水素原子におけるシュレーディンガー方程式の解
ここで、kは
0 ≤ k ≤ m {\displaystyle 0\leq k\leq m} ...(C5)
を満たす整数である。
よって(C4)の解は
u ( ρ ) = L n + l 2 l + 1 ( ρ ) {\displaystyle u(\rho )=L_{n+\ell }^{2\ell +1}(\rho )} ×(規格化定数)
となる。これを変数変換の式(C3)に代入して
R ( ρ ) = ρ l L n + l 2 l + 1 ( ρ ) exp  ( − ρ 2 ) {\displaystyle R(\rho )=\rho ^{\ell }L_{n+\ell }^{2\ell +1}(\rho )\exp \left(-{\frac {\rho }{2}}\right)} ×(規格化定数) ...(C6)
を得る。
ラゲール陪多項式の係数の条件式(C4)から、 h o g e l {\displaystyle hoge\ell } は
0 ≤ l ≤ n − 1 {\displaystyle 0\leq \ell \leq n-1} ...(C7)
を満たす整数でなければならない。
R 1 s = 2 ( Z a 0 ) 3 / 2 exp  ( − Z r a 0 ) R 2 s = 1 2 2 ( Z a 0 ) 3 / 2 ( 2 − Z r a 0 ) exp  ( − Z r 2 a 0 ) R 2 p = 1 2 6 ( Z a 0 ) 3 / 2 Z r a 0 exp  ( − Z r 2 a 0 ) R 3 s = 2 81 3 ( Z a 0 ) 3 / 2 ( 27 − 18 Z r a 0 + 2 Z 2 r 2 a 0 2 ) exp  ( − Z r 3 a 0 ) R 3 p = 4 81 6 ( Z a 0 ) 3 / 2 ( 6 − Z r a 0 ) Z r a 0 exp  ( − Z r 3 a 0 ) R 3 d = 4 81 30 ( Z a 0 ) 3 / 2 Z 2 r 2 a 0 2 exp  ( − Z r 3 a 0 ) R 4 s = 1 768 ( Z a 0 ) 3 / 2 ( 192 − 144 Z r a 0 + 24 Z 2 r 2 a 0 2 − Z 3 r 3 a 0 3 ) exp  ( − Z r 4 a 0 ) R 4 p = 1 256 15 ( Z a 0 ) 3 / 2 ( 80 − 20 Z r a 0 + Z 2 r 2 a 0 2 ) Z r a 0 exp  ( − Z r 4 a 0 ) R 4 d = 1 768 5 ( Z a 0 ) 3 / 2 ( 12 − Z r a 0 ) Z 2 r 2 a 0 2 exp  ( − Z r 4 a 0 ) R 4 f = 1 768 35 ( Z a 0 ) 3 / 2 Z 3 r 3 a 0 3 exp  ( − Z r 4 a 0 ) {\displaystyle {\begin{aligned}R_{1\mathrm {s} }&=2\left({\frac {Z}{a_{0}}}\right)^{3/2}\exp \left(-{\frac {Zr}{a_{0}}}\right)\\R_{2\mathrm {s} }&={\frac {1}{2{\sqrt {2}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}\left(2-{\frac {Zr}{a_{0}}}\right)\exp \left(-{\frac {Zr}{2a_{0}}}\right)\\R_{2\mathrm {p} }&={\frac {1}{2{\sqrt {6}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}{\frac {Zr}{a_{0}}}\exp \left(-{\frac {Zr}{2a_{0}}}\right)\\R_{3\mathrm {s} }&={\frac {2}{81{\sqrt {3}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}\left(27-{\frac {18Zr}{a_{0}}}+{\frac {2Z^{2}r^{2}}{a_{0}^{2}}}\right)\exp \left(-{\frac {Zr}{3a_{0}}}\right)\\R_{3\mathrm {p} }&={\frac {4}{81{\sqrt {6}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}\left(6-{\frac {Zr}{a_{0}}}\right){\frac {Zr}{a_{0}}}\exp \left(-{\frac {Zr}{3a_{0}}}\right)\\R_{3\mathrm {d} }&={\frac {4}{81{\sqrt {30}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}{\frac {Z^{2}r^{2}}{a_{0}^{2}}}\exp \left(-{\frac {Zr}{3a_{0}}}\right)\\R_{4\mathrm {s} }&={\frac {1}{768}}\left({\frac {Z}{a_{0}}}\right)^{3/2}\left(192-{\frac {144Zr}{a_{0}}}+{\frac {24Z^{2}r^{2}}{a_{0}^{2}}}-{\frac {Z^{3}r^{3}}{a_{0}^{3}}}\right)\exp \left(-{\frac {Zr}{4a_{0}}}\right)\\R_{4\mathrm {p} }&={\frac {1}{256{\sqrt {15}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}\left(80-{\frac {20Zr}{a_{0}}}+{\frac {Z^{2}r^{2}}{a_{0}^{2}}}\right){\frac {Zr}{a_{0}}}\exp \left(-{\frac {Zr}{4a_{0}}}\right)\\R_{4\mathrm {d} }&={\frac {1}{768{\sqrt {5}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}\left(12-{\frac {Zr}{a_{0}}}\right){\frac {Z^{2}r^{2}}{a_{0}^{2}}}\exp \left(-{\frac {Zr}{4a_{0}}}\right)\\R_{4\mathrm {f} }&={\frac {1}{768{\sqrt {35}}}}\left({\frac {Z}{a_{0}}}\right)^{3/2}{\frac {Z^{3}r^{3}}{a_{0}^{3}}}\exp \left(-{\frac {Zr}{4a_{0}}}\right)\end{aligned}}}