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Add first draft of projective measurements

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Andreas Tsouchlos
2026-04-18 17:51:13 +02:00
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src/thesis/chapters/2_fundamentals.tex
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@@ -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 $ represents the \ac{pdf} of
Note that $\lvert \psi(x,t) \rvert^2 $ 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{(Q - \braket{Q})\psi \vert (\hat{Q} - \braket{Q}) \psi} \\
&= \lVert (Q - \braket{Q}) \psi \rVert^2 \\[3mm]
&\hspace{-8mm}\Leftrightarrow (\hat{Q} - \braket{Q}) \ket{\psi} = 0 \\
= \braket{\psi \vert (\hat{Q} - \braket{Q})^2 \psi} \nonumber\\
&= \braket{(Q - \braket{Q})\psi \vert (\hat{Q} - \braket{Q})
\psi} \nonumber\\
&= \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