# 歸一條件

（重定向自歸一化

${\displaystyle \int _{-\infty }^{\infty }\psi ^{*}(x)\psi (x)\ dx=1}$ ;

## 歸一化導引

${\displaystyle \Delta P=\mid \psi \mid ^{2}\Delta x}$(1)

${\displaystyle P=\int _{-\infty }^{\infty }\mid \psi \mid ^{2}dx=1}$(2)

## 實例

${\displaystyle \psi (x,\ t)={\begin{cases}Ae^{i(kx-\omega t)},&0\leq x\leq \ell \\0,&elsewhere\end{cases}}}$

${\displaystyle \mid \psi \mid ^{2}=A^{2}e^{i(kx-\omega t)}e^{-i(kx-\omega t)}=A^{2}}$

${\displaystyle \int _{0}^{\ell }A^{2}dx=1}$

${\displaystyle A^{2}\ell =1\qquad \Rightarrow \qquad A=\left({\frac {1}{\sqrt {\ell }}}\right)}$

${\displaystyle \psi (x,t)={\begin{cases}\left({\frac {1}{\sqrt {\ell }}}\right)e^{i(kx-\omega t)},&0\leq x\leq \ell \\0,&{\text{elsewhere}}\end{cases}}}$

## 薛丁格方程的形式不變

${\displaystyle {\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}+V(x)\psi (x)=E\psi (x)}$

${\displaystyle {\frac {-\hbar ^{2}}{2m}}A{\frac {d^{2}\psi }{dx^{2}}}+V(x)A\psi (x)=EA\psi (x)}$
${\displaystyle \Rightarrow A\left({\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}+V(x)\psi (x)\right)=A\left(E\psi (x)\right)}$
${\displaystyle \Rightarrow {\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}+V(x)\psi (x)=E\psi (x)}$

## 歸一化恆定性

${\displaystyle {\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi }{\partial x^{2}}}+V(x)\psi =i\hbar {\frac {\partial \psi }{\partial t}}}$
${\displaystyle P=\int _{-\infty }^{\infty }\psi ^{*}\psi \ dx=1}$

${\displaystyle {\frac {dP}{dt}}={\frac {d}{dt}}\int _{-\infty }^{\infty }\psi ^{*}(x,\ t)\psi (x,\ t)\ dx=\int _{-\infty }^{\infty }{\frac {\partial }{\partial t}}(\psi ^{*}(x,\ t)\psi (x,\ t))\ dx}$

${\displaystyle {\frac {\partial }{\partial t}}(\psi ^{*}\psi )={\frac {\partial \psi ^{*}}{\partial t}}\psi +\psi ^{*}{\frac {\partial \psi }{\partial t}}}$

${\displaystyle {\frac {\partial \psi }{\partial t}}={\frac {i\hbar }{2m}}{\frac {\partial ^{2}\psi }{\partial x^{2}}}-{\frac {i}{\hbar }}V(x)\psi }$

${\displaystyle {\frac {\partial \psi ^{*}}{\partial t}}={\frac {-i\hbar }{2m}}{\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}}+{\frac {i}{\hbar }}V(x)\psi ^{*}}$

${\displaystyle \psi }$${\displaystyle \psi ^{*}}$ 代入被積函數

{\displaystyle {\begin{aligned}{\frac {\partial }{\partial t}}(\psi ^{*}\psi )&=\left({\frac {-i\hbar }{2m}}{\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}}+{\frac {i}{\hbar }}V(x)\psi ^{*}\right)\psi +\psi ^{*}\left({\frac {i\hbar }{2m}}{\frac {\partial ^{2}\psi }{\partial x^{2}}}-{\frac {i}{\hbar }}V(x)\psi \right)\\&=\left({\frac {-i\hbar }{2m}}{\frac {\partial ^{2}\psi ^{*}}{\partial x^{2}}}\right)\psi +\psi ^{*}\left({\frac {i\hbar }{2m}}{\frac {\partial ^{2}\psi }{\partial x^{2}}}\right)\\&=\left({\frac {i\hbar }{2m}}\right){\frac {\partial }{\partial x}}\left(\psi ^{*}{\frac {\partial \psi }{\partial x}}-{\frac {\partial \psi ^{*}}{\partial x}}\psi \right)\\\end{aligned}}}

${\displaystyle {\frac {dP}{dt}}=\left({\frac {i\hbar }{2m}}\right)\left[\left.\left(\psi ^{*}{\frac {\partial \psi }{\partial x}}-{\frac {\partial \psi ^{*}}{\partial x}}\psi \right)\right|_{\infty }-\left.\left(\psi ^{*}{\frac {\partial \psi }{\partial x}}-{\frac {\partial \psi ^{*}}{\partial x}}\psi \right)\right|_{-\infty }\right]}$

${\displaystyle {\frac {dP}{dt}}=0}$

## 參考文獻

• Griffiths, David J. Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. 2004: pp. 12–14. ISBN 0-13-111892-7.