量子力学Ⅰ/球面調和関数/メモ の履歴(No.3)
更新形状†
位相を色分けする
LANG:mathematica With[{lmax=5}, Table[ If[Abs[m]<=l, SphericalPlot3D[Abs[ SphericalHarmonicY[l, m, t, p] 2 ], {t, 0, Pi}, {p, 0, 2 Pi}, PlotRange -> {{-1, 1}, {-1, 1}, {-1.5, 1.5}}, BoxRatios->{1,1,1.5}, PlotPoints->30, Axes->False, Boxed->False, Mesh->False, ColorFunctionScaling -> False, ColorFunction -> Function[{x,y,z,t,p,r}, Blend[{Blue,Yellow}, (Cos[Arg[SphericalHarmonicY[l, m, t, p]]]+1)/2]] ], Null ], {l, 0, lmax}, {m, 0, lmax} ] ]//GraphicsGrid[#,AspectRatio->1.5]&
符号の反転する理由
LANG:mathematica Plot[Sin[x], {x, -3 Pi, 3 Pi}, PlotLegends -> "Expressions", PlotPoints->500,ColorFunctionScaling->False, ColorFunction->Function[{x,y},If[y>0,Blue,Red]],ImageSize->Large,AspectRatio->0.5] Plot[Abs[Sin[x]], {x, -3 Pi, 3 Pi}, PlotLegends -> "Expressions", PlotPoints->500,ColorFunctionScaling->False, ColorFunction->Function[{x,y},If[Sin[x]>0,Blue,Red]],ImageSize->Large,AspectRatio->0.5]
解答:$m$ に関する漸化式†
(1)
&math( l_\pm Y_l{}^{\pm l}(\theta,\phi) &\propto \hbar e^{\pm i\phi} \Big(\pm\frac{\PD}{\PD\theta}+\frac{i}{\tan\theta}\frac{\PD}{\PD\phi}\Big) \sin^l\theta e^{\pm il\phi}\\ &=\hbar e^{\pm i(l+1)\phi} \Big\{\pm(l\sin^{l-1}\theta\cos\theta)+\frac{i\cos\theta}{\sin\theta}\cdot(\pm il\sin^l\theta)\Big\}\\ &=\hbar e^{\pm i(l+1)\phi}\Big(\pm l\sin^{l-1}\theta\cos\theta\mp l\sin^{l-1}\theta\cos\theta\Big)\\ &= 0 );
(2)
&math( &\hat l_-\hat l_+Y_l{}^{m}(\theta,\phi)=\hat l_-\hbar\sqrt{(l-m)(l+m+1)}Y_l{}^{m+1} =\hbar^2(l+m+1)(l-m)Y_l{}^m\\ &\hat l_+\hat l_-Y_l{}^{m}(\theta,\phi)=\hat l_+\hbar\sqrt{(l+m)(l-m+1)}Y_l{}^{m-1} =\hbar^2(l+m)(l-m+1)Y_l{}^{m}\\ );
(3)
&math( \hat l_x^2+\hat l_y^2 &=\frac{1}{2}\left\{ (\hat l_x+i\hat l_y)(\hat l_x-i\hat l_y)+(\hat l_x-i\hat l_y)(\hat l_x+i\hat l_y)\right\}\\ &=\frac{1}{2}\left\{ \hat l_+\hat l_-+\hat l_-\hat l_+\right\}\\ );
(4)
&math( \langle \hat l_x^2+\hat l_y^2\rangle &=\frac{\hbar^2}{2}\big\{(l+m+1)(l-m)+(l+m)(l-m+1)\big\}\\ &=\hbar^2(l^2-m^2+l)\\ );