量子力学Ⅰ/球面調和関数/メモ のバックアップ(No.3)

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)\\ );