Yuanqing Wu’s Blog - 双阱势问题

问题描述

顾名思义, 双阱势是由两个势阱组成的体系. 这样的体系有很多可能的势函数, 例如:

方势阱: \[ V_{\text{sq}} = \Biggr\{ \begin{aligned} & 0 \quad \text{ if } |x-(\frac{a}{2} \pm b)| < \frac{c}{2}\\ & V_0 \quad \text{otherwise} \end{aligned} \]
由两个抛物线势阱构成的双阱势: \[ V_{\text{para}}(x) = \frac{V_{\text{cusp}}}{b^{2}} \Biggl(\Biggl|x-\frac{a}{2}\Biggr| - b \Biggr)^{2} \]
或者, 常见于对称性破缺中的墨西哥帽势: \[ V_{\text{m}}(x) = \frac{V_{\text{max}}}{b^{4}} \Biggl(\Biggl(x-\frac{a}{2}\Biggr)^{2} - b^{2}\Biggr)\]

其中, \(a\) 是体系整体的长度, \(b\) 是体系中心到势阱中心的距离, \(c\) 是势阱的宽度. 图 Figure 1 展示了这几种势的样子.

Code

using Plots, LaTeXStrings
theme(:ggplot2, yticks=nothing, grid=nothing, linewidth=2)
a,b,c = 1.0, 0.25, 0.3
v0,vcusp,vmax = 1.0, 1.5, 0.4
vsq(x) = (abs(x-0.5a+b)<0.5c||abs(x-0.5a-b)<0.5c) ? 0 : v0
vpara(x) = vcusp/(b^2)*(abs(x-0.5a)-b)^2
vm(x) = vmax/(b^4)*((x-a/2)^2-b^2)^2
xticks = (
  [0.0,0.5a-b,0.5a,0.5a+b],
  [L"0", L"\frac{1}{2}-\frac{b}{a}", L"\frac{1}{2}", L"\frac{1}{2}+\frac{b}{a}"]
)

plt = plot(
  [vsq, vpara, vm], 0:0.01:1,
  xlabel = L"x/a", ylabel = L"V(x)",
  label = [L"V_{\rm{sq}}" L"V_{\rm{para}}" L"V_{\rm{m}}"],
  annotation = [
    (0.5, vpara(0.5)+0.1, (L"V_{\rm{cusp}}", 10)),
    (0.5, vm(0.5)+0.1, (L"V_{\rm{max}}", 10)),
    (0.15, v0+0.1, (L"V_0", 10)),
    (0.5a-b, 0.7v0-0.1, (L"c", 10))
  ], xticks=xticks
)
plot!(x->v0, 0:0.5:1, label=nothing, ls=:dot)
plot!(x->0.7v0, 0.5a-b-0.5c:0.1c:0.5a-b+0.5c, label=nothing, c=:black, arrow=(:closed, :both), lw=1)
plt

接下来我们来求解双阱势问题. 不难看出, 当势垒高度趋向无穷时, 无限深方势阱的解也是双阱势问题的解. 因此很直观的, 我们取无限深方式阱的解\(\ket{n}\)作为基:

\[ \psi_{n}(x) = \braket{x|n} = \Biggl\{ \begin{aligned} & \sqrt{\frac{2}{a}}\sin \Biggl(\frac{n \pi x}{a}\Biggr) \quad & \text{if } 0<x<a\\ & 0 & \text{otherwise} \end{aligned}\]

假设双阱势解为:

\[ \ket{a} = c_{m} \ket{m} \]

那么由定态Schrodinger方程, 我们得到:

\[ H_{nm}c_{m} = E c_{n} \]

其中, 矩阵元\(H_{nm}\) 为:

\[ H_{nm} = \bra{n}H\ket{m} = \frac{2}{a}\int _{0}^{a} \sin \Biggl(\frac{n \pi x}{a}\Biggr)V(x)\sin\Biggl(\frac{n \pi x}{a}\Biggr) {\rm d}x \]