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%\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{atomintro}. 
%
%I'll write vectors with bold fonts, to make the assignment easier to read. }


\setcounter{subsection}{2}
\subsection*{The assignment}
\ramme{ The Balmer-$\alpha$ line $n=3 \leftrightarrow 2$ in hydrogen splits into several components due to spin-orbit interaction and other
        relativistic effects. }

\subsubsection*{Question a}\setcounter{thespnr}{1}
\ramme{ List the states for $n=3$ and for $n=2$ in the notation $nl_j$ (with s, p, d, \ldots for $l = 0,1,2,\ldots$). }
For $n=3$, $l$ can be 0, 1 and 2. For each of these three $l$'s, there are two values of $j$, namely $j=l-s$ and $j=l+s$ (but with the restriction that $j$ cannot be negative), where $s$ here is equal to one half. Hence, for $n=3$, I have the five states
\begin{equation*}
 3\text{s}_{1/2} , \quad 3\text{p}_{1/2} , \quad 3\text{p}_{3/2} , \quad 3\text{d}_{3/2} , \quad 3\text{d}_{5/2} .
\end{equation*}
For $n=2$, $l$ can only be 0 and 1, and the conditions on $j$ are the same as before. Hence, for $n=2$, I have the three states
\begin{equation*}
 2\text{s}_{1/2} , \quad 2\text{p}_{1/2} , \quad 2\text{p}_{3/2}  .
\end{equation*}






\subsubsection*{Question b}\setcounter{thespnr}{2}
\ramme{ Evaluate for each of the levels the shift in GHz caused by spin-orbit interaction, by the relativistic mass effect, and by the Darwin effect, as
        well as the combined shift caused by the three effects. 
        
        Enter your results in the table below, and verify that the total shift depends on $n$ and $j$, but not on $l$ and $s$. }
First of all I know that the energy with no of the fine-structure is simply given by the expression from page 25 in \citet{dias090421}, 
\begin{equation}
 E_n = - \frac{m_e M_p}{m_e + M_p} \frac{1}{2 n^2} \left( \frac{e^2}{4\pi\epsilon_0\hbar} \right)^2
 \label{energi}
\end{equation}
for $n = 1, 2, \ldots$. Here I have used $Z=1$, which I will also do with later expressions. When only taking the spin-orbit interaction into account (which only works for states with $l \neq 0$), the energy is changed by the amount
\begin{equation}
 \Delta E_{\text{s-o}} = \left\{ \begin{array}{ll}
                                  - E_n \dfrac{\alpha^2}{2nl(l+1/2)(l+1)} \,l&           \text{for } j = l + s , \\
                                  - E_n \dfrac{\alpha^2}{2nl(l+1/2)(l+1)} \,(-l-1)\quad& \text{for } j = l - s ,
                                 \end{array} \right.
 \label{spinorbit}
\end{equation}
found on page 21 in \citet{dias090505}. When only taking the relativistic mass correction into account (which works on all the states), the energy is changed by the amount
\begin{equation}
 \Delta E_{\text{rel}} = -E_n \frac{\alpha^2}{n^2} \left( \frac{3}{4} - \frac{n}{l+1/2} \right) ,
 \label{relmass}
\end{equation}
found on page 26 in \citet{dias090505}. When only taking the Darwin effect into account (which only works for $l=0$ states), the energy is changed by the amount
\begin{equation}
 \Delta E_{\text{Darwin}} = - E_n \frac{\alpha^2}{n} ,
 \label{darwin}
\end{equation}
found on page 28 in \citet{dias090505}. At last the total energy change, taking account of all three effects, is 
\begin{equation}
 \Delta E_{nj} = E_n \frac{\alpha^2}{n^2} \left( \frac{n}{j+1/2} - \frac{3}{4} \right) ,
 \label{totalenergyshift}
\end{equation}
found by subtracting $E_n$ from the expression on page 29 in \citet{dias090505}. 

To fill in the table, I have to calculate the shift in frequency, so I use the relation $E = h \nu$ on the four found energy shifts, and hence I have the frequency shift. I guess that it's not relevant that I write in all the calculations, so I have just filled in the table with the results:

\begin{table}[htb]
 \centering
 %\begin{center}
 \begin{tabular}{|c|c|S|S|S|S|S|}
  \hline
  \multirow{2}{0mm}{}& \multirow{2}{8mm}{effect}& \multirow{2}{8mm}{$l=0$}& \multicolumn{2}{c|}{$l=1$}& \multicolumn{2}{c|}{$l=2$} \\\cline{4-7}
                     &                           &                          & {$j=l-s$}& {$j=l+s$}&            {$j=l-s$}& {$j=l+s$} \\\hline
  \multirow{4}{10mm}{$n=3$}& Spin-orbit&         0&                         -2.162&      1.081&                -0.648&    0.432 \\\cline{2-7}
                           & Relativistic mass&  -11.35&                    -2.702&      -2.702&               -0.973&    -0.973 \\\cline{2-7}
                           & Darwin&             6.485&                     0&           0&                    0&         0 \\\cline{2-7}
                           & Total&              -4.864&                    -4.864&      -1.621&               -1.621&    -0.540 \\\hline
  \multicolumn{7}{c}{} \\\hline
  \multirow{4}{10mm}{$n=2$}& Spin-orbit&         0&                         -7.296&      3.648&                &          \\\cline{2-7}
                           & Relativistic mass&  -35.57&                    -6.384&      -6.384&               &          \\\cline{2-7}
                           & Darwin&             21.90&                     0&           0&                    &          \\\cline{2-7}
                           & Total&              -13.68&                    -13.68&      -2.736&               &          \\\hline
 \end{tabular}
 %\end{center}
 \caption{The shifts in frequency caused by the different fine-structure effects. All the values are in GHz. The spin-orbit effect is calculated by
          using equation \eqref{spinorbit}, the relativistic mass effect is calculated using equation \eqref{relmass}, the Darwin effect is calculated
          by using equation \eqref{darwin}, and the total effect of the three is calculated using equation \eqref{totalenergyshift}.}
 \label{tab:frekvenser}
\end{table}

If I add the three effects in each row I get the total shift in frequency for the state. Comparing this with the calculated total shift (in the last column of the two tables), I see that this is equal to the sum of the three. So since the expression in \eqref{totalenergyshift}, which I used to calculate the total shift, was independent of $l$ and $s$ and gave the correct result, the total shift in frequency must be independent of $l$ and $s$.


\subsubsection*{Question c}\setcounter{thespnr}{3}
\ramme{ Draw a qualitatively correct diagram similar to figure 2.6 in \citet{atomintro}, and indicate with arrows the allowed transitions from $n=3$ 
        to $n=2$. }
The diagram is shown in figure \vref{fig:diagram}. I have indicated all direct transitions from $n=3$ levels to $n=2$ levels.%, and a couple of transitions from $n=3$ to $n=3$ levels (the transition trough here could be towards an $n=2$ level as part of a decay chain). 

The used selection rules are that $n$ can change arbitrarily, $l$ must change one up or one down ($\Delta l = \pm 1$), and $j$ can change either one up or one down, or just not change ($\Delta j = 0, \pm 1$).

\begin{figure}[htbp]
 \centering
  \psset{xunit=4mm,yunit=50mm}
  \begin{pspicture}(0,-4.95)(32,-1.4)  %\psgrid
   %Energiakse
   \psline[linecolor=black,linewidth=0.5pt]{->}(32,-3)(32,-1.4)      \uput[0](32,-1.4){$E$}
   \pszigzag[linearc=0,coilheight=1.2,coilwidth=0.25,coilarm=0.1,linecolor=black,linewidth=0.5pt,linestyle=solid](32,-3)(32,-3.3)
   \psline[linecolor=black,linewidth=0.5pt]{-}(32,-4.9)(32,-3.3)
   %Niveauer
   \psline[linecolor=black,linewidth=0.5pt,linestyle=dotted](3,-1.511)(29,-1.511)               %n=3,før
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=dotted]{3}{9}{-1.511-3*0.46934}            %n=3,l=0,rel.mass
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{3}{9}{-1.511-3*0.46934+3*0.2682}    %n=3,l=0,rel.mass+darwin
      \uput[0](9,-2.11442){\small 3s$_{1/2}$}
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=dotted]{13}{19}{-1.511-3*0.11175}          %n=3,l=1,rel.mass
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{13}{19}{-1.511-3*0.11175+3*0.0447}  %n=3,l=1,j=l+s,rel.mass+spinorbit
      \uput[0](19,-1.71215){\small 3p$_{3/2}$}
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{13}{19}{-1.511-3*0.11175-3*0.0894}  %n=3,l=1,j=l-s,rel.mass+spinorbit
      \uput[0](19,-2.11445){\small 3p$_{1/2}$}
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=dotted]{23}{29}{-1.511-3*0.0423}           %n=3,l=2,rel.mass
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{23}{29}{-1.511-3*0.0423+3*0.01788}  %n=3,l=2,j=l+s,rel.mass+spinorbit
      \uput[0](29,-1.58426){\small 3d$_{5/2}$}
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{23}{29}{-1.511-3*0.0423-3*0.02682}  %n=3,l=2,j=l-s,rel.mass+spinorbit
      \uput[0](29,-1.71836){\small 3d$_{3/2}$}
   \psline[linecolor=black,linewidth=0.5pt,linestyle=dotted](3,-3.4)(29,-3.4)                   %n=2,før
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=dotted]{3}{9}{-3.4-1.47088}                %n=2,l=0,rel.mass
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{3}{9}{-3.4-1.47088+0.90516}         %n=2,l=0,rel.mass+darwin
      \uput[0](9,-3.96572){\small 2s$_{1/2}$}
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=dotted]{13}{19}{-3.4-0.264}                %n=2,l=1,rel.mass
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{13}{19}{-3.4-0.264+0.15086}         %n=2,l=1,j=l+s,rel.mass+spinorbit
      \uput[0](19,-3.51314){\small 2p$_{3/2}$}
   \psPlot[linecolor=black,linewidth=0.5pt,linestyle=solid]{13}{19}{-3.4-0.264-0.30172}         %n=2,l=1,j=l-s,rel.mass+spinorbit
      \uput[0](19,-3.96572){\small 2p$_{1/2}$}
   %Pile:
   \psset{arrowinset=0.1,arrowsize=0.12}
   \psline[linecolor=ForestGreen,linewidth=0.5pt]{*->}(4,-1.511)(4,-2.91902)      %n=3,l=0,rel.mass
   \psline[linecolor=NavyBlue,linewidth=0.5pt]{*->}(4.5,-2.91902)(4.5,-2.11442)   %n=3,l=0,darwin
   \psline[linecolor=ForestGreen,linewidth=0.5pt]{*->}(14,-1.511)(14,-1.84625)    %n=3,l=1,rel.mass
   \psline[linecolor=BrickRed,linewidth=0.5pt]{*->}(14.5,-1.84625)(14.5,-1.71215) %n=3,l=1,j=l+s,spinorbit
   \psline[linecolor=BrickRed,linewidth=0.5pt]{*->}(14.5,-1.84625)(14.5,-2.11445)  %n=3,l=1,j=l-s,spinorbit
   \psline[linecolor=ForestGreen,linewidth=0.5pt]{*->}(24,-1.511)(24,-1.6379)     %n=3,l=2,rel.mass
   \psline[linecolor=BrickRed,linewidth=0.5pt]{*->}(24.5,-1.6379)(24.5,-1.58426)  %n=3,l=2,j=l+s,spinorbit
   \psline[linecolor=BrickRed,linewidth=0.5pt]{*->}(24.5,-1.6379)(24.5,-1.71836)  %n=3,l=2,j=l-s,spinorbit
   \psline[linecolor=ForestGreen,linewidth=0.5pt]{*->}(4,-3.4)(4,-4.87088)        %n=2,l=0,rel.mass
   \psline[linecolor=NavyBlue,linewidth=0.5pt]{*->}(4.5,-4.87088)(4.5,-3.96572)   %n=2,l=0,darwin
   \psline[linecolor=ForestGreen,linewidth=0.5pt]{*->}(14,-3.4)(14,-3.664)        %n=2,l=1,rel.mass
   \psline[linecolor=BrickRed,linewidth=0.5pt]{*->}(14.5,-3.664)(14.5,-3.51314)   %n=2,l=1,j=l+s,spinorbit
   \psline[linecolor=BrickRed,linewidth=0.5pt]{*->}(14.5,-3.664)(14.5,-3.96572)   %n=2,l=1,j=l-s,spinorbit
   %n-labels:
   \uput[180](2.5,-1.511){$n=3$}
   \uput[180](2.5,-3.4){$n=2$}
   %l-labels:
   \uput[90](6,-1.45){s}
   \uput[90](16,-1.45){p}
   \uput[90](26,-1.45){d}
   %\uput[90](6,-0.85){$l=0$}
   %\uput[90](16,-0.85){$l=1$}
   %\uput[90](26,-0.85){$l=2$}
   %Henfaldslinier:
   \psset{arrowsize=0.2,arrowinset=0.3}
   \psline[linecolor=YellowOrange]{c->}(5.75,-2.11442)(16,-3.96572)                   %3s_1/2 -> 2p_1/2
   \psline[linecolor=YellowOrange]{c->}(6,-2.11442)(15.75,-3.51314)                   %3s_1/2 -> 2p_3/2
   %\psline[linecolor=YellowOrange,linestyle=dashed]{c->}(15.75,-1.71215)(6,-2.11442)  %3p_3/2 -> 3s_1/2
   \psline[linecolor=YellowOrange]{c->}(16,-1.71215)(5.75,-3.96572)                   %3p_3/2 -> 2s_1/2
   \psline[linecolor=YellowOrange]{c->}(16,-2.11445)(6.25,-3.96572)                   %3p_1/2 -> 2s_1/2
      %\psline[linecolor=YellowOrange]{c->}(16,-6.136)(6,-17.07)                      %2p_3/2 -> 2s_1/2
   \psline[linecolor=YellowOrange]{c->}(26.25,-1.71836)(16.25,-3.96572)               %3d_3/2 -> 2p_1/2    
   \psline[linecolor=YellowOrange]{c->}(26,-1.71836)(16.5,-3.51314)                   %3d_3/2 -> 2p_3/2
   \psline[linecolor=YellowOrange]{c->}(26.25,-1.58428)(16,-3.51314)                  %3d_5/2 -> 2p_3/2
   %\psline[linecolor=YellowOrange,linestyle=dashed]{c->}(25.75,-1.71836)(16,-2.11445) %3d_3/2 -> 3p_1/2
   %\psline[linecolor=YellowOrange,linestyle=dashed]{c->}(26,-1.58428)(16,-1.71215)    %3d_5/2 -> 3p_3/2
   %Tegnforklaring:
   \psline[linecolor=ForestGreen,arrowinset=0.1,arrowsize=0.12]{*->}(13,-4.3)(16,-4.3) \uput[0](16,-4.3){\small Relativistic mass correction}
   \psline[linecolor=NavyBlue,arrowinset=0.1,arrowsize=0.12]{*->}(13,-4.4)(16,-4.4)    \uput[0](16,-4.4){\small Darwin term}
   \psline[linecolor=BrickRed,arrowinset=0.1,arrowsize=0.12]{*->}(13,-4.5)(16,-4.5)    \uput[0](16,-4.5){\small Spin-orbit interaction}
   \psline[linecolor=YellowOrange,arrowsize=0.2]{c->}(13,-4.6)(16,-4.6)                \uput[0](16,-4.6){\small Allowed transitions from $n=3$ to $n=2$}
   %\psline[linecolor=YellowOrange,linestyle=dashed,arrowsize=0.2]{c->}(13,-4.7)(16,-4.7)   
   %                                                         \uput[0](16,-4.7){\small Allowed transitions from $n=3$ to $n=3$}
   %                                                         \uput[0](16,-4.8){\small (that could be part of a decay chain)}
  \end{pspicture}
 \caption{ Diagram similar to figure 2.6 in \citet{atomintro} for the $n=2$ and $n=3$ levels in hydrogen (note that this is a sketch, the energy-axis
           scale is not consistent the whole way down). The green, blue and red arrows symbolize the fine-structure corrections to the energy levels,
           while the orange arrows are allowed transitions for decay from $n=3$ to $n=2$ levels. }
 \label{fig:diagram}
\end{figure}


\clearpage
\subsubsection*{Question d}\setcounter{thespnr}{4}
\ramme{ Calculate the frequency of the $3\text{p}_{3/2} \leftrightarrow 2\text{s}_{1/2}$ transition with 8 significant figures. Use physical constants
        from the web site \url{http://physics.nist.gov/cuu/Constants/index.html} (be sure to treat the correction for the reduced mass of the electron
        with the necessary accuracy.) }
Since the total energy with all three relativistic effects is given by
\begin{equation*}
 E_{nj} = E_n \left( 1 + \frac{\alpha^2}{n^2} \left( \frac{n}{j+1/2} - \frac{3}{4} \right) \right) ,
\end{equation*}
from page 29 in \citet{dias090505}, and I have given the energy in equation \eqref{energi} (which already takes the reduced mass into account), %such that
%\begin{equation*}
% E_{nj} = - \frac{m_e M_p}{m_e + M_p} \frac{1}{2n^2} \left( \frac{e^2}{4\pi\epsilon_0\hbar} \right)^2 
%          \left( 1 + \frac{\alpha^2}{n^2} \left( \frac{n}{j+1/2} - \frac{3}{4} \right) \right) ,
%\end{equation*}
the difference in energy of the two levels is
\begin{align*}
 E_{3,3/2} - E_{2,1/2} &= - \frac{m_e M_p}{m_e + M_p} \left( \frac{e^2}{4\pi\epsilon_0\hbar} \right)^2 
                          \bigg[ \frac{1}{2\cdot 3^2} \left( 1 + \frac{\alpha^2}{3^2} \left( \frac{3}{3/2+1/2} - \frac{3}{4} \right) \right) \\
      &\qquad\qquad              - \frac{1}{2\cdot 2^2} \left( 1 + \frac{\alpha^2}{2^2} \left( \frac{2}{1/2+1/2} - \frac{3}{4} \right) \right) \bigg] ,
\end{align*}
and this result I have to divide by $h$ to get it as a frequency. Using constants from NIST (where the ones with the smallest number of significant figures, 9, is $h$ and $m_e$), I get the energy 
\begin{equation*}
 E_{3,3/2} - E_{2,1/2} = \SI{1.88870071}{\eV} ,
\end{equation*}
and hence I get the frequency $\nu_{3,3/2} - \nu_{2,1/2} = \SI{4.5668584e5}{\GHz}$. 


\subsubsection*{Question e}\setcounter{thespnr}{5}
\ramme{ Using information from the NIST web site \url{http://physics.nist.gov/PhysRefData/ASD/lines_form.html} find the experimental frequency, and
        explain the difference between your calculation and the experimental value. }
From the NIST database, I see that the observed wavelength\footnote{Actually it is the indirectly observed wavelength, also called the Ritz wavelength, I use.} corresponding to the $3\text{p}_{3/2} \leftrightarrow 2\text{s}_{1/2}$ transition is $\SI{656.45376}{\nm}$ in vakuum. 

The wavelength corresponding to my result in question d is
\begin{equation*}
 \lambda = \frac{c}{\nu_{3,3/2} - \nu_{2,1/2}} 
         = \frac{\SI[per=frac,fraction=nice]{2.99792458e8}{\meter\per\second}}{\SI{4.5668584e5}{\GHz}}
         %= \SI{6.5645227e-7}{\m}
         = \SI{656.45227}{\nm} ,
\end{equation*}
so the difference is not big - my calculated wavelength is a little smaller than the wavelength from the database (so the calculated energy difference is a little too big). 

This is primarily explained by the Lamb shift, which shifts the energy of the $2\text{s}_{1/2}$ level a little upwards (as explained on page 40 in \citet{atomintro}). If I take this effect into account, my energy gets smaller, and hence the wavelength gets larger -- exactly as wanted. 

Apart from the Lamb shift there is also the hyperfine structure, working with the tiny magnetic dipole moment of the nucleus in the magnetic field generated by the electron -- but this effect is way smaller than the Lamb shift. 


\clearpage
\subsubsection*{Question f}\setcounter{thespnr}{6}
\ramme{ The $2\text{s}_{1/2} \rightarrow 3\text{p}_{3/2}$ transition is excited by a pulse of radiation from a laser tuned to the resonance frequency.
        The laser has an intensity of $\SI[per=frac,fraction=nice]{10}{\watt\per\milli\meter\squared}$. What is the electromagnetic energy density in the beam? }
The electromagnetic energy density $\rho$ is given by the intensity divided by the speed of light, as done in \citet{regneopgave090505}, so
\begin{equation*}
 \rho = \frac{I}{c}
      = \frac{\SI[per=frac,fraction=nice]{10e6}{\watt\per\meter\squared}}{\SI[per=frac,fraction=nice]{3.0e8}{\meter\per\second}}
      = \SI[per=frac,fraction=nice]{3.3e-2}{\joule\per\meter\cubed} .
\end{equation*}


\subsubsection*{Question g}\setcounter{thespnr}{7}
\ramme{ Evaluate the Rabi frequency. }
With the dipole approximation used, the Rabi frequency is given in equation (7.12) in \citet{atomintro}, by
\begin{equation}
 \Omega = \frac{e X_{12} |\vec E_0|}{\hbar} ,
 \label{rabi}
\end{equation}
where $X_{12} = \braket{1}{x}{2}$, for radiation linearly polarized along the $x$-axis. I hence first need to calculate the inner product $X_{12}$ and the electric field strength of the laser $|\vec E_0|$ -- and then I get $\Omega$. 

%To calculate $X_{12}$ I first have to define that $\ket{1}$ is the $2\text{s}_{1/2}$ state, while $\ket{2}$ is the $3\text{p}_{3/2}$ state. 
To calculate $X_{12}$ I find the Einstein $A$-coefficient for the transition $2\text{s}_{1/2} \rightarrow 3\text{p}_{3/2}$ in the NIST database, where it is listed as $\SI{2.245e7}{\per\second}$. I know that this coefficient is related to $D_{12}$ by equation (7.23) in \citet{atomintro} as
\begin{equation*}
 A_{12} = \frac{g_1}{g_2} \frac{4\alpha}{3c^2} \omega^3 |D_{12}|^2 ,
\end{equation*}
which is again related to $X_{12}$ by
\begin{equation*}
 X_{12} = \frac{1}{\sqrt2} D_{12} ,
\end{equation*}
as found in \citet{regneopgave090428}. I know the Einstein coefficient of the transition, I know that both states are non-degenerate (so $g_1 = g_2 = 1$), and I can calculate the angular frequency $\omega$ to be
\begin{equation*}
 \omega = 2\pi \left( \nu_{3,3/2} - \nu_{2,1/2} \right)
        = 2\pi \cdot \SI{4.567e5}{\GHz}
        = \SI{2.869e15}{\per\second} ,
\end{equation*}
and hence
\begin{equation*}
 X_{12} = \sqrt{\frac{3 c^2}{8\alpha \omega^3} A_{12}} 
        = \sqrt{\frac{3\cdot(\SI[per=frac,fraction=nice]{2.998e8}{\meter\per\second})^2}
                     {8\cdot\num{7.297e-3} \cdot (\SI{2.869e15}{\per\second})^3} \SI{2.245e7}{\per\second}} 
        = \SI{6.627e-11}{\meter} .
\end{equation*}

The electric field strength is found by noting that 
\begin{equation*}
 \rho = \frac{1}{2} \,|\vec E_0|^2 \epsilon_0  \qquad\Rightarrow\qquad
 |\vec E_0| = \sqrt{\frac{2\rho}{\epsilon_0}} ,
\end{equation*}
according to the equation at the bottom of page 8 in \citet{dias090424}. Inserting values into this gives me that (note that I have to use a lower number of significant figures, since I only know $\rho$ from question f by two significant figures)
\begin{equation*}
 |\vec E_0| = \sqrt{\frac{2\cdot \SI[per=frac,fraction=nice]{3.3e-2}{\joule\per\meter\cubed}}
                         {\SI[per=frac,fraction=nice]{8.9e-12}{\farad\per\meter}}} 
            = \SI[per=frac,fraction=nice]{8.6e4}{\volt\per\meter} .
\end{equation*}

So at last I can find the Rabi frquency by inserting $X_{12}$ and $|\vec E_0|$ into equation \eqref{rabi}
\begin{equation*}
 \Omega = \frac{\SI{1.60e-19}{\coulomb} \cdot \SI{6.6e-11}{\meter} \cdot \SI[per=frac,fraction=nice]{8.6e4}{\volt\per\meter}}
               {\SI{1.05e-34}{\joule\second}} 
        = \SI{8.6}{\GHz} .
\end{equation*}


\subsubsection*{Question h}\setcounter{thespnr}{8}
\ramme{ What must be the duration of the pulse if all atoms are in the lover level when the pulse starts, and are required to be in the upper level at
        the end of the pulse? }
The laser light is exactly as resonance, so I can use equation (7.29) in \citet{atomintro},
\begin{equation}
 |c_2(t)|^2 = \sin^2{\frac{\Omega t}{2}} 
 \label{prob}
\end{equation}
as the probability of being in the upper of the two states. When $\Omega t = \pi$ the probability becomes $1$, and all the atoms are expected to be in the upper state. This corresponds to the time
\begin{equation*}
 t = \frac{\pi}{\Omega} 
   = \frac{\pi}{\SI{8.6}{\GHz}} 
   = \SI{3.6e-10}{\second} 
   = \SI{0.36}{\nano\second} .
\end{equation*}
So the laser pulse must only be this short. 


\subsubsection*{Question i}\setcounter{thespnr}{9}
\ramme{ What fraction of the atoms is left in the excited state $\SI{10}{\nano\second}$ after the end of the pulse? }
I expect that ``the end of the pulse'' is referring to the pulse being of the length found in question h. When this pulse is over, all the atoms are in the upper level, and there is no laser light to make stimulated emission. Hence the only way for the atoms to decay into the lower level is via spontaneous emission. 

The spontaneous emission is characterized by the Einstein $A$-coefficient, which I found in the NIST database in question g to be $\SI{2.245e7}{\per\second}$. From equation (1.27) in \citet{atomintro} I then know that the number of atoms in the upper state after time $t$ is
\begin{equation*}
 N_2(t) = N_2(0) \exp{(-A_{12} t)} ,
\end{equation*}
where $N_2(0)$ is the number of atoms in the upper state at time $t=0$. So if I simply set $N_2(0) = 1$, I get $N_2(t)$ to be the fraction of the atoms in the state. After $\SI{10}{\nano\second}$ the fraction of atoms in the upper state is hence
\begin{equation*}
 N_2(\SI{10}{\nano\second}) = \exp{(- \SI{2.245e7}{\per\second} \cdot \SI{10}{\nano\second})} 
                            = 0.80
                            \approx \frac{4}{5} .
\end{equation*}




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