量子力学Ⅰ/球面調和関数/メモ のバックアップ(No.5)

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 m=l,\,m=-l 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)\\[-5mm] &\hspace{3mm}\underbrace{\hspace{11mm}}_{=\,0} );

となって、 Y_l{}^{l+1}(\theta,\phi) Y_l{}^{-l-1}(\theta,\phi) 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)

 &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( \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)\\ );


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