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\markright{\sf FY2045/TFY4250 Quantum mechanics I,   exercise  6}{}


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\begin{center}{\Large\bf  EXERCISE 6}\end{center}

\begin{quote}
Problem 6 -- 4 differs somewhat from the Norwegian version, but that shouldn't matter much.
\end{quote}

\subsection*{\underline{Problem 6 -- 1}\h{4mm}{\small Fermi momenta and Fermi energies}}

\pkt{a} Suppose that an ideal gas of (non-interacting) spin-$\hal$-fermions 
 are in the ground state in a volume $V$, such that all states inside a sphere in the three-dimensional momentum space are occupied, while all states outside the sphere are vacant. Use the formula
 \[ dN_{\rm rom}=\frac{V\,d^3p}{h^3}\qquad (\mbox{\# of spatial states})\]
to show that the  radius of the  Fermi sphere is
 \[ p_F=\hs\LT(3\pi^2\frac{N}{V}\RT)^{1/3} \qquad(\mbox{Fermi momentum}),\]
where $N/V$ is the number of fermions per unit volume. [Note that Pauli's exclusion principle allows two spin-$\hal$-fermions to occupy each spatial state.]

\pkt{b} The ideal  Fermi-gas-model gives a  (simplified) description of conduction electrons in metals. Show that the Fermi momentum of these electrons are non-relativistic (in the sense that $p_F&lt;&lt;m_ec$), when you are informed that the number densities ($N/V$) of conduction electrons vary between $\;0.91\cdot10^{22}\,$cm$^{-3}=0.91\cdot10^{28}\,$m$^{-3}$ for cesium and $\,24.7\cdot10^{22}\,$cm$^{-3}\,$ for beryllium. Given: $\hs/(m_ec)=3.862\cdot10^{-13}\,$m.

\pkt{c} Find the  Fermi energy for beryllium in elektron volts, given that $\,\hs^2/(2m_ea_0^2)$(=1\,Ry)$\newline $=13.6\,eV, and $\,a_0=0.529\cdot10^{-10}\,$m.

\pkt{d} The electron density in a white-dwarf star is roughly a factor $10^6$ higher than in ordinary matter. Suppose (for simplicity) that it is a factor $10^6$ higher than the one used  for beryllium in the core of the star. Calculate $p_F/(m_ec)$ in this case, and decide if the Fermi momentum in this case is non-relativistic.

\subsection*{\underline{Problem 6 -- 2}\h{4mm}{\small Fermions in spherical box }}

Consider a spherical box with radius $a$ (with\M{V=0} inside the box and $V=\infty$ outside), containing 18 non-interacting  spin-$\hal$-fermions with mass $m$.

\pkt{a} Suppose that this system is in the ground state, and find the quantum numbers of the occupied one-particle states and the energies of these states in units of  $\hs^2/(2ma^2)$. Show that the energy of the ground state is $473\;\hs^2/(2ma^2)$.

\pkt{b} How much energy does it cost to reduce the radius $a$ of the box by 0.1 \%?

\pkt{c} Explain why the total probability density of the fermions in the ground state is spherically symmetric. Hint:
 \[ \sum_{m=-l}^{+l}|Y_{lm}|^2=\frac{2l+1}{4\pi}.\]
\noindent Given: Table of zeros  $\Pi_n^{(l)}$ of the spherical Bessel functions:
 \ba
\barr{lllll}\qquad &amp;\qquad j_0 &amp; \qquad j_1 &amp; \qquad j_2 &amp; \qquad j_3\\
n_r=0\;\; &amp;\Pi^{(0)}_1=\pi\;\;\;  &amp; \Pi^{(1)}_1=4.4934\;\;\;  &amp; \Pi^{(2)}_1=5.7635\;\;\;  &amp; \Pi^{(3)}_1=6.9879  \\
n_r=1 &amp;\Pi^{(0)}_2=\pi\cdot2  &amp; \Pi^{(1)}_2=7.7253  &amp; \Pi^{(2)}_2=9.0950  &amp; \Pi^{(3)}_2=10.4171  \\
n_r=2 &amp;\Pi^{(0)}_3=\pi\cdot3  &amp; \Pi^{(1)}_3=10.9041  &amp; \Pi^{(2)}_3=12.3229  &amp; \Pi^{(3)}_3=13.6980  \earr\ea

\subsection*{\underline{Oppgave 6 -- 3}\h{4mm}{\small Scattering on two $\D$-barriers}}
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We now return to the problem that was sketched in exercise 3, where we will study scattering of electrons on a potential consisting of two equally strong   $\D$\ barriers:
\[ V(x)=\frac{\hs^2f}{m_ea_0}\LT[\D(x+a/2)+\D(x-a/2)\RT]\qquad (f&gt;0).\]
With the energy $E=\hs^2k^2/(2m_e)$ and a scattering function on the  form 
\[ \psi=\LT\{\barr{lll}{\DS\frac{1}{t}e^{ikx}+Be^{-ikx}} &amp; \mbox{ for } &amp; x&lt;-a/2,\\
Ce^{ikx}+De^{-ikx} &amp; \mbox{ for } &amp; -a/2&lt;x&lt;a/2,\\
e^{ikx} &amp; \mbox{ for } &amp; x&gt;a/2\earr\RT.\]
we can then, as mentioned in exercise 3, show from the matching conditions that
\[ C=1+\frac{if}{ka_0}\OGE{10}D=-\frac{if}{ka_0}\,e^{ika}\]
and that
\[ \frac{1}{t}=\LT(1+\frac{if}{ka_0}\RT)^2+\frac{f^2}{(ka_0)^2}\,e^{2ika}.\]
With this  formula we can study the  transmission coefficient \M{T=|t|^2} as a  function of the various  parameters.

\pkt{a} After some calculation one finds that the transmission coefficient  $T_{(2)}$ of the  double barrier is determined by the formula
\[ \frac{1}{T_{(2)}}=1+\LT[\frac{2f}{ka_0}\LT(\frac{f}{ka_0}\sin ka+\cos ka\RT)\RT]^2.\]
For comparison, we found in exercise 3 for scattering on a single barrier of ``strength'' $f$ that
 \[ \frac{1}{T_{(1)}}=1+\LT(\frac{f}{ka_0}\RT)^2.\]
When the quantity $ka_0$ is small compared to $f$, we obseve that the probability of transmission through the single barrier is small.

\bu  Explain, using the above formula for the double barrier, that we can have   \emph{full} transmission ($T_{(2)}=1$) for certain values of the wavenumber $k$,and find an equation which determines these values.

\pkt{b} This phenomenon is called  tunnelling resonance. 
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The figure shows $T_{(2)}$ and $T_{(1)}$ as functions of $ka_0$ for $f=10$ and $a=2a_0$. \bu Use the figure to find (with reasonable accuracy) a couple of the values of $ka_0$ which give rexonace. \bu Try to use the calculator to find the smallest value of $ka_0$ which fulfills the equation found under point \underline{a}, and compare with the value found from the figure.



\subsection*{\underline{Problem 6 -- 4} \h{4mm} {\small Working with ket and bra
 vectors }}


Let \M{\ket{1},\ket{2}\OGE{1.5}\ket{3}} be three orthonormalized vectors spanning a three-dimensional vector space (a subspace of ``Hilbert'' space).

\pkt{a} Write down relations expressing that the three vectors are normalized and orthogonal.

\pkt{b} Write down the completeness relation for this set of vectors, and use this relation to expand an arbitrary vector $\ket{\psi}$ (in the three-dimensional space) in terms of the basis set $\;\ket{1},\,\ket{2}\,$ and $\,\ket{3}$.


\pkt{c} Show that when the operator \M{P_1\equiv\ket{1}\bra{1}} acts on
$\ket{\psi}$, then it ``projects out'' the component of $\ket{\psi}$ in the 
"$\ket{1}$-direction". Show also that \M{P_1^2=P_1} and that $P_1$ is Hermitian. (Consider the adjoint; see \PKT{h} below.)
Operators of this type are generally called projection operators. Show that also

\M{P_{12}\equiv\ket{1}\bra{1}+\ket{2}\bra{2}} is a projection operator.

\pkt{d} Set \M{\ket{a}=\ket{1}\;} and  \M{\;\ket{b}=(1+i)\ket{1}.} What are then

\M{\;\bra{b},\;\scpr{a}{b},\;\scpr{b}{a},\OGE{2}\scpr{b}{b}?}

\pkt{e} Set \M{\ket{\psi_1}=3^{-1/2}\,\ket{1}+c_1\ket{2}.} 
Which condition must
$|c_1|^2$ satisfy if  $\ket{\psi_1}$ is to be normalized? (Choose $c_1$ real and
positive). Set \M{\ket{\psi_2}=c_2\ket{1}+c_3\ket{2}.} Choose $c_3/c_2$ such that
$\ket{\psi_2}$ becomes orthogonal to $\ket{\psi_1}$ and choose $c_2$ real and positive,
and such that $\ket{\psi_2}$ becomes normalized. Find a third linear combination of the three vectors \M{\ket{1},\ket{2}\OG{1.5}\ket{3}} which is orthogonal to both
$\ket{\psi_1}$ and $\ket{\psi_2}$.

\pkt{f} Let us now leave the three-dimensional vector space above, and return to the ordinary infinite-dimensional Hilbert space. Show that 
  \[  \ket{\psi}=\int dx\, \psi(x)\ket{x}, \]
and that a special case of this formula is
  \[ \ket{p}=\frac{1}{\sqrt{2\pi\hs}}\int dx\,e^{ipx/\hs}\ket{x}. \]



\pkt{g} In a similar manner, find $\ket{x}$ expressed in terms of the  $\ket{p}$ basis.

\pkt{h} Use the definition of the adjoint,
  \[ \bra{a}A^{\dag}\ket{b}=\bra{b}A\ket{a}^{\Ast}, \]
to show that 
  \[ (\ket{c}\bra{d})^{\dag}=\ket{d}\bra{c}. \]

\subsection*{\underline{Problem 6 --5}\h{4mm}{\small More about ket- and bra vectors}}

In the notes we have seen that there is a one-to-one correspondence between the abstract vectors of the Hilbert space and the good old wave functions:
\ba  \psi_a(x)\qquad&amp;\longleftrightarrow&amp;\qquad\ket{\psi_a}\equiv\ket{a},\\
     \psi_b(x)\qquad&amp;\longleftrightarrow&amp;\qquad\ket{\psi_b}\equiv\ket{b},\quad\mbox{etc},\ea
where we are free to choose whichever labels we wish to use. We have also seen that the new scalar products are identical to the old ones:
\[ \LT\langle\;\;\ket{\psi_a}\,,\,\ket{\psi_b}\;\;\right\rangle\equiv\underline{\scpr{\psi_a}{\psi_b}}=\FV{\psi_a,\psi_b}\equiv\underline{\int\psia_a\psi_bd\tau}.\]

\pkt{a} In the good old position representation of quantum mechanics, you will also remember that the eigenvalue equation for the operator $\xop=x$,
\[ x\,\psi_{x'}(x)=x'\,\psi_{x'}(x),\]
has the solution
\[ \psi_{x'}(x)=\D(x-x').\]
(This you can easily check.)
To this wave function there corresponds a vector $\ket{\psi_{x'}}\equiv\ket{x'}$ in  Hilbert space. You may also recall that the eigenvalue equation
\[ \pop_x\,\psi_p(x)=p\,\psi_p(x)\]
for the momentum operator has the solution
\[\psi_p(x)=(2\pi\hs)^{-1/2}\exp(ipx/\hs).\]
To this momentum eigenfunction there corresponds a vector \M{\ket{\psi_p}\equiv\ket{p}} in Hilbert space. Here are the questions: What are $\scpr{x'}{\psi_p}$ and $\scpr{\psi_p}{x'}$?

\pkt{b} The operator $\xop$  representing the observable $x$ is in the  Dirac formalism defined by the eigenvalue equation
 \[ \xop\ket{x'}=x'\ket{x'},\qquad\mbox{where}\quad \xop^{\dag}=\xop,\qquad\mbox{so that}\quad \bra{x'}\xop=x'\bra{x'}.\] 
What is then $\bra{\psi_p}\xop\ket{x'}$? And what is $\bra{x'}\xop\ket{\psi_p}$?

\pkt{c} Let $\ket{n}$ correspond to the eigenfunction $\psi_n(x)$ of a  harmonic oscillator. When the operator $\xop$ acts on the ground state, we get a new vector which we may call $\ket{\rm new}$:
\[ \xop\ket{0}=\ket{\rm new}.\]
What wave function ($\psi_{\rm new}$) does the new vector correspond to? Hint: Consider the scalar product $\scpr{x'}{\rm new}$ and use that $\;\bra{x'}\xop=x'\bra{x'}$. Given: 
\[ x\psi_0(x)=\sqrt{\frac{\hs}{2m\om}}\,\psi_1(x).\]
Show that \M{\ket{\rm new}={\rm constant}\cdot\ket{1},} and find the constant.




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