Appendix C. Statistical states as vectors
https://doi.org/10.48550/arXiv.2403.13981
This appendix shows how the position probability density function ${\pdf}$ of a set of particles can be represented as an element of a Hilbert-Lebesgue space. It is straightforward to adapt it to the probability density function $\pdfO$ of an arbitrary observable, ${\Obs}$.
This appendix deviates from the convention that ${x}$ represents a single coordinate of a single particle’s position vector (see Sec. 2). In this appendix ${x\in\domain}$ is a set of ${\Ndof}$ of coordinates. Therefore it is a specification of the microstructure of a system with $\Ndof$ degrees of freedom.
For simplicity, it will be assumed that ${\domain\subset\realone^{\Ndof}}$, where ${0<\abs{\domain}<\infty}$. It will also be assumed that the precisions of all measurements are finite.
C.1 Statistical states as rays
The position pdf of an arbitrary system of ${\Nparticle=\Ndof/\dimension}$ classical particles confined to ${\domain}$ is \begin{align*}
\pdf:\domain\to\realpos; x\mapsto \pdf(x),
\end{align*}
where ${x\in\domain}$ and the integral of ${\pdf(x)}$ over the set $\domain$ of all microstructures is unity.
The information possessed by $\pdf$ can be specified by a square-integrable function,
\begin{align*}
\Psi(x) = r e^{i\theta(x)}\sqrt{\pdf(x)}\in \lebesgue(\domain),
\end{align*}
where ${r\in\realpos}$ is arbitrary and finite, and ${\theta:\domain\to\realone}$ is arbitrary and may be constant. The position pdf can be retrieved from $\Psi$ as follows: \begin{align*}
\pdf(x) = \frac{\Psi^*(x)\Psi(x)}{\int_\domain\Psi^*(x')\Psi(x')\dd{x'}} = r^{-2}\abs{\Psi(x)}^2.
\tag{106}
\end{align*}
Since $r$ is an arbitrary positive constant and ${\theta}$ is an arbitrary function, there is a one-to-one map between $\Nparticle$-particle position pdfs $\pdf$ and equivalence classes of elements $\Psi$ of ${\lebesgue(\domain)}$.
C.1.1 Exchange symmetry or antisymmetry of $\Psi$
As discussed in Sec. 3.4.2, if ${x}$ specifies the configuration (positions) of a set of identical particles, and if the particles move so fast within the same region of space that it is impossible to follow their individual trajectories, then their identicality makes them indistinguishable. To reflect this indistinguishability, $\pdf$ must be invariant under exchange of any two of the particles’ positions. When that is the case, $\Psi$ must be either exchange symmetric or exchange antisymmetric to preserve the exchange symmetry of $\pdf$.
Under the additional assumption that two or more particles cannot occupy the same point in space, $\Psi$ must vanish at coincidence points, which are points in $\domain$ that represent configurations in which two or more particles coincide.
If $\Psi$ is exchange symmetric, it is not differentiable at a coincidence point unless its derivatives with respect to the positions of the coincident particles vanish. However it can be finite-difference differentiable, while preserving the exchange symmetry of $\pdf$, if it is exchange-antisymmetric and changes sign at coincidence points.
C.2 Statistical states as vectors
A one-dimensional lattice, with spacing ${a\in\realpos}$, is denoted by ${\alattx\integer}$. Therefore ${\discreal\equiv\left(\alattx\integer\right)^{\Ndof}}$ is a ${\Ndof}$-dimensional hypercubic lattice whose ‘volume’ per lattice site will be denoted by ${\hilbv\equiv \alattx^{\Ndof}}$. That is, $\hilbv$ is the measure in ${\realone^{\Ndof}}$ of the set ${\N_x\subset\realone^{\Ndof}}$ of points that are closer in ${\realone^{\Ndof}}$ to the element ${x}$ of ${\discreal}$ than to any other element of $\discreal$.
Now let ${\discreal\equiv (\alattx\integer)^{\Ndof}\cap\domain}$, and let us define a function ${\sqhilbv\tPsiv:\discreal\to\complex}$ whose square modulus divided by $\hilbv$ is
\begin{align*}
\abs{\tPsiv(x)}^2 = \frac{1}{\hilbv}\int_{\N_x} \abs{\Psi(u)}^2 \dd{u}
= \frac{1}{\hilbv}\int_{\N_x} \pdf(u) \dd{u},
\end{align*}
where $\Psi$ is defined as in Appendix C.1. This definition of ${\tPsiv}$ implies that ${\pdf(x)=\lim_{\hilbv\to 0}\abs{\tPsiv(x)}^2}$ and that ${\abs{\sqhilbv\tPsiv(x)}^2}$ is the probability that the true configuration of the system, $x_t$, is closer to $x$ than to any other element of ${\discreal}$. Therefore, ${\sqhilbv\tPsiv}$ is a square root of a probability mass function, and ${\abs{\sqhilbv\tPsi(x)}^2}$ becomes ${\hilbv \pdf(x)}$ in the limit ${\hilbv\to 0}$.
Since $\discreal$ is a finite lattice, ${\sqhilbv\tPsiv}$ can be specified by a finite set of complex numbers, ${\left\{\sqhilbv\tPsiv(u)\right\}_{u\in\discreal}}$. This set is square-summable, because ${\hilbv\abs{\tPsiv(x)}^2}$ is a probability mass function, i.e.,
\begin{align*}
\sum_{u\in\discreal} \abs{\tPsiv(u)}^2\hilbv =\sum_{u\in\discreal}\tPsiv^*(u)\tPsiv(u)\hilbv =1< \infty,
\end{align*}
which implies that ${\sqhilbv\tPsiv\in\seqlebesgue(\discreal)}$, where ${\seqlebesgue(\discreal)=\seqlebesgue(\discreal,\complex)}$ is the Hilbert-Lebesgue space of square-summable functions ${\discreal\to\complex}$.
Any probability mass function, ${\abs{\sqhilbv f}^2}$, for the identity ${x\in\discreal}$ of the neighbourhood ${\N_x}$ containing ${x_t}$ can be represented by one of its square roots, ${\sqhilbv f\in\seqlebesgue(\discreal,\complex)}$. This square root can be represented by a vector ${\ket{\sqhilbv f}\equiv \sqhilbv\ket{f}}$ in a vector space ${\hilbertv}$ whose ‘inner product’ (see Sec. 2.3.5 to interpret the quotes) is defined by
\begin{align*}
\braket{\sqhilbv f}{\sqhilbv g} \equiv \sum_{u\in\discreal}f^*(u)g(u)\epsilon.
\end{align*}
One reason for choosing this form for the inner product is that in the dense-sampling limit ${\alattx\to 0 \implies \hilbv\to 0}$, it becomes the Riemann integral, \begin{align*}
\lim_{\hilbv\to 0}\braket{\sqhilbv f}{\sqhilbv g}=\int_{\domain} f^*(u)g(u)\dd[\Ndof]{u}.
\end{align*}
To achieve this form for the inner product, let us introduce the set \begin{align*}
\left\{\phix(\hilbv)\in\seqlebesgue(\discreal): x\in\discreal\right\},
\end{align*}
where \begin{align*}
\sqhilbv\phix(\hilbv):\discreal\to\complex; u\mapsto \sqhilbv\phix(u;\hilbv),
\end{align*}
and ${\abs{\sqhilbv\phix(\hilbv)}^2}$ is a probability mass function which vanishes everywhere except at ${x\in\discreal}$, i.e., ${\abs{\sqhilbv\phix(u;\hilbv)}^2=\Pr(x_t\in\N_u)}$ vanishes if ${u\neq x}$ and is unity if ${u=x}$. This implies that ${\phix(u;\hilbv)}$ vanishes if ${u\neq x}$ and that ${\phix(x;\hilbv)=e^{i\theta_x}}$, for some ${\theta_x\in\realone}$. For simplicity we choose ${e^{i\theta_x}}$ to be unity (${\implies\theta_x=0}$) for every ${x\in\discreal}$. To simplify notation, I will not make the dependence of $\phix$ on parameter $\hilbv$ explicit in much of what follows, and I will use ${\ket{\sqhilbv x}}$ rather than ${\ket{\sqhilbv \phix(\hilbv)}}$ to denote the element of $\hilbertv$ that represents ${\sqhilbv \phix(\hilbv)}$. I will refer to ${\{\sqhilbv\ket{u}\}_{u\in\discreal}}$ as the microstructure basis set.
The element ${\ket{\sqhilbv f}}$ of ${\hilbertv}$ representing a square-summable function ${\sqhilbv f}$ can be expressed in terms of this basis set as follows,
\begin{align*}
\ket{\sqhilbv f} = \sum_{x\in\discreal}\sqhilbv f(x)\left(\sqhilbv\ket{x}\right) = \sum_{x\in\discreal} f(x)\hilbv\ket{x},
\tag{107}
\end{align*}
where ${\ket{\sqhilbv x}}$ is the element of ${\hilbertv}$ representing ${\sqhilbv\phix}$, i.e., \begin{align*}
\ket{\sqhilbv x} = \sum_{u\in\discreal} \phix(u)\sqhilbv\left(\sqhilbv\ket{u}\right) = \sum_{u\in\discreal}\phix(u)\hilbv\ket{u}.
\end{align*}
This implies that the inner product of ${\ket{\sqhilbv x}}$ and ${\ket{\sqhilbv x'}}$ can be expressed as \begin{align*}
\braket{\sqhilbv x}{\sqhilbv x'}=\hilbv\braket{x}{x'}=\sum_{u,v\in\discreal}\phi_x^*(u)\phi_{x'}(v)\hilbv^2\braket{u}{v},
\end{align*}
where ${\phi_x^*(u)\phi_{x'}(v)}$ vanishes unless ${u=v=x=x'}$, in which case its value is ${1/\hilbv}$. Therefore ${\braket{x}{x'}=0}$ if ${x\neq x'}$, and \begin{align*}
\braket{x}{x}=
\sum_{u\in\discreal}\abs{\phix(u)}^2\hilbv\braket{u}{u} = \abs{\phix(x)}^2\hilbv\braket{x}{x} =\braket{x}{x}.
\end{align*}
These choices do not determine the norm of ${\ket{x}}$, so let us choose it to be \begin{align*}
\norm{\ket{x}}_2\equiv\sqrt{\braket{x}{x}}=\frac{1}{\sqrt{\hilbv}} \implies \braket{x}{x}=\frac{1}{\hilbv},
\end{align*}
which means that the norm of ${\ket{\sqhilbv x}\equiv\sqhilbv\ket{x}}$ is one and that the microstructure basis, ${\{\sqhilbv\ket{u}\}_{u\in\discreal}}$, is orthonormal. In other words, ${\ket{x}}$ is not a unit vector, but ${\sqrt{\hilbv}\ket{x}}$ is a unit vector.
We can express ${\ket{f}}$ in terms of the microstructure basis as follows:
\begin{align*}
\ket{f} = \frac{1}{\sqhilbv}\ket{\sqhilbv f} &=
\frac{1}{\sqhilbv}\sum_{x\in\discreal}\braket{\sqhilbv x}{\sqhilbv f}\sqhilbv\ket{x}
\\
&=\sum_{x\in\discreal}\braket{x}{f}\hilbv\ket{x}.
\end{align*}
By comparison with Eq. (107), this implies that ${\braket{x}{f} = f(x)}$. Therefore, the element of $\hilbertv$ that represents ${\ket{\sqhilbv\tPsiv}\equiv\sqhilbv\ket{\tPsiv}}$ is \begin{align*}
\ket{\sqhilbv \tPsiv}\equiv \sqhilbv \sum_{x\in\discreal}\tPsiv(x)\hilbv\ket{x};
\tag{108}
\end{align*}
and the element that represents ${\ket{\tPsiv}}$ is \begin{align*}
\ket{\tPsiv}= \sum_{x\in\discreal}\tPsiv(x)\hilbv\ket{x}.
\end{align*}
From this point forward let us assume that ${\hilbv}$ is small enough that the variations of ${\Psi}$ and $\pdf$ between any two neighbouring points of $\discreal$ are negligible; and the effects on all theoretical calculations of making it even smaller are negligible. When it is this small, let us denote ${\ket{\tPsiv}}$, ${\tPsiv}$, and ${\hilbertv}$ by ${\ket{\psi}}$, ${\psi}$, and ${\hilbert}$, respectively; and let us express ${\ket{\tPsiv}\equiv \sum_{x\in\discreal}\tPsiv(x)\hilbv\ket{x}}$ as
\begin{align*}
\ket{\psi} \equiv \int_\domain \psi(x)\ket{x}\dd{x},
\tag{109}
\end{align*}
where ${\dd{x}\equiv\hilbv}$ and ${\psi(x)\equiv \braket{x}{\psi}}$. In other words, from now on it is implicit that ${\int_\domain \ldots\dd{x}}$ really means ${\sum_{x\in\discreal}\ldots\hilbv}$. Under this convention, we can express the identity in $\hilbert$ as \begin{align*}
\identity = \sum_{x\in\discreal} \dyad{\sqhilbv x}=\int_\domain \dyad{x}\dd{x}.
\end{align*}
That is, \begin{align*}
\identity\ket{\psi} = \left(\int_\domain\dyad{x}\dd{x}\right)\ket{\psi} = \int_\domain \braket{x}{\psi}\ket{x} \dd{x} = \ket{\psi}
\end{align*}
and \begin{align*}
\identity\ket{x} = \left(\int_\domain\dyad{x'}\dd{x'}\right)\ket{x} = \int_\domain \braket{x'}{x}\ket{x'} \dd{x'} = \ket{x}.
\end{align*}
All of the Hilbert-Lebesgue spaces in this work that are denoted by $\lebesgue$ rather than $\seqlebesgue$, should be regarded as having been constructed in the same way as $\hilbert$. The notation $\lebesgue$ is used despite the fact that the construction of ${\hilbert}$ above implies that ${\hilbv}$ is finite, albeit arbitrarily small; and despite the fact that ${\intdomain\cdots\dd{x}}$ is defined as a Riemann integral. These aspects of the definition of $\hilbert$ imply that, strictly speaking, it is a space of functions whose domain is countable. Therefore it is (an abstract representation of) a ${\seqlebesgue}$-space. Nevertheless, the notation $\lebesgue$ is used throughout this work for spaces defined in this way. A function space with a $2$-norm is only denoted by ${\seqlebesgue}$ when it is important to emphasize that its elements have a countable domain, as was the case for ${\seqlebesgue(\discreal)}$ above.
It may be troubling to some that subtleties appear to have been swept under a rug by using the finite precisions of all measurements to justify quantizing the domain of $\pdf$. A more careful justification of this step is presented in Section 3 of [Tangney, 2024].
C.3 The space of Fourier transforms of elements of ${\lebesgue(\domain)}$
Any function ${\psi\in\lebesgue(\domain)}$ can be Fourier transformed with respect to each of the $\Ndof$ coordinates specified by points ${x\in\domain}$, as follows: \begin{align*}
\ftspsi(k)&\equiv \fourierconst^N \intftsdomain\dd{x} \psi(x) e^{-ikx}
\implies \psi(x)= \fourierconst^N \intdomain\dd{k} \ftspsi(k) e^{ikx},
\end{align*}
where ${\fourierconst\equiv 1/\sqrt{2\pi}}$, and ${\ftsdomain\equiv \domaindual/\fourierconst^2}$, where \begin{align*}
\domain^\ast\equiv \left\{ \adbmal\equiv k/(2\pi):k\in\ftsdomain\right\}
\end{align*}
is the dual space of ${\domain}$. In other words, ${\domain}$ is a vector space over field ${\realone}$, and ${\domain^\ast}$ is the set of all linear maps, \begin{align*}
(\adbmal,\cdot):\domain\to\realone;\; x\mapsto (\adbmal,x)\equiv \;\adbmal x.
\end{align*}
There is nothing within the construction of ${\lebesgue(\domain)}$ and ${\hilbert}$ in Appendix C.2 that would make it inapplicable to space $\ftsdomain$ or space ${\domaindual}$. Therefore let us use an analogous construction to define the Hilbert-Lebesgue space ${\lebesgue(\ftsdomain)}$. Then, since the Fourier transform ${\ftsf}$ of any ${f\in\lebesgue(\domain)}$ exists, and is uniquely defined by $f$; and since Plancherel’s theorem [Iosevich and Liflyand, 2014; Strichartz, 2003] states that
\begin{align*}
\intftsdomain \ftsf^*(k)\ftsf(k)\dd{k}=
\intdomain f^*(x) f(x)\dd{x} =
\norm{f}_2^2<\infty,
\end{align*}
${\ftsf}$ is an element of ${\lebesgue(\ftsdomain)}$ and there is a one to one correspondence between elements of ${\lebesgue(\domain)}$ and elements of ${\lebesgue(\ftsdomain)}$, which implies that there is a one to one correspondence between elements of ${\lebesgue(\ftsdomain)}$ and elements of ${\hilbert}$.
Furthermore, Parseval’s theorem [Iosevich and Liflyand, 2014; Strichartz, 2003] states that the Fourier transforms ${\ftsf,\ftsg\in\lebesgue(\ftsdomain)}$ of any ${f,g\in\lebesgue(\domain)}$ satisfy
\begin{align*}
\braket{f}{g}\equiv\intdomain f^*(x)g(x) \dd{x} = \intftsdomain\ftsf^*(k)\ftsg(k)\dd{k}.
\end{align*}
Therefore, instead of introducing a second abstract Hilbert space, ${\hilbert}$ can play the same role for ${\lebesgue(\ftsdomain)}$ that it plays for ${\lebesgue(\domain)}$:
Recall that each of the elements elements ${\ket{x}}$ of ${\hilbert}$ represents a function ${\phix\in\lebesgue(\domain)}$ which is localized in a neighbourhood of the point ${x\in\domain}$, whose measure is ${\hilbv}$. By following the same procedure in spaces $\ftsdomain$ and ${\lebesgue(\ftsdomain)}$, let us define a set of functions ${\phik\in\lebesgue(\ftsdomain)}$, each of which is localized within a different element of a partition of ${\ftsdomain}$. Let us choose the partition such that the measure of each of its elements is ${\ftshilbv\equiv 1/(\fourierconst^{2\Ndof}\hilbv)}$; and let us denote the element of $\hilbert$ that represents the function ${\phik}$ that is localized within the partition centered at ${k\in\ftsdomain}$ by ${\ket{k}}$, where
\begin{align*}
\norm{\ket{k}}_{\hilbert}=\norm{\phik}_2=\frac{1}{\sqrt{\ftshilbv}} = \fourierconst^\Ndof\sqrt{\hilbv}.
\end{align*}
Since set ${\{\ket{k}\}}$ is a complete orthonormal basis of ${\ftsdomain}$, any ${\ftsf\in\lebesgue(\ftsdomain)}$ is represented in $\hilbert$ by a vector that can be expressed as \begin{align*}
\myket{\ftsf}
&\equiv \intftsdomain \dd{k} \ftsf(k)\ket{k}
= \fourierconst^\Ndof \intftsdomain\dd{k}\intdomain\dd{x} f(x) e^{-ikx} \ket{k}
\\
& = \intdomain\dd{x} f(x) \left(\fourierconst^\Ndof\intftsdomain\dd{k} e^{-ikx}\ket{k}\right).
\tag{110}
\end{align*}
Now, if ${\ftsf}$ and ${f}$ are to be represented by the same element of $\hilbert$, this must be equal to ${\ket{f} = \intdomain\dd{x} f(x) \ket{x}}$. Therefore, \begin{align*}
\ket{x}
&=\fourierconst^\Ndof\intftsdomain\dd{k}e^{-ikx}\ket{k},
\end{align*}
which implies that \begin{align*}
\braket{k}{x} = \fourierconst^\Ndof e^{-ikx} = \frac{e^{-ikx}}{\sqrt{(2\pi)^\Ndof}} = \braket{x}{k}^*.
\end{align*}
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