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\heading{E464: Angular momentum of wave function} 
\auname{Mor Yaakobi}


{\bf The problem:}

\Dn

A particle is described by the wave function ${\psi (x,y,z) = f(r)z(z+1)}$.

\Dn

(1) Express the wave function in terms of ${\ |r, \theta, \phi \rangle }$.

(2) Express the wave function in terms of ${\ |r, l, m \rangle }$.

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

\Dn
 
Let us define  ${\  M_{n} = \int\limits_0^\infty r^n(f(r))^2dr }$

(3) Write a condition for ${\ M_{4} \ , M_{6} }$ so that the wave function will be normalized.

Given ${\ M_{6}= \frac {5}{8\pi}} }$ 

(4) What is the probability of finding the particle to be in a spherical state? ${ (l=0) }$

(5) What are the possible values for measuring ${ \ L_{x} }$ ?

(6) What are the possible values for measuring ${ \ L_{z} }$ ?

(7) What is the probability of finding the particle to be in a polar state ${ (l=0) }$,  for ${ L_{x}=1 }$? 


\Dn\Dn


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{\bf The solution:}

\Dn

(1) 
Converting cartesian to spherical coordinates one gets

\[\psi(r,\theta,\phi) = f(r)r^2cos{^2}\theta + f(r)rcos\theta \]

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\Dn

(2) 
Using the orthonormal basis of the Hilbert space ${ Y_{lm}}$  one gets

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

For simplicity, let us write the above expression schematically:
\[\psi(r,l,m) = f_{0}(r) Y^{00} +f_{1}(r) Y^{10} + f_{2}(r) Y^{20}  \]

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\Dn

(3) 
Normalization requirement (using the schematic expression):
\[\ \langle\psi| \psi \rangle = \sum\limits_{i=0}^2 \int |f_{i}(r)|^2r^2dr 
= P_{0}+P_{1}+P_{2} = \frac{4\pi}{9}M_{6} + \frac{4\pi}{3}M_{4} + \frac{16\pi}{45}M_{6} = 1 \]
 
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\Dn

(4) Let us sum over the probabilities of being in a spherical state $ {(l=0) }$
\[\ P =|\langle f_{0}(r)Y^{00} | \psi \rangle|^2 = P_0 =  \frac{4\pi}{9}M_{6} =  \frac{5}{18} \]

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\Dn

(5) The possible values of ${L_{x}}$ regarding the above wave function are ${-2, -1, 0, 1, 2}$.
In order to find the values whose probability to be measured is bigger than 0, we will check if the aforementioned values are eigenvalues of ${L_{x}}$. 
\[\ L_{x}=\frac{1}{2} (L_{+}+L_{-}) \]
Let us examine each term of ${Y^{lm} }$ separately:
\Dn

${Y^{00} }$ is an eigenstate of ${L_{x} }$, with an eigenvalue of 0, thus is probable to be measured.

Apply ${L_{x} }$ on the states ${Y^{1,m} }$, one gets the following eigenstates:
\[\ {Y_{x}^{11}\mapsto \frac{1}{\sqrt{2}}\left( \amatrix { 1 \cr \sqrt{2} \cr 1 }\right)     
\;\;{Y_{x}^{10}\mapsto \frac{1}{\sqrt{2}}\left( \amatrix { 1 \cr 0 \cr -1 }\right)  
\;\;{Y_{x}^{1-1}\mapsto \frac{1}{\sqrt{2}}\left( \amatrix { 1 \cr -\sqrt{2} \cr 1 }\right) \]
with the eigenvalues of ${1,0,-1}$ accordingly.
\[\ {Y^{10}=\frac{1}{\sqrt{2}}({Y_{x}^{11}-{Y_{x}^{1-1}) \]
Thus ${-1 , 1}$ are probable to be measured.

Remains find the representation of ${Y^{20} }$ in the basis where ${L_{x} }$ is diagonal. Using the same method, one gets
\[\ {Y_{x}^{22}\mapsto \frac{1}{2}\left( \amatrix { 1 \cr 2 \cr \sqrt{6} \cr 2 \cr 1 }\right)     
\;\;{Y_{x}^{20}\mapsto \frac{1}{2}\left( \amatrix 
                                            { 1 \cr 0 \cr -\frac{2}{\sqrt{6}} \cr 0 \cr 1 }\right)
\;\;{Y_{x}^{2-2}\mapsto \frac{1}{{2}}\left( \amatrix { 1 \cr -2 \cr \sqrt{6} \cr -2 \cr 1 }\right) \]
with the eigenvalues of ${2,0,-2}$ accordingly.
\[\ {Y^{20}=c_{1}{Y_{x}^{22}+c_{2}{Y_{x}^{22}+c_{2}{Y_{x}^{2-2} \]
Thus ${-2 , 2}$ are also probable to be measured.

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\Dn

(6)  Using ${ r, l, m }$ representation, ${  L_z}$ is determined to be diagonal. it follows that
${ \ L_{z}|m \rangle  = \m|m \rangle }$ , and therefore the only possible value for measuring 
{$ L_{z} $} is 0.  

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

(7) The probability of measuring ${ L_{x} = 1 }$ when the particle is in a polar state  ${( l=1) }$ is:
\[\  P =|\langle \Uparrow | \Updownarrow \rangle|^2 \times |\langle f_{1}(r)Y^{10} | \psi \rangle|^2 = |\langle \Uparrow | \Updownarrow \rangle|^2 \times P_1 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]


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