拉格朗日乘子法

z 3. the uncertainty principle
ΔpΔq ≥
1 2
h
the wavefunction for a particle at a well-defined location is a
sharply spiked function that has zero amplitude everywhere
p
a particle with high momentum has a wavefunction with a short wavelength(Fig11.14)
z 2. the sharply curved function corresponding to a higher kinetic energy than the less sharply curved function(Fig11.23,11.24,11.25)
11.2(b)
z Describe how a wavefunction determines the dynamical properties of a system and how those properties may be perdicted.
z 1. In quantum mechanics all dynamical properties of a physical system have associated with them a corresponding operator. The system itself is described by a wavefunction.
Hale Waihona Puke Baidu
11.1（b）
z Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation
z 1. explain the energy density distribution of the radiation as a function of wavelength, in particular ,the observed drop to zero as λ 0
z 令其对x，y，z 的三个自变量的偏微熵分别为零，得到三 个新方程式： yz+2C(y+z)=0 xz+2C(x+z)=0 xy+2C(x+y)=0
z 把上面的式子相除得 (x/y)=(x+z)/(y+z) (y/z)=(x+y)/(x+z)
z 再由约束条件得到它们的值是 x=y=z=(a/√6)
∂F ∂x1
=
0
=
∂f ∂x1
+ C1
∂y1 ∂x1
+ C2
∂y2 ∂x1
+ LCm
∂ym ∂x1
L
∂F ∂xn
=
0=
∂f ∂xn
+ C1
∂y1 ∂xn
+ C2
∂y2 ∂xn
+ LCm
∂ym ∂xn
z 很显然，这n 个方程式已经巧妙地把约束条件融合到求解的要求之中 了。拉格朗日就是这样把约束条件的信息放到了求解进程中了。
〈Ω〉 = ∫ψ ∗ Ωψdτ
11.3(b)
z Suggest how the general shape of a wavefunction can be predicted without solving the schro&&dinger equation explicitly.
z 1. the de Broglie relation: λ = h
z 2. When the function representing the state of the system is an eigenfunction of the operator Ω，we solve the eigenvalue equation
Ωψ = ωψ
in order to obtain the observable values,ω ,of the dynamical
z 上面的n 个方程连同约束条件给的m 个方程式已经可以解出n+m 个未 知数。它们就是n 个x（即x1,x2,…,xn ）和m 个C (即C1 ,C2 ,…， Cm )。于是我们就得到了这个函数达到极值时的各个自变量的值。可 以看到约束条件不同，得到的各个x 值也不同。这样就利用拉格朗日 方法解决了问题
拉格朗日乘子法(lagrange mutipliers)
z 欲y1求(x1n,元x2,函…数,xnf()x=10,x，2,…,xn）在如下m个约束条件（m<n） y2(x1,x2,…,xn)=0 … ym(x1,x2,…,xn )=0 下的极值
z 拉乘f格,y朗1 ,y日2 方,…法,y是m,以并1且，把C它1 ,C们2加,…起，来C，m 就这得些到未了知一常个数新（的待函求数的F）顺次 F(x1,x2,…,xn)=f(x1,x2,…,xn) +C1y1(x1,x2,…,xn,…) + C2 y2(x1,x2,…,xn) + … +Cm ym(x1,x2,…,xn )
z 注意到约束条件的各个函数都是零，所以新函数F 达到极值与原函数f 达到极值对自变量（各个x 值）的值是相同的。
z 函数F 既然是各个x 的函数，它达到极值时必然是对各个自变量x 的 偏微商分别等于零（由于是多元函数，所以过去的微分变成了偏微商 了）。根据这个分析，求多个偏微商以后我们就得到n 个新的方程
z 设长方体的三个棱长为x，y，z，则其体积f 为三个边长 的乘积：f(x,y,z)=xyz.要求表面积为a 的平方 于是长方体的6面的面积可以写成2xy+2xz+2yz=a2 即 2xy+2xz+2yz-a2 =0它也就是本问题中仅有一个约束条 件。根据前面介绍的拉格朗日方法制造一个新函数F，并 且放进一个未知的常数C ，于是有 F=xyz+C(2xy+2xz+2yz-a2 )
properties. z 3. When the function is not an eigenfunction of Ω, we can only
find the average or expectation value of dynamical properties by performing the integration
z 2. assume that the energy of the oscillators are quantized E = nhν = nhc / λ we see that at shortwavelength oscillators is very large. This energy is too large for the walls to supply it, so the shortwavelength oscillators remain unexcited.
except at the particle’s position.(Fig11.26,11.27)