量子力学Ⅰ/球面調和関数/メモ の履歴(No.6)
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]&
実数版
LANG:mathematica With[{ lmax=5, F := Function[{l,m,t,p}, If[ m==0, SphericalHarmonicY[l, 0, t, p], SphericalHarmonicY[l, Abs[m], t, p]/Sqrt[2] + If[m<0,-1,1] (-1)^m SphericalHarmonicY[l, -Abs[m], t, p]/Sqrt[2] ] If[m<0,-I,1]] }, Table[ If[Abs[m]<=l, SphericalPlot3D[2 Abs[F[l,m,t,p]], {t, 0, Pi}, {p, 0, 2 Pi}, PlotRange -> {{-1.2, 1.2}, {-1.2, 1.2}, {-1.8, 1.8}}, 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[F[l, m, t, p]]]+1)/2]] ] // Evaluate, Null ], {l, 0, lmax}, {m, -lmax, 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//Darker,Red//Lighter]],ImageSize->Large,AspectRatio->0.3,PlotStyle->Thick,Filling->Axis] Plot[Abs[Sin[x]], {x, -3 Pi, 3 Pi}, PlotLegends -> "Expressions", PlotPoints->500,ColorFunctionScaling->False, ColorFunction->Function[{x,y},If[Sin[x]>0,Blue//Darker,Red//Lighter]],ImageSize->Large,AspectRatio->0.15,PlotStyle->Thick,Filling->Axis]
解答:$m$ に関する漸化式†
(1)
漸化式にそれぞれ を代入すると、
&math( \hat l_+\,Y_l{}^{l}(\theta,\phi)=\hbar&\sqrt{(l-l)(l+l+1)}\,Y_l{}^{l+1}(\theta,\phi)=0\\[-5mm] &\hspace{3mm}\underbrace{\hspace{11mm}}_{=\,0} );
&math( \hat l_-\,Y_l{}^{-l}(\theta,\phi)=\hbar&\sqrt{(l-l)(l+l+1)}\,Y_l{}^{-l-1}(\theta,\phi)=0\\[-5mm] &\hspace{3mm}\underbrace{\hspace{11mm}}_{=\,0} );
となって、 や が得られることはない。
(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)
に注意しつつ、
&math( &\frac{1}{2}\left( \hat l_+\hat l_-+\hat l_-\hat l_+\right)\\ &=\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_x^2+\cancel{i\hat l_y\hat l_x}-\cancel{i\hat l_x\hat l_y}+\hat l_y^2+ \hat l_x^2-\cancel{i\hat l_y\hat l_x}+\cancel{i\hat l_x\hat l_y}+\hat l_y^2\right)\\ &=\hat l_x^2+\hat l_y^2 );
(4)
&math( \big(\hat l_x^2+\hat l_y^2\big)Y_l{}^m &=\frac{\hbar^2}{2}\big\{(l+m+1)(l-m)+(l+m)(l-m+1)\big\}Y_l{}^m\\ &=\hbar^2(l^2-m^2+l)Y_l{}^m\\ );