ヘリウム原子のエネルギー準位と固有関数の空間分布(直交系展開によるエネルギー固有状態の計算結果)
「ヘリウム原子のエネルギー固有状態の計算方法」に基づいて、ヘリウム原子のエネルギー準位と固有関数の空間分布を計算したよ。「ヘリウム原子の基底状態の計算結果」で示したとおり、計算結果はよく知られた精密な実験結果とかなり一致しているよ。
ヘリウム原子のエネルギー準位
次の図は、パラ(対称関数・スピン1重項)とオルト(反対称関数・スピン3重項)の主量子数 $n=1,2,3$ のエネルギー準位だよ。オルトのほうがパラよりも若干小さな値となるね。これは交換相互作用の結果だね。イオン化エネルギーは、電子2個の基底状態から電子1個を引き離すために必要なエネルギーで、「基底状態エネルギー($-79.18[{\rm eV}]$)」から「電子が1個のみのヘリウム原子の基底状態エネルギー( $-54.4[{\rm eV}]$ )」で計算できるよ。
ヘリウム原子の固有状態の空間分布
今回も独立電子近似の場合と同様、粒子の1つが $\varphi_{100}(\boldsymbol{r})|$の最も確率の高い原点( $\boldsymbol{r}_1=0$ あるいは $\boldsymbol{r}_2=0$ )に存在するとして、他方の粒子の空間確率密度を描画するよ。最初の表がパラヘリウム(対称関数・スピン1重項)、次の表がオルトヘリウム(反対称関数・スピン3重項)だよ。
記号 |
$(n_1,l_1,m_1)\times(n_2,l_2,m_2)$ |
展開係数 |
対称関数 |
描画範囲 |
実部 |
虚部 |
XY平面 |
YZ平面 |
ZX平面 |
3面表示 |
$1s$ |
$(1,0,0)\times(1,0,0)$ |
0.921 |
0 |
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|
$L=2[a_B]$ |
$(1,0,0)\times(2,0,0)$ |
-0.384 |
0 |
$2s$ |
$(1,0,0)\times(1,0,0)$ |
-0.114 |
0 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(2,0,0)$ |
-0.606 |
0.001 |
$(1,0,0)\times(3,0,0)$ |
0.766 |
-0.004 |
$2p$ |
$(1,0,0)\times(2,1,-1)$ |
0.194 |
-0.339 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(2,1,0)$ |
-0.155 |
-0.528 |
$(1,0,0)\times(2,1,1)$ |
0.064 |
0.0825 |
$(1,0,0)\times(3,1,-1)$ |
-0.211 |
0.358 |
$(1,0,0)\times(3,1,0)$ |
0.162 |
-0.561 |
$(1,0,0)\times(3,1,1)$ |
-0.0629 |
-0.081 |
$2p$ |
$(1,0,0)\times(2,1,-1)$ |
-0.463 |
-0.058 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(2,1,0)$ |
-0.278 |
0.087 |
$(1,0,0)\times(2,1,1)$ |
-0.379 |
0.101 |
$(1,0,0)\times(3,1,-1)$ |
0.508 |
0.065 |
$(1,0,0)\times(3,1,0)$ |
0.304 |
-0.094 |
$(1,0,0)\times(3,1,1)$ |
0.401 |
-0.116 |
$2p$ |
$(1,0,0)\times(2,1,-1)$ |
0.267 |
0.125 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(2,1,0)$ |
0.241 |
0.134 |
$(1,0,0)\times(2,1,1)$ |
-0.527 |
-0.119 |
$(1,0,0)\times(3,1,-1)$ |
0.290 |
-0.127 |
$(1,0,0)\times(3,1,0)$ |
-0.259 |
-0.151 |
$(1,0,0)\times(3,1,1)$ |
0.565 |
0.136 |
$3s$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3p$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3p$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3p$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=30[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=30[a_B]$ |
記号 |
$(n_1,l_1,m_1)\times(n_2,l_2,m_2)$ |
展開係数 |
反対称関数 |
描画範囲 |
実部 |
虚部 |
XY平面 |
YZ平面 |
ZX平面 |
3面表示 |
$2s$ |
$(1,0,0)\times(2,0,0)$ |
0.875 |
0 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(3,0,0)$ |
-0.469 |
0 |
$2p$ |
$(1,0,0)\times(2,1,-1)$ |
-0.101 |
0.300 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(2,1,0)$ |
0.404 |
-0.309 |
$(1,0,0)\times(2,1,1)$ |
0.005 |
0.492 |
$(1,0,0)\times(3,1,-1)$ |
0.0790 |
-0.245 |
$(1,0,0)\times(3,1,0)$ |
-0.335 |
0.251 |
$(1,0,0)\times(3,1,1)$ |
0.002 |
-0.393 |
$2p$ |
$(1,0,0)\times(2,1,-1)$ |
0.336 |
0.596 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(2,1,1)$ |
-0.254 |
-0.255 |
$(1,0,0)\times(3,1,-1)$ |
-0.268 |
-0.478 |
$(1,0,0)\times(3,1,1)$ |
0.2145 |
0.206 |
$2p$ |
$(1,0,0)\times(2,1,-1)$ |
0.141 |
0.130 |
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|
$L=20[a_B]$ |
$(1,0,0)\times(2,1,0)$ |
0.087 |
0.570 |
$(1,0,0)\times(2,1,1)$ |
0.416 |
0.234 |
$(1,0,0)\times(3,1,-1)$ |
-0.118 |
-0.108 |
$(1,0,0)\times(3,1,0)$ |
-0.0744 |
-0.466 |
$(1,0,0)\times(3,1,1)$ |
-0.336 |
-0.197 |
$3s$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3p$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3p$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3p$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=40[a_B]$ |
$3d$ |
- |
- |
- |
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|
$L=40[a_B]$ |
ちなみにヘリウム原子の場合、直交関数系は $(1,0,0)$ を必ず含むよ。なぜならば、$(1,0,0)$ から次に低い $(2,0,0)$ とした場合の固有エネルギーは約 $-20[{\rm eV}]$ で、イオン化エネルギーよりも高くなるために実質的には先に電離してしまうね。