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\title{MA3441 Problem Set 3}
\author{Professor Fry (typset by Mark Allen)}
\date{Issued: 17th October 2012\\ Due: 24th October 2012 (optional)} % Activate to display a given date or no date (if empty),
         % otherwise the current date is printed 

\begin{document}
\maketitle

\section*{Disclaimer}

Everything I've written might be completely wrong. Check that what I wrote matches with what Prof. Fry wrote, especially if you keep getting the wrong answer.

\section*{Problems}
\begin{enumerate}
\item
Two-state quantum systems.

Examples:
\begin{itemize}
\item
Ammonia molecule NH$^3$
\begin{figure}[htb]
\centering
\includegraphics[width=0.5\textwidth]{qmhw4img1.png}
\label{fig:awesome_image}
\end{figure}
\item
Hydrogen molecular ion H$^+_2$
\begin{figure}[htb]
\centering
\includegraphics[width=0.5\textwidth]{qmhw4img2.png}
\label{fig:awesome_image}
\end{figure}
\pagebreak
\item
Magenta dye

\begin{figure}[!htb]
\centering
\includegraphics[width=0.8\textwidth]{qmhw4img3.png}
\label{fig:awesome_image}
\end{figure}

\item
System with many energy levels, confining our attention to transitions between states $\ket{1}$ and $\ket{2}$ only.
\begin{figure}[!htb]
\centering
\includegraphics[width=0.15\textwidth]{qmhw4img4.png}
\label{fig:awesome_image}
\end{figure}
\end{itemize}
If a two-level system were not allowed to make the transition $\ket{1} \leftrightarrow \ket{2}$ its Hamiltonian operator would be
\[
\left( \begin{matrix}
H_{11} & 0 \\ 0 & H_{22}
\end{matrix} \right)
\]

For the first three examples above, $H_{11} = H_{22} = E_0$ by symmetry. $E_0$ is real. For the last example, $H_{11} = E_1$, $H_{22} = E_2$. For the rest of this problem, we consider the case $H_{11} = H_{22} = E_0$. Then eigenstates of $H$ are 
\[
\ket{1} = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \;\;,\;\; \ket{2} = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right),
\]
and their time dependence is derived from Schr\"{o}dinger equation:
\[
i\hbar \frac{\partial}{\partial t} \ket{1,t} = E_0 \ket{1,t}
\]
\[
\implies \ket{1,t} = e^{-iE_0t/\hbar}\ket{1,0}= e^{-iE_0t/\hbar}\left( \begin{matrix} 1 \\ 0 \end{matrix}\right)
\]
Similarily for $\ket{2,t}$. Then, once in state $\ket{1}$ or $\ket{2}$ the system remains in it forever. Now suppose we allow for the possibility of the transition $\ket{1} \leftrightarrow \ket{2}$ by introducing off-diagonal elements in $H$. Let
\[
H_{12} = -A
\]
Since $H^*_{12} = H_{21}$, $ H_{21} = -A^*$. For simplicity, assume $A^*=A$. Then
\[
H =\left( \begin{matrix} E_0 & -A \\ -A & E_0 \end{matrix}\right)
\]
$A$ corresponds to a tunnelling process. Thus, consider the H$^+_2$ ion: the electron detaches itself from one proton and tunnels through the potential barrier to the other. The business of QM is to calculate $A$.
\begin{figure}[!htb]
\centering
\includegraphics[width=0.8\textwidth]{qmhw4img5.png}
\label{fig:awesome_image}
\end{figure}
\begin{enumerate}
\item
Find the eigenvalues of $H$. Ans: $E_0 \pm A$.
\item
Find the normalized eigenstates of $H$ in terms of $\ket{1}$ and $\ket{2}$. 

Ans: $\ket{\pm} = \frac{1}{\sqrt{2}}( \ket{1} \mp \ket{2})$
\end{enumerate}
Conclusions:
\begin{itemize}
\item
The N atom in NH$_3$ and the electron in H$^+_2$ are not localized. The energy difference 2A is responsible for the ammonia maser and the one-electron bond, respectively.
\item
The dye magenta has two energy states, neither of which is $\ket{1}$ or $\ket{2}$, but a superposition of them. This is the basis for its colour: photons with energy $\hbar \omega = 2A$ are absorbed in the optical region.

More reading: Feynman Lectures, Vol. III, Chapters 8-10.
\end{itemize}
\item
\begin{enumerate}
\item
Consider a spin-$\frac{1}{2}$ particle. It can have `spin up' $(=\frac{\hbar}{2})$ or `spin down' $(=-\frac{\hbar}{2})$ with respect to some axis in space. The following are basis states:
\begin{align*}
\ket{\uparrow}_x &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}   & 
\ket{\uparrow}_y &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}   &
\ket{\uparrow}_z &= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
\intertext{\hphantom{TTTTTTTTTTTTTTTT} or \hphantom{TTTTTTTTTTTTT} or }
\ket{\downarrow}_x &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}   & 
\ket{\downarrow}_y &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}   &
\ket{\downarrow}_z &= \begin{pmatrix} 0 \\ 1 \end{pmatrix}
\end{align*}
We already know that the spin angular momentum operators for the $x$ and $z$ directions are
\[
S_x = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix},\;\; S_z = \frac{\hbar}{2}\begin{pmatrix}1  & 0 \\ 0 & -1 \end{pmatrix}
\]
Find $S_y$ by construction an appropriate unitary transformation.
\item
Show that $[ S_x,S_y] = i\hbar S_z + \text{cyclic perms of } x,y,z$.
\end{enumerate}
\item
\begin{enumerate}
\item
Construct the normalised spin state $\ket{\uparrow}_{\hat{n}}$ with spin $+\frac{\hbar}{2}$ in the direction $\mathbf{\hat{n}}$ (see fig.) where $\mathbf{\hat{n}}$ is a unit vector. Use the results for $S_x$, $S_y$, $S_z$ in problem 2. 

\begin{figure}[!htb]
\centering
\includegraphics[width=0.2\textwidth]{qmhw4img6.png}
\label{fig:awesome_image}
\end{figure}

Hint: You want $\mathbf{S} \cdot \mathbf{\hat{n}} \ket{\uparrow}_{\hat{n}} = + \frac{\hbar}{2}\ket{\uparrow}_{\hat{n}}$.

Ans: $ \ket{\uparrow}_{\hat{n}} = \cos{\left( \frac{\beta}{2} \right)} \ket{\uparrow}_z + e^{i\alpha}\sin{\left( \frac{\beta}{2} \right)}\ket{\downarrow}_z $
\item
Suppose $S_x$ is measured. What is the probability for getting $+\frac{\hbar}{2}$?

Ans: $\frac{1}{2}(1 + \cos{\alpha}\sin{\beta})$
\end{enumerate}
\item
One way to get the basic commutator
\[
[x_i,p_j] = i\hbar \delta_{ij},\;\; i,j = 1,2,3.
\]
Let $\ket{\psi}$ be the state of a system. Suppose the \emph{system} is translated by $\mathbf{a}$ and that its new state is $\ket{\psi'}$.

\begin{figure}[!htb]
\centering
\includegraphics[width=0.4\textwidth]{qmhw4img7.png}
\label{fig:awesome_image}
\end{figure}

Just as for a time translation, require norm preservation:
\[
\braket{\psi'}{\psi'} = \braket{\psi}{\psi}.
\]
This is guaranteed if states are related by a unitary transformation $U$:
\[
U(\mathbf{a})\ket{\psi} = \ket{\psi'},\;\;\; U^+ = U^{-1}.
\]
For an infinitesimal translation let
\[
U(\delta \mathbf{a}) = 1 - i\delta\mathbf{a} \cdot \mathbf{P}/\hbar
\]
where
\[
\mathbf{P}^+ = \mathbf{P}
\]
and by construction $[\mathbf{P}] = $ momentum. This is probably the most basic definition of momentum: the generator of space translations.

Let $\mathbf{X}$ be the position operator. Assume that the expectation value of $\mathbf{X}$ in any state undergoes a translation $\mathbf{a}$ when the state is translated by $\mathbf{a}$. Then
\[
\bra{\psi'}\mathbf{X}\ket{\psi'} = \bra{\psi} \mathbf{X} \ket{\psi} + \mathbf{a}\tag{*}
\]
Specialise (*) to an infinitesimal translation and show that
\[
[X_i,p_j] = i\hbar\delta_{ij},\;\;\; i,j = 1,2,3.
\]
\end{enumerate}
\end{document}
