# 一维谐振子

图1-1 弹簧振子

力:F(x)=kxF(x)=-kx
势函数:V(x)=12kx2=12kmmx2=12ω02mx2V(x)=\frac{1}{2}kx^2=\frac{1}{2}\frac{k}{m}mx^2=\frac{1}{2}\omega_{0}^{2}mx^2
弹簧振子的固有角频率:ω02=km\omega_{0}^{2}=\frac{k}{m}\,

Schrödinger 解法:

一维谐振子势函数:V(x)=12ω02mx2V(x)=\frac{1}{2}\omega_{0}^{2}mx^2
相应 Schrödinger 方程:

itψ(x,t)=22m2x2ψ(x,t)+12ω02mx2ψ(x,t)i\hbar\frac{\partial}{\partial t}\psi(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)+\frac{1}{2}\omega_{0}^{2}mx^2\psi(x,t)

使用分离变量法,把 tt 分离出来

ψ(x,t)=ϕ(x)eiωt\psi\left(x,t\right)=\phi\left(x\right)e^{-i\omega t}ω\omega 为待定系数
Schrödinger 方程改写为

i(iω)ϕ(x)eiωt=[22md2ϕ(x)dx2+12mω02x2ϕ(x)]eiωti\hbar\left(-i\omega\right)\phi\left(x\right)e^{-i\omega t}=\left[-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\phi\left(x\right)}{\mathrm{d}x^2}+\frac{1}{2}m\omega_{0}^{2}x^2\phi\left(x\right)\right]e^{-i\omega t}

ωϕ(x)=Eϕ(x)=22md2ϕ(x)dx2+12mω02x2ϕ(x)\hbar\omega\phi\left(x\right)=E\phi\left(x\right)=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\phi\left(x\right)}{\mathrm{d}x^2}+\frac{1}{2}m\omega_{0}^{2}x^2\phi\left(x\right)

化为解微分方程的定态问题

d2ϕ(x)dx2+2m2(E12mω02x2)ϕ(x)=0\frac{\mathrm{d}^2\phi\left(x\right)}{\mathrm{d}x^2}+\frac{2m}{\hbar^2}\left(E-\frac{1}{2}m\omega_{0}^{2}x^2\right)\phi\left(x\right)=0

奇异点分析:

x\left|x\right|\rightarrow\infty 时,V(x)V\left(x\right)\rightarrow\infty ,粒子为束缚态。limxφ(x)=0\underset{\left|x\right|\rightarrow\infty}{\lim}\varphi\left(x\right)=0

22md2ϕ(x)dx2+12mω02x2ϕ(x)0-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\phi\left(x\right)}{\mathrm{d}x^2}+\frac{1}{2}m\omega_{0}^{2}x^2\phi\left(x\right)\cong 0

令方程解为 ϕ~=eλx\tilde{\phi}=e^{\lambda x} ,则 ϕ=λeλx\phi^{\prime}=\lambda e^{\lambda x}ϕ=λ2eλx\phi^{\prime\prime}=\lambda^2e^{\lambda x}\,

22mλ2eλx+12mω02x2eλx=0-\frac{\hbar^2}{2m}\lambda^2\cancel{e^{\lambda x}}+\frac{1}{2}m\omega_{0}^{2}x^2\cancel{e^{\lambda x}}=0

λ=±mω0x\lambda=\pm\frac{m\omega_0}{\hbar}x

ϕ~=exp(±mω02x2)\tilde{\phi}=\exp\left(\pm\frac{m\omega_0}{2\hbar}x^2\right)

dϕ~(x)dx=exp(mω02x2)(mω0x)\frac{d\tilde{\phi}\left(x\right)}{dx}=\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)\left(-\frac{m\omega_0}{\hbar}x\right)

d2ϕ~(x)dx2=exp(mω02x2)(mω0x)2mω0exp(mω02x2)\frac{d^2\tilde{\phi}\left(x\right)}{dx^2}=\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)\left(-\frac{m\omega_0}{\hbar}x\right)^2-\frac{m\omega_0}{\hbar}\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)

代入渐近方程:

22md2ϕ~(x)dx2+12mω02x2ϕ~(x)=[22m(mω0x)2mω0(22m)]exp(mω02x2)+12mω02x2exp(mω02x2)(12mω02x2+12mω02x2)exp(mω02x2)=0\begin{aligned} -\frac{\hbar^2}{2m}\frac{d^2\tilde{\phi}\left(x\right)}{dx^2}+\frac{1}{2}m\omega_{0}^{2}x^2\tilde{\phi}\left(x\right)&= \left[-\frac{\hbar^2}{2m}\left(\frac{m\omega_0}{\hbar}x\right)^2-\frac{m\omega_0}{\hbar}\left(-\frac{\hbar^2}{2m}\right)\right]\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)+\frac{1}{2}m\omega_{0}^{2}x^2\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)\\ &\cong\left(-\frac{1}{2}m\omega_{0}^{2}x^2+\frac{1}{2}m\omega_{0}^{2}x^2\right)\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)=0 \end{aligned}

推得

22md2ϕ~(x)dx2+12mω02x2ϕ~(x)0-\frac{\hbar^2}{2m}\frac{d^2\tilde{\phi}\left(x\right)}{dx^2}+\frac{1}{2}m\omega_{0}^{2}x^2\tilde{\phi}\left(x\right)\cong 0

解得

ϕ~1=exp(mω02x2)ϕ~2=exp(mω02x2)x,是发散的,舍弃)\tilde{\phi}_1=\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\text{,}\tilde{\phi}_2=\exp\left(\frac{m\omega_{0}}{2\hbar}x^2\right)\text{(}x\rightarrow\infty\text{,是发散的,舍弃)}

我们要求波函数是平方可积的,所以要求 xx\rightarrow\infty ,波函数一定趋近于 00

ϕ(x)=ϕ~(x)χ(x)=exp(mω02x2)χ(x)\phi\left(x\right)=\tilde{\phi}\left(x\right)\chi\left(x\right)=\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi\left(x\right)

ϕ(x)=mω0xexp(mω02x2)χ(x)+exp(mω02x2)χ(x)\phi^{\prime}\left(x\right)=-\frac{m\omega_{0}}{\hbar}x\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi\left(x\right)+\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi^{\prime}\left(x\right)

ϕ(x)=[(mω0x)2mω0]exp(mω02x2)χ(x)2mω0xexp(mω02x2)χ(x)+exp(mω02x2)χ(x)\phi^{\prime\prime}\left(x\right)=\left[\left(\frac{m\omega_{0}}{\hbar}x\right)^2-\frac{m\omega_{0}}{\hbar}\right]\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi\left(x\right)-\frac{2m\omega_{0}}{\hbar}x\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi^{\prime}\left(x\right)+\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi^{\prime\prime}\left(x\right)

代入

22mϕ(x)+12mω02x2ϕ(x)=Eϕ(x)-\frac{\hbar^2}{2m}\phi^{\prime\prime}\left(x\right)+\frac{1}{2}m\omega_{0}^{2}x^2\phi\left(x\right)=E\phi\left(x\right)

22m[(mω0x)2mω0]exp(mω02x2)χ(x)+12mω02x2exp(mω02x2)χ(x)2mω0xexp(mω02x2)χ(x)=Eexp(mω02x2)χ(x)+exp(mω02x2)χ(x)-\frac{\hbar^2}{2m}\left[\left(\frac{m\omega_{0}}{\hbar}x\right)^2-\frac{m\omega_{0}}{\hbar}\right]\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi\left(x\right)+\frac{1}{2}m\omega_{0}^{2}x^2\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi\left(x\right)-\frac{2m\omega_{0}}{\hbar}x\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi^{\prime}\left(x\right)=E\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi\left(x\right)+\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi^{\prime\prime}\left(x\right)

22mχ(x)+ω0xχ(x)+12ω0χ(x)12mω02x2χ(x)+12mω02x2χ(x)=Eχ(x)-\frac{\hbar^2}{2m}\chi^{\prime\prime}\left(x\right)+\hbar\omega_{0}x\chi^{\prime}\left(x\right)+\frac{1}{2}\hbar\omega_{0}\chi\left(x\right)-\frac{1}{2}m\omega_{0}^{2}x^2\chi\left(x\right)+\frac{1}{2}m\omega_{0}^{2}x^2\chi\left(x\right)=E\chi\left(x\right)

22mχ(x)ω0xχ(x)+(E12ω)χ(x)=0(福克斯Fuchs微分方程)\frac{\hbar^2}{2m}\chi^{\prime\prime}\left(x\right)-\hbar\omega_{0}x\chi^{\prime}\left(x\right)+\left(E-\frac{1}{2}\hbar\omega\right)\chi\left(x\right)=0\text{(福克斯}Fuchs\text{微分方程)}

奇异点分析做完,把奇异点扣除,由于方程是二阶的,所以一定有两个线性无关的解在零点处解析(可在零点展成无穷级数)

χ(x)=C0+C1x+C2x2+χ(x)=C1+2C2x+3C3x2+χ(x)=2C2+6C3x+\begin{aligned} \chi\left(x\right)&=\mathrm{C}_0+\mathrm{C}_1x+\mathrm{C}_2x^2+\cdots\\ \chi^{\prime}\left(x\right)&=\mathrm{C}_1+2\mathrm{C}_2x+3\mathrm{C}_3x^2+\cdots\\ \chi^{\prime\prime}\left(x\right)&=2\mathrm{C}_2+6\mathrm{C}_3x+\cdots \end{aligned}

22mχω0xχ+(E12ω)χ=0\frac{\hbar^2}{2m}\chi^{\prime\prime}-\hbar\omega_{0}x\chi^{\prime}+\left(E-\frac{1}{2}\hbar\omega\right)\chi=0

22m(2C2+6C3x+)ω0x(C1+2C2x+3C3x2+)+(E12ω)(C0+C1x+C2x2+)=0\frac{\hbar^2}{2m}\left(2\mathrm{C}_2+6\mathrm{C}_3x+\cdots\right)-\hbar\omega_0x\left(\mathrm{C}_1+2\mathrm{C}_2x+3\mathrm{C}_3x^2+\cdots\right)+\left(E-\frac{1}{2}\hbar\omega\right)\left(\mathrm{C}_0+\mathrm{C}_1x+\mathrm{C}_2x^2+\cdots\right)=0

展开每项

22m2C2+22m6C3x+ω0xC1ω0x2C2x++(E12ω)C0+(E12ω)C1x+(E12ω)C2x2+=0\begin{aligned} \quad\qquad\frac{\hbar^2}{2m}\cdot 2\mathrm{C}_2&+\frac{\hbar^2}{2m}\cdot 6\mathrm{C}_3x+\cdots\\ &-\hbar\omega _0x\mathrm{C}_1-\hbar\omega_{0}x\cdot 2\mathrm{C}_2x+\cdots\\ +\left(E-\frac{1}{2}\hbar\omega\right)\mathrm{C}_0&+\left(E-\frac{1}{2}\hbar\omega\right)\mathrm{C}_1x+\left(E-\frac{1}{2}\hbar\omega\right)\mathrm{C}_2x^2+\cdots=0 \end{aligned}

xx 的一次幂、二次幂、三次幂作为函数是线性无关的,若其线性关系等于 00 ,则所乘系数必须等于 00

x0=22m2C2+(E12ω)C0=0x1=22m6C3ω0xC1+(E12ω)C1=0x2=ω0x2C2+(E12ω)C2=0\begin{aligned} x^0&=\frac{\hbar^2}{2m}\cdot 2\mathrm{C}_2+\left(E-\frac{1}{2}\hbar\omega\right)\mathrm{C}_0=0\\ x^1&=\frac{\hbar^2}{2m}\cdot 6\mathrm{C}_3-\hbar\omega_{0}x\mathrm{C}_1+\left(E-\frac{1}{2}\hbar\omega\right)\mathrm{C}_1=0\\ x^2&=-\hbar\omega_{0}x\cdot 2\mathrm{C}_2+\left(E-\frac{1}{2}\hbar\omega\right)\mathrm{C}_2=0\\ \end{aligned} \vdots

C0\mathrm{C}_0C1\mathrm{C}_1 不确定(有一定自由度),一旦给定,其余系数也都能给定。

  1. C0=1\mathrm{C}_0=1C1=0\mathrm{C}_1=0 ,则 C1=C3=C5==0\mathrm{C}_1=\mathrm{C}_3=\mathrm{C}_5=\cdots =0\,

22m2C2=(E12ω)C0=(E12ω)C2=m(E12ω)C40C60\begin{aligned} \frac{\hbar^2}{2m}\cdot 2\mathrm{C}_2=-\left(E-\frac{1}{2}\hbar\omega\right)&\mathrm{C}_0=-\left(E-\frac{1}{2}\hbar\omega\right)\\ \Downarrow\quad&\\ \mathrm{C}_2=-\frac{m}{\hbar}\left(E-\frac{1}{2}\hbar\omega\right)&\text{,}\mathrm{C}_4\ne 0\text{,}\mathrm{C}_6\ne 0 \end{aligned}

χ(x)=C0+C2x2+C4x4+\chi_{\text{偶}}\left(x\right)=\mathrm{C}_0+\mathrm{C}_2x^2+\mathrm{C}_4x^4+\cdots

xχ(x)=exp(mω0x2)\text{当}\left|x\right|\rightarrow\infty\text{,}\chi_{\text{偶}}\left(x\right)=\exp\left(-\frac{m\omega_{0}}{\hbar}x^2\right)

ϕ(x)=exp(mω02x2)χ(x)x{exp(mω02x2)exp(mω0x2)exp(mω02x2)(发散需截断)\phi\left(x\right)=\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi_{\text{偶}}\left(x\right) \underrightarrow{\left|x\right|\rightarrow\infty} \begin{cases} \exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\exp\left(\frac{m\omega_{0}}{\hbar}x^2\right)\\ \exp\left(\frac{m\omega_{0}}{2\hbar}x^2\right)\text{(发散需截断)} \end{cases}

E12ω=0E-\frac{1}{2}\hbar\omega=0E0=12ωE_0=\frac{1}{2}\hbar\omega ,则 C2=0,C4=0,C6=0,\mathrm{C}_2=0,\mathrm{C}_4=0,\mathrm{C}_6=0,\cdots

χ0(x)=C0+C2x2+C4x4+=C0=1\chi_0\left(x\right)=\mathrm{C}_0+\mathrm{C}_2x^2+\mathrm{C}_4x^4+\cdots =\mathrm{C}_0=1

ϕ(x)=A0exp(mω02x2)χ0(x)=A0exp(mω02x2)A0为归一化系数。\phi\left(x\right)=A_{0}\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\chi_{0}\left(x\right)=A_0\exp\left(-\frac{m\omega_{0}}{2\hbar}x^2\right)\text{,}A_{0}\text{为归一化系数。}

  1. C0=0\mathrm{C}_0=0C1=1\mathrm{C}_1=1C0=C2=C4==0\mathrm{C}_0=\mathrm{C}_2=\mathrm{C}_4=\cdots=0

32mC3=(E32ω)E=32ω(需截断否则发散)\frac{3\hbar^2}{m}\mathrm{C}_3=-\left(E-\frac{3}{2}\hbar\omega\right)\quad\quad E=\frac{3}{2}\hbar\omega\text{(需截断否则发散)}

C3=0,C5=0,\mathrm{C}_3=0,\mathrm{C}_5=0,\cdots

χ1(x)=x\chi_1\left(x\right)=x

结论:当 EE 取值为 En=(n+12)ω0E_n=\left(n+\frac{1}{2}\right)\hbar\omega_0n=0,1,2,n=0,1,2,\cdots 时,χ(x)\chi\left(x\right) 的级数解会在某一阶截断,成为厄密多项式 Hn(y)H_n\left(y\right)(为实函数),χn(x)=Hn(y)=Hn(mω0x)\chi_n\left(x\right)=H_n\left(y\right)=H_n\left(\sqrt{\frac{m\omega_{0}}{\hbar}}x\right)

ϕn(x)=Anexp(mω02x2)Hn(mω0x)An=[mω0π2nn!]12\phi_n\left(x\right)=A_n\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)H_n\left(\sqrt{\frac{m\omega_0}{\hbar}}x\right)\text{,}A_n=\left[\frac{\sqrt{\frac{m\omega_0}{\hbar}}}{\sqrt{\pi}2^nn!}\right]^{\frac{1}{2}}

引入归一化系数 AnA_n ,使得波函数归一化,即

ϕn(x)=Anexp(mω02x2)Hn(mω0x)An=[mω0π2nn!]12\phi_n\left(x\right)=A_n\exp\left(-\frac{m\omega_0}{2\hbar}x^2\right)H_n\left(\sqrt{\frac{m\omega_0}{\hbar}}x\right)\text{,}A_n=\left[\frac{\sqrt{\frac{m\omega_0}{\hbar}}}{\sqrt{\pi}2^nn!}\right]^{\frac{1}{2}}

ϕn(x)ϕn(x)dx=ϕn(x)2dx=1\int_{-\infty}^{\infty}{\phi_{n}^{*}\left(x\right)\phi_{n}\left(x\right)\mathrm{d}x=}\int_{-\infty}^{\infty}{\left|\phi_{n}\left(x\right)\right|^2\mathrm{d}x=}1

(ϕm,ϕn)=ϕm(x)ϕn(x)dx=0(mn)\left(\phi_{m},\phi_{n}\right)=\int_{-\infty}^{\infty}{\phi_{m}^{*}\left(x\right)\phi_{n}\left(x\right)\mathrm{d}x=}0\quad\left(\text{若}m\ne n\right)

ϕm(x)ϕn(x)dx=δmn\int_{-\infty}^{\infty}{\phi_{m}^{*}\left(x\right)\phi_{n}\left(x\right)\mathrm{d}x=}\delta_{mn}

引入记号 α=mω0\alpha=\sqrt{\frac{m\omega_0}{\hbar}}

ϕ0(x)=απ1/4eα2x22,E0=12ω0ϕ1(x)=2απ1/4αxeα2x22,E1=32ω0ϕ2(x)=α/2π1/4(2α2x21)eα2x22,E2=52ω0}En=(n+12)ω0\begin{rcases} \phi_0\left(x\right)=\frac{\sqrt{\alpha}}{\pi^{1/4}}e^{-\frac{\alpha^2x^2}{2}}, & E_0=\frac{1}{2}\hbar\omega_{0}\\ \phi_1\left(x\right)=\frac{\sqrt{2\alpha}}{\pi^{1/4}}\alpha xe^{-\frac{\alpha^2x^2}{2}}, & E_1=\frac{3}{2}\hbar\omega_{0}\\ \phi_2\left(x\right)=\frac{\sqrt{\alpha /2}}{\pi^{1/4}}\left(2\alpha^2x^2-1\right)e^{-\frac{\alpha ^2x^2}{2}}, & E_2=\frac{5}{2}\hbar\omega_{0} \end{rcases} E_n=\left(n+\frac{1}{2}\right)\hbar\omega_{0}

E0=12ω0E_0=\frac{1}{2}\hbar\omega_{0} 为零点能。

估算:

x0xˉ(Δx)2,Px0Px(ΔPx)2x\sim 0\Rightarrow\bar{x}\sim\sqrt{\overline{\left(\varDelta x\right)^2}},\quad P_x\sim 0\Rightarrow\overline{P_x}\sim\sqrt{\overline{\left(\varDelta P_x\right)^2}}

xˉPx(Δx)2(ΔPx)2142=12\bar{x}\cdot\overline{P_x}\sim\sqrt{\overline{\left(\varDelta x\right)^2}\overline{\left(\varDelta P_x\right)^2}}\sim\sqrt{\frac{1}{4}\hbar^2}=\frac{1}{2}\hbar

xˉ=21Px\Rightarrow\bar{x}=\frac{\hbar}{2}\frac{1}{\overline{P_x}}

EP22m+12mω02xˉ2=P22m+12mω02241P2E\cong\frac{\overline{P^2}}{2m}+\frac{1}{2}m\omega_{0}^{2}\bar{x}^2=\frac{\overline{P^2}}{2m}+\frac{1}{2}m\omega_{0}^{2}\frac{\hbar ^2}{4}\frac{1}{\overline{P^2}}

t=P2t=\overline{P^2}

dE(t)dt=12m+18mω022(1t2)=0\frac{dE\left(t\right)}{dt}=\frac{1}{2m}+\frac{1}{8}m\omega_{0}^{2}\hbar^2\left(-\frac{1}{t^2}\right)=0

12m=18mω022(1t2)\frac{1}{2m}=-\frac{1}{8}m\omega_{0}^{2}\hbar^2\left(\frac{1}{t^2}\right)

t2=(12mω0)2tmin=12mω0\Rightarrow t^2=\left(\frac{1}{2}m\omega_{0}\hbar\right)^2\Rightarrow t_{\min}=\frac{1}{2}m\omega_{0}\hbar

Emin=E(tmin)=tmin2m+12mω0224tmin=12mmω02+12mω0224mω02=14ω0+14ω0=12ω0\begin{aligned} E_{\min}&=E\left(t_{\min}\right)\\ &=\frac{t_{\min}}{2m}+\frac{1}{2}m\omega_{0}^{2}\frac{\hbar^2}{4t_{\min}}\\ &=\frac{1}{2m}\frac{m\omega_{0}\hbar}{2}+\frac{1}{2}m\omega_{0}^{2}\frac{\hbar^2}{4\frac{m\omega_{0}\hbar}{2}}\\ &=\frac{1}{4}\hbar\omega_{0}+\frac{1}{4}\hbar\omega_{0}\\ &=\frac{1}{2}\hbar\omega_{0} \end{aligned}

# 一维定态体系的一些基本性质

定理一

设势能函数 V(x)V(x) 为实函数。若 ψ(x)\psi(x) 是本征方程的解,则 ψ(x)\psi^{*}(x) 也是一个解,对应本征值也是 EE(也适用于二维)。

证明:

V(x)=V(x),V(x)为实函数\text{若}V^{*}(x)=V^{*}(x), \text{即}V^{*}(x)\text{为实函数}

Eψ(x)=22md2dx2ψ(x)+V(x)ψ(x)E\psi(x)=-\frac{\hbar^{2}}{2m}\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}\psi(x)+V(x)\psi(x)

[Eψ(x)]=Eψ(x)=[22md2dx2ψ(x)+V(x)ψ(x)]=22md2dx2ψ(x)+V(x)ψ(x)[E\psi(x)]^{*}=E\psi^{*}(x)=[-\frac{\hbar^{2}}{2m}\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}\psi(x)+V(x)\psi(x)]^{*}=-\frac{\hbar^{2}}{2m}\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}\psi^{*}(x)+V(x)\psi^{*}(x)

推论

若方程的本征值 EE 是非简并的,则相应的本征函数可以取作实函数。
非简并就是方程只有一个线性无关的解,一定有 ψ(x)=Cψ(x)\psi^{*}(x)=C\psi(x)

ψ(x)2dx=Cψ(x)2dx=C2ψ(x)2dx\int|\psi^{*}(x)|^{2}\mathrm{d}x=\int|C\psi(x)|^{2}\mathrm{d}x=|C|^{2}\int|\psi(x)|^{2}\mathrm{d}x

假设 ψ(x)2dx=1\int|\psi(x)|^{2}\mathrm{d}x=1 ,因 ψ(x)2dx=1\int|\psi^{*}(x)|^{2}\mathrm{d}x=1 ,则 C2=1|C|^{2}=1C=eiαC=e^{i\alpha}ψ=ψeiα2\psi^{*}=\psi|e^{i\alpha}|^2


定理二

设势能函数 V(x)V(x) 为实函数,则对应于任何方程的本征值 EE ,总可以找到方程的一组实函数解。凡是属于本征值 EE 的任何一个解,都可以表示成这组实函数解的线性叠加。

假设 ψ(x)\psi(x)ψ(x)\psi^{*}(x) 线性无关,即 C1ψ(x)+C2ψ(x)=0C_1\psi(x)+C_2\psi^{*}(x)=0 ,则只有 C1=0,C2=0C_1=0,\;C_2=0
定义 φ(x)=ψ(x)+ψ(x),χ(x)=i[ψ(x)ψ(x)]\varphi(x)=\psi(x)+\psi^{*}(x),\;\chi(x)=-i\left[\psi(x)-\psi^{*}(x)\right](两函数都是实的,取复共轭等于其自身)
假设 D1φ+D2χ=0D_1\varphi+D_2\chi=0

D1[ψ(x)+ψ(x)]+D2{i[ψ(x)ψ(x)]}=0D1ψ+D1ψiD2ψ+iD2ψ=0(D1iD2)ψ+(D1+iD2)ψ=0\begin{aligned} D_1\left[\psi(x)+\psi^{*}(x)\right]+D_2\{-i\left[\psi(x)-\psi^{*}(x)\right]\}&=0\\ D_1\psi+D_1\psi^{*}-iD_2\psi+iD_2\psi^{*}&=0\\ \left(D_1-iD_2\right)\psi+\left(D_1+iD_2\right)\psi^{*}&=0 \end{aligned}

因为 ψ\psiψ\psi^{*} 线性无关,则

{D1iD2=0D1+iD2=0{D1=0D2=0\begin{cases} D_1-iD_2=0\\ D_1+iD_2=0 \end{cases} \Rightarrow \begin{cases} D_1=0\\ D_2=0 \end{cases}


定理三

设势能函数 V(x)V(x) 具有空间反射对称性( V(x)=V(x)V(-x)=V(x) )。若 ψ(x)\psi(x) 是方程的一个解,则 ψ(x)\psi(-x) 亦是该方程的一个解。

证明:令 x=xx^{\prime}=-x

Eψ(x)=22md2dx2ψ(x)+V(x)ψ(x)=22md2dx2ψ(x)+V(x)ψ(x)V(x)具有空间反射对称性。=22md2dx2ψ(x)+V(x)ψ(x)=Eψ(x)\begin{aligned} E\psi(-x)&=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(-x)+V(x)\psi(-x)\\ &=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(x^{\prime})+V(-x^{\prime})\psi(x^{\prime})\\ &\Downarrow V(x)\text{具有空间反射对称性。}\\ &=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(x^{\prime})+V(x^{\prime})\psi(x^{\prime})\\ &=E\psi(x^{\prime}) \end{aligned}

EE 为非简并时,有 P^ψn(x)=ψn(x)=Cψ(x)\hat{P}\psi_n(x)=\psi_n(-x)=\mathrm{C}\psi(x)
经过两次反射后,P^(P^ψ(x))=P^(ψ(x))=P^(Cψ(x))=C2ψ(x)\hat{P}\left(\hat{P}\psi(x)\right)=\hat{P}\left(\psi(-x)\right)=\hat{P}\left(\mathrm{C}\psi(x)\right)=\mathrm{C}^2\psi(x)
因此 C2=1\mathrm{C}^2=1C=±1\mathrm{C}=\pm 1C\mathrm{C} 为波函数的宇称值,P^\hat{P} 称为宇称算符

P^ψn(x)=ψn(x)\hat{P}\psi_n(x)=\psi_{n}(-x) ,则称 ψn(x)\psi_{n}(x) 具有偶宇称;若 P^ψn(x)=ψn(x)\hat{P}\psi_{n}(x)=-\psi_{n}(x) ,则称 ψn(x)\psi_{n}(x) 具有奇宇称。
对于一维谐振子,由于它的每一条能级都是非简并的,因此相应的本征函数都具有确定的宇称。En=(n+12)ω0E_n=(n+\frac{1}{2})\hbar\omega_{0}P^φn(x)=φn(x)=(1)nφn(x)\hat{P}\varphi_n(x)=\varphi_n(-x)=(-1)^{n}\varphi_n(x)


定理四

设势能函数 V(x)V(x) 具有空间反射对称性。则对应于任何一个本征值 EE ,我们总可以找到方程的一组解,它们的每一个都具有确定的宇称。而且,任何对应于本征值 EE 的解 ψ(x)\psi(x) 都可以按照它们来展开。

证明:ψ(x)=Cψ(x)\psi(-x)=\mathrm{C}\psi(x)ψ(x)\psi(x)ψ(x)\psi(-x) 是线性无关的,即 C1ψ(x)+C2ψ(x)=0(C1=0,C2=0)\mathrm{C}_1\psi(x)+\mathrm{C}_2\psi(-x)=0\;(\mathrm{C}_1=0,\mathrm{C}_2=0)
定义

ϕ(x)=ψ(x)+ψ(x)ϕ(x)=ψ(x)+ψ((x))=ψ(x)+ψ(x)=ϕ(x)\begin{aligned} \phi(x)&=\psi(x)+\psi(-x)\\ \phi(-x)&=\psi(-x)+\psi(-(-x))=\psi(-x)+\psi(x)=\phi(x) \end{aligned}

ϕ(x)\phi(x)ϕ(x)\phi(-x) 也是线性无关的,上面看出 ϕ(x)\phi(x) 具有偶宇称。
而在定理二的证明中,χ(x)=ψ(x)ψ((x))=ψ(x)ψ(x)=χ(x)\chi(-x)=\psi(-x)-\psi(-(-x))=\psi(-x)-\psi(x)=-\chi(x),即χ(x)\chi(x) 具有奇宇称。


定理五

对于一维空间的 Schrödinger 方程,若 ψ1(x)\psi_{1}(x)ψ2(x)\psi_{2}(x) 是对应于同一本征值 EE 的两个解,则有 ψ1(x)ψ2(x)ψ1(x)ψ2(x)=常数\psi_{1}(x)\psi_{2}^{\prime}(x)-\psi_{1}^{\prime}(x)\psi_{2}(x)=\text{常数}

证明:

{Eψ1(x)=22md2dx2ψ1(x)+V(x)ψ1(x)(1)Eψ2(x)=22md2dx2ψ2(x)+V(x)ψ2(x)(2)\begin{cases} E\psi_{1}(x)=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_{1}(x)+V(x)\psi_{1}(x) & \text{(1)}\\ E\psi_{2}(x)=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_{2}(x)+V(x)\psi_{2}(x) & \text{(2)} \end{cases}

(1)×ψ2(x)(2)×ψ1(x)\text{(1)}\times\psi_{2}(x)-\text{(2)}\times\psi_{1}(x)

22m[ψ2(x)d2dx2ψ1(x)ψ1(x)d2dx2ψ2(x)]=0-\frac{\hbar^2}{2m}\left[\psi_{2}(x)\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_{1}(x)-\psi_{1}(x)\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_{2}(x)\right]=0

从方程中得

ddx[ψ1(x)ψ2(x)ψ1(x)ψ2(x)]=0\frac{\mathrm{d}}{\mathrm{d}x}\left[\psi_{1}(x)\psi_{2}^{\prime}(x)-\psi_{1}^{\prime}(x)\psi_{2}(x)\right]=0

因此

ψ1(x)ψ2(x)ψ1(x)ψ2(x)=常数(仅一维空间)\psi_{1}(x)\psi_{2}^{\prime}(x)-\psi_{1}^{\prime}(x)\psi_{2}(x)=\text{常数(仅一维空间)}


定理六

假设粒子在一个一维空间中规则(regular)(即没有奇异点)的势函数 V(x)V(x) 中运动,则其束缚态必为非简并的。

证明:势函数 V(x)V(x) 为规则的 \Rightarrow Schrödinger 方程的任何一个解 ψ(x)\psi(x) 及其一阶导数 ψ(x)\psi^{\prime}(x) 在空间的每一点都连续。

V(x)没有太多奇异性(regular){(1)不包含δ函数(2)V(x)有有限个阶跃V(x)\text{没有太多奇异性(regular)}\Rightarrow\begin{cases} \text{(1)不包含}\delta\text{函数}\\ \text{(2)}V(x)\text{有有限个阶跃} \end{cases}

假设 EnE_n 是粒子的一个束缚态的本征值且简并,则可以找到至少两个本征函数 ψ1(x)\psi_{1}(x)ψ2(x)\psi_{2}(x) 与它对应。
另一方面根据定理五,有

ψ1(x)ψ2(x)ψ1(x)ψ2(x)=常数\psi_{1}(x)\psi_{2}^{\prime}(x)-\psi_{1}^{\prime}(x)\psi_{2}(x)=\text{常数}

对于束缚态,有渐进关系

limxψ1(x)=limxψ2(x)=0\lim_{|x|\to\infty}\psi_{1}(x)=\lim_{|x|\to\infty}\psi_{2}(x)=0

则常数为零,方程改写为

ψ1(x)ψ1(x)=ψ2(x)ψ2(x)1ψ1dψ1dx=1ψ2dψ2dxddxlnψ1=ddxlnψ2lnψ1=lnψ2+Cψ1(x)=eCψ2(x)=Aψ2\begin{aligned} \frac{\psi_{1}^{\prime}(x)}{\psi_{1}(x)}&=\frac{\psi_{2}^{\prime}(x)}{\psi_{2}(x)}\\ \frac{1}{\psi_{1}}\frac{\mathrm{d}\psi_{1}}{\mathrm{d}x}&=\frac{1}{\psi_{2}}\frac{\mathrm{d}\psi_{2}}{\mathrm{d}x}\\ \frac{\mathrm{d}}{\mathrm{d}x}\ln\psi_{1}&=\frac{\mathrm{d}}{\mathrm{d}x}\ln\psi_{2}\\ \ln\psi_{1}&=\ln\psi_{2}+C\\ \text{或}\psi_{1}(x)&=e^{C}\psi_{2}(x)=A\psi_{2} \end{aligned}

ψ1(x)\psi_{1}(x)ψ2(x)\psi_{2}(x) 线性相关,这与假设不符,因此得证。

总结:

{(1)定理一、二:势函数为实函数的情况;(2)定理三、四:势函数具有空间反射对称性的情况;(3)定理五、六:仅适用于一维空间,高维空间不成立。\begin{cases} \text{(1)定理一、二:势函数为实函数的情况;}\\ \text{(2)定理三、四:势函数具有空间反射对称性的情况;}\\ \text{(3)定理五、六:仅适用于一维空间,高维空间不成立。} \end{cases}


# 一维方势阱

在量子力学中,一般给定势能函数较复杂,实际中可以近似于方势阱,其结果在定性上与真实出入不大。

图3-1 一维谐振子势函数

V(x)=12kx2V(x)=\frac{1}{2}kx^{2}

图3-2 一维谐振子势函数

V(x)=chx=ex+ex2V(x)=\ch{x}=\frac{e^{x}+e^{-x}}{2}

# 无穷深一维方势阱

对于一维谐振子、双曲余弦函数的近似。
图3-3

V(x)={+,x00,0xa,xaV(x)= \begin{cases} +\infty, & x\leqslant 0\\ 0, & 0\leqslant x\leqslant a\\ -\infty, & x\geqslant a \end{cases}

阱内:

Eψ(x)=22md2dx2ψ(x)E\psi(x)=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(x)

阱外:令 ψ(x=0)=0\psi(x=0)=0ψ(x=a)=0\psi(x=a)=0

Eψ(x)+22md2dx2ψ(x)=0d2ψ(x)dx2+2mE2ψ(x)=0\begin{aligned} E\psi(x)+\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(x)&=0\\ \frac{\mathrm{d}^2\psi(x)}{\mathrm{d}x^2}+\frac{2mE}{\hbar^2}\psi(x)&=0 \end{aligned}

令解 ψ=eλx\psi=e^{\lambda x}ψ=λeλx\psi^{\prime}=\lambda e^{\lambda x}ψ=λ2eλx\psi^{\prime\prime}=\lambda^{2} e^{\lambda x}\,

λ2eλx+2mE2eλx=0λ=2mE2\begin{aligned} \lambda^{2} e^{\lambda x}&+\frac{2mE}{\hbar^2}e^{\lambda x}=0\\ \lambda&=-\frac{2mE}{\hbar^2} \end{aligned}

解得:

ψ1=exp(i2mE2x),ψ2=exp(i2mE2x)\psi_{1}=\exp\left(-i\sqrt{\frac{2mE}{\hbar^2}}x\right)\text{,}\; \psi_{2}=\exp\left(i\sqrt{\frac{2mE}{\hbar^2}}x\right)

因为 ψ1(x=0)=1\psi_{1}(x=0)=1ψ1(x=a)=exp(i2mE2x)0\psi_{1}(x=a)=\exp\left(-i\sqrt{\frac{2mE}{\hbar^2}}x\right)\neq 0
ψ1\psi_{1}ψ2\psi_{2} 线性无关,取其线性组合(原因:Schrödinger 方程是线性的)。

ψ=C1ψ1+C2ψ2=C1exp(i2mE2x)+C2exp(i2mE2x)\psi=C_{1}\psi_{1}+C_{2}\psi_{2}=C_{1}\exp\left(-i\sqrt{\frac{2mE}{\hbar^2}}x\right)+C_{2}\exp\left(i\sqrt{\frac{2mE}{\hbar^2}}x\right)

  1. 由条件 ψ(x=0)=0\psi(x=0)=0 得,C1e0+C2e0=C1+C2C1=C2C_{1}e^{0}+C_{2}e^{0}=C_{1}+C_{2}\implies C_{1}=-C_{2}\,

ψ(x)=C1ψ1+(C1)ψ2=C1(ψ1ψ2)=C1[exp(i2mE2x)exp(i2mE2x)]=2iC1[exp(i2mE2x)exp(i2mE2x)2i]=(2iC1)sin2mE2x=Dsin2mE2x\begin{aligned} \psi(x)&=C_{1}\psi_{1}+(-C_{1})\psi_{2}\\ &=C_{1}(\psi_{1}-\psi_{2})\\ &=C_{1}\left[\exp\left(-i\sqrt{\frac{2mE}{\hbar^2}}x\right)-\exp\left(i\sqrt{\frac{2mE}{\hbar^2}}x\right)\right]\\ &=-2iC_{1}\left[\frac{\exp\left(-i\sqrt{\frac{2mE}{\hbar^2}}x\right)-\exp\left(i\sqrt{\frac{2mE}{\hbar^2}}x\right)}{-2i}\right]\\ &=(-2iC_{1})\sin\sqrt{\frac{2mE}{\hbar^2}}x\\ &=D\sin\sqrt{\frac{2mE}{\hbar^2}}x \end{aligned}

sinx=eixeix2icosx=eix+eix2\begin{aligned} \sin{x}&=\frac{e^{ix}-e^{-ix}}{2i}\\ \cos{x}&=\frac{e^{ix}+e^{-ix}}{2} \end{aligned}

  1. 由条件 ψ(x=a)=0\psi(x=a)=0 得,

ψ(x=a)=Dsin2mE2a\psi(x=a)=D\sin\sqrt{\frac{2mE}{\hbar^2}}a

2mE2a=nπ,n=1,2,3,\sqrt{\frac{2mE}{\hbar^2}}a=n\pi,\;n=1,2,3,\cdots

En=n2π222ma2,ψn(x)=DnsinnπaxE_{n}=\frac{n^2\pi^2\hbar^2}{2ma^2},\quad \psi_{n}(x)=D_{n}\sin\frac{n\pi}{a}x

0nψn(x)2dx=10nDn2sin2nπaxdx=1=Dn20nsin2nπaxdx=Dn20n12(1cos2nπax)dx=Dn2a2\begin{aligned} \int_{0}^{n}\left|\psi_{n}(x)\right|^2\mathrm{d}x&=1\\ \int_{0}^{n}\left|D_{n}\right|^2\sin^2\frac{n\pi}{a}x\mathrm{d}x&=1\\ &=\left|D_{n}\right|^2\int_{0}^{n}\sin^2\frac{n\pi}{a}x\mathrm{d}x\\ &=\left|D_{n}\right|^2\int_{0}^{n}\frac{1}{2}\left(1-\cos\frac{2n\pi}{a}x\right)\mathrm{d}x\\ &=\left|D_{n}\right|^2\frac{a}{2} \end{aligned}

Dn=2a\left|D_{n}\right|=\sqrt{\frac{2}{a}} ,取 Dn=2aD_{n}=\sqrt{\frac{2}{a}}
最终得到,

ψn(x)=2asinnπax\boxed{\psi_{n}(x)=\sqrt{\frac{2}{a}}\sin\frac{n\pi}{a}x}