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Clean up observables section

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Andreas Tsouchlos
2026-04-18 20:33:38 +02:00
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src/thesis/chapters/2_fundamentals.tex
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@@ -593,9 +593,9 @@ decoding of subsequent blocks \cite[Sec.~III.~C.]{hassan_fully_2016}.
Designing codes and decoders for \ac{qec} is generally performed on a
layer of abstraction far removed from the quantum mechanical
processes underlying the actual qubits.
processes underlying the actual physics.
Nevertheless, having a fundamental understanding of the related
quantum mechanical concepts is useful to understand the unique constraints
quantum mechanical concepts is useful to grasp the unique constraints
of this field.
The purpose of this section is to convey these concepts to the reader.
@@ -605,11 +605,12 @@ The purpose of this section is to convey these concepts to the reader.
% Wave functions
In quantum mechanics, the evolution of a state of a particle over tme
and space is described by a \emph{wave function} $\psi(x,t)$.
The connection between this function and the world that we can observe
is the fact that $\lvert \psi (x,t) \rvert^2$ is the \ac{pdf} of
finding a praticle in that particular state.
In quantum mechanics, the state of a particle is described by a
\emph{wave function} $\psi(x,t)$.
The connection between this function and the observable world
is Born's statistical interpretation:
$\lvert \psi (x,t) \rvert^2$ is the \ac{pdf} of finding a praticle at
position $x$ and time $t$ \cite[Sec.~1.2]{griffiths_introduction_1995}.
% Dirac notation
@@ -627,24 +628,32 @@ Their inner product is $\braket{a\vert b}$.
% Expressing wave functions using linear algebra
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)
.%
\end{align*}
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}
.%
\end{align*}
The connection we will make between quantum mechanics and linear
algebra is that we will model the state space of a system as a
\emph{function space}.
We will represent the state of a particle with wave function
$\psi(x,t)$ using the vector $\ket{\psi}$
\cite[Sec.~3.3]{griffiths_introduction_1995}.
% 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)
% .%
% \end{align*}
% 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}
% .%
% \end{align*}
% Operators
Another important notion is that of an \emph{operator}, a component
that takes a function as an input and returns another function as an output.
Another important notion is that of an \emph{operator}, a transformation
that takes a function as an input and returns another function as an
output \cite[Sec.~3.2.2]{griffiths_introduction_1995}.
Operators are useful to describe the relations between different
quantities relating to a particle.
An example of this is the differential operator $\partial x$.
@@ -655,18 +664,22 @@ An example of this is the differential operator $\partial x$.
% Observable quantities
An \emph{observable quantity} $Q$ is \ldots .
Due to the probabilistic nature of quantum mechanics, the result of a
measurement is not deterministic.
Thus, it is useful to consider the \emph{expected value} $\braket{Q}$
of an observable quantity in addition to individual measurement results.
An \emph{observable quantity} $Q(x,p,t)$ is a quantity of a quantum
mechanical system that we can measure, such as the position $x$ or
momentum $p$ of a particle.
In general, such measurements are not deterministic, i.e.,
measurements on identically prepared states can yield different results.
There are some states, however, that are \emph{determinate} for a
specific observable: measuring those will always yield identical
observations \cite[Sec.~3.3]{griffiths_introduction_1995}.
% General expression for expected value of observable quantity
If we know the wave function of a particle, we should be able to
compute $\braket{Q}$ for any observable quantity we wish.
compute the expected value $\braket{Q}$ of any observable quantity we wish.
It can be shown that for any $Q$, we can compute a
corresponding operator $\hat{Q}$ such that%
corresponding operator $\hat{Q}$ such that
\cite[Sec.~3.3]{griffiths_introduction_1995}
\begin{align}
\label{eq:gen_expr_Q_exp}
\braket{Q} = \int_{-\infty}^{\infty} \psi^*(x,t) \hat{Q} \psi(x,t) dx
@@ -675,7 +688,8 @@ corresponding operator $\hat{Q}$ such that%
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%
operator to $\hat{Q} = x$, we can write
\cite[Sec.~1.5]{griffiths_introduction_1995}
\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
@@ -687,14 +701,10 @@ formula simplifies to the direct calculation of the expected value.
% Determinate states and eigenvalues
% TODO: Introduce determinate states above
% TODO: Nicer phrasing
% TODO: Use different symbol for determinate states (not psi)
% TODO: Fix equation
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%
the determinate states of the observable quantity.
We begin by translating \autoref{eq:gen_expr_Q_exp} into linear alebra as
\cite[Eq.~3.114]{griffiths_introduction_1995}
\begin{align}
\label{eq:gen_expr_Q_exp_lin}
\braket{Q} = \braket{\psi \vert \hat{Q}\psi}
@@ -703,53 +713,79 @@ We begin by translating \autoref{eq:gen_expr_Q_exp} into linear alebra as%
\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.%
To relate the two, we note that since determinate states should
always yield the same measurement results, the variance of the
observable should be zero.
We thus compute \cite[Eq.~3.116]{griffiths_introduction_1995}
\begin{align}
0 &\overset{!}{=} \braket{(Q - \braket{Q})^2}
= \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} =
= \braket{e_n \vert (\hat{Q} - \braket{Q})^2 e_n} \nonumber\\
&= \braket{(Q - \braket{Q})e_n \vert (\hat{Q} - \braket{Q})
e_n} \nonumber\\
&= \lVert (Q - \braket{Q}) e_n \rVert^2 \nonumber\\[3mm]
&\hspace{-8mm}\Leftrightarrow (\hat{Q} - \braket{Q}) \ket{e_n} =
0 \nonumber\\
\label{eq:observable_eigenrelation}
&\hspace{-8mm}\Leftrightarrow \hat{Q}\ket{\psi}
= \underbrace{\braket{Q}}_{\lambda_n} \ket{\psi}
&\hspace{-8mm}\Leftrightarrow \hat{Q}\ket{e_n}
= \underbrace{\braket{Q}}_{\lambda_n} \ket{e_n}
.%
\end{align}%
%
Because we have assumed the variance to be zero, $\braket{Q}$ is now
the deterministic measurement value corresponding to the determinate
state $\ket{\psi}$.
Because we have assumed the variance to be zero, the expected value
$\braket{Q}$ is now the deterministic measurement result
corresponding to the determinate state
$\ket{e_n},~n\in \mathbb{N}$.
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} $\lambda_n$.
the observable operator $\hat{Q}$ and that the measurement values are
the corresponding \emph{eigenvalues} $\lambda_n$
\cite[Postulate~3]{griffiths_introduction_1995}.
% 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*}
As we are modelling the wave function $\psi(x,t)$ as a vector
$\ket{\psi}$, we can find a set of basis vectors to decompose it into.
We can use the determinate states for this purpose, expressing the state as%
\footnote{
We are only considering the case of having a \emph{discrete
spectrum} here, i.e., having a discrete set of eigenvalues and vectors.
For continuous spectra, the procedure is analogous.
}
\begin{align}
\label{eq:determinate_basis}
\ket{\psi} = \sum_{n=1}^{\infty} c_n \ket{e_n}, \hspace{3mm}
c_n := \braket{e_n \vert \psi}
.%
\end{align}
Inserting \autoref{eq:determinate_basis} into
\autoref{eq:gen_expr_Q_exp_lin} we obtain
% tex-fmt: off
\cite[Prob.~3.35c)]{griffiths_introduction_1995}
% tex-fmt: on
\begin{align*}
\braket{Q} = \left( \sum_{n=1}^{\infty} c_n \bra{e_n} \right)
\left( \sum_{m=1}^{\infty} c_m\hat{Q}\ket{e_m} \right)
= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} c_n c_m
\lambda_m\braket{e_n \vert e_m}
= \sum_{n=1}^{\infty} \lambda_n \lvert c_n \rvert ^2
.%
\end{align*}
We can thus interpret $\lvert c_n \rvert ^2$ as the probability of
obtaining value $\lambda_n$ from the measurement.
% 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
observable using \autoref{eq:gen_expr_Q_exp_lin}, and describe the
individual determinate states and corresponding measurement values
using \autoref{eq:observable_eigenrelation}.
To summarize, we mathematically model an observable quantity
$Q(x,t,p)$ using a corresponding operator $\hat{Q}$, which allows us
to compute the expected value as $\braket{Q} = \braket{\psi
\vert \hat{Q} \psi}$.
The eigenvectors of $\hat{Q}$ are the determinate states
$\ket{e_n},~n\in \mathbb{N}$ and the eigenvalues are the respective
measurement outcomes.
We can decompose an arbitrary state as $\ket{\psi} = \sum_{n=1}^{\infty} c_n
\ket{e_n}$, where $\lvert c_n \rvert ^2$ represents the probability
of obtaining a certain measurement value.
Note that when we speak of an \emph{observable}, we are usually
refering to the corresponding operator $\hat{Q}$.
%%%%%%%%%%%%%%%%
\subsection{Projective Measurements}
@@ -757,7 +793,7 @@ using \autoref{eq:observable_eigenrelation}.
% Projective measurements
% TODO: Better introduce the collapse of the superposition state
% TODO: Introduce concept of collapsing the wave function onto a basis 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}.