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\markright{\sf FY2045/TFY4250  Kvantemekanikk I, 2012 -  exercise 2}{}


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\begin{center}{\Large\bf  EXERCISE 2}\end{center}
\begin{quote}

\end{quote}



\subsection*{\underline{Problem 2 -- 1} \h{1cm} {\small ``almost'' a  position eigenstate}}

We have seen that the position cannot be measured with \emph{zero} uncertainty. It is, however, in principle possible to prepare an ensemble of particles in a state such that the uncertainty is arbitrarily small (but finite). Imagine that we carry out such a ``measurement'' on an ensemble, leaving it in a state that is immediately afterwards described by the normalized wave function
  \[ \psi(x)=(2\B/\pi)^{1/4}e^{-\B(x-a)^2},\qquad\qquad\B\mbox{ large}.\]

\pkt{a} Find (without calculation) the expectation value $\FV{x}$ of the position. [Hint: Take a look at the distribution of the probability density.]

\pkt{b} Find the uncertainty $\Delta x$ in the position, expressed in terms of $\B$. [Hint: Remember that $\;(\Delta x)^2=\FV{(x-\FV{x})^2}$. You may find it convenient to introduce $\;x'=x-a\;$ as a new integration variable. Some Gaussian integrals are given below.]  Show also that the Gaussian probability distribution $|\psi(x)|^2$ may be written on the form
 \[ |\psi(x)|^2= \frac{1}{\sqrt{2\pi(\Delta x)^2}}\exp\LT[-\frac{(x-a)^2}{2(\Delta x)^2}\RT].\]
Moral: For a Gaussian probability density of this type, we can read out the uncertainty from the exponent. 

\pkt{c} Assume that we choose $\B$  very large, in order to make $\Delta x$ very small, that is, in order to prepare the ensemble in a state with very well-defined position. Calculate $\FV{p_x}$ and $\FV{p_x^2}$, and show that we are ``punished'' with a very large uncertainty $\Delta p_x$ in the momentum. (Check also that the results for $\Delta x$ and $\Delta p_x$ agree with Heisenberg's uncertainty relation.)
\newline
[Hint: You may find it useful to show that $\FV{p_x}$ is always zero for a real wave function.]\newline
Express the expectation value of the kinetic energy in terms of the uncertainty $\Delta x$. What happens if you insist on letting $\Delta x$ go to zero?

\pkt{d} The wave function $\psi(x)$ contains \emph{all} information about the momentum distribution in the prepared state (not only the uncertainty $\Delta p_x$). This information can be extracted from the wave function by calculating the momentum wave function $\phi(p)$, which is the  ``expansion coefficient'' in the  Fourier integral
 \[ \psi(x)=\intuu\phi(p)\psi_p(x)dp\qquad\LT[\psi_p(x)=(2\pi\hs)^{-1/2}e^{ipx/\hs}\RT].\]
As we have seen in the lectures,  $\phi(p)$ is the projection of $\psi(x)$ on the momentum wave function $\psi_p(x)$:
 \[ \phi(p)=\FV{\psi_p,\psi}=\intuu\psia_p(x)\psi(x)dx\qquad\mbox{(the Fourier transform)}.\]
Calculate $\phi(p)$  using the integral given at the end of this exercise, and show that the probability density in momentum space is in this case Gaussian and may be written on the form
 \[ |\phi(p)|^2=\frac{1}{\sqrt{2\pi(\Delta p_x)^2}}\exp\LT[-\frac{p^2}{2(\Delta p_x)^2}\RT],\]
where $\Delta p_x$ is the uncertainty found in \underline{\bf c}. [Hint: Use $x-a$ as integration variable.]\newline What happens with this probability distribution when  $\B$ is chosen very large?\newline
A small challenge: Why can we state (without calculation) that the momentum wave function itself, $\phi(p)$ must depend on $a$?


\pkt{e} What happens \emph{after} the preparation of this state $\psi(x)$ will depend on the potential $V(x)$, and is determined by the Schr\"odinger equation. Instead of solving this problem mathematically, you are invited to speculate. Assume for example that the particle is free ($V(x)=0$) and try to think semi-classically: How far would a particle with mass $m$ and momentum in the range $\;-\Delta p_x&lt;p_x&lt;+\Delta p_x\;$ move in time $t$? How large would you then expect the uncertainty in the position to be after time $t$ (assuming the initial preparation took place at $t=0$)? 

\subsubsection*{Given: Some Gaussian integrals}

\ba I_0(\A) &amp;\equiv &amp; \intuu e^{-\A x^2}dx=\sqrt{\pi/\A},\\
 I_2(\A) &amp; \equiv &amp; \intuu x^2e^{-\A x^2}dx=-\frac{d}{d\A}I_0(\A)=\hal\sqrt{\pi}\,\A^{-3/2}\\
J(A,B) &amp;\equiv&amp; \intuu e^{-Ay^2+By}dy=e^{B^2/4A}\sqrt{\pi/A},\qquad(\RE(A)&gt;0).\ea


\subsection*{\underline{Oppgave 2 -- 2}\h{4mm}{\small Gaussian free-particle wave packet}}

\begin{quote}
{\bf Preliminaries:}{\sf It is easy to check that the Gaussian {\bf probability density}
  \[ \rho(x)=|\psi(x)|^2=(2\pi\si^2)^{-1/2}\,e^{-(x-x_0)^2/2\si^2}\]
is normalized ($\intuu\rho(x)dx=1$), and that the expectation value of $x$ is
 \[ \FV{x}=\intuu x|\psi(x)|^2dx=\intuu x\rho(x)dx=x_0.\]
It is also easy to show that the square of the uncertainty in the position is
 \[ (\Delta x)^2=\FV{(x-\FV{x})^2}=\si^2.\]
Thus, we have the rule
 \[\fbox{$|\psi(x)|^2=\frac{1}{\sqrt{2\pi\si^2}}\,e^{-(x-x_0)^2/2\si^2}\quad\Longrightarrow\quad\Delta x=\si\qquad\mbox{and}\quad\FV{x}=x_0.$}\]
This rule is useful when dealing with Gaussian probability densities, both in $x$- and $p$-space. The results above follow from the results of the previous Problem, and can of course be checked directly using the integrals given above.}


\[ --- *** --- \]
\paragraph {The problem we wish to study} {\sf is a free particle with mass $m$ which is at \M{t=0} prepared in the initial state
 \[ \Psi(x,0)=(2\pi\si^2)^{-1/4}e^{-x^2/4\si^2}e^{ip_0x/\hs}.\]
This has the form of a momentum wave function ($e^{ip_0x/\hs}$) modulated by a Gaussian $e^{-x^2/4\si^2}$,  
and is a special case of the wave function studied in Oppgave 9.
We want to find out how the wave packet  $\Psi(x,t)$ behaves for \M{t&gt;0.}
The figure shows the real part of the initial state:}
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\end{quote}

\pkt{a} Write down the probability density $|\Psi(x,0)|^2$ and read out the initial values $\FV{x}_0$ and $(\Delta x)_0$ using the rules above. You may also want to check these results against those of the previous Problem (substituting $\B=1/4\si^2$.)
\begin{quote}
{\sf The general way to attack a time-dependent problem is to expand the wave function in the stationary states of the system. For the free particle, these are the plane-wave states
 \[ \Psi_p(x,t)=\psi_p(x)\,e^{-i(p^2/2m)t/\hs}=(2\pi\hs)^{-1/2}\,e^{i[px-(p^2/2m)t]/\hs}.\]
Thus (see the lecture notes -- Tillegg 2, section 2.7)}
 \[ (i)\qquad \Psi(x,t)=\intuu\phi(p)\Psi_p(x,t)dp\qquad\Longrightarrow\] 
 \[ \qquad \Psi(x,0)=\intuu\phi(p)\psi_p(x)dp\qquad\Longrightarrow\]
 \[(ii)\qquad\quad\phi(p)=\FV{\psi_p,\Psi(0)}=\intuu\psia_p(x)\Psi(x,0)dx.\]
Thus the Fourier transform $\phi(p)$ is found by projecting the initial state onto the momentum eigenfunction $\psi_p(x)$.
The third step in the recipe is to insert the expression  for $\phi(p)$ into (i), which then gives the wave function at time $t$:
\[ (iii)\qquad \Psi(x,t)=\intuu\FV{\psi_p,\Psi(0)}\Psi_p(x,t)dp.\]
\end{quote}

\pkt{b} It turns out that
 \[ \phi(p)=\LT(\frac{2\si^2}{\pi\hs^2}\RT)^{1/4}e^{-(p-p_0)^2\si^2/\hs^2}.\]
Show this by calculating the integral (ii).
[Hint: The integral is of the type $J(a,b)$ mentioned above. All you have to do is to identify what $a$ and $b$ are in this case.] What is the ``central'' momentum in the wave packet $\Psi(x,t)$?

\pkt{c} According to the physical interpretation of expansion coefficients (see Hemmer, and Tilleg 2),  the probability density in momentum space (``$p$-space'') is now the square of the ``expansion coefficient'' in the Fourier integral (i) above, that is,
$|\phi(p)|^2$. We note that this probability density  is time independent for the free particle.  Use the rule above to read out $\FV{p_x}$ and $\Delta p_x$ from the formula for $|\phi(p)|^2$, and check that $\;(\Delta x)_0\Delta p_x=\hal\hs$. (See previous Problem.)

\pkt{d} To show what happens with the wave packet $\Psi(x,t)$ and the quantities $(\Delta x)_t$ and $\FV{x}_t$
 for \M{t&gt;0,} we can put $\phi(p)$ back into the expansion (i). [This is step (iii) in the procedire outlined above.] \newline
Hint: Show that $\Psi(x,t)$ can be written as
 \[ \Psi(x,t)=\LT(\frac{\si^2}{2\pi^3}\RT)^{1/4}\,e^{i\A}\intuu e^{-ay^2+by}dy,\]
where \M{\A=p_0x/\hs-(p_0^2/2m)t/\hs,\;\;a=\si^2+i\hs t/2m} and \M{b=i(x-p_0t/m).} [Introduce $y=(p-p_0)/\hs$ as integration variable this time.]\newline
Show that the result is
\[  \Psi(x,t)=\frac{(2\pi)^{-1/4}}{\sqrt{\sigma+i\hs t/2m\sigma}}\,\,e^{i[p_0x-(p_0^2/2m)t]/\hs}\, \exp\LT[-\frac{(x-p_0t/m)^2}{4(\sigma^2+i\hs t/2m)}\RT], \]
and check that this gives the correct $\Psi(x,0)$ in the limit \M{t\rpil0.}
Also, show that
\[  \rho(x,t)=|\Psi(x,t)|^2=\frac{1}
  {\sqrt{2\pi(\sigma^2+\hs^2t^2/4m^2\sigma^2)}}\,\exp\LT[-\frac{(x-p_0t/m)^2}
  {2(\sigma^2+\hs^2t^2/4m^2\sigma^2)}\RT]. \]
Hint: For real $\,a,b,c\,$ one has
 \[ |e^{ia}|=1, \;\;|a+ib|^2=a^2+b^2, \]
 \[
 \LT|\exp\LT(\frac{c}{a+ib}\RT)\RT|=\LT|\exp\LT(\frac{c(a-ib)}{a^2+b^2}\RT)\RT|
    =\exp\LT(\frac{ca}{a^2+b^2}\RT). \]

\pkt{e}  Using the rule above, you should now be able to read out from the form of $|\Psi(x,t)|^2$ the values of $\FV{x}_t$ and $(\Delta x)_t$ (at the time $t$). Using the result for $\FV{x}_t$, check that the group velocity  is correct.
Discuss the dispersion of the wave packet, and verify the results of the qualitative discussion in the previous Problem.  

\pkt{f} Show that the uncertainty in $x$ can be written as \M{(\Delta x)_t=\si\sqrt{1+(t/t_1)^2},} where \M{t_1=2m\si^2/\hs.} Then $t_1$ is (roughly) the time it takes the wave packet to double its initial size. Find $t_1$ for an electron with mass 511000 eV/$c^2$ and $\si$ of atomic size ($\sim 10^{-10}\,$m). What is $t_1$ for $\,\si=10$ cm.\footnote{In particle physics it is customary to give the electron mass in units of MeV/c$^2$, as  \M{m_e=0.510999..}MeV/$c^2$, where $c=2.998..$m/s is the velocity of light. Cf Einstein's formula for the rest-mass energy. Thus for the electron we have
 \[ t_1=\frac{2m_e\si^2}{\hs}=\frac{2m_ec^2\,\si^2}{\hs c^2}\sim\frac{2\cdot511999\,{\rm eV}\,(10^{-10}{\rm m})^2}{0.6582\cdot10^{-15}{\rm eVs}\,(2.998\,{\rm m/s})^2}=\cdots.\]
} What is $t_1$ for a ping pong ball with mass 2g and $\,\si\sim10^{-6}\,$m ? [Use $\hs\sim10^{-15}$ eVs\,$\sim10^{-34}$ Nms.]


\subsection*{\underline{Oppgave 2 -- 3}\h{4mm}{\small Matlab program for  wave packets }}

Included (see homepage) you will find the Matlab program, ``fripartikkel.m'', which you should run to see more closely how a free-particle wave packet may behave:

 The program shows the real and imaginart parts of a free-particle wave packet, of the type treated above --- using a three-dimensional animation plot. Run this to see how the packet propagates. Notice in particular the dispersion. Check if you are able to see that the phase velocity is half of the group velocity. 
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