# Finding eigenfunctions using ladder operators

Using $L_1\pm iL_2$ as raising and lowering operators I need to obtain an expression for $Y_{l,l}(\theta,\phi)$ and after this I need to find $Y_{l,m}(\theta,\phi)$ for $(l,m)=(1,1),(1,0),(1,-1),(2,2),(2,1)$ and $(2,0)$ The hint says that $Y_{l,l}(\theta,\phi)$, $Y_{1,1}(\theta,\phi)$, $Y_{2,2}(\theta,\phi)$ should be easy to find as they are in the kernel of $L_+$ and are eigenfunctions of $L_3$. I think I know how to 'lower' these to find the rest but I am struggling to find these. I got the stage where I have $Y_{l,m}(\theta,\phi) = P_{l,m}(\theta)e^{im\phi}$ and I'm not sure where to go from here, or if this is even correct! Any help will be appreciated!

Let $\langle \theta ,\varphi \vert lm\rangle =Y_{l}^{m}(\theta ,\varphi )$, and consider \begin{eqnarray} \hat{L}_{-} &=&\hat{L}_{x}-i\hat{L}_{y} \\ &=&i\hbar \left( \sin \varphi \frac{\partial }{\partial \theta }+\cos \varphi \cos \theta \frac{\partial }{\partial \varphi }\right) +\hbar \left( \cos \varphi \frac{\partial }{\partial \theta }-\sin \varphi \cos \theta \frac{\partial }{\partial \varphi }\right) \\ &=&\hbar \left( e^{-i\varphi }\frac{\partial }{\partial \theta }+\cos \theta (i^{2}\sin \varphi -i\cos \theta )\frac{d}{d\varphi }\right) \\ &=&\hbar \left( -e^{-i\varphi }\frac{\partial }{\partial \theta }+i\cos \theta e^{-i\varphi }\frac{d}{d\varphi }\right) =\hbar e^{-i\varphi }\left( - \frac{\partial }{\partial \theta }+i\cos \theta \frac{\partial }{\partial \varphi }\right) \end{eqnarray}

Then \begin{eqnarray} L_{-}Y_{l}^{-l}(\theta ,\varphi ) &=&0 \\ L_{z}Y_{l}^{-l} &=&l\hbar Y_{l}^{-l}(\theta ,\varphi )\Longrightarrow -i\hbar% \frac{\partial }{\partial \varphi }Y_{l}^{m}(\theta ,\varphi )=m^{2}Y_{l}^{m}(\theta ,\varphi ) \\ &=&Y_{l}^{m}(\theta ,\varphi )=g(\theta )e^{im\varphi } \end{eqnarray}

The equation $L_{-}Y_{l}^{-l}(\theta ,\varphi )=0$ translates into the differential equation $$\left( -\frac{d}{d\theta }+i(-il)\cos \theta \right) g(\theta )=0$$ with solution $g(\theta )=C_{l}(\sin\theta )^{l}$ so that $$Y_{l}^{-l}(\theta ,\varphi )=C_{l}(\sin \theta )^{l}e^{-il\varphi }$$

The integration constant $C_{l}$ is determined from the normalization condition \begin{eqnarray} \int d\Omega |Y_{l}^{-l}(\theta ,\varphi )|^{2} &=&1=\vert C_{l}\vert ^{2}\int_{0}^{2\pi }d\varphi \int_{0}^{\pi}d\theta \sin \theta (\sin \theta )^{2l} \\ &=&2\pi \vert C_{l}\vert ^{2}\int_{0}^{\pi }d\theta \sin \theta (\sin\theta )^{2l} \end{eqnarray}

With this $$Y_{l}^{-l}(\theta ,\varphi )=\sqrt{\frac{(2l+1)!}{4\pi }}\frac{1}{l^{l}2^{l}}% (\sin\theta )^{l}e^{-il\varphi }$$

To obtain the remaining spherical harmonics, consider $\langle \theta ,\varphi |(L_{+})^{l+m}|l-l\rangle$ \ and recall that $$\hat{L}_{+}|lm\rangle =\sqrt{(l-m)(l+m+1)}|l,m+1\rangle$$

Then: \begin{eqnarray} (\theta ,\varphi |(L_{+})^{l+m}|l,-l\rangle &=&\langle \theta ,\varphi |(L_{+})^{l+m-1}|l,-l+1\rangle \sqrt{2l\cdot 1}\hbar \\ &=&\langle \theta ,\varphi |(L_{+})^{l+m-z}|l,-l+2\rangle \sqrt{(2l)\cdot 1}% \cdot \sqrt{(2l-1)\cdot 2}\hbar^{2} \\ &=&\langle \theta ,\varphi |(L_{+})^{l+m-3}|l,-l+3)\sqrt{(2l)(2l-1)(2l-2)% \cdot 1\cdot 2\cdot 3}\hbar \end{eqnarray} and generally: $$\langle \theta ,\varphi |(L_{+})^{l+m}|l,-l\rangle =\sqrt{\frac{(2l)!(l+m)!}{% (l-m)!}}\langle \theta ,\varphi |lm\rangle (\hbar)^{l+m}$$

Hence, $$Y_{l}^{m}(\theta ,\varphi |lm\rangle =\langle \theta ,\varphi |lm\rangle \sqrt{\frac{(l-m)!}{(2l)!(l+m)!}}\langle \theta ,\varphi |(L_{+})^{l+m}|l,-l\rangle \frac{1}{(\hbar)^{l+m}}$$

Using \begin{eqnarray} \hat{L}_{+} &=&-i\hbar e^{i\varphi }\left[ i\frac{\partial }{\partial \theta }-\cos \theta \frac{\partial }{\partial \varphi }\right] \\ &=&\hbar e^{i\varphi }\left[ \frac{\partial }{\partial \theta }+i\cos \theta \frac{\partial }{\partial \varphi }\right] \\ Y_{l}^{-l}(\theta ,\varphi ) &=&\sqrt{\frac{(2l+1)!}{4\pi }}\frac{1}{l!}% 2^{l}(\sin \theta )^{l}e^{-il\varphi } \end{eqnarray} and $$\langle \theta ,\varphi |\hat{L}_{+}^{l+m}|l,-l\rangle =\hbar ^{l+m}e^{i(l+m)\varphi }\left[ \frac{\partial }{\partial \theta }+i\cos \frac{\partial }{\partial \varphi }\right] ^{l+m}Y_{l}^{-l}(\theta ,\varphi )$$ We finally arrive at $$Y_{l}^{m}(\theta ,\varphi )=\sqrt{\frac{(l-m)!}{(2l)!(l+m)!}}\sqrt{\frac{% (2l+1)!}{4\pi }}\frac{1}{l!2^{l}}\left[ e^{i\varphi }\left( \frac{\partial }{% \partial \theta }+i\cos \theta \frac{\partial }{\partial \varphi }\right) % \right] ^{l+m}(\sin \theta )^{l}e^{-il}$$

For example, consider the case where $l=1$ . Then, \begin{eqnarray} Y_{1}^{-1}(\theta ,\varphi ) &=&\sqrt{\frac{(2\cdot 1+1)!}{4\pi }}\frac{1}{1!}\frac{1}{2!}\sin \theta e^{-i\varphi }=\sqrt{\frac{6}{16\pi }}\sin \theta e^{-i\varphi }=\sqrt{\frac{3}{8\pi }}\sin \theta e^{-i\varphi } \\ Y_{1}^{0}(\theta ,\varphi ) &=&\sqrt{\frac{(1-0)!3}{(1+0!4\pi }}\frac{1}{1!} \frac{1}{2^{1}}e^{i(1+0)\varphi }\left( \frac{\partial }{\partial \theta } -i\cos \theta \frac{\partial }{\partial \varphi }\right) ^{1}\sin \theta e^{-i\varphi } \\ &=&\sqrt{\frac{3}{4\pi }}\frac{1}{2}e^{i\varphi }\left[ \left( \cos \theta +i \frac{\cos \theta }{\sin \theta }(-i)\sin \theta \right) e^{-i\varphi } \right] =\sqrt{\frac{3}{4\pi }}\cdot \frac{1}{2}(2\cos \theta )=\sqrt{\frac{3 }{4\pi }}\cos \theta \end{eqnarray}

\begin{eqnarray} Y_{1}^{+1} &=&\sqrt{\frac{(1-1)!}{(1+1)!}\cdot \frac{3}{4\pi }}\frac{1}{1!} \frac{1}{2!}\left[ e^{i\varphi }\left( \frac{\partial }{\partial \theta }% +i\cos \theta \frac{\partial }{\partial \varphi }\right) \right] ^{2}\sin \theta e^{-i\varphi } \\ &=&\sqrt{\frac{3}{2\cdot 4\pi }}\frac{1}{2}e^{-i\varphi }\left( \frac{ \partial }{\partial \theta }+i\cos \theta \frac{\partial }{\partial \varphi } \right) e^{i\varphi }\left( \frac{\partial }{\partial \theta }+i\cos \theta \frac{\partial }{\partial \varphi }\right) \sin \theta e^{-i\varphi } \\ &=&\sqrt{\frac{3}{8\pi }}\frac{1}{2}e^{i\varphi }\left( \frac{\partial }{ \partial \theta }+i\cos \theta \frac{\partial }{\partial \varphi }\right) 2\cos \theta \\ &=&\sqrt{\frac{3}{8\pi }}e^{i\varphi }(-\sin \theta )=-\sqrt{\frac{3}{8\pi }} e^{i\varphi }\sin \theta \end{eqnarray}

• I'm a bit confused how you got the part: $L_{-}Y_{l}^{-l}(\theta ,\varphi ) =0$ and the following lines – Deke Dec 19 '17 at 17:55
• @Deke The state with $m=-l$ is killed by $L_-$ so $\langle \theta\varphi\vert L_-\vert l,-l\rangle = L_-Y_l^{-l}(\theta,\varphi)=0$. I didn't want to solve the problem for you so I started at the bottom of the ladder rather than at the top. – ZeroTheHero Dec 19 '17 at 21:54