贡献者: addis; 切糕糕
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。
$ \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 |
Mathematica 代码(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。
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];
 
 
 
 
 
 
 
 
 
 
 
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