贡献者： addis
本文采用原子单位制。我们要解一维定态薛定谔方程（参考式 9 ）为
\begin{equation}
-\frac{1}{2m} \frac{\mathrm{d}^{2}{\psi}}{\mathrm{d}{x}^{2}} + V\psi= E \psi~.
\end{equation}
令方势垒长度为 $2l$，关于原点对称。势能函数为
\begin{equation}
V(x) =
\begin{cases}
0 & (x < -l)\\
V_0 & (-l \leqslant x < l)\\
0 & (l \leqslant x)~.
\end{cases}
\end{equation}
一些教材把势垒放在区间 $[0, a]$，这时只需令 $a = 2l$，把解出的波函数向右平移 $l$ 即可。以下讨论两种常见的解，一种类似于 $ \sin\left(kx\right) , \cos\left(kx\right) $ 另一种类似于 $ \exp\left( \mathrm{i} kx\right) , \exp\left(- \mathrm{i} kx\right) $。
1. 奇偶函数解
对称势能的好处是存在奇和偶的实函数解，且它们自动正交。
$E \ge V_0$ 的情况
令
\begin{equation}
k = \sqrt{2mE}~,
\qquad
b = \sqrt{2m(E-V_0)}~.
\end{equation}
令对称解和反对称的波函数分别为
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^e(x) =
\begin{cases}
A_1 \cos\left(bx\right) & (0 \leqslant x \leqslant l)\\
C_1 \cos\left(kx\right) + D_1 \sin\left(kx\right) & (l < x)\\
\psi_{ \boldsymbol{\mathbf{k}} }^e(-x) & (x < 0)~,
\end{cases}
\end{equation}
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^o(x) =
\begin{cases}
B_2 \sin\left(bx\right) & (0 \leqslant x \leqslant l)\\
C_2 \cos\left(kx\right) + D_2 \sin\left(kx\right) & (l < x)\\
-\psi_{ \boldsymbol{\mathbf{k}} }^o(-x) & (x < 0)~,
\end{cases}
\end{equation}
其中 $l < x$ 的部分也可以分别记为
\begin{equation}
G_i \sin\left(kx + \phi_i\right) \qquad (i = 1,2)~,
\end{equation}
\begin{equation}
G_i = \sqrt{C_i^2 + D_i^2}
\qquad
\phi_i = \operatorname{Arctan} (C_i, D_i)
\qquad
(i = 1,2)~.
\end{equation}
在 $x = l$ 处匹配波函数和一阶导数，解得
\begin{equation}
\left\{\begin{aligned}
\frac{C_1}{A_1} &= \cos\left(kl\right) \cos\left(bl\right) + \frac{b}{k} \sin\left(kl\right) \sin\left(bl\right) \\
\frac{D_1}{A_1} &= -\frac{b}{k} \cos\left(kl\right) \sin\left(bl\right) + \sin\left(kl\right) \cos\left(bl\right) ~,
\end{aligned}\right.
\end{equation}
\begin{equation}
\left\{\begin{aligned}
\frac{C_2}{B_2} &= \cos\left(kl\right) \sin\left(bl\right) - \frac{b}{k} \sin\left(kl\right) \cos\left(bl\right) \\
\frac{D_2}{B_2} &= \frac{b}{k} \cos\left(kl\right) \cos\left(bl\right) + \sin\left(kl\right) \sin\left(bl\right) ~.
\end{aligned}\right.
\end{equation}
对波函数归一化使 $\int \psi_k'(x)\psi_k(x) = \delta(k'-k)$，令无穷远处振幅为 $1/\sqrt{\pi}$（
定理 1 ），得
\begin{equation}
\begin{aligned}
A_1 = \frac{1}{\sqrt{\pi} \sqrt{ \cos^{2}\left(bl\right) + (b^2/k^2) \sin^{2}\left(bl\right) }}~,\\
B_2 = \frac{1}{\sqrt{\pi} \sqrt{ \sin^{2}\left(bl\right) + (b^2/k^2) \cos^{2}\left(bl\right) }}~.
\end{aligned}
\end{equation}
$0 < E < V_0$ 的情况
令 $\kappa = \sqrt{2m(V_0-E)}$
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^e(x) =
\begin{cases}
A_1 \cosh\left(\kappa x\right) & (0 \leqslant x \leqslant l)\\
C_1 \cos\left(kx\right) + D_1 \sin\left(kx\right) & (l < x)~,
\end{cases}
\end{equation}
\begin{equation}
\psi_{ \boldsymbol{\mathbf{k}} }^o(x) =
\begin{cases}
A_2 \sinh\left(\kappa x\right) & (0 \leqslant x \leqslant l)\\
C_2 \cos\left(kx\right) + D_2 \sin\left(kx\right) & (l < x)~.
\end{cases}
\end{equation}
在 $x = l$ 处匹配波函数和一阶导数，分别得
\begin{equation}
\left\{\begin{aligned}
&A_1 \cosh\left(\kappa l\right) = C_1 \cos\left(kl\right) + D_1 \sin\left(kl\right) \\
&\kappa A_1 \sinh\left(\kappa l\right) = - kC_1 \sin\left(kl\right) + kD_1 \cos\left(kl\right) ~,
\end{aligned}\right. \end{equation}
\begin{equation}
\left\{\begin{aligned}
&A_2 \sinh\left(\kappa l\right) = C_2 \cos\left(kl\right) + D_2 \sin\left(kl\right) \\
&\kappa A_2 \cosh\left(\kappa l\right) = -kC_2 \sin\left(kl\right) + k D_2 \cos\left(kl\right) ~.
\end{aligned}\right. \end{equation}
解得
\begin{equation}
\left\{\begin{aligned}
&\frac{C_1}{A_1} = \cosh\left(\kappa l\right) \cos\left(kl\right) - \frac{\kappa}{k} \sinh\left(\kappa l\right) \sin\left(kl\right) \\
&\frac{D_1}{A_1} = \cosh\left(\kappa l\right) \sin\left(kl\right) + \frac{\kappa}{k} \sinh\left(\kappa l\right) \cos\left(kl\right) ~,
\end{aligned}\right. \end{equation}
\begin{equation}
\left\{\begin{aligned}
&\frac{C_2}{A_2} = \sinh\left(\kappa l\right) \cos\left(kl\right) - \frac{\kappa}{k} \cosh\left(\kappa l\right) \sin\left(kl\right) \\
&\frac{D_2}{A_2} = \sinh\left(\kappa l\right) \sin\left(kl\right) + \frac{\kappa}{k} \cosh\left(\kappa l\right) \cos\left(kl\right) ~.
\end{aligned}\right. \end{equation}
对波函数归一化，同样令无穷远处振幅为 $1/\sqrt{\pi}$ 得
\begin{equation}
A_1 = \frac{1}{\sqrt{\pi}\sqrt{\cosh^2(\kappa l) + (\kappa/k)^2 \sinh^2(kl)}}~,
\end{equation}
\begin{equation}
A_2 = \frac{1}{\sqrt{\pi}\sqrt{\sinh^2(\kappa l) + (\kappa/k)^2 \cosh^2(kl)}}~.
\end{equation}
2. 行波解
令左、中、右三个区间 $\exp$ 项的系数分别为 $A,B$、$C,D$、$E,F$。左边右入射和出射，右边只有出射没有反射
\begin{equation}
\psi_{k,1}(x) = \left\{\begin{aligned}
&A \exp\left( \mathrm{i} kx\right) + B \exp\left(- \mathrm{i} kx\right) &\quad& (x < -L)\\
&C \exp\left( \mathrm{i} bx\right) + D \exp\left(- \mathrm{i} bx\right) && (-L \le x \le L)\\
&F \exp\left(- \mathrm{i} kx\right) &\quad& (x < -L)~,\\
\end{aligned}\right. \end{equation}
\begin{equation}
\psi_{k,2}(x) = \psi_{k,1}(-x)~,
\end{equation}
要归一化只需满足 $A = 1/\sqrt{2\pi}$（
式 19 ）。
$E > V_0$ 的情况
当 $E > V_0$ 时，系数解为
\begin{equation}
\left\{\begin{aligned}
&A = 1/\sqrt{2\pi}\\
&B = \mathcal{N}\cdot (k^2 - b^2) \sin\left(2bl\right) \exp\left(- \mathrm{i} kl\right) \\
&C = \mathcal{N}\cdot \mathrm{i} k (k + b) \exp\left(- \mathrm{i} bl\right) \\
&D = \mathcal{N}\cdot \mathrm{i} k(b-k) \exp\left( \mathrm{i} bl\right) \\
&E = \mathcal{N}\cdot 2 \mathrm{i} kb \exp\left(- \mathrm{i} kl\right) ~,
\end{aligned}\right. \end{equation}
其中
\begin{equation}
\mathcal{N} = \frac{1}{\sqrt{2\pi}}\frac{(k^2 + b^2) \sin\left(2bl\right) - 2 \mathrm{i} kb \cos\left(2bl\right) }{(k^2-b^2)^2\sin^2(2bl) + 4k^2b^2} \exp\left(- \mathrm{i} kl\right) ~.
\end{equation}
可以验证 $\psi_{k,1}, \psi_{k,2}$ 正交的条件（
式 25 ）$ \operatorname{Re} [B^*E] = 0$。
透射率，反射率分别为
\begin{equation}
T = \left\lvert \frac{E}{A} \right\rvert ^2 = \frac{4k^2b^2}{(k^2-b^2)^2\sin^2(2bl) + 4k^2b^2}~,
\end{equation}
\begin{equation}
R = \left\lvert \frac{B}{A} \right\rvert ^2 = \frac{(k^2 - b^2) \sin^{2}\left(2bl\right) }{(k^2-b^2)^2\sin^2(2bl) + 4k^2b^2}~.
\end{equation}
$0 < E < V_0$ 的情况
当 $0 < E < V_0$ 时，系数解为
\begin{equation}
\left\{\begin{aligned}
&A = 1/\sqrt{2\pi}\\
&B = \mathcal{N}\cdot (\kappa ^2 + k^2) \sinh\left(2\kappa l\right) \exp\left(- \mathrm{i} kl\right) \\
&C = \mathcal{N}\cdot k( \mathrm{i} \kappa - k) \exp\left(-\kappa l\right) \\
&D = \mathcal{N}\cdot k( \mathrm{i} \kappa + k) \exp\left(\kappa l\right) \\
&E = \mathcal{N}\cdot 2 \mathrm{i} k\kappa \exp\left(- \mathrm{i} kl\right) ~,
\end{aligned}\right. \end{equation}
\begin{equation}
\mathcal{N} = \frac{1}{\sqrt{2\pi}}\frac{(k^2 - \kappa ^2) \sinh\left(2\kappa l\right) - 2 \mathrm{i} k\kappa \cosh\left(2\kappa l\right) }{(k^2+\kappa^2)^2\sinh^2(2\kappa l) + 4k^2\kappa^2} \exp\left(- \mathrm{i} kl\right) ~,
\end{equation}
\begin{equation}
T = \left\lvert \frac{E}{A} \right\rvert ^2 = \frac{4k^2\kappa ^2}{(k^2+\kappa^2)^2\sinh^2(2\kappa l) + 4k^2\kappa^2}~,
\end{equation}
\begin{equation}
R = \left\lvert \frac{B}{A} \right\rvert ^2 = \frac{(\kappa ^2 + k^2) \sinh^{2}\left(2\kappa l\right) }{(k^2+\kappa^2)^2\sinh^2(2\kappa l) + 4k^2\kappa^2}~.
\end{equation}
未完成：未完成：透射率反射率画图
投影系数
任意平方可积的波函数可以用正交归一化的行波展开为
\begin{equation}
\psi(x) = \int_0^{+\infty} C_1(k)\psi_{k,1}(x) + C_2(k)\psi_{k,2}(x) \,\mathrm{d}{k} ~,
\end{equation}
若 $\psi(x)$ 只在 $x < -l$ 不为零，则有
\begin{equation}
\begin{aligned}
&C_1(k) = \left\langle \psi_{k,1} \middle| \psi \right\rangle = \tilde\psi(k) + \sqrt{2\pi} B^* \tilde\psi(-k)~,\\
&C_2(k) = \left\langle \psi_{k,2} \middle| \psi \right\rangle = \sqrt{2\pi} E^* \tilde\psi(-k)~,
\end{aligned}
\end{equation}
其中 $\tilde\psi(k)$ 是 $\psi(x)$ 的
傅里叶变换
\begin{equation}
\tilde\psi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \psi(x) \mathrm{e} ^{- \mathrm{i} kx} \,\mathrm{d}{x} ~.
\end{equation}
具体例子见 “
方势垒散射数值计算”，我们将把一个高斯波包展开为散射态的积分。