# 歸一條件

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

## 歸一化導引

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

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

## 實例

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

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

$\int_{0}^{\ell} A^2 dx= 1$

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

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

## 薛丁格方程的形式不變

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

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

## 歸一化恆定性

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

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

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

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

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

$\psi$$\psi^*$ 代入被積函數

\begin{align} \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{align}

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

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

## 參考文獻

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

## 外部連結

Middlebury 大學講義：歸一化