\gdef\red#1{{\color{cb8680}{#1}}} \gdef\green#1{{\color{4f8d63}{#1}}} \gdef\gray#1{{\color{gray}{#1}}} \gdef\purple#1{{\color{B189C6}{#1}}} \gdef\orange#1{{\color{dfa04b}{#1}}} \gdef\white#1{{\color{white}{#1}}}
数学家从大量实践中总结出一类 (a,b)\red{\boldsymbol{(a,b)}}(a,b) 上有趣的 Sturm-Liouville 型微分方程
ddx[k(x)dydx] −↓ q(x)y(x)+λρ(x)y(x)=0\frac{d}{dx} \bigg[ k(x) \frac{dy}{dx} \bigg] ~\stackrel{\large\downarrow}{\red{\boldsymbol{-}}} ~ q(x) y(x) + \lambda \rho(x)y(x) = 0 dxd[k(x)dxdy] −↓ q(x)y(x)+λρ(x)y(x)=0
k(x)k(x)k(x)， q(x)q(x)q(x)，ρ(x)\rho(x)ρ(x) 是 已知函数，y(x)y(x)y(x) 是 未知函数，λ\lambdaλ 是常数，
ddx[k(x)y′(x)]−q(x)y(x)+λρ(x)y(x)=0\frac{d}{dx} \bigg[k(x) y'(x) \bigg] - q(x) y(x) + \lambda \rho(x)y(x) = 0 dxd[k(x)y′(x)]−q(x)y(x)+λρ(x)y(x)=0
y′′(x)+P(x)y′(x)+Q(x)y(x)=0y''(x) + P(x) y'(x) + Q(x) y(x) = 0 y′′(x)+P(x)y′(x)+Q(x)y(x)=0
本质是同一个东西
数学家喜欢一个东西变来变去
Sturm-Liouville 标准形式重写
y′′(x)+k′(x)k(x)y′(x)+λρ(x)−q(x)k(x)y(x)=0y''(x) + \frac{k'(x)}{k(x)} y'(x) + \frac{\lambda \rho(x) - q(x)}{k(x)} y(x) = 0 y′′(x)+k(x)k′(x)y′(x)+k(x)λρ(x)−q(x)y(x)=0
与二阶线性常微分方程标准形式比较
P(x)∼k′(x)k(x)=ddxlnx,Q(x)∼λρ(x)−q(x)k(x)P(x) \sim \frac{k'(x)}{k(x)} = \frac{d}{dx}\ln x, \qquad Q(x) \sim \frac{\lambda \rho(x) - q(x)}{k(x)} P(x)∼k(x)k′(x)=dxdlnx,Q(x)∼k(x)λρ(x)−q(x)
kkk 的零点 ⇒\Rightarrow⇒ P(q)P(q)P(q), Q(x)Q(x)Q(x) 的奇点；什么时候是正则奇点？
k(x) > 0,q(x)≥0,ρ(x)≥0,x∈(a,b)k(x) ~ \red{\boldsymbol{>}} ~ 0, \quad q(x) \ge 0, \quad \rho(x) \ge 0, \qquad x \in (a,b) k(x) > 0,q(x)≥0,ρ(x)≥0,x∈(a,b)
k(x)k(x)k(x) 在边界 a,ba, ba,b 有限，可以有零点，最高二阶零点
q(x)q(x)q(x) 在边界 a,ba,ba,b 最高一阶 极点
保证 q(x)/k(x)q(x)/k(x)q(x)/k(x) 极点 不超过二阶
ρ(x)\rho(x)ρ(x) 在边界 a,ba, ba,b 处有限，可以有零点
未知函数 y(x)y(x)y(x) 在开区间 (a,b)(a,b)(a,b) 上连续可微，在闭区间 [a,b][a,b][a,b] 上连续有界
在 [a,b][a,b][a,b] 上连续有界其实就是自然边界条件
后面讨论在边界 a,ba, ba,b 额外添加其他边界条件
这些额外边界条件与自然边界条件相容
无界区间
a,ba,ba,b 可能取正负无穷。此时一般要求 y(x)y(x)y(x) 在该无限区间上面满足平方可积性，
∫∣y(x)∣2dx<∞ \int |y(x)|^2 dx < \infty ∫∣y(x)∣2dx<∞
y′′(x)+λy(x)=0y''(x) + \lambda y(x) = 0 y′′(x)+λy(x)=0
αy′−βy∣x=a=(γy′+δy)∣x=b=0\alpha y' - \beta y \big|_{x = a} = (\gamma y'+\delta y)\big|_{x = b} = 0 αy′−βyx=a=(γy′+δy)x=b=0
k(x)=1>0,q(x)=0≥0,ρ(x)=1>0k(x) = 1 > 0, \qquad q(x) = 0 \ge 0, \qquad \rho(x) = 1 > 0 k(x)=1>0,q(x)=0≥0,ρ(x)=1>0
勒让德方程的本征问题，自然边界条件
y′′(x)−2x1−x2y′+λ1−x2y=0,∣y(±1)∣<∞y''(x) \green{- \frac{2x}{1-x^2}} y' + \green{\frac{\lambda}{1 - x^2}}y = 0, \qquad |y(\pm 1)| < \infty y′′(x)−1−x22xy′+1−x2λy=0,∣y(±1)∣<∞
ddx[(1−x2)y′(x)]+λy(x)=0,∣y(±1)∣<∞\frac{d}{dx}\bigg[(1 - x^2) y'(x)\bigg] + \lambda y(x) = 0, \qquad |y(\pm 1)| < \infty dxd[(1−x2)y′(x)]+λy(x)=0,∣y(±1)∣<∞
拉普拉斯算符在什么坐标系的分离变量法会出现勒让德方程？球坐标
k(x)=1−x2>0,q(x)=0≥0,ρ(x)=1≥0k(x) = 1 - x^2 > 0, \qquad q(x) = 0 \ge 0, \qquad \rho(x) = 1 \ge 0 k(x)=1−x2>0,q(x)=0≥0,ρ(x)=1≥0
Bessel 方程本征值问题，自然边界与第三类边界
ddx(xdydx)−m2xy+λxy=0,0<x<a\frac{d}{dx}\bigg(x \frac{dy}{dx}\bigg) - \frac{m^2}{x} y + \lambda x y = 0, \quad 0 < x < a \\ dxd(xdxdy)−xm2y+λxy=0,0<x<a
∣y(0)∣<∞,αy′(a)+βy(a)=0\purple{|y(0)| < \infty} , \quad \alpha y'(a) + \beta y(a) = 0 ∣y(0)∣<∞,αy′(a)+βy(a)=0
k(x)=x>0,q(x)=m2x≥0,ρ(x)=x≥0k(x) = x > 0, \qquad q(x) = \frac{m^2}{x} \ge 0, \qquad \rho(x) = x \ge 0 k(x)=x>0,q(x)=xm2≥0,ρ(x)=x≥0
ddx[x2dydx]−λ1y+λ2x2y=0,0<x<a\frac{d}{dx}\bigg[ x^2 \frac{dy}{dx} \bigg] - \lambda_1 y + \lambda_2 x^2 y = 0, \quad 0<x<a dxd[x2dxdy]−λ1y+λ2x2y=0,0<x<a
∣y(0)∣<∞,y(a)=0|y(0)| < \infty, \qquad y(a) = 0 ∣y(0)∣<∞,y(a)=0
k(x)=x2>0,q(x)=λ1≥0,ρ(x)=x2≥0k(x) = x^2 > 0, \qquad q(x) = \lambda_1 \ge 0, \qquad \rho(x) = x^2 \ge 0 k(x)=x2>0,q(x)=λ1≥0,ρ(x)=x2≥0
−ℏ22md2ψ(x)dx2+V(x)ψ(x)=Eψ(x)⇒ ddx[dψ(x)dx]−2mℏ2V(x)ψ(x)+2mℏ2Eψ(x)=0\begin{align*} & \ -\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + V(x) \psi(x) = E \psi(x)\\ \Rightarrow & \ \quad \frac{d}{dx}\left[{\frac{d \psi(x)}{dx}}\right] - \frac{2m}{\hbar^2}V(x)\psi(x) + \green{\frac{2m}{\hbar^2}E} \psi(x) = 0 \end{align*} ⇒ −2mℏ2dx2d2ψ(x)+V(x)ψ(x)=Eψ(x) dxd[dxdψ(x)]−ℏ22mV(x)ψ(x)+ℏ22mEψ(x)=0
k(x)=1,q(x)=2mℏ2V(x)≥0,ρ(x)=1>0,λ=2mℏ2Ek(x) = 1, \qquad q(x) = \frac{2m}{\hbar^2} V(x) \ge 0, \qquad \rho(x) = 1 > 0, \quad \lambda = \green{\frac{2m}{\hbar^2} E} k(x)=1,q(x)=ℏ22mV(x)≥0,ρ(x)=1>0,λ=ℏ22mE
讨论三种常见边界条件的使用场景
齐次 第一二三类边界条件
适用于闭区间 [a,b]\red{[a,b]}[a,b] 上全程 k(x) > 0k(x)~ \red{>}~ 0k(x) > 0
αy′(a) − βy(a)=0,γy′(b)+δy(b)=0\alpha y'(a) \ \red{\boldsymbol{-}} \ \beta y(a) = 0, \quad \gamma y'(b) + \delta y(b) = 0 αy′(a) − βy(a)=0,γy′(b)+δy(b)=0
其中 α,β,γ,δ≥0\alpha, \beta, \gamma, \delta \ge 0α,β,γ,δ≥0，α2+β2>0\alpha^2 + \beta^2 > 0α2+β2>0，γ2+δ2>0\gamma^2 + \delta^2 > 0γ2+δ2>0
既然用到 y,y′y, y'y,y′ 在边界上的值，就暗示了 yyy 及其导数 在 [a,b][a,b][a,b] 上 连续且有界，即已经满足自然边界条件
周期性边界条件
适用于正定周期系数 k(a)=k(b) >0k(a) = k(b) \ \red{> 0}k(a)=k(b) >0, 以及周期 q(a)=q(b)q(a) = q(b)q(a)=q(b), ρ(a)=ρ(b)\rho(a) = \rho(b)ρ(a)=ρ(b) 的方程
y(a)=y(b),y′(a)=y′(b)y(a) = y(b), \quad y'(a) = y'(b) y(a)=y(b),y′(a)=y′(b)
∣y(a)∣<∞,∣y(b)∣<∞|y(a)| < \infty, \qquad |y(b)| < \infty ∣y(a)∣<∞,∣y(b)∣<∞
自然边界条件
准确地说应该是 没有添加别的边界条件
更准确来说是 没有办法添加别的边界条件
Why?
核心动机：Sturm-Liouville 本征问题想找到一个本征函数 (保留一个自由常数)
核心原因：k(a)k(a)k(a) 是否为零决定了 aaa 处是常点还是正则奇点
性质 2：本征值无穷多且分立 (可数无穷多)，λ1≤λ2≤⋯\lambda_1 \le \lambda_2 \le \cdotsλ1≤λ2≤⋯,
limn→+∞λn=+∞\lim_{n \to +\infty} \lambda_n = +\infty n→+∞limλn=+∞
证明很困难
性质 3：本征值非简并，即每个 λn\lambda_nλn 只有一个解 yn(x)y_n(x)yn(x)
除了 周期性 边界条件
性质 4：设 λn\lambda_nλn 对应的解为 yny_nyn，则 yn+1(x)y_{n + 1}(x)yn+1(x) 比 yn(x)y_n(x)yn(x) 多一个零点
不作证明
性质 5：不同本征值的本征函数带权正交，即
∫abym(x)‾yn(x)ρ(x)dx=0,n≠m\int_a^b \overline{y_m(x)} y_n(x)\rho(x) dx = 0, \quad n \ne m ∫abym(x)yn(x)ρ(x)dx=0,n=m
y(x)‾\overline{y(x)}y(x) 表示 y(x)y(x)y(x) 的复共轭
万一出现简并，可以对相同本征值的本征函数进行 Schmidt 正交化
本征函数族在 [a,b][a,b][a,b] 上完备
证明本征值非负
ddx[k(x)y′(x)]−q(x)y(x)+λρ(x)y(x)=0\green{\frac{d}{dx} \bigg[k(x) y'(x) \bigg]} \orange{- q(x) y(x)} + \purple{\lambda \rho(x)y(x)} = 0 dxd[k(x)y′(x)]−q(x)y(x)+λρ(x)y(x)=0
λ∫ρ(x)∣y(x)∣2= ∫q(x)∣y(x)∣2−∫y(x)‾ddx(k(x)y′(x))= ∫q(x)∣y(x)∣2 − (y(x)‾k(x)y′(x))∣ab+∫k(x)∣y′(x)∣2\begin{align*} \purple{\lambda \int \rho(x) |y(x)|^2} = & \ \int \orange{q(x) |y(x)|^2} - \green{\int \overline{y(x)} \frac{d}{dx}\left({k(x)y'(x)}\right)} \\ = & \ \int q(x) |y(x)|^2 \ \red{\boldsymbol{-}} \ (\overline{y(x)} k(x) y'(x))\Big|_a^b + \int k(x) |y'(x)|^2 \end{align*} λ∫ρ(x)∣y(x)∣2== ∫q(x)∣y(x)∣2−∫y(x)dxd(k(x)y′(x)) ∫q(x)∣y(x)∣2 − (y(x)k(x)y′(x))ab+∫k(x)∣y′(x)∣2
∫q(x)∣y(x)∣2 − (y(x)‾k(x)y′(x))∣ab+∫k(x)∣y′(x)∣2≥ 0+y(a)‾k(a)y′(a)−y(b)‾k(b)y′(b)+0\begin{align*} & \ \int q(x) |y(x)|^2 \ \red{\boldsymbol{-}} \ (\overline{y(x)} k(x) y'(x))\Big|_a^b + \int k(x) |y'(x)|^2\\ \red{\boldsymbol{\ge}} & \ 0 + \overline{y(a)} k(a)y'(a) - \overline{y(b)} k(b)y'(b) + 0 \end{align*} ≥ ∫q(x)∣y(x)∣2 − (y(x)k(x)y′(x))ab+∫k(x)∣y′(x)∣2 0+y(a)k(a)y′(a)−y(b)k(b)y′(b)+0
y(a)‾k(a)y′(a)−y(b)‾k(b)y′(b)\overline{y(a)} k(a)y'(a) - \overline{y(b)} k(b)y'(b) y(a)k(a)y′(a)−y(b)k(b)y′(b)
y(a)‾k(a)y′(a)=0\overline{y(a)}k(a)y'(a) = 0 y(a)k(a)y′(a)=0
y(a)‾k(a)y′(a)=βαk(a)∣y(a)∣2>0 .\overline{y(a)}k(a)\orange{y'(a)} = \frac{\beta}{\alpha} k(a) |y(a)|^2 > 0 \ . y(a)k(a)y′(a)=αβk(a)∣y(a)∣2>0 .
y(b)‾k(b)y′(b)=0\overline{y(b)}k(b)y'(b) = 0 y(b)k(b)y′(b)=0
− y(b)‾k(b)y′(b)=+δγk(b)∣y(b)∣2>0\red{\boldsymbol{-}} \ \overline{y(b)}k(b)\purple{y'(b)} = + \frac{\delta}{\gamma} k(b) |y(b)|^2 > 0 − y(b)k(b)y′(b)=+γδk(b)∣y(b)∣2>0
y(a)‾k(a)y′(a)−y(b)‾k(b)y′(b)=0\overline{y(a)} k(a)y'(a) - \overline{y(b)} k(b)y'(b) = 0 y(a)k(a)y′(a)−y(b)k(b)y′(b)=0
λ∫ρ(x)∣y(x)∣2≥0+0+0=0\lambda \int \rho(x) |y(x)|^2 \ge 0 + 0 + 0 = 0 λ∫ρ(x)∣y(x)∣2≥0+0+0=0
证明每个本征值都没有简并
y1(x)[ddx(k(x)y2′(x))−q(x)y2(x)+λρ(x)y2(x)]=0y2(x)[ddx(k(x)y1′(x))−q(x)y1(x)+λρ(x)y1(x)]=0\begin{align*} y_1(x) \left[{\frac{d}{dx}\Big(k(x) y'_2(x)\Big) - \red{q(x) y_2(x) + \lambda \rho(x) y_2(x)}}\right] = 0\\ y_2(x) \left[{\frac{d}{dx}\Big(k(x) y'_1(x)\Big) - \red{q(x) y_1(x) + \lambda \rho(x) y_1(x)}}\right] = 0 \end{align*} y1(x)[dxd(k(x)y2′(x))−q(x)y2(x)+λρ(x)y2(x)]=0y2(x)[dxd(k(x)y1′(x))−q(x)y1(x)+λρ(x)y1(x)]=0
y1(x)ddx(k(x)y2′(x))−y2(x)ddx(k(x)y1′(x))=0y_1(x) \frac{d}{dx}\Big({k(x) y'_2(x)}\Big) - y_2(x) \frac{d}{dx}\Big({k(x) y'_1(x)}\Big) = 0 y1(x)dxd(k(x)y2′(x))−y2(x)dxd(k(x)y1′(x))=0
ddx[y1(x)k(x)dy2(x)dx−y2(x)k(x)dy1(x)dx]=0⇒ y1(x)k(x)dy2(x)dx−y2(x)k(x)dy1(x)dx=C\begin{align*} & \ \frac{d}{dx} \left[{y_1(x) k(x) \frac{dy_2(x)}{dx} - y_2(x) k(x) \frac{dy_1(x)}{dx}}\right] = 0\\ \Rightarrow & \ y_1(x) k(x) \frac{dy_2(x)}{dx} - y_2(x) k(x) \frac{dy_1(x)}{dx} = \green{C} \end{align*} ⇒ dxd[y1(x)k(x)dxdy2(x)−y2(x)k(x)dxdy1(x)]=0 y1(x)k(x)dxdy2(x)−y2(x)k(x)dxdy1(x)=C
y1(a)k(a)y2′(a)−y2(a)k(a)y1′(a)=C\begin{align*} y_1(a) k(a) \orange{y'_2(a)} - y_2(a) k(a) y'_1(a) = C \end{align*} y1(a)k(a)y2′(a)−y2(a)k(a)y1′(a)=C
y1(a)k(a)y2′(a)−y2(a)k(a)y1′(a)=C\purple{y_1(a)} \red{k(a)} \orange{y'_2(a)} - \purple{y_2(a)} \red{k(a)} \orange{y'_1(a)} = C y1(a)k(a)y2′(a)−y2(a)k(a)y1′(a)=C
y1,2′(a)=βαy1,2(a)⇒C=y1(a)k(a)βαy2(a)−y2(a)k(a)βαy1(a)=0 y_{1,2}'(a) = \frac{\beta}{\alpha} y_{1,2}(a) \Rightarrow C = y_1(a)k(a) \frac{\beta}{\alpha} y_2(a) - y_2(a) k(a) \frac{\beta}{\alpha} y_1(a) = 0 y1,2′(a)=αβy1,2(a)⇒C=y1(a)k(a)αβy2(a)−y2(a)k(a)αβy1(a)=0
y1(x)dy2(x)dx=y2(x)dy1(x)dx⇒ddxlny1(x)=ddxlny2(x)y_1(x) \frac{dy_2(x)}{dx} = y_2(x) \frac{dy_1(x)}{dx} \quad\Rightarrow\quad \frac{d}{dx} \ln y_1(x) = \frac{d}{dx} \ln y_2(x) y1(x)dxdy2(x)=y2(x)dxdy1(x)⇒dxdlny1(x)=dxdlny2(x)
y1(x)=C′y2(x)y_1(x) = C' y_2(x) y1(x)=C′y2(x)
证明本征函数带权正交
ym‾ddx(kyn′)−qym‾yn+λnρym‾yn=0\overline{y_m} \frac{d}{dx} (k y'_n) - \red{q \overline{y_m}y_n} + \orange{\lambda_n \rho \overline{y_m} y_n} = 0 ymdxd(kyn′)−qymyn+λnρymyn=0
ynddx(kym′‾)−qynym‾+λmρynym‾=0y_n \frac{d}{dx} (k \overline{y'_m}) - \red{q y_n \overline{y_m}} + \orange{\lambda_m \rho y_n \overline{y_m}} = 0 yndxd(kym′)−qynym+λmρynym=0
(λn−λm)∫abρynym‾dx=∫abdx[ym‾ddx(kyn′)−ynddx(kym′‾)]凑全导数= ∫abdxddx(ym‾kyn′−ynkym′‾)=ym‾kyn′−ynkym′‾∣ab = k(b)[yn′(b)ym‾(b)−yn(b)ym′‾(b)] −k(a)[yn′(a)ym‾(a)−yn(a)ym′‾(a)]\begin{align*} & \ \orange{(\lambda_n - \lambda_m) \int_a^b \rho y_n \overline{y_m} dx} = \int_a^b dx \left[{\overline{y_m} \frac{d}{dx}(k y'_n) - y_n \frac{d}{dx}(k \overline{y'_m})}\right]\\ \text{凑全导数} = & \ \int_a^b dx \frac{d}{dx}\left({\overline{y_m}k y'_n - y_nk \overline{y'_m}}\right) = \overline{y_m} k y'_n - y_n k \overline{y'_m} \bigg|_a^b\\ ~\\ = & \ k(b) \Big[\purple{y'_n(b)\overline{y_m}(b) - y_n(b) \overline{y'_m}(b)} \Big] \\ & \ - k(a) \Big[ y'_n(a)\overline{y_m}(a) - y_n(a) \overline{y'_m}(a) \Big]\\ \end{align*} 凑全导数= = (λn−λm)∫abρynymdx=∫abdx[ymdxd(kyn′)−yndxd(kym′)] ∫abdxdxd(ymkyn′−ynkym′)=ymkyn′−ynkym′ab k(b)[yn′(b)ym(b)−yn(b)ym′(b)] −k(a)[yn′(a)ym(a)−yn(a)ym′(a)]
γyn′(b)+δyn(b)=0,γym′(b)‾+δym(b)‾=0\gamma y_n'(b) + \delta y_n(b) = 0, \quad \gamma \overline{y_m'(b)} + \delta \overline{y_m(b)} = 0 γyn′(b)+δyn(b)=0,γym′(b)+δym(b)=0
det(yn′(b)yn(b)ym′(b)‾ym(b)‾)=yn′(b)ym(b)‾−yn(b)ym′(b)‾=0\det \begin{pmatrix} y_n'(b) & y_n(b) \\ \\ \overline{y_m'(b)} & \overline{y_m(b)} \end{pmatrix} = \purple{y_n'(b) \overline{y_m(b)} - y_n(b) \overline{y_m'(b)}} = 0 detyn′(b)ym′(b)yn(b)ym(b)=yn′(b)ym(b)−yn(b)ym′(b)=0
(λn−λm)∫abynym‾ρdx=0⇒∫abynym‾ρdx=0\orange{(\lambda_n - \lambda_m) \int_a^b y_n \overline{y_m} \rho dx} = 0 \quad \Rightarrow \quad \int_a^b y_n \overline{y_m} \rho dx = 0 (λn−λm)∫abynymρdx=0⇒∫abynymρdx=0
p(x)22m+V(x)=E\purple{\frac{p(x)^2}{2m}} + V(x) = E 2mp(x)2+V(x)=E
量子力学：状态由复值波函数 ψ(x)\psi(x)ψ(x) 描述，一维量子力学不含时薛定谔方程
m>0,ℏ>0m > 0, \hbar > 0m>0,ℏ>0, V(x)≥0V(x) \ge 0V(x)≥0，势能是非负实函数
−ℏ22md2dx2ψ(x)+V(x)ψ(x)=Eψ(x)\purple{- \frac{\hbar^2}{2m} \frac{d^2}{dx^2}} \psi(x) + V(x) \psi(x) = E \psi(x) −2mℏ2dx2d2ψ(x)+V(x)ψ(x)=Eψ(x)
ψ(x)\psi(x)ψ(x) 刻画粒子在 xxx 处被探测到的概率幅，概率密度为 ∣ψ(x)∣2|\psi(x)|^2∣ψ(x)∣2，总概率
∫−∞+∞∣ψ(x)∣2dx=1\int_{-\infty}^{+\infty} |\psi(x)|^2 dx = 1 ∫−∞+∞∣ψ(x)∣2dx=1
⇒d2ψ(x)dx2−2mℏ2V(x)ψ(x)+2mℏ2Eψ(x)=0\Rightarrow \quad \frac{d^2 \psi(x)}{dx^2} - \frac{2m}{\hbar^2} V(x) \psi(x) + \frac{2m}{\hbar^2} E \psi(x) = 0 ⇒dx2d2ψ(x)−ℏ22mV(x)ψ(x)+ℏ22mEψ(x)=0
V(x≤a)=+∞,V(x≥b)=+∞,0≤V(a<x<b)<∞V(x \le a) = +\infty, \quad V(x \ge b) = +\infty, \qquad 0 \le V(a < x < b) < \infty V(x≤a)=+∞,V(x≥b)=+∞,0≤V(a<x<b)<∞
方势阱
求解区间 (a,b)(a,b)(a,b)，第一类边界条件
ψ(a)=ψ(b)=0\psi(a) = \psi(b) = 0 ψ(a)=ψ(b)=0
ψ\psiψ 只在 (a,b)(a,b)(a,b) 之间非零，由于势墙无穷高，粒子被束缚在阱内，称为 束缚态 (bound state)
任意束缚态能量本征值 EnE_nEn 没有简并，因此相应本征波函数 ψn(x)\psi_n(x)ψn(x) 可以选为 实连续函数
能级越高的波函数的零点越多，ψn(x)\psi_n(x)ψn(x) 恰好有 n−1n - 1n−1 个零点
∫abψm(x)‾ψn(x)dx=0,n≠m\int_a^b \overline{\psi_m(x)} \psi_n(x) dx = 0, \quad n \ne m ∫abψm(x)ψn(x)dx=0,n=m
本征函数族在 (a,b)(a,b)(a,b) 上完备，即任意波函数 ψ(x)\psi(x)ψ(x) 都可以用本征函数 ψn(x)\psi_n(x)ψn(x) 展开
ψ(x)=∑n=1∞cnψn(x)\psi(x) = {\sum}_{n=1}^\infty c_n \psi_n(x) ψ(x)=∑n=1∞cnψn(x)
全体波函数构成一个 Hilbert 空间，即量子态空间