TITLE: What keeps electrons in an atom from flying away or falling into the nucleus?
QUESTION [13 upvotes]: In atoms, what force or charge, etc. keeps electrons from flying away or into their nucleus? is there a kind of weak-force at work on the atomic scale?
Note I am aware the electron positions are only abstract variables and can be referred to as the electron field and the like. This is not the question.
What is the reason an electron is bound to that nucleus to the point it can sustain an "orbit" or variable probability path, and not fly away or into it's nucleus?

REPLY [4 votes]: The scale of the atom has Planck's constant in it, so it's quantum. The force that keeps the electrons near the nucleus is the electrostatic attraction between the electron and the nucleus. To understand why the electron doesn't fall into the nucleus formally, you can solve the Schrodinger equation, but there are seat-of-the-pants arguments that are correct to give the right order of magnitude.

Uncertainty principle: in order to confine the electron to a box of radius r around the nucleus, you have to give it a momentum of order h/r, which means that it's kinetic energy is roughly ${\hbar^2\over 2m r^2}$, while the potential energy is (negative) ${ke^2\over r}$. The total energy is the difference of these two, and it has a minimum when the size of the box r is of the order of the Bohr radius: $r_B = { \hbar^2\over 2m ke^2}$.
Radiation frequency: The closely related original argument by Bohr is that the frequency of the radiation emitted by the atom should be of the order of the classical orbital frequency. Yet the radiation should also be in quanta. The classical orbital frequency is the reciprocal of the time to go around once, and it obeys Kepler's law ${1\over T}={1\over r^{1.5}}$, but the binding energy goes as ${1\over r}$ so that the level spacing at small r scales to eventually be bigger than the energy magnitude, and there should be a lowest energy level.

For a different power law force, if the potential energy goes to minus infinity with a power faster than $1\over r^2$, the particles attract so much that there is no stable bound state, they end up sitting on top of each other.


text:
\begin{document}
\title{Finite-gap integration of the  $SU(2)$ Bogomolny equations}
\author{H.W. Braden}
\address{School of Mathematics, Edinburgh University, Edinburgh.}
\email{hwb@ed.ac.uk}
\author{V.Z. Enolski}
\address{Department of Mathematics and Statistics, Concordia
University, Montreal.
\newline
On leave: Institute of Magnetism, National Academy of Sciences of
Ukraine.} \email{vze@ma.hw.ac.uk}

\begin{abstract}
The ADHMN construction of magnetic monopoles is given in terms of
the (normalizable) solutions of an associated Weyl equation. We
focus here on solving this equation directly by algebro-geometric
means. The (adjoint) Weyl equation is solved using an ans\"atz of
Nahm in terms of Baker-Akhiezer functions. The solution of Nahm's
equation is not directly used in our development.
\end{abstract}

\maketitle
\section{Introduction}
Consider the dimensional reduction to three dimensions of the four
dimensional Yang-Mills Lagrangian with gauge group $ SU(2)$ under
the assumption that all fields are independent of time. Upon
identifying the $a_4$-component of the gauge field with the Higgs
field $\Phi $ we obtain the three dimensional Yang-Mills-Higgs
Lagrangian
\[L= -\frac12 \mathrm{Tr}\, F_{ij} F^{ij}+\mathrm{Tr}\,D_{i}\Phi\, D^{i}\Phi.\]
Here $F_{ij}=\partial_i a_j -\partial_j a_i+[a_i,a_j]$ is the
curvature of the (spatial) connection of the gauge field
$a_i(\boldsymbol{x})$ and $D_i$  the covariant derivative
$D_i\Phi=\partial_i\Phi+[a_i,\Phi]$,
$\boldsymbol{x}=(x_1,x_2,x_3)\in \mathbb{R}^3$. We are interested
in configurations minimizing the energy of the system. These are
given by the {\it Bogomolny equation}
\begin{equation} D_i\Phi=\pm \sum_{j,k=1}^3\epsilon_{ijk}
F_{jk},\quad i=1,2,3\label{bogomolny}.\end{equation} A solution with the boundary
conditions
\[\left. \sqrt{-\frac12 \mathrm{Tr}\,\Phi(r)^2}\right|_{r\rightarrow\infty}\sim
1-\frac{n}{2r}+O(r^{-2}),\quad r=\sqrt{x_1^2+x_2^2+x_3^2}\] is called a {\it monopole} of  charge $n$.

In this note note we shall follow the
Atiyah-Drinfeld-Hitchin-Manin-Nahm (ADHMN) construction (see the
original papers \cite{nahm82},\cite{hitchin82} and the recent
review \cite{weinbyi06}). This has its origins in the construction
of instanton solutions to the (Euclidean) self-dual Yang-Mills
equations by Atiyah-Drinfeld-Hitchin-Manin (ADHM): here the
self-duality equations, partial differential equation in four
variables, are transformed to an algebraic matrix equation.

In the monopole setting the ADHMN construction reduces the
Bogomolny equation, again a partial differential equation but now
in three variables, to a system of ordinary differential
equations. Our interest here will be to integrate these by
algebro-geometric methods.

The standard approach to the integration of the Bogomolny equation
within the ADHMN construction consists of two stages. First, an
auxiliary equation known as Nahm's equation,
\begin{equation}
\frac{\mathrm{d}T_i(z)}{\mathrm{d}z}
=\frac12\sum_{j,k=1}^3\epsilon_{ijk}[T_j(z),T_k(z)], \quad
z\in[-1,1] \label{nahmeq}\end{equation} is integrated for $n\times n$ matrices $T_i(z)$
subject to
certain boundary conditions. The solution of this equation is then
used to define a differential operator
\begin{align}
\Delta^{\dagger}= \im 1_{2n}\frac{\mathrm{d}}{\mathrm{d}z}
-\sum_{j=1}^3( T_j(z)+\im x_j 1_n) \otimes \sigma_j,
\label{weylequ}\end{align} where $\sigma_j$ are the Pauli
matrices. The Higgs and gauge fields are then expressed  as
certain averages over the normalizable  solutions
$\boldsymbol{{v}}$ to the Weyl equation
$\Delta^{\dagger}\boldsymbol{{v}}(\boldsymbol{x},z)=0$. Nahm's
equation, introduced in the ADHMN construction, plays an important
role in many problems of mathematical physics and the integration
of this equation is of great significance. Its role in the
integration of the Bogomolny equation is nevertheless an auxiliary
one. In this paper we shall concentrate on solving of the Weyl
equation directly by algebro-geometric means using an ansatz
(again of Nahm) that has not been considered previously in this
light. Our new insight is that this ans\"atz  may be solved in
terms of a Baker-Akhiezer function. Although the associated
spectral problem is equivalent to that appearing in the
algebro-geometric integration of the Nahm equation we do not use
solutions of Nahm's equation directly in our development. To
achieve our result we implement the $\theta$-functional
integration of the Nahm equation by Ercolani and Sinha
\cite{ersi89} and our recent analysis \cite{bren06}. The
limitations of space in this volume prevent detailed examples
being given and a fuller exposition will be given elsewhere.

\section{The ADMHN construction}

Set
\begin{equation}
\Delta=\im \dfrac{d}{dz}+x-\im T_4+\boldsymbol{T}\cdot\boldsymbol{\sigma}
=\im \dfrac{d}{dz}-\im R,
\end{equation}
where
\begin{equation}
x=x_4+\im \boldsymbol{x}\cdot\boldsymbol{\sigma},\quad
T=T_4+\im \boldsymbol{T}\cdot\boldsymbol{\sigma},\quad
R=T+ix.
\end{equation}
We will often assume we have chosen a gauge such that $T_4=0$ and
that $x_4=0$. The ADMHN construction may be summarised in  the
following theorem.
\begin{theorem} [\bf ADMHN] The charge $n$
monopole solution of the Bogomolny equation is given by
\begin{align}
\Phi_{ab}(\boldsymbol{x})&=\im\int_{-1}^1dz\,z \boldsymbol{v}_a\sp\dagger(\boldsymbol{x},z)\boldsymbol{v}_b(\boldsymbol{x},z)
,\quad a,b=1,2,\label{higgs}\\
A_{i\, ab}(\boldsymbol{x})&=\int_{-1}^1dz\, \boldsymbol{v}_a\sp\dagger(\boldsymbol{x},z)
\frac{\partial}{\partial x_i} \boldsymbol{v}_b(\boldsymbol{x},z),
\quad
i=1,2,3,\quad a,b=1,2.\label{gauges}
\end{align}
Here the two ($a=1,2$) $2n$-column vectors\footnote{Throughout the
paper vectors are column-vectors and printed in bold, e.g.
$\boldsymbol{a}$; the superscript $\dagger$ means conjugated and
transposed, e.g for vector
$\boldsymbol{a}^{\dagger}=\overline{\boldsymbol{a}}^T$, and this
holds similarly for matrices. }
${{\boldsymbol{v}}}_{a}(\boldsymbol{x},z)
=({v}_1^{(a)}(\boldsymbol{x},z),\ldots,
{v}_{2n}^{(a)}(\boldsymbol{x},z))^T$ form an orthonormal basis on
the interval $z\in[-1,1]$
\begin{equation}\int_{-1}^1dz\, \boldsymbol{v}_a\sp\dagger(\boldsymbol{x},z)
\boldsymbol{v}_b(\boldsymbol{x},z)
=\delta_{ab}
,\label{norm}
\end{equation}
for the normalizable solutions to the the Weyl equation
\begin{align}\Delta\sp\dagger \boldsymbol{v}=0,
\end{align}
where $\Delta^{\dagger}$ is given by (\ref{weylequ}). The
normalizable solutions form a two-dimensional subspace of the
solution space
$(\boldsymbol{v}^{(1)}(\boldsymbol{x},z),\ldots,\boldsymbol{v}^{(2n)}(\boldsymbol{x},z))$.
The $n\times n$-matrices $T_j(z)$, called Nahm data,  satisfy
Nahm's equation (\ref{nahmeq}) and are required to satisfy the
following boundary conditions: they are regular at $z\in(-1,1)$;
have simple poles at $z=\pm1$, the residues of which form the
irreducible $n$-dimensional representation of the $su(2)$ algebra;
further
\begin{equation}
T_i(z)=-T_i^{\dagger}(z),\quad
T_i(z)=T_i^{T}(-z).\label{constraint}\end{equation}
\end{theorem}
A proof consisting of direct verification may be found for example
in the recent exposition by E.Weinberg and P.Yi \cite{weinbyi06}
and references therein.

The integrals in  (\ref{higgs}), (\ref{gauges}) and (\ref{norm})
may be computed in closed form \cite{panagopo83} in the following
way. Denote by
\begin{equation}
\mathcal{H}(\boldsymbol{x})=\boldsymbol{x}\cdot\boldsymbol{\sigma}\otimes
\mathrm{1}_n,\quad \mathcal{T}(z)=\imath
\boldsymbol{\sigma}\otimes\boldsymbol{T}(z),\label{pnotation}
\end{equation}
where $\boldsymbol{\sigma}=(\sigma_1,\sigma_2,\sigma_3)^T$,
$\boldsymbol{T}=(T_1,T_2,T_3)^T$ and
$\boldsymbol{x}\cdot\boldsymbol{\sigma}\otimes
\mathrm{1}_n=\sum_{i=1}\sp3x_i\sigma_i\otimes 1_n$,
$\boldsymbol{\sigma}\otimes\boldsymbol{T}(z)=\sum_{i=1}\sp3\sigma_i\otimes
T_i(z)$. Introduce the $2n\times 2n$ matrix
$\mathcal{Q}(z,\boldsymbol{x})$,
\begin{equation}
\mathcal{Q}(\boldsymbol{x},z)
=\frac{1}{r^2}\mathcal{H}(\boldsymbol{x})\mathcal{T}(z)\mathcal{H}(\boldsymbol{x})-\mathcal{T}(z).
\end{equation}
Then the following formulae of Panagopoulos (see appendix A) are
valid for the normalizable Weyl spinors,
$\boldsymbol{{v}}_{1,2}(\boldsymbol{x},z)$,
\begin{align}
&\int
{\boldsymbol{{v}}_{a}^\dagger}(\boldsymbol{x},z)
\boldsymbol{{v}}_{b}(\boldsymbol{x},z)\mathrm{d}z={\boldsymbol{{v}}_{a}^\dagger}
(\boldsymbol{x},z)
\mathcal{Q}^{-1}(\boldsymbol{x},z)\boldsymbol{{v}}_{b}(\boldsymbol{x},z);
\label{weylnorm}\\
& \int
z{\boldsymbol{{v}}_{a}^\dagger}(\boldsymbol{x},z)
\boldsymbol{{v}}_{b}(\boldsymbol{x},z)\mathrm{d}z={\boldsymbol{{v}}_{a}^\dagger}
(\boldsymbol{x},z)
\mathcal{Q}^{-1}(\boldsymbol{x},z)\left[z+2\mathcal{H}(\boldsymbol{x})\frac{\mathrm{d}}{
\mathrm{d}(r^2)}
 \right]\boldsymbol{{v}}_{b}(\boldsymbol{x},z);
\label{weylhiggs}\\
& \int
{\boldsymbol{{v}}_{a}^\dagger}(\boldsymbol{x},z)
\frac{\partial}{\partial x_i}\boldsymbol{{v}}_{b}(\boldsymbol{x},z)\mathrm{d}z \notag\\
&={\boldsymbol{{v}}_{a}^\dagger}(\boldsymbol{x},z)
\mathcal{Q}^{-1}(\boldsymbol{x},z)\left[\frac{\partial}{\partial
 x_i}+\mathcal{H}(\boldsymbol{x})\frac{zx_i+\imath (\boldsymbol{x}\times \nabla)_i }{ r^2}
 \right]\boldsymbol{{v}}_{b}(\boldsymbol{x},z).
\label{weylgauges}
\end{align}
Therefore only the boundary values of the normalized Weyl spinors
$\boldsymbol{{v}}_{1,2}(\boldsymbol{x},\pm1)$ together with their
derivatives need be computed to find solutions to the Bogomolny
equation. The Nahm data at these boundary values is also needed,
and this involves the residues noted above.

\section{The Nahm Ans\"atz}

Although we wish to solve
$$\Delta\sp\dagger \boldsymbol{v}=0$$ Nahm introduced an ans\"atz  that provides solutions to
$$\Delta \boldsymbol{w}=0$$ that we now recall.
Consider solutions of the form
\begin{equation}\label{Nansatz}
\boldsymbol{w}=(1_2+\boldsymbol{\hat u}(\boldsymbol{x})\cdot\boldsymbol{\sigma})\,e\sp{\im x_4z}|s>\otimes\,
\boldsymbol{\hat w(z)}
\end{equation}
where $|s>$ is an arbitrary normalized spinor not in $\ker
(1_2+\boldsymbol{\hat u}(\boldsymbol{x})\cdot\boldsymbol{\sigma})$
and $\boldsymbol{\hat u}(\boldsymbol{x})$ is (as we shall see) a
unit vector independent of $z$. Substituting in $\Delta
\boldsymbol{w}=0$ we find
$$0=|s>\otimes\left(\im \dfrac{d}{dz}+
\boldsymbol{\hat u}\cdot\boldsymbol{R}\right)\boldsymbol{\hat w(z)}+
\sigma_k|s>\otimes\left(\im {\hat u}\sp{k}\dfrac{d}{dz}+R\sp{k}+
\im(\boldsymbol{R}\times\boldsymbol{\hat u})\sp{k}\right)\boldsymbol{\hat w(z)}$$ and so we require
\begin{align}
0&=\left(\im \dfrac{d}{dz}+
\boldsymbol{\hat u}\cdot\boldsymbol{R}\right)\boldsymbol{\hat w(z)},\label{nahm1}\\
0&=\mathcal{L}_k\boldsymbol{\hat w(z)}:=\left(\im {\hat u}\sp{k}\dfrac{d}{dz}+R\sp{k}+
\im(\boldsymbol{R}\times\boldsymbol{\hat u})\sp{k}\right)\boldsymbol{\hat w(z)}.\label{nahm2a}
\end{align}
The consistency of these equations imposes various constraints.
First consider
\begin{align*}
\left[\mathcal{L}_1,\mathcal{L}_2
\right]&=(\im {\hat u}\sp1+{\hat u}\sp2{\hat u}\sp3)
\left(\dot{T_2}-[T_3,T_1]\right)-(\im {\hat u}\sp2-{\hat u}\sp1{\hat u}\sp3)
\left(\dot{T_1}-[T_2,T_3]\right)
\\
&\qquad-(1-({\hat u}\sp3)\sp2)\left(\dot{T_3}-[T_1,T_2]\right)+(1-\boldsymbol{\hat u}\cdot\boldsymbol{\hat u})\,\dot{T_3}.
\end{align*}
Thus provided $\boldsymbol{\hat u}(\boldsymbol{x})$ is a unit
vector and the $T_i$'s satisfy the Nahm equations we have
consistency of the equations $\mathcal{L}_k\boldsymbol{\hat
w(z)}=0$.

At this stage we introduce a convenient parameterization
(reflected in Hitchin's minitwistor construction). Let
$\boldsymbol{ y}\in \mathbb{C}\sp3$ be a null vector. We may
consider $\boldsymbol{ y}\in \mathbb{P}\sp2$ and parameterize
$\boldsymbol{ y}$ as
\begin{equation}\label{defhy} \boldsymbol{
y}=\left(\frac{1+\zeta^2}{2\im},\frac{1-\zeta^2}2,-\zeta \right).
\end{equation}
Then
$$\boldsymbol{ y}\cdot \boldsymbol{\overline
y}=\frac{(1+|\zeta|\sp2)\sp2}2,\qquad \boldsymbol{
y}\cdot \boldsymbol{ y}=0.$$
The signs here have been chosen so that
$$L(\zeta):=2\im \boldsymbol{ y}\cdot\boldsymbol{T}=
(T_1+\im T_2)-2\im T_3\,\zeta+(T_1-\im T_2)\,\zeta\sp2.$$ In due
course we will see this to be our Lax matrix. Set
\begin{equation}\label{defhu}
\boldsymbol{\hat
u}=\boldsymbol{\hat
u}(\zeta):=\im\,\frac{\boldsymbol{ y}\times \boldsymbol{\overline
y}}{\boldsymbol{ y}\cdot \boldsymbol{\overline
y}}=\frac{1}{1+|\zeta|^2}\left(\im(\zeta-\overline{\zeta}),\;(\zeta+\overline{\zeta}),
1-|\zeta|^2\right).
\end{equation}
Then
$$\boldsymbol{\hat
u}\times\boldsymbol{ y}=-\im\boldsymbol{ y},\qquad
\boldsymbol{\hat u}\times\boldsymbol{\overline
y}=\im\boldsymbol{\overline y}.$$ The three vectors
$\Re(\boldsymbol{ y}),\Im(\boldsymbol{ y})$ and $ \boldsymbol{\hat
u}$ form an orthogonal basis in $\mathbb{R}^3$ with
$|\boldsymbol{\hat u}|=1$, whence any $\boldsymbol{
v}\in\mathbb{R}^3$ may be written as
\begin{equation*}\label{basisuy}\boldsymbol{ v}=\boldsymbol{\hat
u}\,(\boldsymbol{\hat u}\cdot \boldsymbol{v})+
\boldsymbol{\overline y}\left(\frac{\boldsymbol{ y}\cdot
\boldsymbol{v}}{ \boldsymbol{ y}\cdot\boldsymbol{\overline
y}}\right) + \boldsymbol{ y}\left(\frac{\boldsymbol{\overline
y}\cdot \boldsymbol{v}}{ \boldsymbol{ y}\cdot\boldsymbol{\overline
y}}\right).
\end{equation*}
In particular,
\begin{equation}\label{basisucy}\boldsymbol{ v}+\im
\boldsymbol{ v}\times\boldsymbol{\hat u}=\boldsymbol{\hat
u}\,(\boldsymbol{\hat u}\cdot \boldsymbol{v})+2
\boldsymbol{\overline y}\left(\frac{\boldsymbol{ y}\cdot
\boldsymbol{v}}{ \boldsymbol{ y}\cdot\boldsymbol{\overline
y}}\right) .
\end{equation}
 We record that
\begin{align*}
\overline{\boldsymbol{ y}(\zeta)}&=-{\overline\zeta}\sp2\, \boldsymbol{ y}(-1/{\overline
\zeta}),\qquad\boldsymbol{\hat u}(-1/{\overline
\zeta})=-\boldsymbol{\hat u}(\zeta),\\
\boldsymbol{\hat u}&=(-\im\zeta\sp{-1},\zeta\sp{-1},-1)-\frac{2\boldsymbol{ y}}
{\zeta(1+|\zeta|\sp2)}
=(\im\zeta,\zeta,1)+\frac{2{\overline\zeta}\boldsymbol{ y}}
{1+|\zeta|\sp2},\\
\boldsymbol{\hat u}\cdot\boldsymbol{T}&=
-\im\left[(T_1+\im T_2)\zeta\sp{-1}-\im T_3\right]-
\frac{2\boldsymbol{ y}\cdot\boldsymbol{ T}}
{\zeta(1+|\zeta|\sp2)}=
\im\left[(T_1-\im T_2)\zeta-\im T_3\right]-
\frac{2{\overline\zeta}\boldsymbol{ y}\cdot\boldsymbol{ T}}
{1+|\zeta|\sp2}.

\end{align*}

Parameterizing $\boldsymbol{\hat u}$ as above and using
(\ref{basisucy}) we may write
\begin{equation*}
\im \boldsymbol{\hat u}\dfrac{d}{dz}+\boldsymbol{R}+
\im\boldsymbol{R}\times\boldsymbol{\hat u}=\im \boldsymbol{\hat u}\dfrac{d}{dz}+
\boldsymbol{\hat
u}\,(\boldsymbol{\hat u}\cdot \boldsymbol{R})+2
\boldsymbol{\overline y}\left(\frac{\boldsymbol{ y}\cdot
\boldsymbol{R}}{ \boldsymbol{ y}\cdot\boldsymbol{\overline
y}}\right) =\boldsymbol{\hat u}\left(\im \dfrac{d}{dz}+
\boldsymbol{\hat u}\cdot\boldsymbol{R}\right)+2
\boldsymbol{\overline y}\left(\frac{\boldsymbol{ y}\cdot
\boldsymbol{R}}{ \boldsymbol{ y}\cdot\boldsymbol{\overline
y}}\right)
\end{equation*}
and as a consequence (\ref{nahm1}, \ref{nahm2a}) are equivalent to
\begin{align}
0&=\left(\im \dfrac{d}{dz}+
\boldsymbol{\hat u}\cdot\boldsymbol{R}\right)\boldsymbol{\hat w(z)},\label{nahm1b}\\
0&=\left(\boldsymbol{ y}\cdot
\boldsymbol{R}\right)\boldsymbol{\hat w(z)}.\label{nahm2b}
\end{align}
The remaining consistency to be checked is then
\begin{align*}
\left[\im \dfrac{d}{dz}+
\boldsymbol{\hat u}\cdot\boldsymbol{R},\,
\boldsymbol{ y}\cdot
\boldsymbol{R}
\right]
&=
\im {\boldsymbol{y}\cdot{\dot{\boldsymbol{T}}}}
+
\left[\boldsymbol{\hat u}\cdot\boldsymbol{T},\,
{\boldsymbol{y}\cdot{\boldsymbol{T}}}
\right]
=0.
\end{align*}
which upon use of $\boldsymbol{\hat u}\times\boldsymbol{
y}=-\im\boldsymbol{ y}$ is equivalent to Nahm's equations.

Equally from
$$\boldsymbol{\hat u}\cdot\boldsymbol{R}=
-\im\left[(R_1+\im R_2)\zeta\sp{-1}-\im R_3\right]-
\frac{2\boldsymbol{ y}\cdot\boldsymbol{ R}}
{\zeta(1+|\zeta|\sp2)}=
\im\left[(R_1-\im R_2)\zeta-\im R_3\right]-
\frac{2{\overline\zeta}\boldsymbol{ y}\cdot\boldsymbol{ R}}
{1+|\zeta|\sp2}$$ we may write the equations as
\begin{align*}
0&=\left( \dfrac{d}{dz}
+\left[(R_1-\im R_2)\zeta-\im R_3\right]\right)\boldsymbol{\hat w(z)}
=\left( \dfrac{d}{dz}+M
+\im\left[(x_1-\im x_2)\zeta-\im x_3\right]\right)\boldsymbol{\hat w(z)},\\
0&=\left(\boldsymbol{ y}\cdot
\boldsymbol{R}\right)\boldsymbol{\hat w(z)},
\end{align*}
{where} \begin{equation} M=(T_1-\im T_2)\zeta-\im T_3.
\end{equation}

The equations we have obtained are just the Lax equations
\begin{align*}
0&={2\im}\left(\boldsymbol{ y}\cdot
\boldsymbol{R}\right)\boldsymbol{\hat w(z)}
=\left(L(\zeta)-\eta\right)\boldsymbol{\hat w(z)},
\qquad
\eta={2\boldsymbol{y}\cdot{\boldsymbol{x}}},\\
0&=\left(\im \dfrac{d}{dz}+
\boldsymbol{\hat u}\cdot\boldsymbol{R}\right)\boldsymbol{\hat w(z)},
\intertext{and}
\dot L&=[L,M].
\end{align*}
From the first of these we see that
$$0=\det\left(L(\zeta)-\eta\right),$$
which gives the equation of the spectral curve $\mathcal{C}$. Upon
using $\overline{\boldsymbol{ y}(\zeta)}=-{\overline\zeta}\sp2\,
\boldsymbol{ y}(-1/{\overline \zeta})$ we see from
$$0=\det\left(L(\zeta)-\eta\right)\sp\dagger
=\det\left(L(\zeta)\sp\dagger-\overline{\eta}\right)=
\det\left({2\im\overline{\boldsymbol{ y}(\zeta)}\cdot
\boldsymbol{T}}-\overline{\eta}\right)=
\det\left({-2\im{\overline\zeta}\sp2{\boldsymbol{ y}(-1/{\overline
\zeta})}\cdot
\boldsymbol{T}}-\overline{\eta}\right)
$$
that the spectral curve is invariant under
$$(\zeta,\eta)\rightarrow
(-\frac1{\overline\zeta},-\frac{\overline\eta}{{\overline\zeta}\sp2}).$$
The spectral curve then has the form
\begin{equation}
\eta^n+a_1(\zeta)\eta^{n-1}+\ldots+a_n(\zeta)=0, \quad
\mathrm{deg}\, a_k(\zeta) \leq 2k\label{curve1},
\end{equation}
and the genus of $\mathcal{C}$ is $g=(n-1)^2$.

It is worth remarking that Nahm's ans\"atz only yields solutions
of $\Delta \boldsymbol{w}=0$ and does not yield solutions of
$\Delta\sp\dagger \boldsymbol{v}=0$.

\subsection{Strategy of Solution}

The strategy for constructing solutions involves 3 steps. We have
seen that finding solutions to $\Delta \boldsymbol{w}=0$ reduces
to solving
\begin{align}
0&=\left(L(\zeta)-\eta\right)\boldsymbol{\hat w(z)},\label{eqlax1}\\
0&=\left( \dfrac{d}{dz}+M
\right)\boldsymbol{\hat w(z)},\label{eqlax2}
\end{align}
upon using the (slightly modified) ans\"atz
\begin{equation*}
\boldsymbol{w}=(1_2+\boldsymbol{\hat u}(\boldsymbol{x})\cdot\boldsymbol{\sigma})\,
e\sp{-\im z\left[(x_1-\im x_2)\zeta-\im x_3 -x_4\right]}|s>\otimes\,
\boldsymbol{\hat w(z)}.
\end{equation*}
Here  $\boldsymbol{\hat u}(\boldsymbol{x})$ is a unit vector and
$\eta={2\boldsymbol{y}\cdot{\boldsymbol{x}}}$. We might construct
a solution as follows.

\begin{enumerate}
\item Given a spectral curve
    $0=\det\left(L(\zeta)-\eta\right)$ and a position
    $\boldsymbol{x}$ we substitute
    $\eta={2\boldsymbol{y}\cdot{\boldsymbol{x}}}$ using the
    expression for $\boldsymbol{y}$ in terms of $\zeta$. This
    is an equation of degree $2n$ in $\zeta$ which we shall
    refer to as the \emph{Atiyah-Ward} constraint, this
    equation having appeared in their work. The $2n$ solutions
    give us $2n$ associated values $\boldsymbol{\hat
    u}\sp{a}$, $a=1,\dots,2n$. For each of these we solve for
    $\boldsymbol{\hat w(z)}$ yielding a $2n\times 1$ matrix
    $\boldsymbol{w}\sp{a}$. Taking each of the $2n$ solutions
    we obtain a $2n\times 2n$ matrix of solutions $W$.

\item As $0=\Delta W=\im\left(\frac{d}{dz}-R\right)W$, then
\begin{align*}
\frac{d}{dz}W&=RW,\qquad
\frac{d}{dz}W\sp\dagger=W\sp\dagger R,\qquad
\frac{d}{dz}\left(W\sp\dagger\right)\sp{-1}

=-R\left(W\sp\dagger\right)\sp{-1},
\end{align*}
whence
\begin{equation}
0=\Delta \sp\dagger\left(W\sp\dagger\right)\sp{-1}=
\im\left(\frac{d}{dz}+R\right)\left(W\sp\dagger\right)\sp{-1}.
\end{equation}
So given $W$ we may construct
$V=\left(W\sp\dagger\right)\sp{-1}$.

\item To reconstruct the gauge and Higgs fields using the
    formulae of the previous section we must extract from $V$
    the two normalizable solutions.

\end{enumerate}

The new insight that the study of integrable systems brings to
this problem is that $\boldsymbol{\hat w(z)}$ may be understood as
a Baker-Akhiezer function and constructed explicitly. Before
considering this, we conclude the section by noting Nahm's
construction for $\boldsymbol{\hat w(z)}$.

\subsection{Constructing $\boldsymbol{\hat w(z)} $ using the adjoint equation }

We begin with several simple observations. First, assuming $L$ is
invertible, the Lax equation $\dot L=[L,M]$ means also that
\begin{align*}
\frac{d}{dz} L\sp{-1}&=-L\sp{-1}\dot L L\sp{-1}=[L\sp{-1},M],\\
\frac{d}{dz} \adj L&=\frac{d}{dz}\left(\det(L)\,L\sp{-1}\right) =[
\adj L,M].
\end{align*}
Second, suppose $\lambda_i$ is an eigenvalue of $L$ with
    associated eigenvector $f_i$, $Lf_i=\lambda_i f_i$. Then
    $f_i$ is only determined up to a scale $f_i\rightarrow f_i
    h_i(z)$ which may differ from eigenvector to eigenvector.
    Set
$$F=(f_1,\dots,f_n),\qquad
\Lambda=\diag(\lambda_1,\ldots,\lambda_n).$$ Then
\begin{equation*}LF=F\Lambda \label{eqLF}
\end{equation*}
is compatible with the Lax equation if and only if $F=F(z)$ is
governed by
\begin{equation}\label{eqldta}
\left(\frac{d}{dt}+M\right)F=F\diag(\alpha_1,\ldots,\alpha_n),
\end{equation}
for some $\alpha_i(z)$. Conversely, given a solution of this
equation we may reconstruct $L$ satisfying $\dot L=[L,M]$ via
$L=F\Lambda F\sp{-1}$. Third, if $\lambda_i$ is an eigenvalue of
$L$ we may construct a corresponding eigenvectors $f_i$ via
\begin{equation}
\label{nevw}
f_i=\adj(L-\lambda_i)\nu h_i(z),
\end{equation} where $\nu$ is any
constant vector. This follows as
$$(L-\lambda_i )f_i=(L-\lambda_i
)\adj(L-\lambda_i)\nu h_i(z)=\det(L-\lambda_i)\nu h_i(z)=0.$$ With
such eigenvectors $f_i$  we see that \begin{align*} \dot
F&=[\adj(L-\lambda_i),M]\nu h_i(z)+\adj(L-\lambda_i)\nu \dot
h_i(z)\\&=-MF+F h_i\sp{-1}\dot h_i+\adj(L-\lambda_i)M\nu
h_i(z)\end{align*} and for this to be of the form (\ref{eqldta})
we require
$$0=\adj(L-\lambda_i)\nu(\dot h_i-h_i\alpha_i)+\adj(L-\lambda_i)M\nu
h_i(t).$$ Taking the inner product with an arbitrary vector
$\mu$ then yields the differential equation
$$h_i\sp{-1}\frac{dh_i}{dz}=\alpha_i(z)-
\frac{\mu\sp{T}\adj(L-\lambda_i)M\nu}{\mu\sp{T}\adj(L-\lambda_i)\nu}.$$
Therefore requiring the differential equation for $F$ leads to a
differential equation for $h_i$. Suppose we write
$$h_i(z)=\frac{\exp\left[-\theta_i(z)+\int\sp{z}\alpha_i(z)\,dz
\right]}{\sqrt{\mu\sp{T}\adj(L-\lambda_i)\nu}}
$$
then
$$h_i\sp{-1}\frac{dh_i}{dz}=\alpha_i(z)-
\frac{d\theta_i}{dz}-\frac12
\frac{\mu\sp{T}\left[\adj(L-\lambda_i),M\right]\nu}{\mu\sp{T}\adj(L-\lambda_i)\nu}$$
which provides a solution if $$ \frac{d\theta_i}{dz}=\frac12
\frac{\mu\sp{T}\left\{M,\adj(L-\lambda_i)\right\}\nu}{\mu\sp{T}\adj(L-\lambda_i)\nu}
.$$

Nahm's approach to construct $\boldsymbol{\hat w(z)} $ was to
express this in the form (\ref{nevw}) together with the one
dimensional differential equations for $\theta_i$. This method has
only been implemented  in the charge $2$ case and we now propose
an alternative approach.

\section{A Spectral Problem}
We have identified $\boldsymbol{\hat w(z)} $ with the
Baker-Akhiezer function and now must ask whether this can be
constructed. Set
\begin{align*}
A_{-1}&=T_1+iT_2,\
  A_0   =-2iT_3,\
  A_{1} =T_1-iT_2,\\
\intertext{and so}
  L(\zeta)&=A_{-1}+A_0\zeta+A_1\zeta\sp2,\qquad
  M=\frac{1}{2}A_0+A_1\zeta.
\end{align*} Viewing equation (\ref{eqlax2}) as a spectral problem
$$
\left( \dfrac{d}{dz}+\frac12 A_0(z)
\right)\boldsymbol{\hat w(z)}=-\zeta A_1(z)\boldsymbol{\hat w(z)},
$$
we seek to solve this. The $z$-dependence of the right hand side
means however this is not a standard eigenvalue problem, but it
may be reduced to such using a trick of \cite{ersi89}. With the
notation introduced, Nahm's equation yield
\begin{equation}\frac{{d}}{{d}
z}A_1(z)=\frac12[A_0(z),A_1(z)]\end{equation} and so by
introducing the matrix $C(z)$ with \begin{equation*}
\frac{{d}}{{d} z}C(z)=\frac12 A_0(z)C(z), \qquad
C(0)=1_n\end{equation*}
we may write
$$A_1(z)=C(z)A_1(0)C(z)\sp{-1}.$$
Then upon performing a gauge transformation
$$Q_0(z)=C(z)A_0(z)C(z)\sp{-1},\qquad
\Phi(z)=C(z)\sp{-1}\hat w(z)$$ we obtain the spectral problem
\begin{align} \label{standard}
\left( \dfrac{d}{dz}+Q_0(z)
\right)\boldsymbol{\Phi}(z)&=-\zeta A_1(0)\boldsymbol{\Phi}(z).
\end{align}
Here
$\boldsymbol{\Phi}(z)=\boldsymbol{\Phi}(\zeta,\eta,z)=\boldsymbol{\Phi}(P,z)$
is given by the Baker-Akhiezer function on the curve,
$P=(\zeta,\eta)\in\mathcal{C}$. Then
$$\boldsymbol{\hat w(z)}=C(z)\boldsymbol{\Phi}(\zeta,\eta,z)$$
and
\begin{align*}
\boldsymbol{w}&=(1_2+\boldsymbol{\hat u}(\boldsymbol{x})\cdot\boldsymbol{\sigma})\,
e\sp{-\im z\left[(x_1-\im x_2)\zeta-\im x_3 -x_4\right]}|s>\otimes\,
C(z)\boldsymbol{\Phi}(\zeta,\eta,z),\\
&=1_2\otimes C(z) \left(
(1_2+\boldsymbol{\hat u}(\boldsymbol{x})\cdot\boldsymbol{\sigma})\,
e\sp{-\im z\left[(x_1-\im x_2)\zeta-\im x_3 -x_4\right]}|s>\otimes\,
\boldsymbol{\Phi}(\zeta,\eta
=\frac{2\boldsymbol{y}\cdot{\boldsymbol{x}}}{\zeta},z)
\right)
\end{align*}
Again, if we group all $2n$ solutions $\boldsymbol{\Phi}$ together
into a $n\times 2n$ matrix $\boldsymbol{\hat\Phi}$ we then obtain
$$W=\left(1_2\otimes C(z)\right) \varphi,
\qquad \varphi=(1_2+\boldsymbol{\hat u}(\boldsymbol{x})\cdot\boldsymbol{\sigma})\,
e\sp{-\im z\left[(x_1-\im x_2)\zeta-\im x_3 -x_4\right]}|s>\otimes \boldsymbol{\hat \Phi}
$$
where $\varphi$ is a $2n\times 2n$ matrix. Then
$$V\sp\dagger =W\sp{-1}=\varphi\sp{-1} (1_2\otimes C(z)\sp{-1})$$
will be in terms of the Baker-Akhiezer function. It remains then
to construct $\boldsymbol{\Phi}$.

\subsection{The Baker-Akhiezer function}
By a constant gauge transformation we may assume that $A_1(0)$ is
diagonal. Its behaviour may be read from the spectral curve
(\ref{curve1}),
$$A_1(0)=\mathrm{Diag}\left(\rho_1,\ldots,
\rho_m\right),\quad
\rho_m=\mathrm{Res}_{P\rightarrow\infty_m}\frac{\eta}{\zeta},
$$
where $\infty_m$ ($m=1,\ldots,n$) are the $n$ points above
$\zeta=\infty$. Thus the integration of the Adjoint Weyl equation
reduces to the matrix spectral problem (\ref{standard}). The same
problem appeared in \cite{ersi89} and \cite{bren06} when focussing
on the algebro-geometric integration of the Nahm equation and we
shall use the results of our recent paper \cite{bren06} for the
integration of the Weyl equation.

Let $\theta$ be the canonical $\theta$-function of the curve
$\mathcal{C}$ and let $\tau$ be its period matrix. The period
lattice is then generated by $\Lambda=(1_g,\tau)$ and
\[ \theta(\boldsymbol{w})=\sum_{\boldsymbol{k}\in\mathbb{Z}^g} \mathrm{exp}\left\{
\imath\pi\boldsymbol{k}^T \tau \boldsymbol{k}+2\imath\pi
\boldsymbol{w}^T\boldsymbol{k} \right\}.
\]
Denote by $\Theta=\{\boldsymbol{w}|\theta(\boldsymbol{w})=0\}$ the
$\theta$-divisor in the Jacobi variety of the curve $\mathcal{C}$,
$\mathbb{C}^g/\Lambda$.

\begin{theorem} Let $\boldsymbol{\Phi}(P,z)=(\Phi_1(P,z),\ldots,
\Phi_n(P,z))^T$ be the eigenfunction (or Baker-Akhiezer function)
of the standard spectral problem (\ref{standard}). The components
$\Phi_j(P,z)$  are given by
\begin{equation}\label{newbafnch}
\Phi _{j }\left( P,z\right) =g_{j }(P)\, \frac{ \theta \left(
\boldsymbol{\phi} (P)-\boldsymbol{\phi}(\infty_{j
})+(z+1)\,\boldsymbol{U}-\widetilde{\boldsymbol{K}}\right) \theta
\left(-\widetilde{\boldsymbol{K}} \right) } { \theta
\left(\boldsymbol{\phi} (P)-\boldsymbol{\phi}(\infty_{j })
-\widetilde{\boldsymbol{K}}\right) \theta \left(
(z+1)\,\boldsymbol{U}-\widetilde{\boldsymbol{K}}\right) }\,
e^{z\,\int\limits_{P_0}^{P}\gamma_{\infty}-z\,\nu_j}.
\end{equation}
Here $\boldsymbol{\phi}(P)$ is the Abel map, $z\in(-1,1)$, and
$P\in\mathcal{C}$. The vector $\widetilde{\boldsymbol{K}}$ is
defined by
$$\widetilde{\boldsymbol{K}}=
\boldsymbol{K}+\boldsymbol{\phi}\left((n-2)
\sum_{k=1}\sp{n}\infty_k\right),$$ where $\boldsymbol{K}$ is the
vector of Riemann constants. We have that
\begin{enumerate}\item $\widetilde{\boldsymbol{K}}$ is independent of the choice
of base point of the Abel map; \item $
\theta(\widetilde{\boldsymbol{K}})=0$;
\item $2\widetilde{\boldsymbol{K}} \in \Lambda$; \item for $n\ge3$ we
have $ \widetilde{\boldsymbol{K}}\in \Theta_{\rm singular}$.
\end{enumerate}
For each $j$ the function $g_j(P)$ is meromorphic on
$\mathcal{C}$, $g_j(\infty_j)=1$, and has a zero-divisor of degree
$g+n-1$ that includes the $n-1$ points
$(\infty_1,\ldots,\widehat{\infty_j},\ldots,\infty_n)$.

The matrix $Q_0(z)$ (which has poles of first order at $z=\pm1$)
is given by
\begin{equation}\label{ourq0}
Q_{0}(z)_{jl}  = \epsilon_{jl}\,\frac{\rho_{j}-
\rho_{l}}{\mathcal{E}(\infty_j,\infty_l)}\,e\sp{i\pi\boldsymbol{\tilde
q}\cdot(\boldsymbol{\phi}(\infty_l)-\boldsymbol{\phi}(\infty_j))}\,
    \,\frac{\theta(\boldsymbol{\phi}(\infty_{l}) -\boldsymbol{\phi}(
    \infty_{j}) + (z+1)\boldsymbol{U} - \widetilde{\boldsymbol{K}})}{\theta(
    (z+1)\boldsymbol{U}
    - \widetilde{\boldsymbol{K}})}\,e^{z(\nu_{l} - \nu_{j})}.
\end{equation}
Here $E(P,Q)=\mathcal{E}(P,Q)/\sqrt{\mathrm{d}x(P)\mathrm{d}x(Q)}$
is the Schottky-Klein prime form, $\boldsymbol{U} -
\widetilde{\boldsymbol{K}}=\frac12\boldsymbol{\tilde
p}+\frac12\tau\boldsymbol{\tilde q}$ ($\boldsymbol{\tilde p}$,
$\boldsymbol{\tilde q}\in\mathbb{Z}\sp{g}$) is a non-singular even
$\theta$-characteristic, and $\epsilon_{jl}=\epsilon_{lj}=\pm1$ is
determined (for $j<l$)
 by
$\epsilon_{jl}=\epsilon_{jj+1}\epsilon_{j+1j+2}\dots\epsilon_{l-1l}$.
The $n-1$ signs $\epsilon_{jj+1}=\pm1$ are arbitrary.
\end{theorem}
In passing we note that a formula with similar features was
obtained by Dubrovin \cite{dubrovin77} when giving a
$\theta$-functional solution to the Euler equation describing
motion of the $n$-dimensional rigid body. The essential difference
is that the curve $\mathcal{C}$ here should be subjected to the
the following three constraints
$\mathbf{H1},\mathbf{H2},\mathbf{H3}$ of Hitchin who showed a
bijection between such curves and magnetic monopoles
\cite{hitchin83}:

$\mathbf{H1}$ $\mathcal{C}$ admits the involution:
$(\zeta,\eta)\longrightarrow(-1/\overline{\zeta},-\overline{\eta}/\overline{\zeta}^2)$.

$\mathbf{H2}$ Let $\gamma_{\infty}(P)$ be the unique differential
 of the second kind on $\mathcal{C}$ defined by the conditions
\[ \gamma_{\infty}(P)\vert_{P\to\infty_i}=\left(\frac{\rho_i}{\xi^2}+O(1)\right)\mathrm{d}\xi,\quad
\oint_{\mathfrak{a}_k}\gamma_{\infty}(P)=0,\quad i, k=1,\ldots,g,
\] where $\xi$ is a local coordinate and
$\rho_i=\mathrm{Res}_{P\rightarrow\infty_i} \eta/\zeta$. Then
$\mathfrak{b}$-periods defining the winding vector
$\boldsymbol{U}$ are to be half-periods,
\begin{equation}
\boldsymbol{U}=\frac{1}{2 \pi\imath} \left(
\oint_{\mathfrak{b}_1}\gamma_{\infty},\ldots,
\oint_{\mathfrak{b}_g}\gamma_{\infty}
\right)^T=\frac12\boldsymbol{n}+\frac12\tau\boldsymbol{m}.
\end{equation}
The vectors $\boldsymbol{n}, \boldsymbol{m}\in \mathbb{Z}^g$ are
called  {\it Ercolani-Sinha vectors}. They should be {\it
primitive}, i.e. $s\boldsymbol{U}$ belongs to the period lattice
$\Lambda$ if and only if $s=0$ or $s=2$ (equivalently,
$z=s-1=\pm1$). (Hitchin's original constraint was  reformulated to
this form in \cite{bren06}.)

$\mathbf{H3}$ All components of the Baker-Akhiezer function
$\Phi_j(P,z)$ are real and smooth for $z\in (-1,1)$.

Bringing the previous results together then yields:
\begin{proposition}
Let $\boldsymbol{w}^{(k)}(\boldsymbol{x},z)$, $k=1,\ldots,2n$ be
the column vectors
\begin{equation}
\boldsymbol{w}^{(k)}(\boldsymbol{x},z)=
(1_2+\boldsymbol{\hat u}(P_k)\cdot\boldsymbol{\sigma})\,
e\sp{-\im z\left[(x_1-\im x_2)\zeta-\im x_3 -x_4\right]}|s>\otimes\,
C(z)\boldsymbol{\Phi}(P_k,z)
\end{equation}
where $P_k=(\zeta_k,\eta_k)\in\mathcal{C}$ are solutions to the
Atiyah-Ward constraint, $C(z)^{-1}$ is the  fundamental solution
to the ODE
\[\frac{\mathrm{d}}{\mathrm{d} z} C(z)^{-1}+\frac12 Q_0(z) C(z)^{-1}=0 \]
normalized by the condition $C(0)=1_n$, and the $n\times n$-matrix
$Q_0(z)$ and $n$-vector $\boldsymbol{\Phi}(P,z)$ are given by the
$\theta$-functional formulae (\ref{ourq0}) and (\ref{newbafnch})
respectively. Then the $2n\times 2n$ matrix
\begin{equation}V(\boldsymbol{x},z)=
\left\{\left(\boldsymbol{w}^{(1)}(\boldsymbol{x},z),\boldsymbol{w}^{(2)}(\boldsymbol{x},z),\ldots,
\boldsymbol{w}^{(2n)}(\boldsymbol{x},z)\right)^{-1}\right\}^{\dagger}
\end{equation}
defines the fundamental solution to the Weyl equation,
$\Delta^{\dagger}V=0$.

\end{proposition}

\section{Conclusions}

Although nonabelian magnetic monopoles have been objects of
fascination for some decades now, very few explicit solutions are
known. This note fits into our longer programme of seeing how far
the techniques from integrable systems will allow us to construct
such solutions. Here we have considered the explicit construction
of magnetic monopoles using algebro-geometric constructions coming
from integrable systems. Previous studies along these lines have
focussed on the construction of solutions to Nahm's equation which
is an auxiliary problem to that of the explicit integration of the
Bogomolny equations. Although the ADHM construction is based upon
normalizable solutions of the equation
$\Delta^{\dagger}\boldsymbol{v}=0$ an ans\"atz of Nahm naturally
gives solutions of the adjoint equation $\Delta\boldsymbol{w}=0$:
the matrices of fundamental solutions of these equations are
related by $V=\left(W\sp\dagger\right)\sp{-1}$. Here we have
expressed $\boldsymbol{w}$ in terms of a Baker-Akhiezer function
and given explicit expressions for this. Assuming one has a
spectral curve these expressions may be evaluated algorithmically.
Unfortunately the curves characterizing magnetic monopoles are
often restricted by transcendental constraints (\textbf{H2},
\textbf{H3} above), but this is a separate and interesting story
to the one presented here. Finally we have not addressed here the
remaining problem of extracting the normalizable solutions from
this data. Details and examples of this approach will be given
elsewhere.

\section{Acknowlegments} Both authors wish to acknowledge the partial
support of the European Science Foundation Programme MISGAM
(Methods of Integrable System, Geometry and Applied Mathematics)
and V.Z.E further thanks the ENIGMA network for additional support
while at Imperial College.

\appendix

\section{The Panagopolous formulae}

We have
$$
\Delta=\im \dfrac{d}{dz}+x_4+\im\boldsymbol{x}\cdot\boldsymbol{\sigma} -\im
T_4+\boldsymbol{T}\cdot\boldsymbol{\sigma}.$$ Thus
$$\Delta\sp\dagger=
\im
\dfrac{d}{dz}+x_4-\im\boldsymbol{x}\cdot\boldsymbol{\sigma}-\im
T_4-\boldsymbol{T}\cdot\boldsymbol{\sigma}.$$ Set
\[ \Delta^{\dagger}=\im\left[1_{2n}\frac{d}{dz}-\im x_4-T_4+\mathcal{H}+\mathcal{F}\right]\]
with Hermitian
\[  \mathcal{H}=-\sum_{j=1}^3 x_j \sigma_j\otimes 1_n,\qquad  \mathcal{F}=\imath \sum_{j=1}^3 \sigma_j\otimes T_j .\]
Then if $\boldsymbol{v}$ is any solution of
$\Delta^{\dagger}\boldsymbol{v}=0$ we have that
\begin{align*}
1_{2n}\frac{d}{dz}\boldsymbol{v}=\left[\im x_4+T_4-(\mathcal{H}+\mathcal{F})\right]\boldsymbol{v}.
\end{align*}

For completeness we prove here the Panagopolous formulae
\cite{panagopo83} using the method described in that reference
(extending very slightly to the case $x_4,T_4$ possibly nonzero).
These integral formulae reduce to the problem of finding for any
given operator $\mathcal{A}$ and any two solutions
$\boldsymbol{{v}}_{a,b}$ of $\Delta^{\dagger}\boldsymbol{v}=0$ an
operator $\mathcal{B}$ such that
\begin{equation}
\boldsymbol{{v}}_a\sp\dagger
\mathcal{A}\boldsymbol{{v}}_b=\frac{d}{dz} \left(\boldsymbol{{v}}_a\sp\dagger
\mathcal{B}\boldsymbol{{v}}_b\right).
\end{equation}
In this case
\begin{align*}
\boldsymbol{{v}}_a\sp\dagger\mathcal{A}\boldsymbol{{v}}_b
&=\frac{d \boldsymbol{{v}}_a\sp\dagger}{dz}
\mathcal{B}\boldsymbol{{v}}_b
+\boldsymbol{{v}}_a\sp\dagger
\frac{d\mathcal{B}}{dz}\boldsymbol{{v}}_b
+\boldsymbol{{v}}_a\sp\dagger\mathcal{B}
\frac{d\boldsymbol{{v}}_b}{dz}
=\boldsymbol{{v}}_a\sp\dagger\left(
 \frac{d\mathcal{B}}{dz} -(\mathcal{H}+\mathcal{F})\mathcal{B}-\mathcal{B}(\mathcal{H}+\mathcal{F})
  \right)\boldsymbol{{v}}_b.
\end{align*}
and thus we seek to relate the operators $\mathcal{A}$ and
$\mathcal{B}$ by
\[\mathcal{A}=
\frac{d\mathcal{B}}{dz}
-(\mathcal{H}+\mathcal{F})\mathcal{B}-\mathcal{B}(\mathcal{H}+\mathcal{F}).
\]
Introduce the operator $\mathcal{D}$ by
\[  \mathcal{D} (\mathcal{R})= \frac{d\mathcal{R}}{dz}
-(\mathcal{H}+\mathcal{F})\mathcal{R}-\mathcal{R}(\mathcal{H}+\mathcal{F}).\]
We shall use the following relations
\begin{equation}
\mathcal{F}^2=-1_2\otimes
\sum_{i=1}^3T_iT_i-\im\sum_{i,j,k=1}^3\epsilon_{ijk}\,\sigma_k\otimes
T_iT_j\label{rel1}
\end{equation}
and
\begin{equation}
\frac{d\mathcal{F}}{dz}=\im\sum_{i,j,k=1}^3\epsilon_{ijk}\, \sigma_k\otimes
T_iT_j\label{rel2}
\end{equation}
Therefore
\[  \mathcal{F}^2+\frac{d\mathcal{F}}{dz}=-1_2\otimes\sum_{i,j=1}^3 T_iT_j \]
and
\[   \left[\mathcal{F}^2+\frac{d\mathcal{F}}{dz},\mathcal{H}\right]=0. \]

\begin{proposition} \label{panagopo1} Let
\[  \mathcal{Q}=\frac{1}{r^2} \mathcal{H}\mathcal{F}\mathcal{H}-\mathcal{F} \]
Then we have the antiderivative
\begin{equation}
\int dz\, \boldsymbol{{v}}_a\sp\dagger \boldsymbol{{v}}_b
= \boldsymbol{{v}}_a\sp\dagger \mathcal{Q}^{-1}
\boldsymbol{{v}}_b.
\end{equation}
\end{proposition}
\begin{proof}
In this case $\mathcal{A}=1_{2n}$ and we must show that
\begin{align}
\frac{d\mathcal{Q}^{-1}}{dz}-(\mathcal{H}+\mathcal{F})\mathcal{Q}^{-1}
-\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F}) =1_{2n} \label{panrel}
\end{align}
The left-hand side  of (\ref{panrel}) may be rewritten as follows
\begin{align*}
\mathcal{Q}^{-1} &\left[ -\frac{d}{dz}\left(
\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}-\mathcal{F}
\right)-(\mathcal{H}+\mathcal{F})\left(
\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}-\mathcal{F} \right)
-\left(
\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}-\mathcal{F}
\right)(\mathcal{H}+\mathcal{F}) \right]\mathcal{Q}^{-1}\\
=&\mathcal{Q}^{-1}
\left[-\frac{1}{r^2}\mathcal{H}\frac{d\mathcal{F}}{dz}
\mathcal{H}+\frac{d\mathcal{F}}{dz}
-\frac{1}{r^2}(\mathcal{H}+\mathcal{F})\mathcal{H}\mathcal{F}\mathcal{H}
-\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}(\mathcal{H}+\mathcal{F})\right.\\
&\qquad \qquad
+(\mathcal{H}+\mathcal{F})\mathcal{F}+\mathcal{F}(\mathcal{H}+\mathcal{F})\bigg]\mathcal{Q}^{-1}\\
=&\mathcal{Q}^{-1}
\left[\frac{1}{r^2}\mathcal{H}\left(\mathcal{F}^2+1_2\otimes\sum_{i,j=1}^3
T_iT_j\right)\mathcal{H}-\left(\mathcal{F}^2+1_2\otimes\sum_{i,j=1}^3T_iT_j\right)\right.\\
&\left.\qquad\ -\frac{1}{r^2}\mathcal{H}^2\mathcal{F}\mathcal{H}-\frac{1}{r^2}
\mathcal{F}\mathcal{H}\mathcal{F}\mathcal{H}+\mathcal{H}\mathcal{F}+\mathcal{F}^2
-\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}^2-\frac{1}{r^2}
\mathcal{H}\mathcal{F}\mathcal{H}\mathcal{F}+\mathcal{F}\mathcal{H}+\mathcal{F}^2
\right]\mathcal{Q}^{-1}
\end{align*}
Now $\mathcal{H}^2=r^21_{2n}$ and
\begin{align*}
\mathcal{Q}^2&=\left(\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}-\mathcal{F}\right)^2
=\frac{1}{r^4}\mathcal{H}\mathcal{F}\mathcal{H}^2\mathcal{F}\mathcal{H}+\mathcal{F}^2
-\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}\mathcal{F}
-\frac{1}{r^2}\mathcal{F}\mathcal{H}\mathcal{F}\mathcal{H}\\
&=\frac{1}{r^2}\mathcal{H}\mathcal{F}^2\mathcal{H}
-\frac{1}{r^2}\mathcal{H}\mathcal{F}\mathcal{H}\mathcal{F}
-\frac{1}{r^2}\mathcal{F}\mathcal{H}\mathcal{F}\mathcal{H}+\mathcal{F}^2
\end{align*}
Performing the appropriate cancellations we obtain the necessary
result.

\end{proof}

\begin{proposition}\label{panagopo2} Let $\mathcal{Q}$ be as in the
Proposition \ref{panagopo1} and
\[  \mathcal{S}=\mathcal{Q}^{-1} \left( z+2\mathcal{H}\frac{d}{d r^2}   \right). \]
Then we have the antiderivative
\begin{equation}
\int dz\, z \boldsymbol{{v}}_a\sp\dagger \boldsymbol{{v}}_b
= \boldsymbol{{v}}_a\sp\dagger \mathcal{S}
\boldsymbol{{v}}_b.\label{parel2}
\end{equation}
\end{proposition}
\begin{proof}
Denote
\[ \mathcal{S}_1=\mathcal{Q}^{-1} z,\qquad  \mathcal{S}_2=\mathcal{Q}^{-1} 2
\mathcal{H}\frac{d}{dr^2}.
\]
Then
\begin{align*}
\mathcal{D}(\mathcal{S}_1)&=\frac{d}{dz}\left(z\mathcal{Q}^{-1}\right)-z
(\mathcal{H}+\mathcal{F})\mathcal{Q}^{-1}-z\mathcal{Q}^{-1}(\mathcal{H}
+\mathcal{F})
=z\mathcal{D}(\mathcal{Q}^{-1})+\mathcal{Q}^{-1}=z1_{2n}+\mathcal{Q}^{-1}.
\end{align*}
Further, using (\ref{panrel}),
\begin{align*}
\mathcal{D}(\mathcal{S}_2)&=\frac{d}{dz}\left(\mathcal{Q}^{-1}
2\mathcal{H}
\frac{d}{dr^2}\right)-(\mathcal{H}+\mathcal{F})\mathcal{Q}^{-1}2\mathcal{H}
\frac{d}{dr^2}
-\mathcal{Q}^{-1}\,2\mathcal{H}\,
\frac{d}{dr^2}(\mathcal{H}+\mathcal{F})\\
&=\mathcal{D}(\mathcal{Q}^{-1})\,2\mathcal{H}\frac{d}{dr^2}
\;\;+\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F})2\mathcal{H}
\frac{d}{dr^2}
-\mathcal{Q}^{-1}\,2\mathcal{H}
\frac{d}{dr^2}(\mathcal{H}+\mathcal{F})\\
&=2\mathcal{H}\frac{d}{dr^2}+\mathcal{Q}^{-1}(2r^2+2\mathcal{F}\mathcal{H})\frac{d}{dr^2}
-\mathcal{Q}^{-1}(2r^2+2\mathcal{H}\mathcal{F})\frac{d}{dr^2}
-\mathcal{Q}^{-1}2\mathcal{H}\frac{d\mathcal{H}}{dr^2}.
\end{align*}
The last term may be expressed as
\[-\mathcal{Q}^{-1}2\mathcal{H}\frac{d\mathcal{H}}{dr^2}
=-\mathcal{Q}^{-1}\frac{d\mathcal{H}^2}{dr^2}=-\mathcal{Q}^{-1}\]
whence
\begin{align*}\mathcal{D}(\mathcal{S})=
\mathcal{D}(\mathcal{S}_1+\mathcal{S}_2)=z1_{2n}
+\left(
2\mathcal{H}+2\mathcal{Q}^{-1}\mathcal{F}\mathcal{H}-2\mathcal{Q}^{-1}\mathcal{H}\mathcal{F}
\right)\frac{d}{dr^2}.
\end{align*}
Now the expression in brackets vanishes as a consequence of
\begin{align*}
2\mathcal{H}&+2\mathcal{Q}^{-1}\mathcal{F}\mathcal{H}-2\mathcal{Q}^{-1}\mathcal{H}\mathcal{F}
=2\mathcal{Q}^{-1}\left[\left( \frac{1}{r^2} \mathcal{H}\mathcal{F}\mathcal{H} -
\mathcal{F}\right)\mathcal{H}+\mathcal{F}\mathcal{H}-\mathcal{H}\mathcal{F}\right]\\
&=2\mathcal{Q}^{-1}\left[\mathcal{H}\mathcal{F}-\mathcal{F}\mathcal{H}
+\mathcal{F}\mathcal{H} -\mathcal{H}\mathcal{F}  \right]=0
\end{align*}
and the result follows.
\end{proof}

\begin{proposition}\label{panagopo3} Let $\mathcal{Q}$ be as in the
Proposition \ref{panagopo1} Then the antiderivative
\begin{align}
\int \boldsymbol{{v}}_a\sp\dagger\frac{\partial}{\partial
x_i}\boldsymbol{{v}}_b dz
&=\boldsymbol{{v}}_a\sp\dagger\mathcal{Q}^{-1} \left[
\frac{\partial}{\partial x_i}+\mathcal{H}\frac{z}{r^2}\,
x_i+\mathcal{H}\frac{\imath}{r^2} \left(
\boldsymbol{x}\times\boldsymbol{\nabla}\right)_i \right]\boldsymbol{{v}}_b
.\label{parel3}
\end{align}
\end{proposition}

\begin{proof}
Let $L=L_1+L_2+L_3$ with
\[ L_1=\mathcal{Q}^{-1}\frac{\partial}{\partial x_i},\quad
L_2= \mathcal{Q}^{-1}\mathcal{H}\frac{z}{r^2}\,x_i,\quad
L_3=\mathcal{Q}^{-1}\mathcal{H}\frac{\imath}{r^2}\left( \boldsymbol{x}\times
\boldsymbol{\nabla} \right)_i
\]
We compute $\mathcal{D}(L_i)$, $i=1,2,3$. First
\begin{align*}
\mathcal{D}(L_1)&=\frac{d}{dz}\left( \mathcal{Q}^{-1} \frac{\partial
}{\partial x_i} \right)
-(\mathcal{H}+\mathcal{F})\mathcal{Q}^{-1}\frac{\partial}{\partial
x_i}-\mathcal{Q}^{-1}\frac{\partial }{\partial
x_i}(\mathcal{H}+\mathcal{F})\\
&=\left[
\frac{d\mathcal{Q}^{-1}}{dz}-(\mathcal{H}+\mathcal{F})\mathcal{Q}^{-1}
-\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F})
\right]\frac{\partial}{\partial x_i}-\mathcal{Q}^{-1}\frac{\partial
\mathcal{H}}{\partial x_i}\\
&=1_{2n}\frac{\partial}{\partial x_i}-\mathcal{Q}^{-1}\frac{\partial
\mathcal{H}}{\partial x_i}
\end{align*}
where we use Proposition \ref{panagopo1}. Next
\begin{align*}
\mathcal{D}(L_2)&=\frac{d}{dz}\left( \mathcal{Q}^{-1}
\mathcal{H}\frac{z}{r^2}x_i \right)-(\mathcal{H}+\mathcal{F})
\mathcal{Q}^{-1} \mathcal{H}\frac{z}{r^2}x_i-\mathcal{Q}^{-1}
\mathcal{H}\frac{z}{r^2}x_i(\mathcal{H}+\mathcal{F})\\
&=\left[
\frac{d\mathcal{Q}^{-1}}{dz}-(\mathcal{H}+\mathcal{F})\mathcal{Q}^{-1}
-\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F})
\right]\mathcal{H}\frac{z}{r^2}x_i\\
&\qquad +\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F})\mathcal{H}\frac{z}{r^2}x_i
-\mathcal{Q}^{-1}\mathcal{H}\frac{z}{r^2}x_i(\mathcal{H}+\mathcal{F})
+\mathcal{Q}^{-1}\mathcal{H}\frac{x_i}{r^2}\\
&=\mathcal{H}\frac{z}{r^2}x_i+\mathcal{Q}^{-1}(r^2+\mathcal{F}\mathcal{H})\frac{z}{r^2}x_i
-\mathcal{Q}^{-1}(r^2+\mathcal{H}\mathcal{F})\frac{z}{r^2}x_i+\mathcal{Q}^{-1}\mathcal{H}\frac{x_i}{r^2}\\
&=\mathcal{Q}^{-1}\mathcal{H}\frac{x_i}{r^2}+\frac{z}{r^2}x_i(
\mathcal{H}+\mathcal{Q}^{-1}\mathcal{F}\mathcal{H}-\mathcal{Q}^{-1}\mathcal{H}\mathcal{F})
\end{align*}
Now the expression in brackets vanishes (see the proof of
Proposition \ref{panagopo2}). Therefore
\[\mathcal{D}(L_2)=\mathcal{Q}^{-1}\mathcal{H}\frac{x_i}{r^2}.\]
Next we calculate $\mathcal{D}(T_3)$ for $i=1$. We have
\begin{align*}
\mathcal{D}(L_3)&=\frac{d}{dz}\left[ \mathcal{Q}^{-1}
\mathcal{H}\frac{\imath}{r^2}\left( x_2\frac{\partial}{\partial
x_3}-x_3\frac{\partial}{\partial x_2} \right) \right]\\
&\quad -(\mathcal{H}+\mathcal{F})\left[ \mathcal{Q}^{-1}
\mathcal{H}\frac{\imath}{r^2}\left( x_2\frac{\partial}{\partial
x_3}-x_3\frac{\partial}{\partial x_2} \right) \right]
-\left[ \mathcal{Q}^{-1} \mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right) \right](\mathcal{H}+\mathcal{F})\\
&=\left[  \frac{d\mathcal{Q}^{-1}}{dz}
-(\mathcal{H}+\mathcal{F})\mathcal{Q}^{-1}
-\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F})\right]\mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)\\
&\quad +\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F})\mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)
-\mathcal{Q}^{-1}\mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)(\mathcal{H}+\mathcal{F})\\
&=\mathcal{H}\frac{\imath}{r^2}\left( x_2\frac{\partial}{\partial
x_3}-x_3\frac{\partial}{\partial x_2}
\right)+\mathcal{Q}^{-1}(\mathcal{H}+\mathcal{F})\mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)\\
&\qquad-\mathcal{Q}^{-1}\mathcal{H}(\mathcal{H}+\mathcal{F})\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)-\mathcal{Q}^{-1}\mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)\mathcal{H}\\
&=\left[\mathcal{H}+\mathcal{Q}^{-1}(\mathcal{H}^2+\mathcal{F}\mathcal{H})
-\mathcal{Q}^{-1}(\mathcal{H}^2+\mathcal{H}\mathcal{F})\right]
\frac{\imath}{r^2}\left( x_2\frac{\partial}{\partial
x_3}-x_3\frac{\partial}{\partial x_2} \right)\\
&\quad-\mathcal{Q}^{-1}\mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)\mathcal{H}.
\end{align*}
Finally we obtain
\[  \mathcal{D}(L_3)= -\mathcal{Q}^{-1}\mathcal{H}\frac{\imath}{r^2}\left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)\mathcal{H}. \] Altogether we have
\begin{align*}
\mathcal{D}(T)=1_{2n}\frac{\partial}{\partial
x_1}+\mathcal{Q}^{-1}\left\{-\frac{\partial \mathcal{H}}{\partial
x_1}+\mathcal{H}\frac{x_1}{r^2}-\mathcal{H}\frac{\imath}{r^2} \left(
x_2\frac{\partial}{\partial x_3}-x_3\frac{\partial}{\partial x_2}
\right)\mathcal{H}\right\}
\end{align*}
Multiplying the expression in the parentheses by $-r^2$ gives
\begin{align*}
-\left\{\cdot\right\}r^2&=-\sigma_1\otimes 1_2(x_1^2+x_2^2+x_3^2)
+\sigma_1\otimes 1_2x_1^2+\sigma_2\otimes 1_2x_1
x_2+\sigma_3\otimes 1_2x_1 x_3\\
&\quad
-\imath(\sigma_1\otimes 1_2 x_1 +\sigma_2\otimes 1_2 x_2+\sigma_3\otimes 1_2 x_3  )
\times (x_2\sigma_3\otimes 1_2-x_3\sigma_2\otimes 1_2)
\end{align*}
which vanishes by standard relations, proving the result.

\end{proof}

\providecommand{\bysame}{\leavevmode\hbox
to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

\providecommand{\MRhref}[2]{
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}

config: arxiv


TITLE: How to calculate projected value?
QUESTION [2 upvotes]: I am trying to figure out how to calculate a projected value for our goals.
Each of our sales persons is assigned an annual goal for 2014, say 60 units. Let's say, as of today, one sales person has sold 5 units.
How would I calculate, assuming everything trends the way it currently is, their projected actual value for year end?
I'd like to show them how they are trending.

REPLY [2 votes]: Pace = 5 units completed / 16 days
Projected Yearly Outcome = pace * 365
*Note: this example is counting physical days
If you want more accurate results then you need to track this over time (months) and take the average of their paces then just simply multiply it to get the yearly estimate. Keep in mind this is just math so there is no intelligence such as Christmas and New Years where the office could be closed for a week or more. It may also make more sense to look at this in terms of working days then physical (calendar) days.


TITLE: If $f$ is holomorphic function then are the zeros of $\Re(f)$ isolated?
QUESTION [1 upvotes]: Let $f$ be a holomorphic function.
We know that the the zeros of $f$ are isolated
I would like to know if is it true that the zeros of $\Re(f)$ are isolated.
Thanks.

REPLY [6 votes]: Hint: consider $f(z) = z$ and the set $\{ z \in \mathbb{C} : \Re(z) = 0 \}$.
In fact, zeros of $\Re(f)$ are rarely isolated, because if $f(x+iy) = u(x, y) + i v(x, y)$ is holomorphic (where $u$ and $v$ are real), then under some natural assumption* the solution to the equation $u(x, y) = 0$ is a one-dimensional submanifold of $\mathbb{R}$, which is clearly not a discrete set.
*When $0$ is a regular value of $u(x, y)$ and $0 \in \operatorname{im} u$.


TITLE: Do protons and electrons actually precess?
QUESTION [2 upvotes]: I know that protons and electrons do not actually spin, although they have the property of spin.
I was learning about MRI. I was introduced to the idea that you have the spin of the proton in the external magnetic field. Since they are not perfectly aligned, it will cause precession just like in a spin top. I understand how precession occurs in spin tops and bike wheels, but that's dependent on the object actually spinning and having the velocity to change.
If protons don't spin, do they actually precess too? I would imagine not, because that would imply that they're actually spinning. If not actually precessing, then what is this precession?

REPLY [0 votes]: These particles have half-integer spin, so their angular momentum can't be explained as if they're physical objects spinning rigidly. That would produce only integer values of angular momentum.
They do "really" precess in the following sense. Suppose you prepare an electron in state A, with its spin pointing along the positive x axis. Then you subject it to a magnetic field along the z axis and let it act for an amount of time that would, classically, cause a precession by some angle. For example, let's say it's 90 degrees counterclockwise about the z axis. Now it's in some state B. If you then measure the component of the electron's spin along the y axis, you will find what you expected, with 100% probability: the spin is aligned with the y axis.
But the results of other measurements on B will not give the results you would expect classically. For example, if you measure the component of the spin along the x axis, you will not get zero. (That's not a possible value for a spin-1/2 particle.) You will get either +1/2 or -1/2, with equal probability.
The answer by anna v is wrong. She says, "The (x,y,z) of particles in the quantum mechanical frame are described by orbitals , probability loci where a measurement will give the non-sequential (x,y,z) of a particle." The spin-1/2 of an electron has nothing to do with the x, y, and z degrees of freedom.


text:
\section{Generating Fraction for Lucas Numbers}
Tags: Lucas Numbers

\begin{theorem}
The [[Definition:Fraction|fraction]]:
:$\dfrac {199} {9899}$
has a [[Definition:Decimal Expansion|decimal expansion]] which contains within it the start of the [[Definition:Lucas Number|Lucas sequence]]:
:$0 \cdotp 02010 \, 30407 \, 11 \ldots$
\end{theorem}

\begin{proof}
By [[Definition:Long Division|long division]]:
<pre>
         0.0201030407111829
     ----------------------
9899 ) 199.0000000000000000
       197 98
       --- --
         1 0200
           9899
         - ----
            30100
            29697
            -----
              40300
              39596
              -----
                70400
                69293
                -----
                 11070
                  9899
                 -----
                  11710
                   9899
                  -----
                   18110
                    9899
                   -----
                    82110
                    79192
                    -----
                     29180
                     19798
                     -----
                      93820
                      89091
                      -----
                       4729
</pre>
By [[Generating Function for Lucas Numbers]]:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} L_k 10^{-2 k - 2}
      | r = \frac {2 - 10^{-2} } {1 - 10^{-2} - 10^{-4} } \times 10^{-2}
}}
{{eqn | r = \frac {2 \times 10^2 - 1} {10^4 - 10^2 - 1}
}}
{{eqn | r = \frac {199} {9899}
}}
{{end-eqn}}
The first few terms are contained in the decimal expansion, as long as $L_{k + 1} < 100$, where there is no carry.
{{qed}}
\end{proof}

config: wiki


TITLE: Identities For Generalized Harmonic Number
QUESTION [7 upvotes]: I have been searching for identities involving generalized harmonic numbers
\begin{equation*}H_n^{(p)}=\sum_{k=1}^{n}\frac{1}{k^p}\end{equation*}
I found several identities in terms of $H_n^{(1)}$, but I am looking for some interesting identities for $H_n^{(2)}$. Does anyone know of any identities know of any nontrivial identities for $H_n^{(2)}$?  I found some listed on Wikipedia, but this list is not comprehensive. Thanks for your help.

integral identities
summation identities
recursive identities
in terms of another function

REPLY [1 votes]: I discovered a Generalized Harmonic Summation Identity some time ago.

$$ \sum_{r=1}^{n} \dfrac{H_{r} ^{(m)}}{r^m} = \dfrac{1}{2} \left( [H_{n}^{(m)}]^2 + H_{n}^{(2m)} \right) \quad ; \quad m \geq 1 $$

I have posted a proof here.


TITLE: matrix singular proof
QUESTION [2 upvotes]: Let A, B be n×n matrices. Show that if AB = A and B≠I then A must be singular.
I was thikning to prove it by contradiction, showing if A is nonsingular then we have
thta  AB=BA=A, therefore B is the inverse of A. Since B≠I then A need to be singular.
Is this proof complete and correct?

REPLY [2 votes]: This is a direct proof: Assuming that $B\ne I$ then there is $x,y\in\Bbb R^n$ such that $\boxed{x\ne y}$ and $Bx=y$. Now we have
$$ABx=\boxed{Ay=Ax}\implies A\;\text{isn't injective}\iff A\;\text{isn't bijective}$$


TITLE: Group of order $p^2$ with Center of order $p$
QUESTION [0 upvotes]: I am sure this question was asked here before, maybe even multiple times. But I can't find it right now and it really bugs me:

Let $G$ be a group with $|G|=p^2$. Then $G$ is abelian.

For the proof we have 3 cases: $|Z(G)|\in\{1,p,p^2\}$, two of which I understand. Let $|Z(G)|=p$, then $G/Z(G)\cong\mathbb{Z}_p$, so $G/Z(G)$ is cyclical and therefore $G$ is abelian.
Question: This would mean that $Z(G)=G$, so there it's ultimately a contradiction although it proofs that $G$ is abelian. How do I interpret this proof?

REPLY [1 votes]: As you have mentioned, $|Z(G)|=1,p  \text{ or } p^2$. If $|Z(G)|=p$, then by this argument, $|G/Z(G)|=p$ which implies $G/Z(G)$ is cyclic. But as the proof mentions, this cannot be the case because of this result. Hence $|Z(G)|=p^2=|G|$, so that $Z(G)=G$ and $G$ is abelian.


text:
\setcounter{figure}{0}
\renewcommand{\thefigure}{B\arabic{figure}}

\section{Additional examples for reconstructing images as summation of rank-1 images} \label{sec:appx_rank1}

Here, we provide an example of writing a testing image as summation of rank-1 images obtained from the training set (in the wavelet space). Hence the same common basis obtained from all classes of training set, works for efficient decomposition of testing images, as well.

Figure~\ref{fig_reconstruction_te2} shows the reconstruction of the 3rd testing image of the Cat class in CIFAR-10.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_100.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_200.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_300.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_1500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_1800.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_2000.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_3000.png}\\
       \vspace{0.5cm}
   \includegraphics[width=0.7\linewidth]{fig_cd_te3_residual.png}
  \caption{3rd testing image of the Cat class in CIFAR-10. We have used the common right basis obtained from training set to decompose this image into rank-1 images.}
  \label{fig_reconstruction_te2}
\end{figure}

Figure~\ref{fig_reconstruction_te331} shows the reconstruction of a testing image of Cat class that is misclassified by state of art model on CIFAR-10.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_100.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_200.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_300.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_1000.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_1500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_2000.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_3000.png}\\
       \vspace{0.5cm}
   \includegraphics[width=0.7\linewidth]{fig_cd_te331_residual.png}
  \caption{331st testing image of the Cat class which is misclassified by state of art model on CIFAR-10. For this decomposition,we have used the same $V$ obtained from training set.}
  \label{fig_reconstruction_te331}
\end{figure}

file: 2006.09879/patterns in image classification/a2-examples.tex


text:
\section{Introduction}
\label{sec:intro}

Nonnegative Matrix Factorization (NMF) is a low-rank model widely used for feature extraction in applications such as multispectral imaging, text mining, or blind source separation; see~\cite{gillis_why_2014,fu2019nonnegative} and the references therein.
Given a nonnegative data matrix $M \in \mathbb{R}_+^{m \times n}$ and a factorization rank $r$, NMF consists in finding two nonnegative matrices $W \in \mathbb{R}^{m \times r}_+$ and $H \in \mathbb{R}^{r \times n}_+$ such that $M \approx W H$.
NMF can be formalized as the following optimization problem:
\begin{equation}
    \min_{W \geq 0, H \geq 0} \| M - WH \|_F^2 .
\end{equation}
In this paper, we use the Frobenius norm to measure the quality of the approximation.
Although other measures are possible, the Frobenius norm is by far the most commonly used, because it assumes Gaussian noise (which is reasonable in many real-life applications) and allows for efficient computations~\cite{gillis_why_2014}.

One of the advantages of NMF over similar methods such as principal component analysis (PCA) is that the nonnegativity constraint favors a part-based representation~\cite{lee1999learning}, which is to say that the factors are more easily interpretable, in particular when they have a physical meaning.
If each column of $M$ represents a data point, then each corresponding column of $H$ contains the coefficients to reconstruct it from the $r$ atoms represented by the columns of $W$, since $M(:,j) \approx WH(:,j)$ for all $j$.
Every data point is therefore expressed as a linear combination of atoms.
For example, when using NMF for multispectral unmixing, a data point is a pixel, an atom is a specific material, and each column of $H$ contains the abundance of these materials in the corresponding pixel; see Section~\ref{sec:xp-real} for more details.
Geometrically, the atoms (columns of $W$) can be seen as $r$ vertices whose convex hull contains the data points (columns of $M$), under appropriate scaling.

\subsection{Separability}

In general, computing NMF is NP-hard~\cite{vavasis_complexity_2010}.
However, Arora et al.~\cite{arora_computing_2012} proved that NMF is solvable in polynomial time under the \emph{separability}  assumption on the input matrix.
\begin{definition}\label{def:separability}
A matrix $M$ is $r$-separable if there exists a subset of $r$ columns of $M$, indexed by $\mathcal{J}$, and a nonnegative matrix $H \geq 0$, such that $M = M(:,\mathcal{J}) H$.
\end{definition}
Equivalently, $M$ is $r$-separable if $M$ has the form
$M = W [ I_r , H' ] \Pi$, where $I_r$ is the identity matrix of size $r$, $H'$ is a nonnegative matrix, and $\Pi$ is a permutation.
Separable NMF consists in selecting the right $r$ columns of $M$ such that $M$ can be reconstructed perfectly.
In other words, it consists in finding the atoms (columns of $W$) \emph{among} the data points (columns of $M$).

\begin{problem}[Separable NMF]
Given a $r$-separable matrix $M$, find $W = M(:,\mathcal{J})$ with $|\mathcal{J}| = r$  and $H \geq 0$ such that $M = W H$.
\end{problem}
Note that, if $W$ is known, the computation of $H$ is straightforward: it is a convex problem that can be solved using any nonnegative least squares (NNLS) solver (for example, it can be solved with the Matlab function \texttt{lsqnonneg}).
However, the solution is not necessarily unique, unless $W$ is full rank.

In the presence of noise, which is typically the case in real-life applications, this problem is called near-separable NMF and is also solvable in polynomial time given that the noise level is sufficiently small~\cite{arora_computing_2012}. In this case, we are given a near-separable matrix $M \approx M(:,\mathcal{J}) H$ where $|\mathcal{J}| = r$ and $H \geq 0$.

\subsection{Successive Nonnegative Projection Algorithm}

Various algorithms have been developed to tackle the (near-)separable NMF problem.
Some examples are the successive projections algorithm (SPA)~\cite{araujo_successive_2001}, the fast canonical hull algorithm~\cite{kumar_fast_2013}, or the successive nonnegative projections algorithm (SNPA)~\cite{gillis_successive_2014}.
Such algorithms start with an empty matrix $W$ and a residual matrix $R = M$, and then alternate between two steps: a greedy selection of one column of $R$ to be added to $W$, and an update of $R$ using $M$ and the  columns extracted so far.
As SNPA was shown, both theoretically and empirically, to perform better and to be more robust than its competitors~\cite{gillis_successive_2014}, it is the one we study here in detail. Moreover, SNPA is able to handle the underdetermined case when $\rank(W) < m$ which will be key for our problem setting (see below for more details).

SNPA is presented in \cref{alg:snpa}.
\input{21pseudocode-snpa}
SNPA selects, at each step, the column of $M$ maximizing a function $f$ (which can be any strongly convex function such that $f(0)=0$, and $f = ||.||_2^2$ is the most common choice).
Then, the columns of $M$ are projected onto the convex hull of the origin and the columns extracted so far, see step 8 where we use the notation
\[
\Delta = \Big\{ h \ \big| \ h \geq 0, \sum_i h_i \leq 1 \Big\},
\]
whose dimension is clear from the context.

After $r$ steps, given that the noise is sufficiently small and that the columns of $W$ are vertices of $\conv(W)$, SNPA is guaranteed to identify $W$.
An important point is that SNPA requires the columns of $H$ to satisfy $\|H(:,j)\|_1 \leq 1$ for all $j$, where $\|x\|_1 = \sum_i |x_i|$ is the $\ell_1$ norm. This assumption can be made without loss of generality by properly scaling the columns of the input matrix to have unit $\ell_1$ norm; see the discussion in~\cite{gillis_successive_2014}.

\subsection{Model Limitations}

Unfortunately, some data sets cannot be handled successfully by separable NMF, even when all data points are linear combinations of a subset of the input matrix.
In fact, in some applications, the columns of the basis matrix $W$, that is, the atoms, might not be vertices of $\conv(W)$.
This may happen when one seeks a matrix $W$ which is not full column rank.
For example, in multispectral unmixing, $m$ is the number of spectral bands which can be smaller than $r$, which is the number of materials  present in the image; see Section~\ref{sec:xp-real} for more details.
Therefore, it is possible for some columns of $W$ to be contained in the convex hull of the other columns, that is, to be additive linear combinations of others columns of $W$; see \cref{fig:examples} for illustrations in three dimensions (that is, $m = 3$).

\begin{figure}[t]
    \centering
    \subfloat[\label{fig:allext}]{
        \input{img/allext}
    }\hfil
    \subfloat[\label{fig:intvertex}]{
        \input{img/oneint}
    }
    \caption{On the left (a), all vertices are exterior, and SNPA is assured to identify them all. On the right (b), the data points are 2-sparse combinations of 4 points, one of which (vertex 4) is ``interior'' hence it cannot be identified with separable NMF.}
    \label{fig:examples}
\end{figure}

These difficult cases cannot be handled with separable NMF, because it assumes the data points to be linear combinations of vertices, so an ``interior vertex'' cannot be distinguished from another data point.
However, if we assume the \emph{sparsity} of the mixture matrix $H$, we may be able to identify these interior vertices.
To do so, we introduce a new model, extending the approach of SNPA using additional sparsity constraints.

We introduce in \cref{sec:ssnmf} a proper definition of this new problem, which we coin as sparse separable NMF (SSNMF). Before doing so, let use recall the literature on sparse NMF.

\subsection{Sparse NMF}

A vector or matrix is said to be \emph{sparse} when it has few non-zero entries.
Sparse NMF is one of the most popular variants of NMF, as it helps producing more interpretable factors.
In this model, we usually consider column-wise sparsity of the factor $H$, meaning that a data point is expressed as the combination of only a few atoms.
For example, in multispectral unmixing, the column-wise sparsity of $H$ means that a pixel is composed of fewer materials than the total number of materials present in the image.
When sparsity is an a priori knowledge on the structure of the data, encouraging sparsity while computing NMF is likely to reduce noise and produce better results.

Sparse NMF is usually solved by extending standard NMF algorithms with a regularization such as the $\ell_1$ penalty~\cite{hoyer2002non,kim_sparse_2007}, or constraints on some sparsity measure, like the one introduced in~\cite{hoyer2004non}.
Recently, exact $k$-sparse methods based on the $\ell_0$-``norm'' have been used for NMF, using a brute-force approach~\cite{cohen_nonnegative_2019-1}, or a dedicated branch-and-bound algorithm~\cite{nadisicexact}.
They allow the explicit definition of a maximum number (usually noted $k$) of non-zero entries per column of $H$.
These approaches leverage the fact that, in most NMF problems, the factorization rank $r$ is small, hence it is reasonable to solve the $k$-sparse NNLS subproblems exactly.

\subsection{Contributions and Outline}

In this work, we study the SSNMF model from a theoretical and a  pratical point of view.
Our contributions can be summarized as follows:
\begin{itemize}
    \item In \cref{sec:ssnmf}, we introduce the SSNMF model. We prove that, unlike separable NMF, SSNMF is NP-complete.

    \item In \cref{sec:algo}, we propose an algorithm to tackle SSNMF, based on SNPA and an exact sparse NNLS solver.

    \item In \cref{sec:analysis}, we prove that our algorithm is correct under reasonable assumptions, in the noiseless case.

    \item In \cref{sec:xp}, experiments on both synthetic and real-world data sets illustrate the relevance and efficiency of our algorithm.
\end{itemize}

file: 2006.07553/arXiv SSNMF/20intro.tex


TITLE: Why is do no complete graphs have genus 7?
QUESTION [2 upvotes]: My teacher mentioned this in class, but he rushed his explanation due to time restraints. Why is it that complete graphs ($K_n$) never have genus $7$?

REPLY [7 votes]: It is a theorem of Ringel and Youngs that for all $n \geq 3$, the genus of the complete graph $K_n$ -- i.e., the minimal genus of a compact orientable surface into which $K_n$ embeds -- is $\lceil \frac{(n-3)(n-4)}{12} \rceil$.  Granting this result, what you ask about is very straightforward: the given function is weakly increasing.  For $n = 12$ it takes the value $6$.  For $n = 13$ it takes the value $8$.  Thus it never takes the value $7$ (the first of infinitely many values that it skips).
Not being a graph theorist, I confess that I don't know the proof of the Ringel-Youngs Theorem, and the paper was not freely available to me on the internet.  For all I know it may be possible to avoid this result and give an argument more specific to your precise question...but I will guess that it will probably be more fruitful to learn the proof of the general result.

Ringel, G. and Youngs, J. W. T. "Solution of the Heawood Map-Coloring Problem." Proc. Nat. Acad. Sci. USA 60, 438-445, 1968.


TITLE: Finding absorption time of a markov chain
QUESTION [1 upvotes]: I have the following problem.

Consider $N$ cells in a line. The first one is set to $1$ and the rest are  set to $0$. At each step one of the elements that is $0$ is randomly picked. If it is adjacent to one that has a $1$, then it is set to $1$; otherwise it remains $0$. The process is repeated until all cells are filled. Draw the possible states and the transition rates. What is the expected absorption time?

I think the transition probabilities are pretty easy to figure out, but how do you find the expected absorption time? If it was a numeric problem I know I could do that using the fundamental matrix. But I am not sure how you can do that here. I also tried to derive a difference equation but didn't really work out
Any ideas will be appreciated
Thanks

REPLY [2 votes]: Let $t_i$ denote the absorption time if we start from state $i$.
There is only one absorption state $M$.
Hence $$t_M=0.$$
For $i \neq M$, by law of total expectation,
$$t_i = \sum_{k=0}^Np_{ik}t_k+1$$
Now we just have to solve a linear system to compute the absorption time.


TITLE: Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers
QUESTION [1 upvotes]: Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers

Clearly $k$ is a rational number, so I let $k=\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$, this gives $\frac{12b}{a+b}$ and $\frac{6b}{a-b}$ are positive integers. I tried different things here but couldn't make progress. I want to verify that $2, 3$ and $\frac{7}{5}$ are the only possible values for $k$. How to proceed from here? Thanks in advance.

REPLY [0 votes]: Starting from $\frac{6}{k-1}=n \in \mathbb N$, we can generate all possible rational $k=\frac{n+6}{n}$ without regard to the second condition.
Considering the second condition, $\frac{12}{k+1}=m \in \mathbb N$, we substitute to get $m=\frac{12}{\frac{n+6}{n}+1}=\frac{12n}{2n+6}=\frac{6n}{n+3}$. We need to find positive integers $n$ that yield positive integers $m$.
Note that $\frac{6n}{n+3}$ asymptotically  approaches $6$, so we only have to look at values of $n$ such that $\frac{6n}{n+3} \le 5$, which by algebraic rearrangement yields $n \le 15$. We find $m\in \mathbb N \Rightarrow n= 3,6,15$, which in turn yields $k=3,2,\frac{7}{5}$, which is exactly what you found by trial and error.