%------------------------------------------------------------------------------------------------------------------------------------%
\typeout{<----------- PREAMBLE OG EKSTRA KOMMANDOER ----------->}
\newcommand{\theaflnr}{1}                %Definition af afleveringsnummer, til sektionsnummerering, forside og sidehoved
\newcommand{\theopgnr}{6.31}             %Definition af opgavenummer, til sidehoved
\input{../preamble.tex}                  %Input af preamble
%\usepackage[notref,notcite]{showkeys}   %Hvis aktiv vises labels i margin - bør kun anvendes i draft
\linespread{1.2}
%------------------------------------------------------------------------------------------------------------------------------------%
\typeout{<----------- START PÅ DOKUMENT MED FORSIDE ----------->}
\begin{document}                         % Starter dokumentet efter preamble
\pagestyle{empty}                        % Sidehoved og sidefod for forsiden
\input{../forside.tex}                   % Input af forsiden, der ligger i forside.tex
%------------------------------------------------------------------------------------------------------------------------------------%
\typeout{<---------- START PÅ AFLEVERINGSOPGAVEN SELV ---------->}
\newpage                                 % Ny side efter forsiden er indsat
\pagestyle{fancy}                        % Sidehoved og sidefod for resten af dokumentet
\setcounter{page}{2}                     % Resten af afleveringen har sidenummer 2 og frem
\setcounter{section}{\theaflnr}          % reset tælleren til at nummerere sections (ugeopgavenummer)

%------------------------------------------------------------------------------------------------------------------------------------%
%\subsubsection*{Notice}
%{\footnotesize In this assignment I'll refer to equations, theorems, definitions and so forth. When I do this, I don't make any references to where they're coming from, but -- unless I state anything else -- they're from the book used in the course, \cite{griffithskvant}. }


\setcounter{subsection}{1}
\subsection*{Problem 6.31}
\ramme{{\bfseries Van der Waals interaction}\\
       Consider two atoms a distance $R$ apart. Because they are electrically neutral you might suppose there would be no force between them, but 
       if they are polarizable there is in fact a weak attraction. 
       \begin{center}\psset{unit=5mm}\begin{pspicture}(0,-2)(17,1) %\psgrid
        %Coils and their labels:
        \rput(0,0){\psline{c-c}(0,0)(0.46,0)
                   \rput(0.6,0){\psCoil[coilaspect=-45,coilwidth=3mm,coilheight=1.32,arrows=c-c]{90}{1710}}
                   \psline{c-c}(3.54,0)(4,0)}
        \uput{9pt}[-90](2,0){$x_1$}
        \rput(13,0){\psline{c-c}(0,0)(0.46,0)
                   \rput(0.6,0){\psCoil[coilaspect=-45,coilwidth=3mm,coilheight=1.32,arrows=c-c]{90}{1710}}
                   \psline{c-c}(3.54,0)(4,0)}
        \uput{9pt}[-90](15,0){$x_2$}
        %Dots and their labels:
        \pscircle*[linecolor=Periwinkle](0,0){2pt}      \uput[90](0,0){$+e$}
        \pscircle*[linecolor=Periwinkle](4,0){2pt}      \uput[90](4,0){$-e$}
        \pscircle*[linecolor=Periwinkle](13,0){2pt}     \uput[90](13,0){$+e$}
        \pscircle*[linecolor=Periwinkle](17,0){2pt}     \uput[90](17,0){$-e$}
        %Brace:
        \psbrace[rot=90,braceWidthInner=1mm,braceWidth=1pt,braceWidthOuter=1mm,nodesepB=9pt,nodesepA=-4pt](0,-1.2)(13,-1.2){$R$}
       \end{pspicture}\end{center}
       To model this system, picture each atom as an electron (mass $m$, charge $-e$)
       attached by a spring (spring constant $k$) to the nucleus (charge $+e$), as in the figure above. We'll assume the nuclei are heavy, and 
       essentially motionless. The Hamiltonian for the unpertubed system is
       \begin{equation}
        H^0 = \frac{1}{2m} p_1^2 + \frac{1}{2} k x_1^2 + \frac{1}{2m} p_2^2 + \frac{1}{2} k x_2^2 .
        \label{Hnul}
       \end{equation}
       The Coulumb interaction between the atoms is 
       \begin{equation}
        H' = \frac{1}{4 \pi \epsilon_0} \left( \frac{e^2}{R} - \frac{e^2}{R+x_1} - \frac{e^2}{R - x_2} + \frac{e^2}{R+x_1-x_2} \right) .
        \label{Hpert}
       \end{equation}\vspace*{-0.4cm}}

\subsubsection*{Question a}
\ramme{ Explain equation \eqref{Hpert}. Assuming that $|x_1|$ and $|x_2|$ are both much less than $R$, show that 
        \begin{equation}
         H' \approx - \frac{e^2 x_1 x_2}{2 \pi \epsilon_0 R^3} .
         \label{Hpertapprox}
        \end{equation}\vspace*{-0.4cm}}
To explain equation \eqref{Hpert}, I will explain the four different terms separately. First I generally know that the Coulomb potential between two charges $q_1$ and $q_2$ with a distance $r$ from each other is given by
\begin{equation*}
 V = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} ,
\end{equation*}
according to \cite{griffithsel}. Hence I get the following explanations for the four terms in \eqref{Hpert}:
\begin{itemize}\ns
 \item The first term, $e^2/R$, is corresponding to the potential between the two nuclei.
 \item The second term, $-e^2/(R+x_1)$, is corresponding to the potential between the electron of the first atom and the nucleus of the second atom.
 \item The third term, $-e^2/(R-x_2)$, is corresponding to the potential between the electron of the second atom and the nucleus of the first atom.
 \item And at last, the fourth term, $e^2/(R+x_1-x_2)$, is corresponding to the potential between the two electrons.
\end{itemize}
Actually though, it seems as if $x_1$ and $x_2$ should have had different signs (or should have been switched), such that $x_1$ in the second and the fourth term should be subtracted from $R$ instead of being added, and $x_2$ in the third and the fourth term should have been added to $R$ instead of being subtracted. This doesn't make any difference later -- so I'll just let it be as it is. 

Now I can move on to showing the approximation in equation \eqref{Hpertapprox}. I first note that, according to equation (22.8) in \cite{schaums}, an expression of the form $(1 \pm y)^{-1}$, for a $y \ll 1$, can be approximated (to second order) by
\begin{equation*}
 (1 \pm y)^{-1} \approx 1 \mp y + y^2 .
\end{equation*}
Hence, if I assume that $x \ll k$, such that the fraction $x/k$ is much smaller than one, I have, to second order, that 
\begin{equation*}
 \frac{1}{k \pm x} = \frac{1}{k} \left( 1 \pm \frac{x}{k} \right)^{-1}
                   \approx \frac{1}{k} \left[ 1 \mp \frac{x}{k} + \frac{x^2}{k^2} \right] .
\end{equation*}
Since I could actually assume that $|x_1|$ and $|x_2|$ was much smaller than $R$ according to the text in this question, I can use this to get an approximation of my pertubation Hamiltonian $H'$ in \eqref{Hpert}, simply by seeing that
\begin{align*}
 H' &= \frac{1}{4 \pi \epsilon_0} e^2 \left[ \frac{1}{R} - \frac{1}{R + x_1} - \frac{1}{R - x_2} + \frac{1}{R + x_1 - x_2} \right] \\
    &= \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R} \left[ 1 - \left( 1 + \frac{x_1}{R} \right)^{-1} - \left( 1 - \frac{x_2}{R} \right)^{-1} 
                                                         + \left( 1 + \frac{x_1-x_2}{R} \right)^{-1} \right] \\
    &\approx \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R} \left[ 1 - \left( 1 - \frac{x_1}{R} + \frac{x_1^2}{R^2} \right) 
                                                               - \left( 1 + \frac{x_2}{R} + \frac{x_2^2}{R^2} \right) 
                                                               + \left( 1 - \frac{x_1-x_2}{R} + \frac{(x_1-x_2)^2}{R^2} \right) \right] \\
    &= \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R} \left[ 1 - 1 + \frac{x_1}{R} - \frac{x_1^2}{R^2} 
                                                         - 1 - \frac{x_2}{R} - \frac{x_2^2}{R^2}
                                                         + 1 - \frac{x_1-x_2}{R} + \frac{(x_1-x_2)^2}{R^2} \right] \\
    &= \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R} \left[ - \frac{x_1^2}{R^2} - \frac{x_2^2}{R^2} + \frac{(x_1-x_2)^2}{R^2} \right] \\
    &= \frac{1}{4 \pi \epsilon_0} \frac{e^2}{R} \frac{-2x_1x_2}{R^2} \\
    &= - \frac{e^2 x_1 x_2}{2 \pi \epsilon_0 R^3} ,
\end{align*}
exactly as I was supposed to get.


\clearpage
\subsubsection*{Question b}
\ramme{ Show that the total Hamiltonian (equation \eqref{Hnul} plus equation \eqref{Hpertapprox}) separates into two harmonic oscillator Hamiltonians:
        \begin{equation}
         H = \left[ \frac{1}{2m} p_+^2 + \frac{1}{2} \left( k - \frac{e^2}{2\pi\epsilon_0 R^3} \right) x_+^2 \right] 
             + \left[ \frac{1}{2m} p_-^2 + \frac{1}{2} \left( k + \frac{e^2}{2\pi\epsilon_0 R^3} \right) x_-^2 \right] 
         \label{Hharmonic}
        \end{equation}
        under the change of variables
        \begin{equation}
         x_\pm \equiv \frac{1}{\sqrt 2} (x_1 \pm x_2) , \qquad\text{which entails } p_\pm = \frac{1}{\sqrt 2} (p_1 \pm p_2) .
         \label{hamonicvariables}
        \end{equation}\vspace*{-0.6cm}}
This is easiest done by simply ``calculating backwards'', by inserting the new variables defined in \eqref{hamonicvariables} back into equation \eqref{Hharmonic}. I then get that
\begin{align*}
 H &= \frac{1}{2m} p_+^2 + \frac{1}{2} \left( k - \frac{e^2}{2\pi\epsilon_0 R^3} \right) x_+^2 
      + \frac{1}{2m} p_-^2 + \frac{1}{2} \left( k + \frac{e^2}{2\pi\epsilon_0 R^3} \right) x_-^2 \\
   &= \frac{1}{4m} (p_1^2 + p_2^2 + 2p_1p_2) + \frac{1}{4} \left( k - \frac{e^2}{2\pi\epsilon_0 R^3} \right) (x_1^2 + x_2^2 + 2x_1x_2) \\
   &\qquad   + \frac{1}{4m} (p_1^2 + p_2^2 - 2p_1p_2) + \frac{1}{4} \left( k + \frac{e^2}{2\pi\epsilon_0 R^3} \right) (x_1^2 + x_2^2 - 2x_1x_2) \\
   &= \frac{1}{2m} p_1^2 + \frac{1}{2} k x_1^2 + \frac{1}{2m} p_2^2 + \frac{1}{2} k x_2^2 + \frac{1}{4} \frac{e^2}{2\pi \epsilon_0 R^3} (-4 x_1 x_2) \\
   &= \frac{1}{2m} p_1^2 + \frac{1}{2} k x_1^2 + \frac{1}{2m} p_2^2 + \frac{1}{2} k x_2^2 - \frac{e^2}{2\pi \epsilon_0 R^3} x_1 x_2 ,
\end{align*}
which is exactly $H = H^0 + H'$, where $H^0$ is defined in \eqref{Hnul} and $H'$ is the approximation defined in \eqref{Hpertapprox}. Hence it is certainly possible to write the Hamiltonian into two seperate harmonic oscillator Hamiltonians with the given change of variables. 

I could also have done it the other way round, simply by noting that $x_\pm^2 = ( x_1^2 + x_2^2 \pm 2 x_1 x_2 )/2$ and $p_\pm^2 = ( p_1^2 + p_2^2 \pm 2 p_1 p_2 )/2$, and hence that
\begin{equation*}
 p_+^2 + p_-^2 = p_1^2 + p_2^2 ,  \qquad
 x_+^2 + x_-^2 = x_1^2 + x_2^2 ,  \qquad
 x_-^2 - x_+^2 = -2 x_1 x_2 .
\end{equation*}
%Then I can see that the following three expressions can be used:
%\begin{align*}
% \frac{1}{2m} (p_+^2 + p_-^2) &= \frac{1}{2m} (p_1^2 + p_2^2) , \\
% \frac{1}{2} k (x_+^2 + x_-^2) &= \frac{1}{2} k (x_1^2 + x_2^2) , \\
% \frac{1}{2} \frac{e^2}{2\pi \epsilon_0 R^3} (x_-^2 - x_+^2) &= - \frac{e^2}{2\pi \epsilon_0 R^3} x_1 x_2 ,
%\end{align*}
By using these I can rewrite my hamiltonian operator $H$ with the new variables $x_\pm$ and $p_\pm$ as
\begin{align*}
 H &= H^0 + H' \\
   &= \frac{1}{2m} p_1^2 + \frac{1}{2} k x_1^2 + \frac{1}{2m} p_2^2 + \frac{1}{2} k x_2^2 - \frac{e^2}{2\pi \epsilon_0 R^3} x_1 x_2 \\
   &= \frac{1}{2m} (p_1^2 + p_2^2) + \frac{1}{2} k (x_1^2 + x_2^2) - \frac{e^2}{2\pi \epsilon_0 R^3} x_1 x_2 \\
   &= \frac{1}{2m} (p_+^2 + p_-^2) + \frac{1}{2} k (x_+^2 + x_-^2) + \frac{1}{2} \frac{e^2}{2\pi \epsilon_0 R^3} (x_-^2 - x_+^2) \\
   &= \left[ \frac{1}{2m} p_+^2 + \frac{1}{2} k x_+^2 - \frac{1}{2} \frac{e^2}{2\pi \epsilon_0 R^3} x_+^2 \right]
      + \left[ \frac{1}{2m} p_-^2 + \frac{1}{2} k x_-^2 + \frac{1}{2} \frac{e^2}{2\pi \epsilon_0 R^3} x_-^2 \right] \\
   &= \left[ \frac{1}{2m} p_+^2 + \frac{1}{2} \left( k - \frac{1}{2} \frac{e^2}{2\pi \epsilon_0 R^3} \right) x_+^2 \right]
      + \left[ \frac{1}{2m} p_-^2 + \frac{1}{2} \left( k + \frac{1}{2} \frac{e^2}{2\pi \epsilon_0 R^3} \right) x_-^2 \right] ,
\end{align*}
exactly as I wanted to show. 



\clearpage
\subsubsection*{Question c}
\ramme{ The ground state energy for this Hamiltonian is evidently
        \begin{equation}
         E = \frac{1}{2} \hbar (\omega_+ + \omega_-), \qquad\text{where } \omega_\pm = \sqrt{\frac{k \mp \dfrac{e^2}{2\pi\epsilon_0 R^3}}{m}} .
         \label{angvel}
        \end{equation}
        Without the Coulomb interaction it would have been $E_0 = \hbar \omega_0$, where $\omega_0 = \sqrt{k/m}$. Assuming that 
        $k \gg (e^2/2\pi\epsilon_0R^3)$, show that
        \begin{equation*}
         \Delta V \equiv E - E_0 
                  \approx - \frac{\hbar}{8 m^2 \omega_0^3} \left( \frac{e^2}{2\pi\epsilon_0} \right)^2 \frac{1}{R^6} .
        \end{equation*}
        {\slshape Conclusion:} There is an attractive potential between the atoms, proportional to the inverse sixth power of their separation. 
        This is the van de Waals interaction between two neutral atoms. }
To get an approximation, I first note that, according to equation (22.12) in \cite{schaums}, an expression of the form of the general angular velocities given in \eqref{angvel}, can be approximated (to second order) by
\begin{equation*}
 (1 \pm y)^{1/2} \approx 1 \pm \frac{1}{2} y - \frac{1}{8} y^2 ,
\end{equation*}
and hence the general angular velocities $\omega_\pm$ are, after a little re-writing by setting $\omega_0 = \sqrt{k/m}$ ouside the square root, approximated by
\begin{align*}
 \omega_\pm = \sqrt{\frac{k \mp \dfrac{e^2}{2\pi\epsilon_0 R^3}}{m}} 
            = \omega_0 \left( 1 \mp \frac{e^2}{2\pi\epsilon_0 R^3 k} \right)^{1/2} 
            \approx \omega_0 \left[ 1 \mp \frac{1}{2} \frac{e^2}{2\pi\epsilon_0 R^3 k} 
                                    - \frac{1}{8} \left( \frac{e^2}{2\pi\epsilon_0 R^3 k} \right)^2 \right] .
\end{align*} 
With this I now see that the first order of these terms in respectively $\omega_+$ and $\omega_-$ cancel each other in $\omega_+ + \omega_-$, and hence I get that the potential difference is approximated by
\begin{align*}
 \Delta V &\approx \frac{1}{2} \hbar \omega_0 \left[ 2 - \frac{1}{4} \left( \frac{e^2}{2\pi\epsilon_0 R^3 k} \right)^2 \right] - \hbar \omega_0 \\
          &= \hbar \omega_0 \left[ 1 - \frac{1}{8} \left( \frac{e^2}{2\pi\epsilon_0 R^3 k} \right)^2 - 1\right] \\ %%%
          &= - \frac{\hbar \omega_0}{8} \left( \frac{e^2}{2\pi\epsilon_0 R^3 k} \right)^2 \\
          &= - \frac{\hbar \omega_0}{8} \left( \frac{e^2}{2\pi\epsilon_0 R^3 \omega_0^2 m} \right)^2 \\
          &= - \frac{\hbar}{8 m^2 \omega_0^3} \left( \frac{e^2}{2\pi\epsilon_0} \right)^2 \frac{1}{R^6} ,
\end{align*}
just as I had to show (note that I used the definition of $\omega_0 = \sqrt{k/m}$ to substitute for $k$ between the third and fourth line). 


\clearpage
\subsubsection*{Question d}
\ramme{ Now do the same calculation using second-order pertubation theory. } %{\slshape Hint:} The unpertubed states are of the form
        %$\psi_{n_1}(x_1)\psi_{n_2}(x_2)$, where $\psi_n(x)$ is a one-particle oscillator wave function with mass $m$ and spring constant $k$; 
        %$\Delta V$ is the second-order correction to the ground state energy, for the pertubation in equation \eqref{Hpertapprox} (notice that the
        %first-order correction is zero). }
First of all, I will write the unpertubed states as $\psi_{n_1n_2}(x_1,x_2) = \psi_{n_1}(x_1) \psi_{n_2}(x_2)$, where $\psi_n(x)$ is a one-particle oscillator wave function with mass $m$ and spring constant $k$, such that its angular velocity is $\omega_0 = \sqrt{k/m}$. When writing the wave functions as bra's or ket's however, I will simply write $\bra{n_1 n_2}$ or $\ket{n_1 n_2}$. 

With this notation, the first-order correction to the ground state energy is, according to equation (6.9) in \cite{griffithskvant}, given by
\begin{align*}
 E_{00}^{(1)} &= \braket{00}{H'}{00} \\
              &= - \frac{e^2}{2\pi\epsilon_0 R^3} \braket{00}{x_1x_2}{00} \\
              &= - \frac{e^2}{2\pi\epsilon_0 R^3} \braket{0}{x_1}{0} \braket{0}{x_2}{0} \\
              &= - \frac{e^2}{2\pi\epsilon_0 R^3} \left( \sqrt{\frac{\hbar}{2m\omega}} \braket{0}{a^\dag + a}{0} \right)^2 \\
              &= - \frac{e^2}{2\pi\epsilon_0 R^3} \frac{\hbar}{2m\omega} \left( \braket{0}{a^\dag}{0} + \braket{0}{a}{0} \right)^2 \\
              &= 0 ,
\end{align*}
where I used that the two wave functions $\psi_{n_1}(x_1)$ and $\psi_{n_2}(x_2)$ are totally independent on each other (and hence I could split up the inner product between the two), that the position operator can be written in terms of the lowering and raising operators $a$ and $a^\dag$ by equation (2.69) in \cite{griffithskvant} by
\begin{equation*}
 x = \sqrt{\frac{\hbar}{2m\omega}} (a^\dag + a) ,
\end{equation*}
and that the states of the harmonic oscillator is an orthonormal set. 

To find the second-order correction to the ground state energy, I use equation (6.15), which gives me that
\begin{align*}
 E_{00}^{(2)} &= \sum_{n_1 \neq 0} \sum_{n_2 \neq 0} \frac{\left| \braket{00}{H'}{n_1n_2} \right|^2}{E_{00}^{(0)} - E_{n_1n_2}^{(0)}} \\
              &= \left( \frac{e^2}{2\pi\epsilon_0 R^3} \right)^2 
                 \sum_{n_1 \neq 0} \sum_{n_2 \neq 0} \frac{\left| \braket{00}{x_1x_2}{n_1n_2} \right|^2}{E_{00}^{(0)} - E_{n_1n_2}^{(0)}} \\
              &= \left( \frac{e^2}{2\pi\epsilon_0 R^3} \right)^2 
                 \sum_{n_1 \neq 0} \sum_{n_2 \neq 0} 
                 \frac{\left| \braket{0}{x_1}{n_1} \braket{0}{x_2}{n_2} \right|^2}{E_{00}^{(0)} - E_{n_1n_2}^{(0)}} \\
              &= \left( \frac{e^2}{2\pi\epsilon_0 R^3} \right)^2 \left( \frac{\hbar}{2m\omega_0} \right)^2
                 \sum_{n_1 \neq 0} \sum_{n_2 \neq 0} 
                 \frac{\left| \braket{0}{a^\dag + a}{n_1} \braket{0}{a^\dag + a}{n_2} \right|^2}{E_{00}^{(0)} - E_{n_1n_2}^{(0)}} ,
\end{align*}
where I again used that I could split up the inner product and used the expression of the position operator in terms of the lowering and raising operators. According to equation (2.66) in \cite{griffithskvant} and the orthonormality of the states of the harmonic oscillator, I further see that
\begin{align*}
 \braket{0}{a^\dag + a}{n_i} &= \braket{0}{a^\dag}{n_i} + \braket{0}{a}{n_i} \\
                             &= \sqrt{n_i+1} \twobraket{0}{n_i+1} + \sqrt{n_i} \twobraket{0}{n_i-1} \\
                             &= \sqrt{n_i+1} \,\delta_{0,n_i+1} + \sqrt{n_i} \,\delta_{0,n_i-1} ,
\end{align*}
for $i \in \{1, 2\}$. Hence the terms in the sum is only non-zero for $n_1 = n_2 = 1$ (remembering that neither $n_1$ nor $n_2$ was allowed to be zero), so moving on, knowing that $E_{n_1n_2}^{(0)} = \hbar \omega_0 (1 + n_1 + n_2)$, I get that
\begin{align*}
 E_{00}^{(2)} %&= \left( \frac{e^2}{2\pi\epsilon_0 R^3} \right)^2 \left( \frac{\hbar}{2m\omega_0} \right)^2
              %   \sum_{n_1 \neq 0} \sum_{n_2 \neq 0} 
              %   \frac{\left| \left( \sqrt{n_1 + 1} \delta_{0,n_1+1} + \sqrt{n_1} \delta_{0,n_1-1} \right) 
              %                \left( \sqrt{n_2 + 1} \delta_{0,n_2+1} + \sqrt{n_2} \delta_{0,n_2-1} \right) 
              %                \right|^2}{E_{00}^{(0)} - E_{n_1n_2}^{(0)}} \\
              &= \left( \frac{e^2}{2\pi\epsilon_0 R^3} \right)^2 \left( \frac{\hbar}{2m\omega_0} \right)^2
                 \frac{\left| \left( 0 + \sqrt{1} \cdot 1 \right) \left( 0 + \sqrt{1} \cdot 1 \right) \right|^2}{E_{00}^{(0)} - E_{n_1n_2}^{(0)}} \\
              &= \left( \frac{e^2}{2\pi\epsilon_0 R^3} \right)^2 \left( \frac{\hbar}{2m\omega_0} \right)^2 \frac{1}{\hbar\omega_0 - 3 \hbar\omega_0} \\
              &= \left( \frac{e^2}{2\pi\epsilon_0 R^3} \right)^2 \left( \frac{\hbar}{2m\omega_0} \right)^2 \frac{1}{- 2 \hbar\omega_0} \\
              &= - \frac{\hbar}{8m^2 \omega_0^3} \left( \frac{e^2}{2\pi\epsilon_0} \right)^2 \frac{1}{R^6} ,
\end{align*}
exactly the result I found as $\Delta V$ in question c, just as expected. 






\vspace{5cm}
\bibliography{../bibliography}

\label{lastpage}
\end{document}