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\heading{Ex464: Angular momentum of a wave function} 
\auname{Yaki Ben-Yakar, corrections by DC}


{\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 the ${| r  \theta  \phi \rangle }$ basis.

(2) Write the wave function in the ${| r  l  m \rangle }$ basis.
  
\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 
    
(3) write a condition on ${M_6}$ and ${M_4}$ so that the wave function will be normalized,Given ${M_{n}=\int^{\infty}_{0}r^n(f(r))^2dr}$ and ${M_6=\frac{5}{8\pi}}$. find what is ${M_4}$.

\Dn


(4) Find the probability of the particle to be in a spherical state (l=0).

(5) What are the possible values in the measurement of ${L_x}$?   

(6) What are the possible values in the measurement of ${L_z}$? 

(7)  Find the probability of the particle to be in a polar state (l=1) with ${L_x=1}$.
 

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
By changing variables from cartesian cordinates to spherical cordinates one gets:

\[  \Psi(r,\theta,\phi) = f(r)r^2 \cos^2(\theta)+f(r)r \cos(\theta) \]


%%%%%%%%%%%%%%%
\Dn

(2)
Using the ${Y^{lm}}$ for the basis of Hilbert space one get:


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



\Dn

{\LARGE Note by DC:} Let us write it schemtically as  

\[  
\Psi(r,\theta,\phi)=  f_{0}(r) Y^{00} +  f_{1}(r) Y^{10} + f_{2}(r)  Y^{20}
\]


%%%%%%%%%%%%%%
\Dn

(3)
Normalizing the given wave function one get the wanted condition:
\[1= \langle \Psi | \Psi \rangle = \int_{all space}\Psi^*(r,\theta,\phi)\Psi(r,\theta,\phi)r^2drd\Omega= \frac{4\pi}{5}\int^{\infty}_{0}r^6(f(r))^2dr+ \frac{4\pi}{3}\int^{\infty}_{0}r^4(f(r))^2dr=\frac{4\pi}{5}M_6+\frac{4\pi}{3}M_4 \]
\Dn

If substitute ${M_6}$ in the above equation it is easy to find that: ${ M_4=\frac{3}{8\pi} }$.           


\Dn

{\LARGE Note by DC:} Schematically the calculation here is  

\[  
\mbox{Normalization}  
\ \ = \ \  \sum_{\ell=0}^{3}  \int |f_{\ell}(r)|^2 r^2dr  
\ \ =  \ \ p_0 + p_1 + p_2  \ \ = \ \ \frac{4\pi}{9}M_6 + \frac{4\pi}{3}M_4 + \frac{16\pi}{45}M_6   \ \ = \ \ 1 
\]



\newpage
%%%%%%%%%%%%%
\Dn

(4)
Finding the probability is as finding the projection of the given state on the wantet state to find, i.e in our case ${\Phi=Y^{00}}$. so we get: 

\[ P= |\langle \Phi | \Psi \rangle|^2 =\int_{all space} |\Phi^*(\theta,\phi)\Psi(r,\theta,\phi)|^2r^2drd\Omega=\frac{4\pi}{9}M_6 =\frac{5}{18}  \]


\Dn

{\LARGE Note by DC:} The above way of writing is meaningless.
what we have here is simply a sum over probabilities: 

\[ 
P \ \ = \ \ \sum_r |\langle r , \ell{=}0 | \Psi \rangle|^2 \ \ = \ \ p_0 
\]



%%%%%%%%%%%%%
\Dn

(5)
Using the same idea as in the last section we want to find the projection of our wave function on ${L_x}$. The given wave function consist three angular parts ${Y^{00}}$,${Y^{10}}$ and
${Y^{20}}$. It is obvious that the ${Y^{00}}$ can contribute only the value of ${L_x=0}$ becouse ${l}$ is equal to zero. Moving to the next term ${Y^{10}}$, we can express this term in the linear basis, so ${Y^{10}=\frac{1}{\sqrt{2}}(-Y_{x}^{11}+Y_{x}^{1-1})}$, from here we see that the possible values of ${L_x}$ is -1 and 1. The last term is quite difficult to handle becouse it take a lot of algebra to express ${Y^{20}}$ in another basis just as we did before,but we know that it cannot exceed the value of ${l=2}$ and we allready got the values -1,0,1 so all we have to check is if it is possible to get also the values 2 and -2.
We can proof that by negation (i.e "stira") assuming that the valuse 2 and -2 are not possible. 
We do know that on the one hand ${\langle L^2 \rangle=2(2+1)=4}$ while on the other hand ${\langle L^2 \rangle=\langle L_x^2 \rangle+\langle L_y^2 \rangle+\langle L_z^2 \rangle}$.   


Becouse of symmetry ${\langle L_x^2 \rangle=\langle L_y^2 \rangle}$ so we get ${\langle L^2 \rangle = 2\langle L_x^2 \rangle}$ which equel 4,in our problem ${\langle L_z^2 \rangle=0}$, and from our assuming ${\langle L_x^2 \rangle}$ can get the values 0 or 1,therefore ${\langle L^2 \rangle\leq2}$ in contra to ${\langle L^2 \rangle=4}$ and we get a negation.

\Dn
So the possible values in the system is ${L_x=-2,-1,0,1,2}$.

\Dn

{\LARGE Note by DC:} The argumentation why for $Y^{20}$  there is non-zero probability 
to measure $L_x=\pm2$ is with typos. The correct phrasing is: 
For this state we have  ${\langle L^2 \rangle= (2+1)2=6}$. 
On the other hand from XY symmetry consideration  
${\langle L^2 \rangle=\langle L_x^2 \rangle+\langle L_y^2 \rangle+\langle L_z^2 \rangle = 2 \langle L_x^2 \rangle}$.  
The latter implies ${\langle L_x^2 \rangle= 3}$,  
which cannot be true unless $L_x=\pm2$ has non-zero probability. 


\Dn
\Dn

(6) It is obviouse that the given wave function is an  eigenvector of the operator ${L_z}$ so we can see that the only Possible value that ${L_z}$ can get is:

\[ L_z=0 \]

\Dn

(7) 
The probability of finding the particle in ${l=1}$ and ${L_x=1}$ is:

\[ 
P(l=1,L_x=1)=|\langle l=1, L_x=1 | \Psi \rangle|^2
=|\langle l=1,L_x=1 | l=1,m=0 \rangle\langle l=1,m=0 | \Psi \rangle|^2= \] 

\[ 
=|\langle \Uparrow | \Leftrightarrow \rangle|^2\times\int|(Y^{10})^*\Psi|^2 d^3r=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4} 
\]


\Dn

{\LARGE Note by DC:} The calculation here is 

\[ 
P \ \ = \ \ |\langle \Rightarrow  | \Updownarrow \rangle|^2 \times p_1
\]

and it is useful to observe that  $|\langle \Rightarrow | \Updownarrow \rangle|^2 = |\langle \Uparrow | \Leftrightarrow \rangle|^2 =1/2$, 
where the latter equality is implied by the standard representation of linear polarization along X.   



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