Add first draft of projective measurements

src/thesis/chapters/2_fundamentals.tex

@@ -90,7 +90,6 @@ We can arrange the coefficients of these equations in a
 \textit{parity-check matrix} (\acs{pcm}) $\bm{H} \in
 \mathbb{F}_2^{(n-k) \times n}$ and equivalently define the code as
 \cite[Sec.~3.1.1]{ryan_channel_2009}
-%
 \begin{align*}
     \mathcal{C} = \left\{ \bm{x} \in \mathbb{F}_2^n :
     \bm{H}\bm{x}^\text{T} = \bm{0} \right\}
@@ -630,7 +629,6 @@ Their inner product is $\braket{a\vert b}$.
 
 We can model a wave function $\psi(x,t)$ as a linear combination of different
 \emph{basis functions} $e_n(x,t),~n\in \mathbb{N}$ as%
-%
 \begin{align*}
     \psi(x,t) = \sum_{n=1}^{\infty} c_n \cdot e_n(x,t)
     .%
@@ -638,7 +636,6 @@ We can model a wave function $\psi(x,t)$ as a linear combination of different
 To express this relation using linear algebra, we represent
 $\psi(x,t)$ and $e_n(x,t)$ as vectors $\ket{\psi}$ and $\ket{e_n}$.
 We write%
-%
 \begin{align*}
     \ket{\psi} = \sum_{n=1}^{\infty}  c_n \ket{e_n}
     .%
@@ -670,25 +667,21 @@ If we know the wave function of a particle, we should be able to
 compute $\braket{Q}$ for any observable quantity we wish.
 It can be shown that for any $Q$, we can compute a
 corresponding operator $\hat{Q}$ such that%
-%
 \begin{align}
     \label{eq:gen_expr_Q_exp}
     \braket{Q} = \int_{-\infty}^{\infty} \psi^*(x,t) \hat{Q} \psi(x,t) dx
     .%
 \end{align}%
-%
 While the derivation of this relationship is out of the scope of this
 work, we can at least look at an example to illustrate it.
 Considering the position $Q = x$ of a particle and setting the observable
 operator to $\hat{Q} = x$, we can write%
-%
 \begin{align*}
     \braket{x} = \int_{-\infty}^{\infty} \psi^*(x,t) \cdot x \cdot \psi(x,t) dx
     = \int_{-\infty}^{\infty} x \lvert \psi(x,t) \rvert ^2 dx
     .%
 \end{align*}
-%
+Note that $\lvert \psi(x,t) \rvert^2 $ represents the \ac{pdf} of
-Note that $\lvert \psi(x,t) \rvert $ represents the \ac{pdf} of
 finding a particle in a specific state. We immediately see that the
 formula simplifies to the direct calculation of the expected value.
 
@@ -702,40 +695,55 @@ Let us now examine how the observable operator $\hat{Q}$ relates to
 the determinate states that make up the overall superposition state
 of the particle.
 We begin by translating \autoref{eq:gen_expr_Q_exp} into linear alebra as%
-%
 \begin{align}
     \label{eq:gen_expr_Q_exp_lin}
     \braket{Q} = \braket{\psi \vert \hat{Q}\psi}
     .%
 \end{align}
-%
 \autoref{eq:gen_expr_Q_exp_lin} expresses an inherently probabilistic
 relationhip.
 The determinate states are inherently deterministic.
 To relate the two, we look at those states $\ket{\psi}$, where the
 variance of the measurements of $Q$ is zero. These are exactly the
 determinate states.%
-%
+\begin{align}
-\begin{align*}
     0 &\overset{!}{=} \braket{(Q - \braket{Q})^2}
-    = \braket{\psi \vert (\hat{Q} - \braket{Q})^2 \psi} \\
+    = \braket{\psi \vert (\hat{Q} - \braket{Q})^2 \psi} \nonumber\\
-    &= \braket{(Q - \braket{Q})\psi \vert (\hat{Q} - \braket{Q}) \psi} \\
+    &= \braket{(Q - \braket{Q})\psi \vert (\hat{Q} - \braket{Q})
-    &= \lVert (Q - \braket{Q}) \psi \rVert^2 \\[3mm]
+    \psi} \nonumber\\
-    &\hspace{-8mm}\Leftrightarrow (\hat{Q} - \braket{Q}) \ket{\psi} = 0 \\
+    &= \lVert (Q - \braket{Q}) \psi \rVert^2 \nonumber\\[3mm]
+    &\hspace{-8mm}\Leftrightarrow (\hat{Q} - \braket{Q}) \ket{\psi} =
+    0 \nonumber\\
     \label{eq:observable_eigenrelation}
-    &\hspace{-8mm}\Leftrightarrow \hat{Q}\ket{\psi}  = \braket{Q} \ket{\psi}
+    &\hspace{-8mm}\Leftrightarrow \hat{Q}\ket{\psi}
+    = \underbrace{\braket{Q}}_{\lambda_n} \ket{\psi}
     .%
-\end{align*}%
+\end{align}%
 %
 Because we have assumed the variance to be zero, $\braket{Q}$ is now
 the deterministic measurement value corresponding to the determinate
 state $\ket{\psi}$.
 We can see that the determinate states are the \emph{eigenstates} of
 the observable operator $\hat{Q}$ and that the corresponding
-(deterministic) measurement values are the corresponding \emph{eigenvalues}.
+(deterministic) measurement values are the corresponding
+\emph{eigenvalues} $\lambda_n$.
+
+% Determinate states as a basis
+
+% TODO: Rephrase
+% TODO: Show that |c_n|^2 is the probability of finding a particle in
+% a given state
+% In particular, using the determinate states $\ket{e_n}$ as a basis to
+% write the superimposed state
+% \begin{align*}
+%     \ket{\psi} = \sum_{n=1}^{\infty} c_n \ket{e_n}
+%     ,
+% \end{align*}
 
 % Recap
 
+% TODO: Mention that `observable` is used to refer to the observable operator
+% TODO: Mention eigenstates and eigenvalues again
 To summarize, we can mathematically express any observable quantity
 $Q$ using a corresponding operator $\hat{Q}$.
 This operator allows us to both compute the expected value of the
@@ -747,22 +755,67 @@ using \autoref{eq:observable_eigenrelation}.
 \subsection{Projective Measurements}
 \label{subsec:Projective Measurements}
 
+% Projective measurements
+
+% TODO: Better introduce the collapse of the superposition state
+The measurements we considered in the previous section, for which
+\autoref{eq:gen_expr_Q_exp_lin} holds, belong to the category of
+\emph{projective measurements}.
+For these, certain restrictions such as repeatability apply: after
+measuring a quantum state and thus collapsing it onto one of the
+determinate states, futher measurements should yield the same value.
+More general methods of modelling measurements exist, e.g., describing
+destructive measurements, but they are not relevant to us here
+\cite[Box~2.5]{nielsen_quantum_2010}.
+
+% Projection operators
+
+% TODO: Fix notational issues related to e_n
+We can model the collapse of the original state onto one of the
+superimposed basis states as a \emph{projection}.
+To see this, we insert \autoref{eq:determinate_basis} into
+\autoref{eq:observable_eigenrelation}, obtaining%
+\begin{align*}
+    \hat{Q}\ket{\psi} = \sum_{n=1}^{\infty} c_n \hat{Q} \ket{e_n}
+    = \sum_{n=1}^{\infty} \lambda_n c_n \ket{e_n}
+    .%
+\end{align*}%
+We see that $\hat{Q}$ has the effect of multiplying the component
+along each basis vector with the corresponding eigenvalue.
+We decompose $\hat{Q}$ into its constituent parts that act on each of
+the separate components as
+\begin{align*}
+    \hat{Q} = \sum_{n=1}^{\infty} \lambda_n \hat{P}_n
+\end{align*}
+using \emph{projection operators}
+\begin{align*}
+    \hat{P}_n := \ket{e_n}\bra{e_n}, \hspace{3mm} n\in \mathbb{N}
+    .
+\end{align*}%
+These project a vector onto the subspace spanned by $\ket{e_n}$.
+
+% Using projection operators to measure if a state has a component
+% along a basis vector
+
+A particularly interesting property of projection operators is that
+\begin{align*}
+    \hat{P}_n (\hat{P}_n \ket{\psi}) = \hat{P}_n^2 \ket{\psi}
+    = \hat{P}_n \ket{\psi},
+\end{align*}%
+and the only way this can hold for any $\ket{\psi}$ is if $\hat{P}_n$
+only has the eigenvalues $0$ or $1$.
+As explained in the previous section, the eigenvalues are the results
+of performing a measurement.
+We can thus use the projection operator as an observable and treat
+the eigenvalue as an indicator of the state having a component along
+the related basis vector.
+
 %%%%%%%%%%%%%%%%
 \subsection{Qubits and Multi-Qubit States}
 \label{subsec:Qubits and Multi-Qubit States}
 
 \red{
     \begin{itemize}
-        \item Projective measurements
-            \begin{itemize}
-                \item Using the determinate states as a basis for the
-                    superposition state
-                \item The effect of using the obserrvable operator on
-                    the superposition state
-                \item Modelling the observable operator as a series
-                    of projections
-                \item The projection operator and its eigenvalues
-            \end{itemize}
         \item Qubits and multi-qubit states
             \begin{itemize}
                 \item The qubit