３次元調和振動子/メモ

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量子力学Ⅰ/３次元調和振動子

解答†

(1)

&-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}rR(r)+ \left\{\frac{1}{2}Kr^2+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right\} rR(r)=\varepsilon rR(r)\\ &-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}\underbrace{rR(r)}_{X(\xi)}+ \left\{\frac{1}{2}m\omega^2r^2+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right\} rR(r)=\varepsilon rR(r)\\ &-\frac{\hbar\omega}{2}\frac{d^2}{d\xi^2}X(\xi)+ \left\{\frac{1}{2}\hbar\omega\xi^2+\frac{\hbar\omega}{2}\frac{l(l+1)}{\xi^2}\right\} X(\xi)=\varepsilon X(\xi)\\ &-\frac{d^2}{d\xi^2}X(\xi)+ \left\{\xi^2+\frac{l(l+1)}{\xi^2}\right\} X(\xi)=\frac{2\varepsilon}{\hbar\omega} X(\xi)\\ &\frac{d^2}{d\xi^2}X(\xi)= \left\{\xi^2-\frac{2\varepsilon}{\hbar\omega} +\frac{l(l+1)}{\xi^2}\right\} X(\xi)\\

(2)

\frac{d^2}{d\xi^2}\left(\xi^k e^{-\xi^2/2}\right) &=\frac{d}{d\xi}\left(k\xi^{k-1} e^{-\xi^2/2}-\xi^{k+1} e^{-\xi^2/2}\right)\\ &=(k-1)k\xi^{k-2} e^{-\xi^2/2}-k\xi^k e^{-\xi^2/2}-(k+1)\xi^k e^{-\xi^2/2}+\xi^{k+2} e^{-\xi^2/2}\\ &=\left(\xi^2-(2k+1)+\frac{(k-1)k}{\xi^2}\right)\xi^k e^{-\xi^2/2}

(3)

X''(\xi)&=\left\{\xi^2-\{2(l+1)+1\}+\frac{\{(l+1)-1\}(l+1)}{\xi^2}\right\}X(\xi)\\ &=\left\{\xi^2-(2l+3)+\frac{l(l+1)}{\xi^2}\right\}X(\xi)\\ &=\left\{\xi^2-\frac{2\varepsilon}{\hbar\omega}+\frac{l(l+1)}{\xi^2}\right\} X(\xi)

この式は 2l+3=\frac{2\varepsilon}{\hbar\omega} すなわち、 \varepsilon=\left(l+\frac{3}{2}\right)\hbar\omega のとき成り立つ。

(4)

X''(\xi)&=\left\{ \left(\xi^2-(2l+3)+\frac{l(l+1)}{\xi^2}\right)\xi^{l+1}- \left(\xi^2-(2l+7)+\frac{(l+2)(l+3)}{\xi^2}\right)\frac{2}{2l+3}\xi^{l+3}\right\}e^{-\xi^2/2}\\ &=\left\{ \left(\xi^2-(2l+7)+\frac{l(l+1)}{\xi^2}\right)\xi^{l+1}- \left(\xi^2-(2l+7)+\frac{l^2+5l+6}{\xi^2}-\frac{2(2l+3)}{\xi^2}\right)\frac{2}{2l+3}\xi^{l+3}\right\}e^{-\xi^2/2}\\ &=\left\{ \left(\xi^2-(2l+7)+\frac{l(l+1)}{\xi^2}\right)\xi^{l+1}- \left(\xi^2-(2l+7)+\frac{l(l+1)}{\xi^2}\right)\frac{2}{2l+3}\xi^{l+3}\right\}e^{-\xi^2/2}\\ &=\left(\xi^2-\underbrace{(2l+7)}_{2\varepsilon/\hbar\omega}-\frac{l(l+1)}{\xi^2}\right) \underbrace{\left(\xi^{l+1}-\frac{2}{2l+3}\xi^{l+3}\right)e^{-\xi^2/2}}_{X(\xi)}\\

(5) (発展)

　 X''(\xi)=&\left(\sum_{k=0}^n 2k(2k-1)c_k\xi^{2(k-1)}\right)\xi^{l+1}e^{-\xi^2/2}\\ &+2\left(\sum_{k=0}^n 2kc_k\xi^{2k-1}\right)\Big\{(l+1)\xi^{l}e^{-\xi^2/2}-\xi^{l+2}e^{-\xi^2/2}\Big\}\\ &+\left(\sum_{k=0}^n c_k\xi^{2k}\right)\left\{\xi^2-2(l+3/2)+\frac{l(l+1)}{\xi^2}\right\}\xi^{l+1}e^{-\xi^2/2}\\ =&\left(\sum_{k=0}^{n-1} 2(k+1)(2k+1)c_{k+1}\xi^{2k}\right)\xi^{l+1}e^{-\xi^2/2}\\ &+\left(\sum_{k=0}^n 4k(l+1)c_k\xi^{2k-2}\right)\xi^{l+1}e^{-\xi^2/2}\\ &+\left(\sum_{k=0}^n -4kc_k\xi^{2k}\right)\xi^{l+1}e^{-\xi^2/2}\\ &+\left(\sum_{k=0}^n +4nc_k\xi^{2k}\right)\xi^{l+1}e^{-\xi^2/2}\\ &+\left(\sum_{k=0}^n c_k\xi^{2k}\right)\left\{\xi^2-2(l+2n+3/2)+\frac{l(l+1)}{\xi^2}\right\}\xi^{l+1}e^{-\xi^2/2}\\ =&\left[\sum_{k=0}^{n-1}\Big\{2(k+1)(2k+1+2l+2)c_{k+1}-4(k-n)c_k\Big\}\xi^{2k}\right]\xi^{l+1}e^{-\xi^2/2}\\ &+\left\{\xi^2-2(l+2n+3/2)+\frac{l(l+1)}{\xi^2}\right\}X(\xi)

したがって、

　 2(k+1)(2k+2l+3)c_{k+1}=4(k-n)c_k

であれば条件を満たす。

すなわち、適当な c_0 に対して

　 c_{k+1}=-\frac{2(k-n)}{(k+1)(2l+2k+3)}c_k

を満たすよう c_k を定めれば与式を満たす。

グラフ†

LANG:Mathematica
ListPlot[Table[{l, l + 2 n + 3/2}, {n, 0, 20}, {l, 0, 8}] // Flatten[#, 1] &, 
  PlotRange -> {{-0.2, 7.2}, {0, 10}}, 
  GridLines -> {Table[i, {i, 1, 8}], Table[i, {i, 1, 16}]}, 
  GridLinesStyle -> LightGray, PlotStyle -> {PointSize[0.02]}, 
  ImageSize -> 400]
LANG:Mathematica
c[k_, n_, l_] := 2 (k - 1 - n)/((k - 1 + 1) (2 l + 2 (k - 1) + 3)) c[k - 1, n, l];
c[0, n_, l_] := 1

X[n_, l_] := Sum[c[k, n, l] x^(2 k), {k, 0, n}] x^(l + 1) Exp[-x^2/2]

Table[ D[X[n, l], {x, 2}] == (x^2 - 2 (l + 2 n + 3/2) + l (l + 1)/x^2) X[n, l], 
       {n, 0, 3}, {l, 0, 3}] // Simplify // Flatten // Apply[And, #] &

Table[Plot[
   Table[(x X[n, l])^2/
      Integrate[(x X[n, l])^2 /. x -> y, {y, 0, 10}], {l, 0, 3}] // 
    Evaluate, {x, 0, 6}, PlotRange -> Full, 
   PlotLabel -> ("n = " <> ToString[n]), 
   PlotLegends -> 
    If[n == 3, Table["l = " <> ToString[l], {l, 0, 3}], None]], {n, 
   0, 3}] // Partition[#, 2] & // TableForm

$N=2$ の場合†

　 \varphi_{200}\propto \left(-1+2\frac{r^2}{r_0^2}\cos^2\phi\sin^2\theta\right)e^{-\frac{r^2}{2r_0^2}}

　 \varphi_{020}\propto \left(-1+2\frac{r^2}{r_0^2}\sin^2\phi\sin^2\theta\right)e^{-\frac{r^2}{2r_0^2}}

　 \varphi_{002}\propto \left(-1+2\frac{r^2}{r_0^2}\cos^2\theta\right)e^{-\frac{r^2}{2r_0^2}}

　 \varphi_{110}\propto \left(\frac{r^2}{r_0^2}\sin\phi\cos\phi\sin^2\theta\right)e^{-\frac{r^2}{2r_0^2}}

　 \varphi_{101}\propto \left(\frac{r^2}{r_0^2}\sin\phi\sin\theta\cos\theta\right)e^{-\frac{r^2}{2r_0^2}}

　 \varphi_{011}\propto \left(\frac{r^2}{r_0^2}\cos\phi\sin\theta\cos\theta\right)e^{-\frac{r^2}{2r_0^2}}

であり、

　 Y_0^0\propto 1

　 Y_3^3\propto 3\cos^2\theta-1

　 Y_2^{\pm 1}\propto e^{\pm i\phi}\sin\theta\cos\theta

　 Y_2^{\pm 2}\propto e^{\pm 2i\phi}\sin^2\theta

であるから、

　 &\varphi_{200}\propto \left\{ 2 \sqrt{\pi}\left(-1+\frac{2}{3}\frac{r^2}{r_0^2}\right) Y_{0}^{0} -\frac{4}{3} \sqrt{\frac{\pi }{5}} \frac{r^2}{r_0^2} Y_{2}^{0} +2 \sqrt{\frac{2 \pi }{15}} \frac{r^2}{r_0^2} (Y_{2}^{-2}+Y_{2}^{2}) \right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{020}\propto \left\{ 2 \sqrt{\pi}\left(-1+\frac{2}{3}\frac{r^2}{r_0^2}\right) Y_{0}^{0} -\frac{4}{3} \sqrt{\frac{\pi }{5}} \frac{r^2}{r_0^2} Y_{2}^{0} -2 \sqrt{\frac{2 \pi }{15}} \frac{r^2}{r_0^2} (Y_{2}^{-2}+Y_{2}^{2}) \right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{002}\propto \left\{ 2 \sqrt{\pi}\left(-1+\frac{2}{3}\frac{r^2}{r_0^2}\right) Y_{0}^{0} +\frac{8}{3} \sqrt{\frac{\pi }{5}} \frac{r^2}{r_0^2} Y_{2}^{0} \right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{110}\propto \left\{ -i 2 \sqrt{\frac{\pi }{30}} \frac{r^2}{r_0^2}(Y_{2}^{2}-Y_{2}^{-2}) \right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{101}\propto \left\{ i 2 \sqrt{\frac{\pi }{30}} \frac{r^2}{r_0^2}(Y_{2}^{-1}+Y_{2}^{1}) \right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{011}\propto \left\{ 2 \sqrt{\frac{\pi }{30}} \frac{r^2}{r_0^2}(Y_{2}^{-1}-Y_{2}^{1}) \right\}e^{-\frac{r^2}{2r_0^2}}\\

のように、

　 \left(-1+\frac{2}{3}\frac{r^2}{r_0^2}\right)e^{-\frac{r^2}{2r_0^2}}\,Y_{0}^{0}

と、

　 \frac{r^2}{r_0^2}e^{-\frac{r^2}{2r_0^2}}\,Y_{2}^{m}

との線形結合で \varphi_{abc} を表せる。係数比は、

  • \varphi_{200},\ \varphi_{020}
    • Y_0^0:Y_2^0:Y_2^{+2}:Y_2^{-2}=1:1/2:3/4:3/4 より、
    • l=0:l=2 は 1:2
    • m=0:m=2:m=-2 は 2:1:1
  • \varphi_{002}
    • Y_0^0:Y_2^0=1:2 より、
    • l=0:l=2 は 1:2
    • m=0 は固有値

これを逆に解けば \left(-1+\frac{2}{3}\frac{r^2}{r_0^2}\right)e^{-\frac{r^2}{2r_0^2}}\,Y_{0}^{0} と、 \frac{r^2}{r_0^2}e^{-\frac{r^2}{2r_0^2}}\,Y_{2}^{m} とを \varphi_{abc} で表すこともできる。

LANG:mathematica
Y[l_, m_] := SphericalHarmonicY[l, m, \[Theta], \[Phi]]
-1 + 2 ( \[Xi] Cos[\[Phi]] Sin[\[Theta]])^2 == 2 Sqrt[\[Pi]] (-1 + (2 \[Xi]^2)/3) Y[0, 0] + 2 Sqrt[(2 \[Pi])/15] \[Xi]^2 (Y[2, 2] + Y[2, -2]) - 4/3 Sqrt[\[Pi]/5] \[Xi]^2 Y[2, 0] // FullSimplify
-1 + 2 ( \[Xi] Sin[\[Phi]] Sin[\[Theta]])^2 == 2 Sqrt[\[Pi]] (-1 + (2 \[Xi]^2)/3) Y[0, 0] - 2 Sqrt[(2 \[Pi])/15] \[Xi]^2 (Y[2, 2] + Y[2, -2]) - 4/3 Sqrt[\[Pi]/5] \[Xi]^2 Y[2, 0] // FullSimplify
-1 + 2 ( \[Xi] Cos[\[Theta]])^2 == 2 Sqrt[\[Pi]] (-1 + (2 \[Xi]^2)/3) Y[0, 0] + 8/3 Sqrt[\[Pi]/5] \[Xi]^2 Y[2, 0] // FullSimplify
Sin[\[Phi]] Cos[\[Phi]] Sin[\[Theta]]^2 == -I 2 Sqrt[\[Pi]/30] (Y[2, 2] - Y[2, -2]) // FullSimplify
Sin[\[Phi]] Sin[\[Theta]] Cos[\[Theta]] == I 2 Sqrt[\[Pi]/30] (Y[2, 1] + Y[2, -1]) // FullSimplify
Cos[\[Phi]] Sin[\[Theta]] Cos[\[Theta]] == 2 Sqrt[\[Pi]/30] (-Y[2, 1] + Y[2, -1]) // FullSimplify

-1+2 (\xi \cos \theta)^2 &= 2 \sqrt{\pi } \left(-1+\frac{2 \xi^2}{3}\right) Y_{0}^{0} +\frac{8}{3} \sqrt{\frac{\pi }{5}} \xi^2 Y_{2}^{0} \\ -1+2 (\xi \sin \phi \sin \theta)^2 &= 2 \sqrt{\pi } \left(-1+\frac{2 \xi^2}{3}\right) Y_{0}^{0} -\frac{4}{3} \sqrt{\frac{\pi }{5}} \xi^2 Y_{2}^{0} -2 \sqrt{\frac{2 \pi }{15}} \xi^2 (Y_{2}^{-2}+Y_{2}^{2}) \\ -1+2 (\xi \cos \phi \sin \theta)^2 &= 2 \sqrt{\pi } \left(-1+\frac{2 \xi^2}{3}\right) Y_{0}^{0} -\frac{4}{3} \sqrt{\frac{\pi }{5}} \xi^2 Y_{2}^{0} +2 \sqrt{\frac{2 \pi }{15}} \xi^2 (Y_{2}^{-2}+Y_{2}^{2}) \\ \sin \phi \cos \phi \sin ^2\theta &=-i 2 \sqrt{\frac{\pi }{30}} (Y_{2}^{2}-Y_{2}^{-2}) \\ \sin \phi \sin \theta \cos \theta &=i 2 \sqrt{\frac{\pi }{30}} (Y_{2}^{-1}+Y_{2}^{1}) \\ \cos \phi \sin \theta \cos \theta &=2 \sqrt{\frac{\pi }{30}} (Y_{2}^{-1}-Y_{2}^{1}) \\

$N=3$ の場合†

　 &\varphi_{300}\propto e^{-\frac{r^2}{2r_0^2}}\left\{\frac{r}{r_0}\cos\phi\sin\theta -\frac{2}{3}\left(\frac{r}{r_0}\cos\phi\sin\theta\right)^3\right\}\\ &\varphi_{030}\propto e^{-\frac{r^2}{2r_0^2}}\left\{\frac{r}{r_0}\sin\phi\sin\theta -\frac{2}{3}\left(\frac{r}{r_0}\sin\phi\sin\theta\right)^3\right\}\\ &\varphi_{003}\propto e^{-\frac{r^2}{2r_0^2}}\left\{\frac{r}{r_0}\cos\theta -\frac{2}{3}\left(\frac{r}{r_0}\cos\theta\right)^3\right\}\\ &\varphi_{210}\propto e^{-\frac{r^2}{2r_0^2}}\frac{r}{r_0}\left\{1-2\left(\frac{r}{r_0}\cos\phi\sin\theta\right)^2\right\}\sin\phi\sin\theta\\ &\varphi_{021}\propto e^{-\frac{r^2}{2r_0^2}}\frac{r}{r_0}\left\{1-2\left(\frac{r}{r_0}\sin\phi\sin\theta\right)^2\right\}\cos\theta\\ &\varphi_{102}\propto e^{-\frac{r^2}{2r_0^2}}\frac{r}{r_0}\left\{1-2\left(\frac{r}{r_0}\cos\theta\right)^2\right\}\cos\phi\sin\theta\\ &\varphi_{120}\propto e^{-\frac{r^2}{2r_0^2}}\frac{r}{r_0}\left\{1-2\left(\frac{r}{r_0}\sin\phi\sin\theta\right)^2\right\}\cos\phi\sin\theta\\ &\varphi_{012}\propto e^{-\frac{r^2}{2r_0^2}}\frac{r}{r_0}\left\{1-2\left(\frac{r}{r_0}\cos\theta\right)^2\right\}\sin\phi\sin\theta\\ &\varphi_{201}\propto e^{-\frac{r^2}{2r_0^2}}\frac{r}{r_0}\left\{1-2\left(\frac{r}{r_0}\cos\phi\sin\theta\right)^2\right\}\cos\theta\\ &\varphi_{111}\propto e^{-\frac{r^2}{2r_0^2}}\left(\frac{r}{r_0}\right)^3\cos\phi\sin\theta\sin\phi\sin\theta\cos\theta\\

がんばって計算すると、

　 &\varphi_{300}\propto \left\{\sqrt{\frac{2\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}(-Y_1^1+Y_1^{-1}) +\frac{2}{5}\sqrt{\frac{\pi }{21}}\frac{r^3}{r_0^3}(-Y_3^1+Y_3^{-1}) -\frac{2}{3}\sqrt{\frac{\pi }{35}}\frac{r^3}{r_0^3}(-Y_3^3+Y_3^{-3})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{030}\propto \left\{i\sqrt{\frac{2\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}(Y_1^1+Y_1^{-1}) +i\frac{2}{5}\sqrt{\frac{\pi }{21}}\frac{r^3}{r_0^3}(Y_3^1+Y_3^{-1}) +i\frac{2}{3}\sqrt{\frac{\pi }{35}}\frac{r^3}{r_0^3}(Y_3^3+Y_3^{-3})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{003}\propto \left\{2 \sqrt{\frac{\pi }{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}Y_1^0-\frac{8}{15}\sqrt{\frac{\pi }{7}}\frac{r^3}{r_0^3}Y_3^0\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{210}\propto \left\{i\sqrt{\frac{2\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}(Y_1^1+Y_1^{-1}) +i\frac{2}{5}\sqrt{\frac{\pi }{21}}\frac{r^3}{r_0^3}(Y_3^1+Y_3^{-1}) -i2 \sqrt{\frac{\pi }{35}}\frac{r^3}{r_0^3}(Y_3^3+Y_3^{-3})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{021}\propto \left\{2\sqrt{\frac{\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}Y_1^0 +\frac{4}{5}\sqrt{\frac{\pi }{7}}\frac{r^3}{r_0^3}Y_3^0 +2 \sqrt{\frac{2 \pi }{105}}\frac{r^3}{r_0^3}(Y_3^2+Y_3^{-2})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{102}\propto \left\{\sqrt{\frac{2\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}(-Y_1^1+Y_1^{-1}) -\frac{8}{5}\sqrt{\frac{\pi }{21}}\frac{r^3}{r_0^3}(-Y_3^1+Y_3^{-1})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{120}\propto \left\{\sqrt{\frac{2\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}(-Y_1^1+Y_1^{-1}) +\frac{2}{5}\sqrt{\frac{\pi }{21}}\frac{r^3}{r_0^3}(-Y_3^1+Y_3^{-1}) +2 \sqrt{\frac{\pi }{35}}\frac{r^3}{r_0^3}(-Y_3^3+Y_3^{-3})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{012}\propto \left\{i\sqrt{\frac{2\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}(Y_1^1+Y_1^{-1}) -i\frac{8}{5}\sqrt{\frac{\pi }{21}}\frac{r^3}{r_0^3}(Y_3^3+Y_3^{-3})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{201}\propto \left\{2\sqrt{\frac{\pi}{3}}\left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}Y_1^0 +\frac{4}{5}\sqrt{\frac{\pi }{7}}\frac{r^3}{r_0^3}Y_3^0 -2 \sqrt{\frac{2 \pi }{105}}\frac{r^3}{r_0^3}(Y_3^2+Y_3^{-2})\right\}e^{-\frac{r^2}{2r_0^2}}\\ &\varphi_{111}\propto \left\{i\sqrt{\frac{2 \pi }{105}}\frac{r^3}{r_0^3}(-Y_3^2+Y_3^{-2})\right\}e^{-\frac{r^2}{2r_0^2}}\\

のように、 \left(1-\frac{2}{5}\frac{r^2}{r_0^2}\right)\frac{r}{r_0}e^{-\frac{r^2}{2r_0^2}}\,Y_1^m と、 \frac{r^3}{r_0^3}e^{-\frac{r^2}{2r_0^2}}\,Y_3^m との線形結合で \varphi_{abc} を表せる。

LANG:mathematica
Y[l_, m_] := SphericalHarmonicY[l, m, \[Theta], \[Phi]]
\[Xi] Cos[\[Phi]] Sin[\[Theta]] - 2/3 (\[Xi] Cos[\[Phi]] Sin[\[Theta]])^3 == -(1/ 3) (2 Sqrt[\[Pi]/35] \[Xi]^3 (Y[3, -3] - Y[3, 3])) + 2/5 Sqrt[\[Pi]/21] \[Xi]^3 (Y[3, -1] - Y[3, 1]) + Sqrt[(2 \[Pi])/ 3] (1 - (2 \[Xi]^2)/5) \[Xi] (Y[1, -1] - Y[1, 1]) // FullSimplify
\[Xi] Cos[\[Phi]] Sin[\[Theta]] - 2/3 (\[Xi] Cos[\[Phi]] Sin[\[Theta]])^3 == -(1/ 3) (2 Sqrt[\[Pi]/35] \[Xi]^3 (Y[3, -3] - Y[3, 3])) + 2/5 Sqrt[\[Pi]/21] \[Xi]^3 (Y[3, -1] - Y[3, 1]) + Sqrt[(2 \[Pi])/ 3] (1 - (2 \[Xi]^2)/5) \[Xi] (Y[1, -1] - Y[1, 1]) // FullSimplify
\[Xi] Sin[\[Phi]] Sin[\[Theta]] - 2/3 (\[Xi] Sin[\[Phi]] Sin[\[Theta]])^3 == I 2/3 Sqrt[\[Pi]/35] \[Xi]^3 (Y[3, -3] + Y[3, 3]) + I 2/5 Sqrt[\[Pi]/21] \[Xi]^3 (Y[3, -1] + Y[3, 1]) + I Sqrt[(2 \[Pi])/ 3] (1 - (2 \[Xi]^2)/5) \[Xi] (Y[1, -1] + Y[1, 1]) // FullSimplify
\[Xi] Cos[\[Theta]] - 2/3 (\[Xi] Cos[\[Theta]])^3 == 2 Sqrt[ \[Pi]/3] (1 - (2 \[Xi]^2)/5) \[Xi] (Y[1, 0]) - 8/15 Sqrt[\[Pi]/7] \[Xi]^3 Y[3, 0] // FullSimplify
(1 - 2 (\[Xi] Cos[\[Phi]] Sin[\[Theta]])^2) Sin[\[Phi]] Sin[\[Theta]] == I Sqrt[( 2 \[Pi])/3] (1 - (2 \[Xi]^2)/5) (Y[1, 1] + Y[1, -1]) + I 2/5 Sqrt[\[Pi]/21] \[Xi]^2 (Y[3, 1] + Y[3, -1]) - I 2/1 Sqrt[\[Pi]/35] \[Xi]^2 (Y[3, 3] + Y[3, -3]) // FullSimplify
(1 - 2 (\[Xi] Sin[\[Phi]] Sin[\[Theta]])^2) Cos[\[Theta]] == 2 Sqrt[ \[Pi]/3] (1 - (2 \[Xi]^2)/5) (Y[1, 0]) + 4/5 Sqrt[\[Pi]/7] \[Xi]^2 (Y[3, 0]) + 2/1 Sqrt[(2 \[Pi])/105] \[Xi]^2 (Y[3, 2] + Y[3, -2]) // FullSimplify
(1 - 2 (\[Xi] Cos[\[Theta]])^2) Cos[\[Phi]] Sin[\[Theta]] == Sqrt[(2 \[Pi])/3] (1 - (2 \[Xi]^2)/5) (-Y[1, 1] + Y[1, -1]) - 8/5 Sqrt[\[Pi]/21] \[Xi]^2 (-Y[3, 1] + Y[3, -1]) // FullSimplify
(1 - 2 (\[Xi] Sin[\[Phi]] Sin[\[Theta]])^2) Cos[\[Phi]] Sin[\[Theta]] == Sqrt[(2 \[Pi])/3] (1 - (2 \[Xi]^2)/5) (-Y[1, 1] + Y[1, -1]) + 2/5 Sqrt[\[Pi]/21] \[Xi]^2 (-Y[3, 1] + Y[3, -1]) + 2/1 Sqrt[\[Pi]/35] \[Xi]^2 (-Y[3, 3] + Y[3, -3]) // FullSimplify
(1 - 2 (\[Xi] Cos[\[Theta]])^2) Sin[\[Phi]] Sin[\[Theta]] == I Sqrt[(2 \[Pi])/3] (1 - (2 \[Xi]^2)/5) (Y[1, 1] + Y[1, -1]) - (I 8)/ 5 Sqrt[\[Pi]/21] \[Xi]^2 (Y[3, 1] + Y[3, -1]) // FullSimplify
(1 - 2 (\[Xi] Cos[\[Phi]] Sin[\[Theta]])^2) Cos[\[Theta]] == 2 Sqrt[ \[Pi]/3] (1 - (2 \[Xi]^2)/5) (Y[1, 0]) + 4/5 Sqrt[\[Pi]/7] \[Xi]^2 (Y[3, 0]) - 2/1 Sqrt[(2 \[Pi])/105] \[Xi]^2 (Y[3, 2] + Y[3, -2]) // FullSimplify
Cos[\[Phi]] Sin[\[Phi]] Sin[\[Theta]]^2 Cos[\[Theta]] == I Sqrt[( 2 \[Pi])/105] (-Y[3, 2] + Y[3, -2]) // FullSimplify

\xi \cos \phi \sin \theta-\frac{2}{3} (\xi \cos \phi \sin \theta)^3&= \left(1-\frac{2 \xi^2}{5}\right) \sqrt{\frac{2 \pi }{3}} (Y_{1}^{-1}-Y_{1}^{1}) \xi +\frac{2}{5} \sqrt{\frac{\pi }{21}} (Y_{3}^{-1}-Y_{3}^{1}) \xi^3 -\frac{2}{3} \sqrt{\frac{\pi }{35}} \xi^3 (Y_{3}^{-3}-Y_{3}^{3}) \\ \xi \cos \phi \sin \theta-\frac{2}{3} (\xi \cos \phi \sin \theta)^3&= \left(1-\frac{2 \xi^2}{5}\right) \sqrt{\frac{2 \pi }{3}} (Y_{1}^{-1}-Y_{1}^{1}) \xi +\frac{2}{5} \sqrt{\frac{\pi }{21}} (Y_{3}^{-1}-Y_{3}^{1}) \xi^3 -\frac{2}{3} \sqrt{\frac{\pi }{35}} \xi^3 (Y_{3}^{-3}-Y_{3}^{3}) \\ \xi \sin \phi \sin \theta-\frac{2}{3} (\xi \sin \phi \sin \theta)^3&= \left(1-\frac{2 \xi^2}{5}\right) i \sqrt{\frac{2 \pi }{3}} (Y_{1}^{-1}+Y_{1}^{1}) \xi +\frac{2}{5} i \sqrt{\frac{\pi }{21}} (Y_{3}^{-1}+Y_{3}^{1}) \xi^3 +\frac{2}{3} i \sqrt{\frac{\pi }{35}} (Y_{3}^{-3}+Y_{3}^{3}) \xi^3 \\ \xi \cos \theta-\frac{2}{3} (\xi \cos \theta)^3&= \left(1-\frac{2 \xi^2}{5}\right) 2 \sqrt{\frac{\pi }{3}} Y_{1}^{0} \xi -\frac{8}{15} \sqrt{\frac{\pi }{7}} Y_{3}^{0} \xi^3 \\ \left(1-2 (\xi \cos \phi \sin \theta)^2\right) \sin \phi \sin \theta&= \left(1-\frac{2 \xi^2}{5}\right) i \sqrt{\frac{2 \pi }{3}} (Y_{1}^{-1}+Y_{1}^{1}) +\frac{2}{5} i \sqrt{\frac{\pi }{21}} (Y_{3}^{-1}+Y_{3}^{1}) \xi^2 -i 2 \sqrt{\frac{\pi }{35}} (Y_{3}^{-3}+Y_{3}^{3}) \xi^2 \\ \left(1-2 (\xi \sin \phi \sin \theta)^2\right) \cos \theta&= \left(1-\frac{2 \xi^2}{5}\right) 2 \sqrt{\frac{\pi }{3}} Y_{1}^{0} +2 \sqrt{\frac{2 \pi }{105}} (Y_{3}^{-2}+Y_{3}^{2}) \xi^2 +\frac{4}{5} \sqrt{\frac{\pi }{7}} Y_{3}^{0} \xi^2 \\ \left(1-2 (\xi \cos \theta)^2\right) \cos \phi \sin \theta&= \left(1-\frac{2 \xi^2}{5}\right) \sqrt{\frac{2 \pi }{3}} (Y_{1}^{-1}-Y_{1}^{1}) -\frac{8}{5} \sqrt{\frac{\pi }{21}} (Y_{3}^{-1}-Y_{3}^{1}) \xi^2 \\ \left(1-2 (\xi \sin \phi \sin \theta)^2\right) \cos \phi \sin \theta&= \left(1-\frac{2 \xi^2}{5}\right) \sqrt{\frac{2 \pi }{3}} (Y_{1}^{-1}-Y_{1}^{1}) +2 \sqrt{\frac{\pi }{35}} (Y_{3}^{-3}-Y_{3}^{3}) \xi^2 +\frac{2}{5} \sqrt{\frac{\pi }{21}} (Y_{3}^{-1}-Y_{3}^{1}) \xi^2 \\ \left(1-2 (\xi \cos \theta)^2\right) \sin \phi \sin \theta&= \left(1-\frac{2 \xi^2}{5}\right) i \sqrt{\frac{2 \pi }{3}} (Y_{1}^{-1}+Y_{1}^{1}) -i\frac{8}{5} \sqrt{\frac{\pi }{21}} (Y_{3}^{-1}+Y_{3}^{1}) \xi^2 \\ \left(1-2 (\xi \cos \phi \sin \theta)^2\right) \cos \theta&= \left(1-\frac{2 \xi^2}{5}\right) 2 \sqrt{\frac{\pi }{3}} Y_{1}^{0} -2 \sqrt{\frac{2 \pi }{105}} (Y_{3}^{-2}+Y_{3}^{2}) \xi^2 +\frac{4}{5} \sqrt{\frac{\pi }{7}} Y_{3}^{0} \xi^2 \\ \cos \phi \sin \phi \sin ^2\theta \cos \theta&= i \sqrt{\frac{2 \pi }{105}} (Y_{3}^{-2}-Y_{3}^{2}) \\

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