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

   本文使用原子单位制

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 $ \left\langle \psi_{n',l',m'} \middle| z \middle| \psi_{n,l,m} \right\rangle $,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 $\psi_{n,l,m}$ 与 $Z$ 成反比进行缩放(保持归一化),导致 $ \left\lvert \psi_{n,l,m} \right\rvert ^2$ 和 $z$ 的平均值也是如此。另外注意 $ \left\langle \psi_{n',l',m'} \middle| z \middle| \psi_{n,l,m} \right\rangle = \left\langle \psi_{n',l',-m'} \middle| z \middle| \psi_{n,l,-m} \right\rangle $。

   根据式 3 可以用 3j 系数表示角向积分,进而化简为式 20

\begin{equation} \mathcal C_{l,m} = \left\langle Y_{l,m} \middle| \cos\theta \middle| Y_{l+1,m} \right\rangle = \sqrt{\frac{(l+1)^2-m^2}{(2l+1)(2l+3)}}~. \end{equation}
仅留下径向积分:
\begin{equation} \begin{aligned} &\quad \left\langle \psi_{n',l',m'} \middle| r\cos\theta \middle| \psi_{n,l,m} \right\rangle \\ &= (-1)^m\sqrt{(2l+1)(2l'+1)} \begin{pmatrix}l' & 1 & l\\ 0 & 0 & 0\end{pmatrix} \begin{pmatrix}l' & 1 & l\\ -m' & 0 & m\end{pmatrix} \int_0^\infty R_{n',l'}(r)R_{n,l}(r)r^2 \,\mathrm{d}{r} \\ &= \delta_{m,m'}(\delta_{l+1,l'}\mathcal C_{l,m} + \delta_{l,l'+1}C_{l',m'})\int_0^\infty R_{n',l'}(r)R_{n,l}(r)r^3 \,\mathrm{d}{r} ~. \end{aligned} \end{equation}

表1:$ \left\langle \psi_{n',l',0} \middle| z \middle| \psi_{n,l,0} \right\rangle $(实对称矩阵)的下半三角,$Z=1$
$ \left\lvert n,l,m \right\rangle $ $ \left\lvert 1,0,0 \right\rangle $ $ \left\lvert 2,0,0 \right\rangle $ $ \left\lvert 2,1,0 \right\rangle $ $ \left\lvert 3,0,0 \right\rangle $ $ \left\lvert 3,1,0 \right\rangle $ $ \left\lvert 3,2,0 \right\rangle $ $ \left\lvert 4,0,0 \right\rangle $ $ \left\lvert 4,1,0 \right\rangle $ $ \left\lvert 4,2,0 \right\rangle $ $ \left\lvert 4,3,0 \right\rangle $
$ \left\lvert 1,0,0 \right\rangle $ 0
$ \left\lvert 2,0,0 \right\rangle $ 0 0
$ \left\lvert 2,1,0 \right\rangle $ $\frac{128\sqrt 2}{243}$ $-3$ 0
$ \left\lvert 3,0,0 \right\rangle $ 0 0 $\frac{3456\sqrt 6}{15625}$ 0
$ \left\lvert 3,1,0 \right\rangle $ $\frac{27}{64\sqrt 2}$ $\frac{27648}{15625}$ 0 $-3\sqrt 6$ 0
$ \left\lvert 3,2,0 \right\rangle $ 0 0 $\frac{110592\sqrt 3}{78125}$ 0 $-3 \sqrt 3$ 0
$ \left\lvert 4,0,0 \right\rangle $ 0 0 $\frac{1024\sqrt 2}{6561}$ 0 $\frac{5750784 \sqrt 2}{5764801}$ 0 0
$ \left\lvert 4,1,0 \right\rangle $ $\frac{6144}{15625 \sqrt 5}$ $\frac{512\sqrt{10}}{2187}$ 0 $\frac{4700160 \sqrt{15}}{5764801}$ 0 $\frac{3538944}{5764801}\sqrt{\frac 65}$ $-6\sqrt 5$ 0
$ \left\lvert 4,2,0 \right\rangle $ 0 0 $\frac{4096\sqrt 2}{6561}$ 0 $\frac{15925248 \sqrt 2}{5764801}$ 0 0 $-\frac{24}{\sqrt 5}$ 0
$ \left\lvert 4,3,0 \right\rangle $ 0 0 0 0 0 $\frac{191102976}{40353607}\sqrt{\frac 65}$ 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 $ \left\langle 3,1,0 \middle| z \middle| 3,2,0 \right\rangle $ 和 $ \left\langle 3,1,1 \middle| z \middle| 3,2,1 \right\rangle $ 是不一样的。因为球谐函数中 $m$ 不光决定相位因子 $ \mathrm{e} ^{ \mathrm{i} m\phi}$ 也会决定连带勒让德函数 $P_l^m(\cos\theta)$。

   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{equation} \left\lvert 2,0,0 \right\rangle , \left\lvert 2,1,0 \right\rangle \Rightarrow-\frac{3}{Z} \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{3}{Z} \Rightarrow \frac{1}{\sqrt{2}}(1, \pm1)~. \end{equation}

   【$n=3, m=0$】

\begin{equation} \left\lvert 3,0,0 \right\rangle , \left\lvert 3,1,0 \right\rangle , \left\lvert 3,2,0 \right\rangle \Rightarrow -\frac{3\sqrt{3}}{Z} \begin{pmatrix}0 & \sqrt{2} & 0\\ \sqrt{2} & 0 & 1\\ 0 & 1 & 0\end{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{9}{Z} \Rightarrow \left(\frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{6}} \right) ~. \end{equation}
\begin{equation} \frac{0}{Z} \Rightarrow \left(\frac{1}{\sqrt{3}}, 0, -\sqrt{\frac{2}{3}} \right) ~. \end{equation}

   【$n=3, m=1$】

\begin{equation} \left\lvert 3,1,\pm1 \right\rangle , \left\lvert 3,2,\pm1 \right\rangle \Rightarrow -\frac{9}{2Z} \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{9}{2Z} \Rightarrow \frac{1}{\sqrt 2}(1, \pm 1)~. \end{equation}

   【$n=4, m=0$】

\begin{equation} \left\lvert 4,0,0 \right\rangle , \left\lvert 4,1,0 \right\rangle , \left\lvert 4,2,0 \right\rangle , \left\lvert 4,3,0 \right\rangle \Rightarrow -\frac{6}{\sqrt{5}Z} \begin{pmatrix}0 & 5 & 0 & 0\\ 5 & 0 & 4 & 0\\ 0 & 4 & 0 & 3\\ 0 &0 &3 &0\end{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{18}{Z} \Rightarrow \frac{1}{2\sqrt{5}}(\sqrt{5},3,\pm\sqrt{5},1)~. \end{equation}
\begin{equation} \mp\frac{6}{Z} \Rightarrow \frac{1}{3\sqrt{14}} \left(\pm\sqrt{5}, 1,\mp\sqrt{5},-3 \right) ~. \end{equation}

   【$n=4, m=1$】

\begin{equation} \left\lvert 4,1,\pm1 \right\rangle , \left\lvert 4,2,\pm1 \right\rangle , \left\lvert 4,3,\pm1 \right\rangle \Rightarrow -\frac{12}{\sqrt{5}Z} \begin{pmatrix}0 & \sqrt{3} & 0\\ \sqrt{3} & 0 & \sqrt{2}\\ 0 & \sqrt{2} &0\end{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{12}{Z} \Rightarrow \frac{1}{2\sqrt{5}} \left(\sqrt{3},\pm\sqrt{5},\sqrt{2} \right) ~. \end{equation}
\begin{equation} \frac{0}{Z} \Rightarrow \frac{1}{\sqrt{5}} \left(-\sqrt{2},0,\sqrt{3} \right) ~. \end{equation}

   【$n=4, m=2$】

\begin{equation} \left\lvert 4,2,\pm2 \right\rangle , \left\lvert 4,3,\pm2 \right\rangle \Rightarrow -\frac{6}{Z} \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{6}{Z} \Rightarrow \frac{1}{\sqrt{2}}(1,\pm 1)~. \end{equation}

   【$n=5, m=0$】

\begin{equation} \left\lvert 5,0,0 \right\rangle ,\dots, \left\lvert 5,4,0 \right\rangle \Rightarrow -\frac{3}{\sqrt{7}} \begin{pmatrix} 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{pmatrix} ~. \end{equation}
\begin{equation} \pm30 \Rightarrow \frac{1}{\sqrt{70}} \left(\sqrt{14},\pm 2\sqrt{7},2 \sqrt{5},\pm\sqrt{7},1 \right) ~. \end{equation}
\begin{equation} \mp 15 \Rightarrow \frac{1}{\sqrt{70}} \left(-\sqrt{14},\mp\sqrt{7},\sqrt{5},\pm 2\sqrt{7},4 \right) ~. \end{equation}
\begin{equation} 0 \Rightarrow \frac{1}{\sqrt{35}} \left(\sqrt{7},0,-\sqrt{10},0,3 \sqrt{2} \right) ~. \end{equation}

   【$n=5, m=1$】

\begin{equation} \left\lvert 5,1,1 \right\rangle ,\dots, \left\lvert 5,4,1 \right\rangle \Rightarrow -\frac{3\sqrt{5}}{2\sqrt{7}} \begin{pmatrix} 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{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{45}{2}\Rightarrow\frac{1}{\sqrt{70}} \left(\pm\sqrt{14},\sqrt{30},\pm\sqrt{21},\sqrt{5} \right) ~. \end{equation}
\begin{equation} \mp\frac{15}{2}\Rightarrow\frac{1}{\sqrt{70}} \left(\mp\sqrt{21},-\sqrt{5},\pm\sqrt{14},\sqrt{30} \right) ~. \end{equation}

   【$n=5, m=2$】

\begin{equation} \left\lvert 5,2,2 \right\rangle ,\dots, \left\lvert 5,4,2 \right\rangle \Rightarrow -\frac{15}{\sqrt{7}} \begin{pmatrix} 0 & 2 & 0 \\ 2 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \end{pmatrix} ~. \end{equation}
\begin{equation} \mp 15 \Rightarrow \frac{1}{\sqrt{14}} \left(2,\sqrt{7},\sqrt{3} \right) ~. \end{equation}
\begin{equation} 0 \Rightarrow \frac{1}{\sqrt{14}} \left(-\sqrt{6},0,2 \sqrt{2} \right) ~. \end{equation}

   【$n=5, m=3$】

\begin{equation} \left\lvert 5,3,3 \right\rangle ,\dots, \left\lvert 5,4,3 \right\rangle \Rightarrow -\frac{15}{2} \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix} ~. \end{equation}
\begin{equation} \mp\frac{15}{2} \Rightarrow \frac{1}{\sqrt{2}}(1,\pm 1)~. \end{equation}

                     

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