人思考的。又比如这里,我们就应该问为什么需要Separation of variables? 而不是什么是Separation of variables. 概念是在回答了为什么之后自然而然出来的东西。 不要给我炫概念,告诉我为什么我需要知道这个概念。网络上可以查到概念,但是为什么会出现这个概念却需要自己去领悟。 教书授课也要秉承这些理念,学生不是来看一本字典的,而是要弄清楚为什么是这样。

Step 1: To solve Γ(t)\Gamma(t),

dΓ(t)dt=GiΓ(t)\frac{d \Gamma(t)}{dt} = \frac{G}{i\hbar} \Gamma(t)

Γ(t)=keGit=keiGt=eiGt,k=1\Gamma(t) = ke^{\frac{G}{i\hbar}t}= ke^{\frac{-iG}{\hbar}t} = e^{\frac{-iG}{\hbar}t} , k = 1

Now, the question is, what is GG? To find it out, let't operate on the wavefunction ψ(x,t)\psi(x,t) with the energy operator iti\hbar \frac{\partial}{\partial t}.

iψt=i(χ(x)Γ(t))t=iχ(x)Γ(t)t=iχ(x)(iG)Γ(t)=Gχ(x)Γ(t)=Gψi\hbar \frac{\partial \psi}{\partial t} = i \hbar \frac{\partial \left( \chi(x) \Gamma (t)\right)}{\partial t} = i\hbar\chi (x) \frac{\partial \Gamma (t)}{\partial t} = i \hbar \chi(x) \left( \frac{-iG}{\hbar} \right) \Gamma(t) = G\chi(x)\Gamma(t) = G\psi

As for iψt=Eψi\hbar\frac{\partial \psi}{\partial t} = E \psi, we see that the separation constant GG is the total energy EE of the sysytem. Thus the time-dependent part of the wavefunction ψ(x,t)\psi(x,t) is,

Γ(t)=eiEt\Gamma(t) = e^{\frac{-iE}{\hbar}t}

Step2: To solve χ(x)\chi(x),

22m1χ(x)d2χ(x)dx2+Ep(x)=G=E-\frac{\hbar^{2}}{2 m} \frac{1}{\chi(x)} \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x)= G = E

Then we get

22md2χdx2+Ep(x)χ=Eχ-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}+E_{p}(x) \chi = E \chi

This equation is called time-independent Schrödinger equation. For a given Ep(x)E_p(x), the equation has to be solved to find the possible χ(x)\chi(x)'s that can be associated with the system.

都说薛定谔方程难解

到底是哪里在哪里?

So far, there is no restrictions on the values of the total energy EE of the system. Thus, for different EE's we expect to get different functions χ(x)\chi(x). We will see, however, that χ(x)\chi(x) must satisfy certain requirements. That requirements will, in some cases, restrict the number of physically acceptable χ(x)\chi(x)'s and consequently the values of EE that the system may have.

The solutions χ\chi are called the eigenfunctions or eighenstates; the corresponding values of EE are called the eigenvalues. (In German, "eigen" means "proper")

TIP

Different eigenvalues correspond to different eigenstates, different wavefunctions. That is amazing.

# Required Properties of the Eighenfunction and its Derivative

χ(x)\chi(x) must be well-behaved. This means that the eighenfunciton χ(x)\chi(x) and it derivative dχ/dxd\chi/dx must have the following properties:

  1. They must be finite everywhere.
  2. They must be single-valued everywhere.
  3. They must be continuous everywhere.(In some special cases dχ/dxd\chi/dx is not continuous.)

When these conditions are satisfied, the eigenfunction (χ(x)\chi(x)) and the associated wavefunction (ψ(x,t)=χ(x)Γ(t)\psi(x, t) = \chi(x) \Gamma(t)) are said to be well-behaved. Why must these conditions be satisfied?

Recall that

x¯=ψxψdx\bar{x} = \int_{-\infty}^{\infty} \psi^{*}x\psi dx

p¯=ψ(ix)ψdx\bar{p} = \int_{-\infty}^{\infty} \psi^* ( -i \hbar \frac{\partial}{\partial x}) \psi dx

If χ\chi or dχ/dxd\chi/dx are not finite, the average value of the momentum will be infinite; infinite momentum means infinite energy; this is not physically possible.

If χ\chi or dχ/dxd\chi/dx were not single-valued, the average position and the average momentum would not be defined.

The requirement of contunuity is closely tied to the requrement of finiteness. If χ\chi is not continuous at a given point in space, then dχ/dxd\chi/dx would be infinite at that point. This, as we have indicated, is not allowed. If dχ/dxd\chi/dx is discontinuous at some point, then d2χ/dx2d^2 \chi / d x^2 would be infinite at that point. From the time-independent Schrödinger equation,

22md2χdx2+Ep(x)χ=Eχ-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}+E_{p}(x) \chi = E \chi

this would imply ahat either EE or EpE_p is infinite, which is not physically possible.

TIP

Let A=a+biA = a + bi, B=c+diB= c + di.

A=abi,B=cdiA^* = a - bi, B^* = c - di

AB=(abi)(cdi)=ac(bc+ad)i+bidi=acbd(bc+ad)iA^* B^* = (a - bi) (c - di) = ac -(bc + ad) i + bidi = ac - bd - (bc + ad) i

Then

AB=(a+bi)(c+di)=ac+(bc+ad)i+bidi=acbd+(bc+ad)iAB = (a + bi) ( c + di) = ac + (bc + ad)i + bidi = ac - bd + (bc + ad) i

(AB)=acbd(bc+ad)i(AB)^* = ac - bd -(bc + ad)i

Therefore

(AB)=AB(AB)^* = A^* B^*