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\heading{Ex464: Angular momentum of wave function} 
\auname{Alexander Elikashvili}


{\bf The problem:}

\Dn

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

\Dn

(1) Write the wave function in ${| r,  \theta,  \phi \rangle }$ representation.

(2) 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

${M_n=\int_0^\infty r^{n}(f(r))^2 dr}$, \, ${M_6=\frac{5}{8\pi}}$
	
\Dn 
    
(3) Write a condition on ${M_6}$ and ${M_4}$ so that the wave function will be normalized.

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

(5) What are the possible non-zero probability values of ${L_x}$?   

(6) What are the possible non-zero probability values of ${L_z}$? 

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

\Dn\Dn


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{\bf The solution:}

\Dn

(1) 
Writing  the wave function in spherical coordinates one gets:

\[  \langle r,\theta, \phi| \Psi \rangle = f(r)r^2 \cos^2(\theta)+f(r)r \cos(\theta)= \sqrt{\frac{16 \pi}{45}}r^2f(r)Y^{20}(\theta, \phi)- \sqrt{\frac{4 \pi}{9}}r^2f(r)Y^{00}(\theta, \phi) + \sqrt{\frac{4 \pi}{3}}rf(r)Y^{10}(\theta, \phi) \]


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\Dn

(2)
\[ \langle r, \ell,m| \Psi \rangle=\int \langle r,\ell,m|r',\theta,\phi \rangle \langle r', \theta, \phi| \Psi \rangle r'^{2}dr'd\Omega= \sqrt{\frac{16 \pi}{45}}r^{3}f(r) \delta_{\ell 2} \delta_{m0}- \sqrt{\frac{4 \pi}{9}} r^{3}f(r) \delta_{\ell 0} \delta_{m0}+ \sqrt{\frac{4 \pi}{3}} r^{2}f(r) \delta_{\ell 1} \delta_{m0} \] 


\Dn

Or in abstract form:

\[ |\Psi \rangle = \sum_{r}( \psi_{20}(r) |r,2,0 \rangle + \psi_{00}(r) |r,0,0 \rangle + \psi_{10}(r) |r,1,0 \rangle ) \]
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\Dn

(3)
\[ \langle \Psi | \Psi \rangle = \sum_{r, \ell,m}| \langle r, \ell, m | \Psi \rangle|^2= \frac{16 \pi}{45} M_{6}+\frac{4 \pi}{9} M_{6}+\frac{4 \pi}{3} M_{4}=1 \]
\Dn

By substituting the value of ${M_6}$ in the above expression one finds: ${ M_4=\frac{3}{8\pi} }$.           


\Dn



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\Dn

(4)

\[ P= \sum_{r} | \langle r,0,0| \Psi \rangle|^{2}= \frac{4 \pi}{9} \int_{0}^{\infty} r^6 f^2(r)dr=\frac{4 \pi}{9} M_{6}= \frac{5}{18}  \]


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\Dn

(5)
 This section will be solved by examining each sub-space (${ \ell=0, \ell=1, \ell=2}$) separately. 
\Dn

 In ${ \ell=0}$ sub-space:
\Dn

It is trivial case when ${L_{x}=0}$ is the only value.
\Dn
\Dn

 In ${ \ell=1}$ sub-space:
\Dn

Notations used:
\Dn

{\bf Standart basis:} ${|m_{x}=-1 \rangle,|m_{x}=0 \rangle,|m_{x}=1 \rangle}$ \; { \bf Linear basis:} ${|m_{x}=0 \rangle,|m_{y}=0 \rangle,|m_{z}=0 \rangle}$

\[ | \Psi \rangle \propto |m_{z}=0 \rangle= \langle m_{x}=-1 | m_{z}=0 \rangle | m_{x}=-1 \rangle + \langle m_{x}=1 | m_{z}=0 \rangle | m_{x}=1 \rangle \]

\Dn

If ${\langle m_{x}=-1 | m_{z}=0 \rangle=0}$ it would mean that ${|m_{z}=0 \rangle \propto | m_{x}=1 \rangle }$
\Dn

If ${\langle m_{x}=1 | m_{z}=0 \rangle=0}$ it would mean that ${|m_{z}=0 \rangle \propto | m_{x}=-1 \rangle}$
\Dn

If ${\langle m_{x}=-1 | m_{z}=0 \rangle=0}$ and ${\langle m_{x}=1 | m_{z}=0 \rangle=0}$ it would mean that ${|m_{z}=0 \rangle=0}$
\Dn

All three cases above are impossible, hence ${ \langle m_{x}=-1| \Psi \rangle \neq 0}$ and ${\langle m_{x}=1| \Psi \rangle \neq 0}$, or in other words ${L_{x}= \pm 1}$ has non-zero probability. 
\Dn
\Dn

 In ${ \ell=2}$ sub-space:
\[ \langle L^{2} \rangle = \langle L_{x}^{2} \rangle + \langle L_{y}^{2} \rangle + \langle L_{z}^{2} \rangle = \ell (\ell + 1)=6 \]
Notice that ${\langle L_{z}^2 \rangle = 0}$ because ${\hat{L}_{z} | \Psi \rangle = 0| \Psi \rangle}$ and ${\langle L_{x}^{2} \rangle = \langle L_{y}^{2} \rangle}$ because of symmetry considerations, therefore ${ \langle L^{2} \rangle =2 \langle L_{x}^{2} \rangle }$, which implies that ${\langle L_{x}^{2} \rangle=3}$.
\Dn

Now suppose that getting ${L_{x}= \pm 2}$ has zero probability, then one would get:
\[ \langle L_{x}^{2} \rangle=(-1)^2P(L_{x}=-1)+1^2P(L_{x}=1)<3 \]
Therefore values ${L_{x}= \pm 2}$ must have non-zero probabilities.
Final conclusion implies that ${L_{x}}$ can get values -2,-1,0,1,2.


\Dn

(6) 
\[ \hat{L}_{z} | \Psi \rangle =\sum_{r}( \psi_{20}(r) \hat{L}_{z} |r,2,0 \rangle + \psi_{00}(r) \hat{L}_{z} |r,0,0 \rangle + \psi_{10}(r) \hat{L}_{z} |r,1,0 \rangle ) = 0| \Psi \rangle \]
\Dn

(7) 
\[ \langle r, \ell=1, m_{x}=1 | \Psi \rangle = \sum_{r',\ell',m_{z}} \langle r, \ell=1, m_{x}=1|r', \ell', m_{z} \rangle \langle r', \ell', m_{z}| \Psi \rangle=\psi_{10}(r) \langle m_{x}=1|m_{z}=0 \rangle \]
One can express ${ \langle m_{x}=1|= \frac{1}{2}(1, \sqrt{2}, 1)}$ and ${ |m_{z}=0 \rangle = \left( \amatrix { 0 \cr 1 \cr 0 }\right)}$, which implies ${\langle m_{x}=1|m_{z}=0 \rangle = \frac{1}{\sqrt{2}}}$, also from section (2) it is clear that ${\psi_{10}(r) = \sqrt{\frac{4 \pi}{3}} r^{2}f(r)}$.
\[ P= \sum_{r}^{\infty}|\langle r, \ell=1, m_{x}=1| \Psi \rangle|^2= \int_{0}^{\infty} \frac{4 \pi}{3}}r^4 f^2(r) |\langle m_{x}=1|m_{z}=0 \rangle|^2 dr= \frac{2 \pi}{3}M_{4}= \frac{1}{4} \] 
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