# 量子力学讨论班（五）

[left{begin{array}{c}frac{partialpsi}{partial t}=frac{ihbar}{2m}frac{partial^2psi}{partial x^2}\psi(x,0)=psi_0(x)end{array}right.]

[left{begin{array}{c}langle Xrangle_{psi(t)}=langle Xrangle_{psi(0)}+frac{t}{m}langle Prangle_{psi_0}\langle Prangle_{psi(t)}=langle Prangle_{psi_0}end{array}right.]

$S1$. Fourier变换法求解：

[psi(x,t)=e^{i(kx-omega(k)t)}=expleft[ikleft(x-frac{omega(k)}{k}tright)right]]

[psi(x,t)=frac{1}{sqrt{2pi}}int_{-infty}^inftyhat{psi}_0(k)e^{i(kx-omega(k)t)}dk]

[int_{mathbb{R}^2}psi(x,t)left[frac{partialchi}{partial t}+frac{ihbar}{2m}frac{partial^2chi}{partial t^2}right]dxdt=0]

[mathcal{F}(psi)(mathbf{k})=(2pi)^{-n/2}lim_{Arightarrowinfty}int_{|mathbf{x}|leq A}e^{-imathbf{k}cdotmathbf{x}}psi(mathbf{x})dmathbf{x}]

[(mathcal{F}^{-1}f)(mathbf{x})=(2pi)^{-n/2}lim_{Arightarrowinfty}int_{|mathbf{k}|leq A}e^{imathbf{k}cdotmathbf{x}}f(mathbf{k})dmathbf{k}]

$S2$. 卷积法求解：

[psi(x,t)=sqrt{frac{m}{2pi ithbar}}int_{-infty}^inftyexpleft{ifrac{m}{2thbar}(x-y)^2right}psi_0(y)dy]

$S3$. 波包传播法（第一种方法）：

$$psi(x,t)=A(x,t)e^{itheta(x,t)}$$

$$left{begin{array}{l}frac{partial A}{partial t}=-frac{hbar}{m}frac{partial A}{partial x}frac{partialtheta}{partial x}-frac{hbar}{2m}Afrac{partial^2theta}{partial x^2}\\frac{partialtheta}{partial t}=frac{hbar}{2m}frac{1}{A}frac{partial^2A}{partial x^2}-frac{hbar}{2m}left(frac{partialtheta}{partial x}right)^2end{array}right.$$

$$frac{1}{A}frac{partial^2A}{partial x^2}llleft(frac{partialtheta}{partial x}right)^2$$

$$left{begin{array}{l}frac{partial A}{partial t}=-frac{hbar}{m}frac{partial A}{partial x}frac{partialtheta}{partial x}-frac{hbar}{2m}Afrac{partial^2theta}{partial x^2}\\frac{partialtheta}{partial t}=-frac{hbar}{2m}left(frac{partialtheta}{partial x}right)^2end{array}right.$$

$$theta(x,t)=frac{p_0}{hbar}left(x-frac{p_0}{2m}tright),qquad A(x,t)=A_0left(x-frac{p_0}{m}tright)$$

$$psi(x,t)=A_0left(x-frac{p_0}{m}tright)expleft[ifrac{p_0}{hbar}left(x-frac{p_0}{2m}tright)right]$$

$S4$. 波包传播法（第二种方法）：

$$psi(x,t)=frac{1}{sqrt{2pi}}int_{-infty}^inftyhat{psi}_0(k)exp[i(kx-omega(k)t)]dk$$

$$omega(k)approxomega(k_0)+omega'(k_0)(k-k_0)$$

$$psi(x,t)approx e^{iomega'(k_0)k_0t-iomega(k_0)t}psi_0(x-omega'(k_0)t)$$

$$psi(x,t)=A_0left(x-frac{p_0}{m}tright)expleft[ifrac{p_0}{hbar}left(x-frac{p_0}{2m}tright)right]$$

$$|psi(x,t)-phi(x,t)|_{L^2(mathbb{R})}leqfrac{|t|hbarkappa^2}{2m}$$

$$kappa=left[int_{-infty}^infty|hat{psi}_0(k)|^2(k-k_0)^4dkright]^{1/4}$$

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GMT+8, 2024-2-26 15:11