氢原子的跃迁偶极子矩阵元计算（束缚态之间）

贡献者： addis; 切糕糕

预备知识　氢原子的跃迁偶极子矩阵元和选择定则

1. $z$ 方向电场，束缚态到束缚态

　　表 1 给出了核电荷数 $Z = 1$ 时的 $⟨ ψ_{n^{'}, l^{'}, m^{'}} | z | ψ_{n, l, m} ⟩$ ，由于这是一个实对称矩阵，只给出矩阵的下半三角。当 $Z > 1$ 时把表中每个矩阵元除以 $Z$ 即可。这是因为 $ψ_{n, l, m}$ 与 $Z$ 成反比进行缩放（保持归一化），导致 ${| ψ_{n, l, m} |}^{2}$ 和 $z$ 的平均值也是如此。另外注意 $⟨ ψ_{n^{'}, l^{'}, m^{'}} | z | ψ_{n, l, m} ⟩ = ⟨ ψ_{n^{'}, l^{'}, - m^{'}} | z | ψ_{n, l, - m} ⟩$ 。

　　根据式 3 可以用 3j 系数表示角向积分，进而化简为式 20

\begin{matrix} (1) & C_{l, m} = ⟨ Y_{l, m} | \cos θ | Y_{l + 1, m} ⟩ = \sqrt{\frac{(l + 1)^{2} - m^{2}}{(2 l + 1) (2 l + 3)}} . \end{matrix}

仅留下径向积分：

\begin{matrix} (2) & \begin{aligned} ⟨ ψ_{n^{'}, l^{'}, m^{'}} | r \cos θ | ψ_{n, l, m} ⟩ \\ = (- 1)^{m} \sqrt{(2 l + 1) (2 l^{'} + 1)} (\begin{array}{c} l^{'} & 1 & l \\ 0 & 0 & 0 \end{array}) (\begin{array}{c} l^{'} & 1 & l \\ - m^{'} & 0 & m \end{array}) \int_{0}^{\infty} R_{n^{'}, l^{'}} (r) R_{n, l} (r) r^{2} d r \\ = δ_{m, m^{'}} (δ_{l + 1, l^{'}} C_{l, m} + δ_{l, l^{'} + 1} C_{l^{'}, m^{'}}) \int_{0}^{\infty} R_{n^{'}, l^{'}} (r) R_{n, l} (r) r^{3} d r . \end{aligned} \end{matrix}

表1：

⟨ ψ_{n^{'}, l^{'}, 0} | z | ψ_{n, l, 0} ⟩

（实对称矩阵）的下半三角，

Z = 1

$\| n, l, m ⟩$	$\| 1, 0, 0 ⟩$	$\| 2, 0, 0 ⟩$	$\| 2, 1, 0 ⟩$	$\| 3, 0, 0 ⟩$	$\| 3, 1, 0 ⟩$	$\| 3, 2, 0 ⟩$	$\| 4, 0, 0 ⟩$	$\| 4, 1, 0 ⟩$	$\| 4, 2, 0 ⟩$	$\| 4, 3, 0 ⟩$
$\| 1, 0, 0 ⟩$	0
$\| 2, 0, 0 ⟩$	0	0
$\| 2, 1, 0 ⟩$	$\frac{128 \sqrt{2}}{243}$	$- 3$	0
$\| 3, 0, 0 ⟩$	0	0	$\frac{3456 \sqrt{6}}{15625}$	0
$\| 3, 1, 0 ⟩$	$\frac{27}{64 \sqrt{2}}$	$\frac{27648}{15625}$	0	$- 3 \sqrt{6}$	0
$\| 3, 2, 0 ⟩$	0	0	$\frac{110592 \sqrt{3}}{78125}$	0	$- 3 \sqrt{3}$	0
$\| 4, 0, 0 ⟩$	0	0	$\frac{1024 \sqrt{2}}{6561}$	0	$\frac{5750784 \sqrt{2}}{5764801}$	0	0
$\| 4, 1, 0 ⟩$	$\frac{6144}{15625 \sqrt{5}}$	$\frac{512 \sqrt{10}}{2187}$	0	$\frac{4700160 \sqrt{15}}{5764801}$	0	$\frac{3538944}{5764801} \sqrt{\frac{6}{5}}$	$- 6 \sqrt{5}$	0
$\| 4, 2, 0 ⟩$	0	0	$\frac{4096 \sqrt{2}}{6561}$	0	$\frac{15925248 \sqrt{2}}{5764801}$	0	0	$- \frac{24}{\sqrt{5}}$	0
$\| 4, 3, 0 ⟩$	0	0	0	0	0	$\frac{191102976}{40353607} \sqrt{\frac{6}{5}}$	0	0	$- \frac{18}{\sqrt{5}}$	0

　　这可以用于计算类氢原子斯塔克效应以及跃迁率等。

　　注意 $m$ 值不同矩阵元也会不同，例如 $⟨ 3, 1, 0 | z | 3, 2, 0 ⟩$ 和 $⟨ 3, 1, 1 | z | 3, 2, 1 ⟩$ 是不一样的。因为球谐函数中 $m$ 不光决定相位因子 $e^{i m ϕ}$ 也会决定连带勒让德函数 $P_{l}^{m} (\cos θ)$ 。

　　 Mathematica 代码如下（请自行修改矩阵尺寸和循环范围），HydrogenR 函数见类氢原子的束缚态。

代码 1：DipoleZ.m

DipoleZ[Z_, n1_, l1_, m1_, n2_, l2_, 
   m2_] := (-1)^
    m1 Sqrt[(2 l1 + 1) (2 l2 + 1)] ThreeJSymbol[{l1, 0}, {1, 0}, {l2, 
     0}] Integrate[
    HydrogenR[Z, n1, l1, r]\[Conjugate] HydrogenR[Z, n2, l2, 
      r] r^3, {r, 0, +\[Infinity]}] ThreeJSymbol[{l1, -m1}, {1, 
     0}, {l2, m2}];
d = ConstantArray[0, {10, 10}];
i1 = 0; i2 = 0;
For[n1 = 1, n1 <= 4, n1++, For[l1 = 0, l1 < n1, l1++,
  ++i1; i2 = 0;
  For[n2 = 1, n2 <= 4, n2++, For[l2 = 0, l2 < n2, l2++,
    ++i2;
    d[[i1, i2]] = DipoleZ[1, n1, l1, 0, n2, l2, 0];
  ]]
]];
Print[d // MatrixForm];

顺便给出

x, y, z

三个方向的矩阵元的代码

代码 2：Dipole.m

(* eq_SelRul_1 *)
Dipole[Z_, n1_, l1_, m1_, n2_, l2_, m2_] :=\
(-1)^m1 Sqrt[(2 l1 + 1) (2 l2 + 1)]\
  ThreeJSymbol[{l1, 0}, {1, 0}, {l2, 0}] Integrate[\
  HydrogenR[Z, n1, l1, r]\[Conjugate] HydrogenR[Z, n2, l2, r] r^3, {r,\
    0, +∞}] {(ThreeJSymbol[{l1, -m1}, {1, -1}, {l2, m2}] - \
     ThreeJSymbol[{l1, -m1}, {1, 1}, {l2, m2}])/Sqrt[\
   2], (ThreeJSymbol[{l1, -m1}, {1, -1}, {l2, m2}] + \
     ThreeJSymbol[{l1, -m1}, {1, 1}, {l2, m2}])/Sqrt[2], \
  ThreeJSymbol[{l1, -m1}, {1, 0}, {l2, m2}]}

　　 Matlab 代码如下（请自行修改矩阵尺寸和循环范围），hydrogen_Rnl 函数见类氢原子的束缚态。

代码 3：hydrogen_trans_dipole_z.m

% hydrogen <n1,l1,m1|z|n2,l2,m2>
% r_max is upper bound of r integral
function d_z = hydrogen_trans_dipole_z(Z,n1,l1,m1,n2,l2,m2,r_max)
if (n1 <= 0 || l1 < 0 || l1 >= n1 || abs(m1) > l1)
    error('illegal n1,l1,m1');
end
if (n2 <= 0 || l2 < 0 || l2 >= n2 || abs(m2) > l2)
    error('illegal n2,l2,m2');
end
if (abs(l1-l2) ~= 1 || m1 ~= m2)
    d_z = 0; return;
end
integrand = @(r) hydrogen_Rnl(Z,n1,l1,r) ...
    .* hydrogen_Rnl(Z,n2,l2,r) .* r.^3;
% radial integral
% change of variable r = tan(x) to avoid infinite upperbound
I_r = integral(integrand, 0, r_max);
if isnan(I_r) || isinf(I_r)
    error('integral failed!');
end
% angular integral
I_th = (-1)^m2 * sqrt((2*l1+1)*(2*l2+1)) ...
    * ThreeJ(l1,0,1,0,l2,0) * ThreeJ(l1,-m1,1,0,l2,m2);
d_z = I_r * I_th;
end

　　以下函数需要 parallel toolbox 加速，如果没有可能会较慢（parfor 自动变为 for）。

代码 4：hydrogen_trans_dipole_z_mat.m

% hydrogen <n1,l1,m1|z|n2,l2,m2> matrix
% n_l(i,:) = [n, l] for each basis
function [d_z, n_l] = hydrogen_trans_dipole_z_mat(Z, m, n_max, r_max)
if nargin == 0
% === params ===
Z = 1;
m = 0;
n_max = 5;
% ==============
end
Ndim = (n_max-abs(m))*(n_max-abs(m)+1)/2;
n_l = zeros(Ndim, 2);
n1l1_n2l2_k1k2 = zeros(Ndim*(Ndim-1)/2, 6);
k1 = 0; k = 0;
for n1 = 1:n_max
    for l1 = m:n1-1
        k1 = k1 + 1; k2 = 0;
        n_l(k1, :) = [n1, l1];
        for n2 = 1:n_max
            for l2 = m:n2-1
                k2 = k2 + 1;
                if (k2 >= k1)
                    continue;
                end
                k = k + 1;
                n1l1_n2l2_k1k2(k, :) = [n1, l1, n2, l2, k1, k2];
            end
        end
    end
end
if k ~= Ndim*(Ndim-1)/2 || k1 ~= Ndim
    error('error!')
end

Nlist = size(n1l1_n2l2_k1k2, 1);
d_val = zeros(1, Nlist);
parfor i = 1:Nlist
    n1 = n1l1_n2l2_k1k2(i,1); l1 = n1l1_n2l2_k1k2(i,2);
    n2 = n1l1_n2l2_k1k2(i,3); l2 = n1l1_n2l2_k1k2(i,4);
    disp(n1l1_n2l2_k1k2(i,1:4));
    d_val(i) = hydrogen_trans_dipole_z(Z,n1,l1,m,n2,l2,m,r_max);
end

d_z = zeros(Ndim, Ndim);
for i = 1:Nlist
    k1 = n1l1_n2l2_k1k2(i,5); k2 = n1l1_n2l2_k1k2(i,6);
    d_z(k1, k2) = d_val(i);
end
% copy to upper triangle
d_z = d_z + d_z.';
end

2. 好本征态列表（z 方向电场）

　　注意第 $n$ 能级有 $n^{2}$ 基底。

　　【 $n = 2, m = 0$ 】

\begin{matrix} (3) & | 2, 0, 0 ⟩, | 2, 1, 0 ⟩ \Rightarrow - \frac{3}{Z} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (4) & \mp \frac{3}{Z} \Rightarrow \frac{1}{\sqrt{2}} (1, \pm 1) . \end{matrix}

　　【 $n = 3, m = 0$ 】

\begin{matrix} (5) & | 3, 0, 0 ⟩, | 3, 1, 0 ⟩, | 3, 2, 0 ⟩ \Rightarrow - \frac{3 \sqrt{3}}{Z} (\begin{matrix} 0 & \sqrt{2} & 0 \\ \sqrt{2} & 0 & 1 \\ 0 & 1 & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (6) & \mp \frac{9}{Z} \Rightarrow (\frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{6}}) . \end{matrix}

\begin{matrix} (7) & \frac{0}{Z} \Rightarrow (\frac{1}{\sqrt{3}}, 0, - \sqrt{\frac{2}{3}}) . \end{matrix}

　　【 $n = 3, m = 1$ 】

\begin{matrix} (8) & | 3, 1, \pm 1 ⟩, | 3, 2, \pm 1 ⟩ \Rightarrow - \frac{9}{2 Z} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (9) & \mp \frac{9}{2 Z} \Rightarrow \frac{1}{\sqrt{2}} (1, \pm 1) . \end{matrix}

　　【 $n = 4, m = 0$ 】

\begin{matrix} (10) & | 4, 0, 0 ⟩, | 4, 1, 0 ⟩, | 4, 2, 0 ⟩, | 4, 3, 0 ⟩ \Rightarrow - \frac{6}{\sqrt{5} Z} (\begin{matrix} 0 & 5 & 0 & 0 \\ 5 & 0 & 4 & 0 \\ 0 & 4 & 0 & 3 \\ 0 & 0 & 3 & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (11) & \mp \frac{18}{Z} \Rightarrow \frac{1}{2 \sqrt{5}} (\sqrt{5}, 3, \pm \sqrt{5}, 1) . \end{matrix}

\begin{matrix} (12) & \mp \frac{6}{Z} \Rightarrow \frac{1}{3 \sqrt{14}} (\pm \sqrt{5}, 1, \mp \sqrt{5}, - 3) . \end{matrix}

　　【 $n = 4, m = 1$ 】

\begin{matrix} (13) & | 4, 1, \pm 1 ⟩, | 4, 2, \pm 1 ⟩, | 4, 3, \pm 1 ⟩ \Rightarrow - \frac{12}{\sqrt{5} Z} (\begin{matrix} 0 & \sqrt{3} & 0 \\ \sqrt{3} & 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (14) & \mp \frac{12}{Z} \Rightarrow \frac{1}{2 \sqrt{5}} (\sqrt{3}, \pm \sqrt{5}, \sqrt{2}) . \end{matrix}

\begin{matrix} (15) & \frac{0}{Z} \Rightarrow \frac{1}{\sqrt{5}} (- \sqrt{2}, 0, \sqrt{3}) . \end{matrix}

　　【 $n = 4, m = 2$ 】

\begin{matrix} (16) & | 4, 2, \pm 2 ⟩, | 4, 3, \pm 2 ⟩ \Rightarrow - \frac{6}{Z} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (17) & \mp \frac{6}{Z} \Rightarrow \frac{1}{\sqrt{2}} (1, \pm 1) . \end{matrix}

　　【 $n = 5, m = 0$ 】

\begin{matrix} (18) & | 5, 0, 0 ⟩, \dots, | 5, 4, 0 ⟩ \Rightarrow - \frac{3}{\sqrt{7}} (\begin{matrix} 0 & 5 \sqrt{14} & 0 & 0 & 0 \\ 5 \sqrt{14} & 0 & 7 \sqrt{5} & 0 & 0 \\ 0 & 7 \sqrt{5} & 0 & 6 \sqrt{5} & 0 \\ 0 & 0 & 6 \sqrt{5} & 0 & 10 \\ 0 & 0 & 0 & 10 & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (19) & \pm 30 \Rightarrow \frac{1}{\sqrt{70}} (\sqrt{14}, \pm 2 \sqrt{7}, 2 \sqrt{5}, \pm \sqrt{7}, 1) . \end{matrix}

\begin{matrix} (20) & \mp 15 \Rightarrow \frac{1}{\sqrt{70}} (- \sqrt{14}, \mp \sqrt{7}, \sqrt{5}, \pm 2 \sqrt{7}, 4) . \end{matrix}

\begin{matrix} (21) & 0 \Rightarrow \frac{1}{\sqrt{35}} (\sqrt{7}, 0, - \sqrt{10}, 0, 3 \sqrt{2}) . \end{matrix}

　　【 $n = 5, m = 1$ 】

\begin{matrix} (22) & | 5, 1, 1 ⟩, \dots, | 5, 4, 1 ⟩ \Rightarrow - \frac{3 \sqrt{5}}{2 \sqrt{7}} (\begin{matrix} 0 & 7 \sqrt{3} & 0 & 0 \\ 7 \sqrt{3} & 0 & 8 \sqrt{2} & 0 \\ 0 & 8 \sqrt{2} & 0 & 5 \sqrt{3} \\ 0 & 0 & 5 \sqrt{3} & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (23) & \mp \frac{45}{2} \Rightarrow \frac{1}{\sqrt{70}} (\pm \sqrt{14}, \sqrt{30}, \pm \sqrt{21}, \sqrt{5}) . \end{matrix}

\begin{matrix} (24) & \mp \frac{15}{2} \Rightarrow \frac{1}{\sqrt{70}} (\mp \sqrt{21}, - \sqrt{5}, \pm \sqrt{14}, \sqrt{30}) . \end{matrix}

　　【 $n = 5, m = 2$ 】

\begin{matrix} (25) & | 5, 2, 2 ⟩, \dots, | 5, 4, 2 ⟩ \Rightarrow - \frac{15}{\sqrt{7}} (\begin{matrix} 0 & 2 & 0 \\ 2 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (26) & \mp 15 \Rightarrow \frac{1}{\sqrt{14}} (2, \sqrt{7}, \sqrt{3}) . \end{matrix}

\begin{matrix} (27) & 0 \Rightarrow \frac{1}{\sqrt{14}} (- \sqrt{6}, 0, 2 \sqrt{2}) . \end{matrix}

　　【 $n = 5, m = 3$ 】

\begin{matrix} (28) & | 5, 3, 3 ⟩, \dots, | 5, 4, 3 ⟩ \Rightarrow - \frac{15}{2} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) . \end{matrix}

\begin{matrix} (29) & \mp \frac{15}{2} \Rightarrow \frac{1}{\sqrt{2}} (1, \pm 1) . \end{matrix}

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$\| n, l, m ⟩$	$\| 1, 0, 0 ⟩$	$\| 2, 0, 0 ⟩$	$\| 2, 1, 0 ⟩$	$\| 3, 0, 0 ⟩$	$\| 3, 1, 0 ⟩$	$\| 3, 2, 0 ⟩$	$\| 4, 0, 0 ⟩$	$\| 4, 1, 0 ⟩$	$\| 4, 2, 0 ⟩$	$\| 4, 3, 0 ⟩$
$\| 1, 0, 0 ⟩$	0
$\| 2, 0, 0 ⟩$	0	0
$\| 2, 1, 0 ⟩$	$\frac{128 \sqrt{2}}{243}$	$- 3$	0
$\| 3, 0, 0 ⟩$	0	0	$\frac{3456 \sqrt{6}}{15625}$	0
$\| 3, 1, 0 ⟩$	$\frac{27}{64 \sqrt{2}}$	$\frac{27648}{15625}$	0	$- 3 \sqrt{6}$	0
$\| 3, 2, 0 ⟩$	0	0	$\frac{110592 \sqrt{3}}{78125}$	0	$- 3 \sqrt{3}$	0
$\| 4, 0, 0 ⟩$	0	0	$\frac{1024 \sqrt{2}}{6561}$	0	$\frac{5750784 \sqrt{2}}{5764801}$	0	0
$\| 4, 1, 0 ⟩$	$\frac{6144}{15625 \sqrt{5}}$	$\frac{512 \sqrt{10}}{2187}$	0	$\frac{4700160 \sqrt{15}}{5764801}$	0	$\frac{3538944}{5764801} \sqrt{\frac{6}{5}}$	$- 6 \sqrt{5}$	0
$\| 4, 2, 0 ⟩$	0	0	$\frac{4096 \sqrt{2}}{6561}$	0	$\frac{15925248 \sqrt{2}}{5764801}$	0	0	$- \frac{24}{\sqrt{5}}$	0
$\| 4, 3, 0 ⟩$	0	0	0	0	0	$\frac{191102976}{40353607} \sqrt{\frac{6}{5}}$	0	0	$- \frac{18}{\sqrt{5}}$	0

氢原子的跃迁偶极子矩阵元计算（束缚态之间）

1. z 方向电场，束缚态到束缚态

2. 好本征态列表（z 方向电场）

1. $z$ 方向电场，束缚态到束缚态