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\exnumber{4640}
\heading{Angular momentum of a wave function} 
\auname{Dan Shaked}


{\bf The problem:}

\Dn

The wave function of a particle is  ${ \psi(x,y,z)=f(r)z(z+1)}$. 

\Dn
\begin{enumerate}
\item Write the wave function in ${| r,  \theta,  \varphi \rangle }$ representation.

\item Write the wave function in ${| r,  \ell,  m \rangle }$ representation.

\Dn


Given that:

$Y^{00}=\sqrt{\frac{1}{4\pi}}$,  \,  
$Y^{10}=\sqrt{\frac{3}{4\pi}}cos(\theta)$,
  \,  $Y^{20}=\sqrt{\frac{5}{16\pi}}(3cos^2(\theta)-1)$,  

\Dn

${M_n=\int_0^\infty r^{n}(f(r))^2 dr}$
	
\Dn 
    
\item Write a condition on ${M_6}$ and ${M_4}$ so that the wave function will be normalized.

\Dn
Following it is given: ${M_6=\frac{5}{8\pi}}$

\item What is a probability of the particle to be on spherical state ${(\ell=0)}$?

\item What are the possible non-zero probability values of ${L_z}$? 

\item What are the possible non-zero probability values of ${L_x}$?   

\item Find the probability of the particle to be on polar state ${(\ell=1)}$ with ${L_x=1}$.
\end{enumerate} 
\Dn\Dn


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{\bf The solution:}

\Dn

\begin{enumerate}

\item $\psi(r,\theta,\varphi)=f(r)r^2\cos^2\theta+f(r)r\cos\theta$
\item $\psi(r,\ell,m)=f(r)r^2\sqrt{\frac{16\pi}{45}}Y^{20}+f(r)r\sqrt{\frac{4\pi}{3}}Y^{10}+f(r)r^2\sqrt{\frac{4\pi}{9}}Y^{00}$
\item The normalization condition is \[\|\langle r\eel m|r\eel m \rangle\|^2=(\frac{16\pi}{45}+\frac{4\pi}{9})\int^\infty_0{r^6f(r)}dr+(\frac{4\pi}{3})\int^\infty_0{r^4f(r)}dr=1\]
placing the definitions of $M_n$ we get \[\frac{4\pi}{5}M_6+\frac{4\pi}{3}M_4=1\]
\item \(\|\langle r00|\psi\rangle\|^2=\frac{4\pi}{9}M_6=\frac{5}{18}\)
\item $L_z=0$ with probability 1.
\item I'll show that the possible values are $L_x=0,\pm1,\pm2$:\\
In order to find the possible values of $L_x$ we'll go through the subspaces $\ell=0,1,2$.\\
\begin{itemize}
	\item For $\ell=0$ it is clear that we can get the value $L_x=0$ from the possibility to get $Y^{00}$.
\item In the subspace $\ell=1$ we know that\[Y^{10}=\frac{1}{\sqrt{2}}(Y_x^{11}-Y_x^{1-1})\]
Hence we get that it is possible to get $L_x=\pm1$.
\item The case of $\ell=2$ is more complicated, therefore we sall prove by contridiction that it is possible to get $L_x=\pm2$:\\
Let us assume that the only possible values of $L_x$ are $-1,0,1$. We know that $L^2=L_x^2+L_y^2+L_z^2=2L_x^2$. Then we get $L^2=2$ is the maximum value.
On the other hand we know that $\|\langle r20|\psi\rangle\|^2>0$ and in the $\ell=2$ subspcae $L^2=\ell(\ell+1)=6$. Therefore we must conclude that the our first assumption is false and that the values $L_x=\pm2$ are possible. 
\end{itemize}
Q.E.D.
\item $\ell=1$: 
\[\|\langle L_x=1|\psi\ranlge\|^2=\|\langle \vec{e_x}|\psi\ranlge\|^2=\frac{4\pi}{3}M_4\|\langle\vec{e_x}|\Updownarrow\ranlge\|^2=\frac{1}{4}\]
\end{enumerate}
\end{document} 
