A template and examples for upper level physics students taking quantum mechanics at St. Mary's College of Maryland. Includes examples of kets and column vectors.
\begin
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\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{depreequantum}
\title{Quantum Homework Template}
\author{Erin De Pree}
\date{Fall 2019}
\usepackage{natbib}
\usepackage{graphicx}
\begin{document}
\maketitle
\section*{Suppressing Section Numbering}
If you are going to label your sections by problem number, then it is a good idea to use \verb+\section*+ instead of \verb+\section+ so you can suppress the section numbering. This way you won't write ``3 Problem 2''!
Here, I'm going to work through problem 1.5 from our text. This will show you how to write kets and such. I've also included a section on matrices so you can see how to write them up.
Be sure to include the \verb+depreequantum.sty+ style file as it loads the necessary packages.
\section*{Problem 1.5}
A beam of spin-1/2 particles is prepared in the state
\[ \ket{\psi} = \tfrac{2}{\sqrt{13}} \ket{+} + i \tfrac{3}{\sqrt{13}} \ket{-} \]
\paragraph{Part a} What are the possible results of a measurement of the spin component $S_z$, and with probabilities would they occur?
You will either measure $S_z = +\hbar/2$ (spin up) or $S_z = - \hbar/2$ (spin down).
The probability of measuring $S_z = +\hbar/2$ is
\begin{align*}
\mathcal{P}_+ &= \abs{ \braket{+}{\psi} }^2
= \abs{ \bra{+} \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{2}{\sqrt{13}} \braket{+} + i \frac{3}{\sqrt{13}} \braket{+}{-} }^2\\
&= \abs{ \frac{2}{\sqrt{13}} + 0 }^2
= \boxed{ \frac{4}{13} \simeq 31\% }
\end{align*}
While the probability of measuring $S_z = - \hbar/2$ is
\begin{align*}
\mathcal{P}_- &= \abs{ \braket{-}{\psi} }^2
= \abs{ \bra{-} \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{2}{\sqrt{13}} \braket{-}{+} + i \frac{3}{\sqrt{13}} \braket{-} }^2 \\
&= \abs{ 0 + i \frac{3}{\sqrt{13}} }^2
= \qty( -i\frac{3}{\sqrt{13}} ) \qty( i \frac{3}{\sqrt{13}} ) \\
&= -i^2 \frac{9}{13}
= \boxed{ \frac{9}{13} \simeq 69\% }
\end{align*}
\paragraph{Part b} What are the possible results of a measurement of the spin component $S_x$, and with probabilities would they occur?
Measure $S_x = \pm \hbar/2$.
Probability of measuring $S_x = +\hbar/2$ is
\begin{align*}
\mathcal{P}_{+x} &= \abs{ {}_x\braket{+}{\psi} }^2
= \abs{ \qty( \frac{1}{\sqrt{2}} \bra{+} + \frac{1}{\sqrt{2}} \bra{-} ) \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{1}{\sqrt{2}} \frac{2}{\sqrt{13}} + i \frac{1}{\sqrt{2}} \frac{3}{\sqrt{13}} }^2
= \abs{ \frac{2 + 3i}{\sqrt{26}} }^2 \\
&= \frac{1}{26} \qty( 2 - 3i ) \qty( 2+3i )
= \frac{ 4 - 6i + 6i - 9i^2}{26} \\
&= \frac{4 + 9}{26}
= \frac{13}{26}
= \boxed{ \frac{1}{2} = 50\% }
\end{align*}
Probability of measuring $S_x = -\hbar/2$ is
\begin{align*}
\mathcal{P}_{-x} &= \abs{ {}_x\braket{-}{\psi} }^2
= \abs{ \qty( \frac{1}{\sqrt{2}} \bra{+} -\frac{1}{\sqrt{2}} \bra{-} ) \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{1}{\sqrt{2}} \frac{2}{\sqrt{13}} - i \frac{1}{\sqrt{2}} \frac{3}{\sqrt{13}} }^2
= \abs{ \frac{2 - 3i}{\sqrt{26}} }^2 \\
&= \frac{1}{26} \qty( 2 + 3i ) \qty( 2-3i )
= \frac{ 4 - 6i + 6i - 9i^2}{26} \\
&= \frac{4 + 9}{26}
= \frac{13}{26}
= \boxed{ \frac{1}{2} = 50\% }
\end{align*}
\paragraph{Part c} Histograms! See \cref{fig:HistrogramsProblem1.5} for the histograms. Also note the figure appears on the next page because \LaTeX{} is automatically placing it where it makes most sense (after it is referenced).
\begin{figure}
\begin{center}
\begin{tikzpicture}[>=latex]
\begin{scope}
\draw[->] (0,0) -- (6,0) node[right]{$S_z$};
\draw[->] (0,0) -- (0,5) node[right]{$\mathcal{P}$};
\foreach \x / \spin / \prob in {2/-/0.31, 4/+/0.69}
{ \draw[fill=Grey] (\x,0) ++(-0.5, 0) rectangle ++(1,4*\prob) ++(-0.5,0) node[above]{$\prob$};
\draw (\x, -0.25) node[below]{$\spin \hbar/2$} -- ++(0, 0.5);
}
\foreach \y / \prob in {1/0.25, 2/0.50, 3/0.75, 4/1}
{ \draw (-0.2, \y) node[left]{$\prob$} -- ++(0.4,0); }
\end{scope}
\begin{scope}[xshift=9cm]
\draw[->] (0,0) -- (6,0) node[right]{$S_x$};
\draw[->] (0,0) -- (0,5) node[right]{$\mathcal{P}$};
\foreach \x / \spin / \prob in {2/-/0.5, 4/+/0.5}
{ \draw[fill=Grey] (\x,0) ++(-0.5, 0) rectangle ++(1,4*\prob) ++(-0.5,0) node[above]{$\prob$};
\draw (\x, -0.25) node[below]{$\spin \hbar/2$} -- ++(0, 0.5);
}
\foreach \y / \prob in {1/0.25, 2/0.50, 3/0.75, 4/1}
{ \draw (-0.2, \y) node[left]{$\prob$} -- ++(0.4,0); }
\end{scope}
\end{tikzpicture}
\end{center}
\caption{Probabilities of measuring various spins in problem 1.5. You may find it easier to make your histograms in Excel or Google Sheets or by hand and include them as a picture.}
\label{fig:HistrogramsProblem1.5}
\end{figure}
\section*{Matrix Notation}
Going to use $\dot{=}$ to mean \emph{represented by}. In \LaTeX\,, use the notation: \verb+\dot{=}+. The quantum state isn't a column vector, but it can be represented by a column vector
\begin{align}
\ket{+} &\dot{=} \mqty( 1 \\ 0 )
& \ket{-} &\dot{=} \mqty( 0 \\ 1)
\label{eq:basismatrix}
\end{align}
This is the $z$-basis set.
Let's use the column vector notation to represent a state $\ket{\psi}$ as a projection of the state on the $\ket{+}$ and the projection on the $\ket{-}$:
\[ \ket{\psi} \dot{=} \mqty( \braket{+}{\psi} \\ \braket{-}{\psi} ) \]
\[ \ket{\psi} = a \ket{+} + b \ket{-} \]
\[ \ket{\psi} \dot{=} \mqty( a \\ b ) \]
\[ \bra{\psi} \dot{=} \mqty( a^\ast & b^\ast ) \]
We can also write the $x$ and $y$ kets in the $z$-basis:
\begin{align*}
\ket{+}_x &\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ 1)
& \ket{+}_y&\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ i ) \\
\ket{-}_x &\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ -1 )
& \ket{-}_y &\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ -i )
\end{align*}
\end{document}