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\markright{\sf  TFY4250 Atom- \&amp; molekylfysikk/FY2045 Kvantefysikk,  løsning øving 10}{}


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\begin{center}{\Large\bf LØSNING ØVING 10}\end{center}


\subsection*{\underline{Løsning oppgave 28}\h{4mm}{\small }}

\pkt{a}  The eigenfunctions of $\hatl_z$ and $\hhat$ are 
 \[ \Phi^{(m)}(\phi)=A\,e^{im\phi},\qquad m=0,\pm1,\pm1,\cdots,\]
where $A$ is a normalisation constant. ($A=1/\sqrt{2\pi}$ gives a normalised wave function.) It is the requirement of a continuous wave function that leads to integer quantum numbers $m$, and thereby the quantization of both $L_z$ and the energy:
 \[ L_z=m\hs,\qquad E^{(m)}=\frac{\hs^2m^2}{2\mu r_0^2}.\]
We note that the degeneracy is equal to  2, except for the ground state, which has \M{m=0} and \M{E^{(0)}=0,}  and for which the wave function is \M{\Phi^{(0)}(\phi)=1/\sqrt{2\pi}. } 

\pkt{b}   Noting that
\[ \hatl_z^2\cos2\phi =-\hs^2\frac{d^2}{d\phi^2}\cos2\phi=-\hs^2(-4\cos2\phi),\]
we see that $\Phi(\phi)$ is an eigenfunction of $\;\hhat=\hatl_z^2/(2mr_0^2)\,$ with the eigenvalue $4\hs^2/(2mr_0^2)$:
 \[ \hhat\Phi(\phi)=\frac{\hatl_z^2}{2mr_0^2}\,\Phi(\phi)=\frac{4\hs^2}{2mr_0^2}\,\Phi(\phi).\]  
The possible measured values of  $L_z$ are (in general) the eigenvalues, i.e., integer multiples of $\hs$. Since $L_z^2$ has the sharp value $4\hs^2$ for the state in question, the values that can be measured here are $L_z=\pm2\hs$. The probabilities are found very simply, by writing the wave function as a superposistion of two eigenfunctions of $\hatl_z$: \M{\Phi^{(\pm2)}=(2\pi)^{-1/2}e^{\pm2i\phi }:} 
 \[\Phi(\phi)=\frac{1}{\sqrt{\pi}}\,\hal(e^{2i\phi}+e^{-2i\phi})=\frac{1}{\sqrt{2}}\Phi^{(+2)}+\frac{1}{\sqrt{2}}\,\Phi^{(-2)}.\]
The coefficients $1/\sqrt{2}$ then are the probability \emph{amplitudes} to measure \M{L_z=\pm2\hs,} respectively, and the probabilities are both equal to  $\hal$. After such a measurement, the system will be left in the eigenstate corresponding to the measured eigenvalue of $L_z$. Thus, if we measure \M{L_z=2\hs,} then the system will be left in the state
 \[ \Phi^{(2)}\equiv\frac{1}{\sqrt{2\pi}}\,e^{2i\phi}.\]

\pkt{c} In the ground state, the energy associated withe the motion along the ring is equal to zero, corresponding to the function \M{\Phi^{(0)}=1/\sqrt{2\pi}.}
For a particle which is confined to a finite region of space, where  \M{V=0,} it should really not be possible to have the energy \M{E=0} (which we get from the  formula above for $m=0$).
On the contrary, it follows from the uncertainty relation that the smaller we make the cross section of the ring, the ``wilder'' the particle gets. The energy then is far from being zero:

(i) With a quadratic cross section \M{L^2=(r_0/100)^2,}  the two-dimensional motion ``along'' this cross section (i.e., perpendicular to the azimuthal direction) is that of a two-dimensional box. This ``perpendicular'' motion contributes to the energy eigenvalues  with an amount
 \[ E^{\perp}_{n_1n_2}=\frac{\hs^2\pi^2}{2\mu L^2}(n_1^2+n_2^2)
=\frac{\hs^2\pi^2}{2\mu r_0^2}(n_1^2+n_2^2)\cdot10^4,\qquad n_1=1,2,\cdots,\quad n_2=1,2,\cdots.\]
The ground state corresponds to \M{n_1=n_2=1} and \M{m=0,}
 and the energy of this state is
\[ E=2\cdot10^4\cdot\frac{\hs^2\pi^2}{2\mu r_0^2}.\]
This energy level is succeded by a long sequence of levels which have \M{n_1=n_2=1,}  while $|m|$ increases. This goes on, to rather large values of $|m|$, before ``perpendicular'' excitations  ($n_1=2,\,n_2=1$ f.ex.) enter the picture. For low energies, the system therefore  effectively is one-dimensional.

(ii) The circular cross section, with radius $r_0/200$, 
 is a little smaller than the quadratic one. From the uncertainty relation we would then expect a ground state energy that is a little larger. This is confirmed by the results from Tillegg 11 for the circular box; the lowest energy for the ``perpendicular'' motion is 17 percent higher than for the quadratic cross section,
\[ \frac{[\hs\Pi^{(0)}_1]^2}{2\mu (r_0/200)^2}=\frac{\hs^2\pi^2}{2\mu r_0^2}\cdot10^4\LT(\frac{2\Pi^{(0)}_1}{\pi}\RT)^2=2.34\cdot10^4\cdot\frac{\hs^2\pi^2}{2\mu r_0^2},\]
since \M{\Pi_1^{(0)}=2.4048.}


\pkt{d} The eigenfunctions of $\hatl_z$ obviuosly are the functions from \PKT{a}:
 \[ \Phi^{(m)}(\phi)=\frac{1}{\sqrt{2\pi}}\,e^{im\phi},\qquad\hatl_z\Phi^{(m)}=\hs m\,\Phi^{(m)},\qquad m=0,\pm1,\cdots.\]
The eigenfunctions of $\hhat_z$ are box solutions for a one-dimensional box with width $2a$, which requires that $k_{n_z}\cdot2a=n_z\pi$:
 \[ \psi_{n_z}(z)=\sqrt{\frac{1}{a}}\sin(n_z\pi z/2a),\qquad E_{n_z}=\frac{\hs^2\pi^2}{2\mu(2a)^2}\,n_z^2.\]
Inserting into \M{(\hhat-E)\psi=(\hhat-E)R(r)\Phi^{(m)}(\phi)\psi_{n_z}(z)=0,} we get
 \ba 0 &amp;=&amp; \LT[-\frac{\hs^2}{2\mu}\LT(\pfrac{^2}{r^2}+\frac{1}{r}\,\pfrac{}{r}\RT)+\frac{\hatl_z^2}{2\mu r^2}- \frac{\hs^2}{2\mu}\,\pfrac{^2}{z^2}-E\RT]R(r)\Phi^{(m)}(\phi)\psi_{n_z}(z)\\
 &amp;=&amp; \LT[-\frac{\hs^2}{2\mu}\LT(\pfrac{^2}{r^2}+\frac{1}{r}\,\pfrac{}{r}\RT)+\frac{\hs^2m^2}{2\mu r^2}+E_{n_z}-E \RT]R(r)\Phi^{(m)}(\phi)\psi_{n_z}(z).\ea
Setting \M{E=E_{n_z}+\hs^2k^2/2\mu} and multiplying with $-2\mu r^2/\hs^2$, we find the radial equation 
 \[ (kr)^2\,\frac{d^2R}{d(kr)^2}+kr\,\frac{dR}{d(kr)}+[(kr)^2-m^2]R=0,\qquad R(a)=0.\]
This is the Bessel equation, which was found page 5 in Tillegg 11. As expected, the energy eigenfunctions therefore are products of the eigenfunctions of $\hhat_z$ and the eigenfunctions we found in Tillegg 11 for the circular two-dimensional box:
 \[ \psi^{(m)}_{nn_z}(r,\phi,z)=A_{nm}J_m(\Pi^{(m)}_nr/a)\,\frac{1}{\sqrt{2\pi}}\,e^{im\phi}\cdot\frac{1}{\sqrt{a}}\sin(n_z\pi z/2a).\]
Here, $A_{nm}$  normalisation constants. 
For each of these states, the energy is a sum of $E_{n_z}$ and the energy related to the perpendicular motion: 
\[ E^{(m)}_{nn_z}=\frac{\hs^2}{2\mu a^2}\LT((\Pi^{(m)}_n)^2+\frac{\pi^2n_z^2}{4}\RT).\]
Here,  $\Pi_n^{(m)}$ are the zeroes of the Bessel function $J_m$. From the table page 6 in Tillegg 11, we then find, in order of increasing energy:
 \[ \barr{llllll}E=8.25(\hs^2/(2\mu a^2))\;\; &amp;\;\; n=1 \;\;&amp;\;\; n_r=0\;\; &amp;\;\; m=0\;\; &amp;\;\; n_z=1\;\; &amp; \;\;\mbox{(ground state)}\\
E=15.65(\hs^2/(2\mu a^2))\;\; &amp;\;\; n=1 \;\;&amp;\;\; n_r=0\;\; &amp;\;\; m=0\;\; &amp;\;\; n_z=2\;\; &amp; \;\;\mbox{(1. excited level)}\\

E=17.15(\hs^2/(2\mu a^2))\;\; &amp;\;\; n=1 \;\;&amp;\;\; n_r=0\;\; &amp;\;\; m=\pm1\;\; &amp;\;\; n_z=1\;\; &amp; \;\;\mbox{(2. excited level)}\\
E=24.55(\hs^2/(2\mu a^2))\;\; &amp;\;\; n=1 \;\;&amp;\;\; n_r=0\;\; &amp;\;\; m=\pm1\;\; &amp;\;\; n_z=2\;\; &amp; \;\;\mbox{(3. excited level)}\\
E=27.99(\hs^2/(2\mu a^2))\;\; &amp;\;\; n=1 \;\;&amp;\;\; n_r=0\;\; &amp;\;\; m=0\;\; &amp;\;\; n_z=3\;\; &amp; \;\;\mbox{(4. excited level)}\\
E=28.84(\hs^2/(2\mu a^2))\;\; &amp;\;\; n=1 \;\;&amp;\;\; n_r=0\;\; &amp;\;\; m=\pm2\;\; &amp;\;\; n_z=1\;\; &amp; \;\;\mbox{(5. excited level)}\\
E=32.94(\hs^2/(2\mu a^2))\;\; &amp;\;\; n=2 \;\;&amp;\;\; n_r=1\;\; &amp;\;\; m=0\;\; &amp;\;\; n_z=1\;\; &amp; \;\;\mbox{(6. excited level)}
\earr\]
Here, $n_r$ is the radial quantum number, i.e., the number of zeroes in the radial function.


\subsection*{\underline{Løsning oppgave 29}\h{4mm}{\small Coulomb problem in ``half-space''}}

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In the half-space \M{z&gt;0} the time-independent Schr\"odinger equation has the same form as for a hydrogenlike system,
  \[ \LT[-\frac{\hs^2}{2m}\,\nabla^2-\frac{e^2}{4\pi\epsilon r}\RT]
  \psi(r,\theta,\phi)=E\,\psi(r,\theta,\phi). \]
We know the solutions $\psi_{nlm}$ of this equation, and we know that it has no other square-integrable solutions for \M{E&lt;0.} What is new is only the boundary condition \M{\psi=0} for \M{z\leq0.} This means that not all the solutions $\psi_{nlm}$ are applicable. Only those that are zero in the $xy$-plane, that is, for \M{\theta=\hal\pi,} can be used.
By consulting a table of spherical harmonics, we find at once that all $s$-states ($l=0$) are ruled out, like $\psi_{100}$. However, 
\M{Y_{10},\,Y_{2,\pm1},\,
Y_{3,\pm2},} etc are zero for \M{\theta=\hal\pi.}
Thus the smallest  $l$-value giving an acceptable solution is \M{l=1.} 
Therefore we can only use solutions with principal quantum number $\;n\geq2$.
Among these solutions the one with the lowest energy is
  \[ \psi_{210}=R_{21}(r)Y_{10}(\theta,\phi)
  =(32\pi a^5)^{-1/2}\cos\theta\;re^{-r/2a}\;\;;
  \;\;\;\;a="a_0"=\frac{\epsilon}{\epsilon_0}\,a_0. \]
This is the ground state in the present potential, and its energy is
  \[ E_{\rm gr.st.}=-\hal("\alpha")^2\frac{m_ec^2}{2^2}\equiv
  -\hal\LT(\frac{e^2}{4\pi\epsilon\hs c}\RT)^2\frac{m_ec^2}{4}
  =(-13.6\,eV)\cdot\frac{1}{4}\frac{\epsilon_0^2}{\epsilon^2}, \]
if we forget the complications due to the effective mass of the electron; see Hemmer p 111.

Note that if we want a normalized wave function, we must add an extra factor  $\sqrt{2}$ in the formula above, since the wave function is restricted to \M{z&gt;0.}




\subsection*{\underline{Løsning oppgave 30A}\h{4mm}{\small Compatible and incompatible observables}}

\pkt{a} We want to show the operator identity \M{[\hat{A},\hat{B}]=0,} or in other words, that the commutator gives zero applied to an arbitrary function $\psi$:
 \[ [\hat{A},\hat{B}]\psi=(\hat{A}\hat{B}-\hat{B}\hat{A})\sum_nc_n\psi_n=\sum_nc_n(a_nb_n-b_na_n)\psi_n=0,\qed\]

\pkt{b} See Hemmer p 88, or Griffiths p 146, how to derive the {\bf angular momentum algebra},
   \[ [\hat{L}_x,\hat{L}_y]=i\hs\hat{L}_z,\]
   \[ [\hat{L}_y,\hat{L}_z]=i\hs\hat{L}_x,\]
   \[ [\hat{L}_z,\hat{L}_x]=i\hs\hat{L}_y.\]
Since these operators do not commute, they cannot have simultaneous eigenfunctions. Thus the observables $\,L_x,\,L_y$ and $L_z$ are not compatible; they cannot have sharp values simultaneously.

\pkt{c}  The eigenvalue \M{\ve{L}^2=2\hs^2} corresponds to an angular-momentum quantum number \M{l=1.} A good indication is obtained by considering the special case \M{\hatn=\hatz.} Then the possible values (of  $\,\vel\prikk\hatn=\vel\prikk\hatz=L_z\,$) are $\,0,\,\pm\hs$. The next thing we have to realize is that there is really nothing special about the $\hatz$-direction. So, if we choose instead to measure $L_x$ or $L_y$, the results still will be $\,0,\,\pm\hs$. \footnote{For each of these eigenvalues, it is also possible to find the respective eigenfunctions. A way to find them is to express the three eigenfunctions $Y_{1m}$ of $L_z$ in terms of $x,y,z$ and then change variables in a ``cyclic'' manner. Try it, and check that they are eigenfunctions!}
And the same holds for any component $\ve{L}\prikk\hatn$ of $\ve{L}$;
the possible eigenvalues of \emph{any} component of $\ve{L}$  are the same as for $L_z$, namely \M{\hatn\prikk\ve{L}=0,\pm\hs.}
 If you think this is strange, I remind you that the systems we  consider (for example the hydrogen atom) is  described by a  spherically symmetric potential. Then there can be no preferred directions in space.



\subsection*{\underline{Løsning Oppgave 30B}\h{4mm}{\small Angular momentum $1$}}

\pkt{a} With \M{\exp(\pm i\phi)=\cos\phi\pm i\sin\phi} we have
  \[ Y_{1\pm1}=\mp\sqrt{\frac{3}{8\pi}}\,\sin\theta(\cos\phi\pm i\sin\phi)
= \mp\sqrt{\frac{3}{8\pi}}\,\frac{x\pm iy}{r}.\]

\pkt{b} With the substitution \M{ z\rpil x,\quad x\rpil y,\quad y\rpil z,} we get the three functions
 \[ X_{10}=\sqrt{\frac{3}{4\pi}}\,\frac{x}{r}=\sqrt{\frac{3}{4\pi}}\,\sin\theta\cos\phi\quad(\equiv p_x),\qquad X_{1\pm1}=\mp\sqrt{\frac{3}{8\pi}}\,\frac{y\pm iz}{r}=\mp\sqrt{\frac{3}{8\pi}}\,(\sin\theta\sin\phi\pm i\cos\theta).\]

\pkt{c} We find
 \ba \hatl_xX_{10} &amp;=&amp; \frac{\hs}{i}\LT(-\sin\phi\,\pfrac{}{\theta}-\cot\theta\,\cos\phi\,\pfrac{}{\phi}\RT)\sqrt{\frac{3}{4\pi}}\,\sin\theta\cos\phi\\
 &amp;=&amp; \sqrt{\frac{3}{4\pi}}\;\frac{\hs}{i}\LT[-\sin\phi\,\cos\theta\,\cos\phi-\frac{\cos\theta}{\sin\theta}\,\cos\phi\,\sin\theta (-\sin\phi)\RT]=0,\ea
and
 \ba \hatl_xX_{1\pm1} &amp;=&amp; \mp\sqrt{\frac{3}{8\pi}}\;\frac{\hs}{i}\LT(-\sin\phi\,\pfrac{}{\theta}-\cot\theta\,\cos\phi\,\pfrac{}{\phi}\RT)(\sin\theta\sin\phi\pm i\cos\theta)\\
&amp;=&amp;  \mp\sqrt{\frac{3}{8\pi}}\;\frac{\hs}{i}
\LT[-\sin\phi(\cos\theta\,\sin\phi\pm i(-\sin\theta))-\frac{\cos\theta}{\sin\theta}\,\cos\phi\,\sin\theta\,\cos\phi\RT]\\
&amp;=&amp; \pm\hs\LT[\mp\sqrt{\frac{3}{8\pi}}\,(\sin\theta\sin\phi\pm i\cos\theta)\RT]
=\pm\hs\,X_{1,\pm1}.\ea
Thus as expected we find the same eigenvalues as for the set $Y_{1m}$. 

[Comment:  A rotation by 45 degrees would give a similar set of function with the same form with respect to an axis $\hatn$ pointing midway between $\hatx$ and $\hatz$. These functions of course are eigenfunctions of $\ve{S}\prikk\hatn$ with the same eigenvalues as above.]

\pkt{d}  Measurements of $\ve{L}^2$ and $L_x$ will leave the system in eigenstates of the corresponding operators, according to the measurement postulate. If such measurements are made on an ensemble of systems, we can single out those systems for which we find \M{\vel^2=2\hs^2} and \M{L_x=0,} respectively. The resulting ensemble will then be in the state $X_{10}$. 

\pkt{e}  Since $X_{10}$ is a superposition of $Y_{11}$ and $Y_{1-1}$, a measurement of $L_z$ on this ensemble will give either \M{L_z=\hs} or \M{L_z=-\hs.} The probability amplitudes and the probabilities  for these two results are
 \[ c_{\pm}=\FV{Y_{1\pm1},X_{10}}=\mp1/\sqrt{2}\quad\mbox{ and }\quad P_{\pm \hs}=|c_{\pm}|^2=\hal,\]
according to the rule (T3.31) in Tillegg 3. See also section 2.5.2 in Hemmer. 
 

\pkt{f} A measurement giving \M{L_z=\hs} leaves the system in $Y_{11}$. The probability to measure \M{L_x=\hs} then is the square of the projection of $Y_{11}$ on  $X_{11}$ (according to the rule just mentioned),
\[ P_{+\hs}=|\FV{X_{11},Y_{11}}|^2=|\FV{Y_{11},X_{11}}|^2=|-i/2|^2=1/4.\]
(Here we have used the fact that $\;\FV{\psi_2,\psi_1}=\FV{\psi_1,\psi_2}^{\Ast}$ for a scalar product.)



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\subsection*{\underline{Løsning Oppgave 30}\h{4mm}{\small Ladder operators for angular momentum}}

\pkt{a} Calculating the commutator
 \ba [\hatl_z,\hatl_{\pm}] &amp;=&amp; [\hatl_z,\hatl_x\pm i\hatl_y]=[\hatl_z,\hatl_x]\pm i[\hatl_z,\hatl_y]\\
 &amp;=&amp; i\hs\hatl_y\pm i(-i\hs\hatl_x)=\pm\hs\hatl_{\pm},\ea
we have
 \[ \hatl_z\hatl_{\pm} = \hatl_{\pm}\hatl_z\pm\hs\hatl_{\pm},\qed\]
It is also easy to show that
 \ba \hatl_{\pm}\hatl_{\mp} &amp;=&amp; (\hatl_x\pm i\hatl_y)(\hatl_x\mp i\hatl_y)\\
 &amp;=&amp; \hatl_x^2+\hatl_y^2\mp i(\hatl_x\hatl_y-\hatl_y\hatl_x)=\hllve-\hatl_z^2\mp i(i\hs\hatl_z)\\
 &amp;=&amp; \hllve-\hatl_z^2\pm\hs\hatl_z,\qed\ea

\pkt{b} Let us start with the bottom rung of the ladder, \M{Y_{1-1}=\sqrt{3/8\pi}\,\sin\theta \exp(-i\phi)} and illustrate how $\hatl_+$ works:
 \ba \hatl_+Y_{1-1} &amp;=&amp; \frac{\hs}{i}\,e^{i\phi}\LT(i\,\pfrac{}{\theta}-\cot\theta\,\pfrac{}{\phi}\RT)\sqrt{\frac{3}{8\pi}}\,\sin\theta\,e^{-i\phi}\\
 &amp;=&amp; \frac{\hs}{i}\,e^{i\phi}\sqrt{\frac{3}{8\pi}}\LT[i\cos\theta-\cot\theta\,\sin\theta(-i)\RT]e^{-i\phi}=\hs\sqrt{2}\,Y_{10}.\ea
This agrees with the general formula (for $l=1,\,m=-1$). Next,
 \ba \hatl_+Y_{10} &amp;=&amp; \frac{\hs}{i}\,e^{i\phi}\LT(i\,\pfrac{}{\theta}-\cot\theta\,\pfrac{}{\phi}\RT)\sqrt{\frac{3}{4\pi}}\,\cos\theta\\
 &amp;=&amp; \frac{\hs}{i}\,e^{i\phi}\sqrt{\frac{3}{4\pi}}\LT[i(-\sin\theta)\RT]=\hs\sqrt{2}\,Y_{11},\ea
which is also agrees with the general formula. Let us also try
    \ba \hatl_+Y_{11} &amp;=&amp; \frac{\hs}{i}\,e^{i\phi}\LT(i\,\pfrac{}{\theta}-\cot\theta\,\pfrac{}{\phi}\RT)\LT(-\sqrt{\frac{3}{8\pi}}\,\sin\theta\,e^{i\phi}\RT)\\
 &amp;=&amp; \frac{\hs}{i}\,e^{i\phi}\LT(-\sqrt{\frac{3}{8\pi}}\RT)\LT[i\,\cos\theta-\cot\theta\,\sin\theta\cdot i\RT]e^{i\phi}=0.\ea
The same thing happens if you apply $\hatl_-$ to $Y_{1-1}$. If you apply $\hatl_-$ to $Y_{11}$, you get $\hs\sqrt{2}$ times $Y_{10}$, etc.






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