TITLE: Gauge invariance of non-Abelian theories under Pauli-Villars-Regularisation
QUESTION [3 upvotes]: Under the ordinary Pauli -Villars Regularisation one introduces a heavy mass ($\Lambda$) term  $$\frac{1}{p^2-m^2+i\epsilon} \rightarrow \frac{1}{p^2-m^2+i\epsilon} - \frac{1}{p^2-\Lambda^2+i\epsilon}.$$
Now Kugo in his book "Gauge theories" says that the Pauli-Villars Regularisation term violates gauge invariance for non-Abelian case.
For the usual propagator term before regularisation one could argue that the gauge invariance is manifestly preserved since the propagator in its form is derived directly from the gauge-invariant Lagrangian.
The regularisation process doesn't take place on the Lagrangian level, so the gauge invariance can not be proved by simple gauge transformations of the fields. What is the argument for the gauge invariance violation of the $\Lambda$ - term. How can one "see" this? Or similarly, how can one see the gauge invariance of the regularisation in the abelian case?

REPLY [2 votes]: In abelian case, the photon propagator  under Pauli- Villars regularization goes as $$\frac{1}{k^2}\rightarrow\frac{1}{k^2}- \frac{1}{k^2-{\Lambda^2}}$$ which is basically $1/(k^2- \Lambda^{-2}k^4)$. The presence of the extra term in the propagator can be obtained by supplementing the Lagrangian with an extra term $$\frac{1}{2\Lambda^2}\Box A_\mu\Box A^\mu$$ which using Lorenz gauge can be converted into $$\frac{1}{2\Lambda^2}\partial^\nu F_{\nu\mu}\partial_\sigma F^{\sigma\mu}$$ as $$\partial_\sigma F^{\sigma\mu}=\partial_\sigma(\partial^\sigma A^\mu-\partial^\mu A^\sigma)$$ and using Lorenz gauge second term can be dropped. Now the expression $\frac{1}{2\Lambda^2}\partial^\nu F_{\nu\mu}\partial_\sigma F^{\sigma\mu}$ is U(1) invariant as $F^{\nu\mu}$ is U(1) invariant so Pauli -Villars regularization preserves gauge invariance in abelian case.
In Yang -Mills case, a term like $\frac{1}{2\Lambda^2}\Box A_\mu\Box A^\mu$ violates gauge invariance due to presence of simple derivative instead of covariant derivatives. It is not possible to convert this term by any gauge transformation to a form which has an explicit gauge $covariance$. Gauge invariance can only be obtained by modifying the Lagrangian appropriately for Yang-Mills case which does not follow from Pauli- Villars procedure.
The requirement of covariant derivative in non-Abelian case for gauge invariance comes form the fact that in non-Abelian case the Yang-Mills field strength transform homogeneously under gauge transformation as $F\rightarrow UFU^{-1}$ as opposed to the Abelian case where it is gauge invariant. Covariant derivative has the same transformation law as the field under gauge transformation as $D\rightarrow UDU^{-1}$, so the combination $D_\mu F_{\mu\nu}$ will transform homogeneously which will not be the case if we use a partial derivative instead of covariant derivative. Under a trace term in the Lagrangian which includes a couple of terms like this one can use cyclic property of the trace to show that the term (like the one mentioned below) is unaffected by gauge transformation.
A requirement of gauge invariant term in the Lagrangian  will involve higher covariant derivative term like $tr(\frac{1}{\Lambda^2}D_\alpha F_{\mu\nu}D_\alpha F_{\mu\nu})$ .
This modified form of Pauli- Villars regularization was used by A. A. Slavnov in Nucl. Phys. B, 31, 301 (1971).


TITLE: $\pi$-separable group and its homomorphic image
QUESTION [0 upvotes]: Show that every subgroup and every homomorphic image of a $\pi $-separable group $G$ is also $\pi $-separable. I find a normal series of a subgroup or homomorphic image but I can not show that every quotient group is $\pi $ or $\pi '$ group.

REPLY [2 votes]: Le $N/M$ be a quotient of the normal series of your group $G$ with factors that are either $\pi$- or $\pi'$-groups.
Consider first the case of subgroups.
If $H$ is a subgroup of $G$, the induced factor in the series for $H$ will be
$$
\frac{H \cap N}{H \cap M}
=
\frac{H \cap N}{(H \cap N) \cap M}
\cong
\frac{(H \cap N)M}{M}
\le
\frac{N}{M}.
$$
So $\dfrac{H \cap N}{H \cap M}$ is isomorphic to a subgroup of $N/M$.
For quotients, let $K$ be a normal subgroup of $G$. The induced factor in the series for $G/K$ will be
$$
\frac{N K}{M K}
=
\frac{N (M K)}{MK}
\cong
\frac{N}{N \cap M K}
\cong
\frac{N / M}{(N \cap M K)/M}.
$$
So $\dfrac{N K}{M K}$ is isomorphic to a quotient of $N/M$.


text:
\begin{document}

\begin{frontmatter}

\title{The 2D $\kappa$-Dirac oscillator}

\author[uepg]{Fabiano M. Andrade}
\ead{fmandrade@uepg.br}
\author[ufma]{Edilberto O. Silva}
\ead{edilbertoo@gmail.com}

\address[uepg]{
  Departamento de Matem\'{a}tica e Estat\'{i}stica,
  Universidade Estadual de Ponta Grossa,
  84030-900 Ponta Grossa-PR, Brazil
}
\address[ufma]{
  Departamento de F\'{i}sica,
  Universidade Federal do Maranh\~{a}o,
  Campus Universit\'{a}rio do Bacanga,
  65085-580 S\~{a}o Lu\'{i}s-MA, Brazil
}

\begin{abstract}
In this Letter, 2D Dirac oscillator in the quantum deformed framework
generated by the $\kappa$-Poincar\'{e}-Hopf algebra is considered.
The problem is formulated using the $\kappa$-deformed Dirac equation.
The resulting theory reveals that the energies and wave functions of the
oscillator are modified by the deformation parameter.
\end{abstract}

\begin{keyword}
$\kappa$-Poincar\'{e}-Hopf algebra \sep Dirac oscillator
\end{keyword}

\end{frontmatter}

\date{\today}

\section{Introduction}
\label{sec:introduction}

The Dirac oscillator, established in 1989 by Moshinsky and Szczepaniak
\cite{JPA.1989.22.817,Book.Moshinsky.1996}, is considered a natural
framework to access the relativistic quantum properties of quantum
harmonic oscillator-like systems.
This model has inspired a great deal of investigations in recent years.
These studies have allowed the exploration of new models in theoretical
and experimental physics.
In the context of recent investigations, the interest in this issue
appears, for example, in quantum optics
\cite{PRA.2007.76.041801,PRA.2008.77.033832}, deformed Kempf algebra
\cite{SR.2013.3.3221}, graphene physics \cite{arXiv:1311.2021},
noncommutative space
\cite{JMP.2014.55.032105,IJMPA.2011.26.4991,IJTP.2012.51.2143}, quantum
phase transition \cite{PRA.2008.77.063815,arXiv:1312.5251} and
topological defects \cite{AP.2013.336.489,PRA.2011.84.32109}.
Among several recent contributions on the Dirac oscillator, we refer to
its first experimental verification \cite{PRL.2013.111.170405}.
For a more detailed approach of the Dirac oscillator see Refs.
\cite{JPA.1997.30.2585,AIPCP.2011.1334.249,Book.1998.Strange,
PRA.1994.49.586,EPJB.2001.22.31,MPLA.2004.19.2147,arXiv:1403.4113}.

The dynamics of the Dirac oscillator is governed by the Dirac equation
with the nonminimal prescription
\begin{equation}\label{eq:prescription}
  \mathbf{p}\rightarrow \mathbf{p}-im\omega\tilde{\beta}\mathbf{r},
\end{equation}
where $\mathbf{p}$ is the momentum operator, $m$ is the mass, $\omega$
the frequency of the oscillator and $\mathbf{r}$  is the position
vector.
In the same context, another usual framework where one can study the
dynamics of the Dirac oscillator is that in connection with the theory
of quantum deformations.
These quantum deformations are realized based on the
$\kappa$-Poincar\'{e}-Hopf algebra
\cite{PLB.1991.264.331,PLB.1992.293.344,PLB.1994.329.189,
PLB.1994.334.348}
and has direct implications on the quantum dynamics of relativistic and
nonrelativistic quantum systems.
The deformation parameter $\kappa$ appearing in the theory is usually
interpreted as being the Planck mass $m_{P}$ \cite{PLB.2012.711.122}.
Some important contributions on $\kappa$-deformation have been studied
in Refs.
\cite{AoP.1995.243.90,CQG.2010.27.025012,
NPB.2001.102-103.161,EPJC.2003.31.129,PLB.2002.529.256,
PRD.2011.84.085020,JHEP.2011.1112.080,EPJC.2013.73.2472,
PRD.2009.79.045012,EPJC.2006.47.531,EPJC.2008.53.295,
PRD.2013.87.125009,PRD.2012.85.045029,PRD.2009.80.025014}.

The physical properties of $\kappa$-deformed relativistic quantum
systems can be accessed by solving the $\kappa$-deformed Dirac equation
\cite{PLB.1993.302.419,PLB.1993.318.613}.
Recently, some studies involving $\kappa$-deformation have also been
reported with great interest.
Some theoretical contributions in this
context can be found, for example, in Refs.
\cite{PLB.2013.719.467,PLB.1994.339.87,PRD.2007.76.125005,
PLB.1995.359.339,MPLA.1995.10.1969}.
The 3D Dirac oscillator has been discussed in connection with the theory
of quantum deformations in Ref. \cite{PLB.2014.731.327}.
However, it is well known that the 2D Dirac oscillator exhibits a
dynamics completely different from that of 3D one.
In this context, the main goal of this Letter is study the
dynamics of the 2D Dirac oscillator in the quantum deformed framework
generated by the $\kappa$-Poincar\'{e}-Hopf algebra and after compare
with the usual (undeformed) 2D Dirac oscillator.

This Letter is organized as follow.
In Section \ref{sec:2ddiraco}, we revise the 2D Dirac oscillator and
determine the energy levels and wave functions.
In Section \ref{sec:kappadirac}, the 2D Dirac oscillator in the
framework of the quantum deformation is discussed.
A brief conclusion is outlined in Section \ref{sec:conclusion}.

\section{The 2D Dirac oscillator}
\label{sec:2ddiraco}

In this section, we briefly discuss the usual 2D Dirac oscillator
for later comparison with the deformed one.
One begins by writing the Dirac equation for the four-component spinor
$\Psi$
\begin{equation}\label{eq:dirac}
  \left(
    \tilde{\beta}\tilde{\boldsymbol{\gamma}} \cdot
    \mathbf{p}+\tilde{\beta} m
  \right)\Psi=E\Psi.
\end{equation}
The 2D Dirac oscillator is obtained throught the nonmininal prescription
in Eq. \eqref{eq:prescription} where $\mathbf{r}=(x,y)$ is the position
vector.
We shall now make use of the underlying symmetry of the system to reduce
the four-component Dirac equation to a two-component spinor equation.
We use the following representation for the $\tilde{\gamma}$ matrices
\cite{PRD.1978.18.2932,NPB.1988.307.909,PRL.1989.62.1071}
\begin{align}
  \tilde{\beta}=\tilde{\gamma}_{0}=
  \left(
    \begin{array}{cc}
      \sigma_{z} & 0 \\
      0          & -\sigma_{z}
    \end{array}
  \right),
   & \qquad
  \tilde{\gamma}_{1}=
  \left(
    \begin{array}{cc}
      i\sigma_{y} & 0 \\
      0          & -i\sigma_{y}
    \end{array}
  \right),\\
  \tilde{\gamma}_{2}=
  \left(
    \begin{array}{cc}
     - i\sigma_{x} & 0 \\
      0          & i\sigma_{x}
    \end{array}
  \right),
  & \qquad
  \tilde{\gamma}_{3}=
  \left(
    \begin{array}{cc}
      0          & 1 \\
      -1          & 0
    \end{array}
  \right),
\end{align}
where $\sigma_{i}$ are the Pauli matrices.

The 2D Dirac oscillator being independent of the $z$ direction, allow us
decouple the usual four-component Dirac equation \eqref{eq:dirac} into
two two-component equations
\begin{equation}\label{eq:dirac_oscillator}
  H_{0}\psi=\left[
    \beta \gamma \cdot
    (\mathbf{p}-i m\omega \beta\mathbf{r})+\beta m
  \right]\psi=E\psi,
\end{equation}
where $\psi$ is a two-component spinor and the three-dimensional
$\gamma$ matrices are \cite{PRL.1990.64.2347,IJMPA.1991.6.3119}
\begin{equation}
  \beta = \gamma_{0} = \sigma_{z}, \qquad
  \gamma_{1}=i\sigma_{y}, \qquad
  \gamma_{2}=-is\sigma_{x},
\end{equation}
where the parameter $s$, which is twice the spin value, can be
introduced to characterize the two spin states \cite{EPJC.2014.74.2708},
with $s=+1$ for spin ``up'' and $s=-1$ for spin ``down''.
By squaring Eq. \eqref{eq:dirac_oscillator}, one obtains
\begin{equation}\label{eq:2ddiracoscillator}
  \left[
    p^2+m^2\omega^2r^2-2m\omega(\sigma^{3}+s L_{3})
  \right]\psi=
  (E^2-m^{2})\psi.
\end{equation}
If one adopts the following decomposition
\begin{equation}\label{eq:ansatz}
  \psi=
  \left(
    \begin{array}{c}
      \psi_{1} \\
      \psi_{2}
    \end{array}
  \right)=
  \left(
    \begin{array}{c}
      f(r)\;e^{i l \phi} \\
      i g(r)\;e^{i(l+s)\phi}
    \end{array}
  \right),
\end{equation}
for the spinor, from Eq. \eqref{eq:2ddiracoscillator}, it is possible to
obtain the energy spectrum \cite{arXiv:1403.4113}:
\begin{equation}\label{eq:energy2ddo}
E=\pm\sqrt{m^{2}+2m\omega\left(2n+|l|-sl\right)},
\end{equation}
and unnormalized wave functions
\begin{equation}\label{eq:eigenfunction_2d_dirac}
  f(\rho)
  =\rho^{(|l|+1)/2} e^{-\rho/2}\;
  M\left(-n,1+|l|,\rho\right).
\end{equation}
Here, $n=0,1,2,\ldots$, $l=0,\pm 1, \pm 2,\ldots$, $\rho=m\omega r^2$
and $M(z)$ is the confluent hypergeometric function of the first kind
\cite{Book.1972.Abramowitz}.
It should be noted that the spectrum is spin dependent, and for $sl>0$
the energy eigenvalues are independent of $l$ as depicted in the
Fig. \ref{fig:fig1}.

\begin{figure}
  \centering
  \includegraphics*[width=\columnwidth]{fig1}
  \caption{ \label{fig:fig1}
    (Color online) The positive energy spectrum,
    Eq. \eqref{eq:energy2ddo}, for the 2D Dirac oscillator for different
    values of $n$ and $l$ with $m=\omega=1$ and $s=1$.
    Notice that levels with quantum numbers $n \pm q$ have the same
    energy as levels with $l \pm q$, with $q$ an integer.}
\end{figure}

\section{The 2D $\kappa$-Dirac Oscillator}
\label{sec:kappadirac}

In this section, we address the 2D Dirac oscillator in the framework of
$\kappa$-Poincar\'{e}-Hopf algebra.
We begin with the $\kappa$-Dirac equation defined in
\cite{PLB.1993.302.419,PLB.1993.318.613,PLB.1995.359.339} when the third
spatial coordinate is absent.
So, using the same reasoning of the previous section we have
\begin{equation}
  \left\{
    \gamma_{0} P_{0}-\gamma_{i}P_{i}+
    \frac{\varepsilon}{2}
    \left[
      \gamma_{0}(P_{0}^{2}-P_{i}P_{i})-m P_{0}
    \right]
  \right\}\psi=m\psi,
\end{equation}
with $i=1,2$.
Identifying $P_{0}=H=E$ and $P_{i}=\pi_{i}=p_{i}-im\omega\beta r_{i}$,
we have
\begin{equation}\label{eq:def_dirac}
  H\psi=
  \left[
    H_{0}-\frac{\varepsilon}{2}(H^{2}-\pi_{i}\pi_{i}-m\beta H)
  \right]
  \psi=E\psi,
\end{equation}
By iterating Eq. \eqref{eq:def_dirac}, we have up to
$\mathcal{O}(\varepsilon)$ ($\varepsilon^{2}\approx 0$)
\begin{equation}\label{eq:def_dirac_oe}
  H\psi=
  \left[
    H_{0}-\frac{\varepsilon}{2}(H_{0}^{2}-\pi_{i}\pi_{i}-m\beta H_{0})
  \right]
  \psi=E\psi,
\end{equation}
with $H_{0}$ given in Eq. \eqref{eq:dirac_oscillator}.
By using the same representation of the previous section for the
$\gamma$ matrices, Eq. \eqref{eq:def_dirac_oe} assumes the form
\begin{multline}\label{eq:ddiracocart}
  \left(1+\frac{m\varepsilon}{2}\sigma_{z}\right)
  (\sigma_{x}\pi_{x}+s\sigma_{y}\pi_{y}+m\sigma_{z})\psi=\\
  \left\{
    E+\varepsilon
    \left[
      m^2\omega^2r^2-m\omega sL_{z}+
      im\omega\sigma_{z}(\mathbf{r}\cdot\mathbf{p})
    \right]
  \right\}\psi.
\end{multline}
Equation \eqref{eq:ddiracocart}, in polar coordinates $(r,\phi)$, reads
\begin{multline}\label{eq:ddiracopolar}
  e^{is\sigma_{z}\phi}
  \left(1+\frac{m\varepsilon}{2}\sigma_{z}\right)
  \left[
    \sigma_{x}\partial_{r}+
    \sigma_{y}\left(\frac{s}{r}\partial_{\phi}-im\omega r \right)
  \right]\psi=\\
  i\left[
    E-m\sigma_{z}+\varepsilon
    \left(
      m^{2}\omega^{2}r^{2}+im\omega s\partial_{\phi}+
      m\omega\sigma_{z}r\partial_{r}
    \right)
  \right]\psi.
\end{multline}

Our task now is solve Eq. \eqref{eq:ddiracopolar}.
As the deformation does not break the angular symmetry, then we can use a
similar ansatz as for the usual (undeformed) case
\begin{equation}\label{eq:ansatzeps}
  \psi=
   \left(
    \begin{array}{c}
      f_{\varepsilon}(r)\;e^{i l \phi} \\
      i g_{\varepsilon}(r)\;e^{i(l+s)\phi}
    \end{array}
  \right),
\end{equation}
but now with the radial part labeled by the deformation parameter.
So, by using the ansatz \eqref{eq:ansatzeps} into
Eq. \eqref{eq:ddiracopolar}, we find a set of two coupled radial
differential equations of first order
\begin{subequations}
\begin{multline}\label{eq:ddiracopolarcomp1}
\left(1+\frac{m\varepsilon}{2}\right)
  \left[
    \frac{d}{dr}+
    \frac{s(l+s)}{r}-m\omega r
  \right]g_{\varepsilon}(r)= (E-m)f_{\varepsilon}(r)\\
  +    \varepsilon
    \left(
      m^{2}\omega^{2}r^{2}-m\omega s l +
      m\omega r \frac{d}{dr}
    \right)f_{\varepsilon}(r),
\end{multline}
\begin{multline}\label{eq:ddiracopolarcomp2}
\left(1-\frac{m\varepsilon}{2}\right)
  \left[
    \frac{d}{dr}-
    \frac{sl}{r}+m\omega r
  \right]f_{\varepsilon}(r)= -(E+m)g_{\varepsilon}(r)\\
  -\varepsilon
    \left(
      m^{2}\omega^{2}r^{2}-m\omega s(l+s)-
      m\omega\rho \frac{d}{dr}
    \right)g_{\varepsilon}(r).
\end{multline}
\end{subequations}
The above system of equations can be decoupled yielding a single second
order differential equation for $f_{\varepsilon}(r)$,
\begin{multline} \label{eq:edof}
  -f_{\varepsilon}''(r)-
  \left(
    \frac{1}{r}+2m^{2}\varepsilon \omega r
  \right)f_{\varepsilon}'(r)\\
  + \left[
    \frac{l^{2}}{r^{2}}+
    \left(1-2 E\varepsilon \right) m^{2}\omega^{2}r^{2}-
    k_{\varepsilon}^{2}
  \right] f_{\varepsilon}(r) =0,
\end{multline}
where
\begin{equation}\label{eq:mue}
  k_{\varepsilon}^{2}=E^2-m^2+
    2m\omega\left[(sl+1)(1-\varepsilon E)+m\varepsilon\right].
\end{equation}
A similar equation for $g_{\varepsilon}$ there exists.
The regular solution for Eq. \eqref{eq:edof} is
\begin{equation}\label{eq:def_eigen}
    f_{\varepsilon}(\rho)
  =(\lambda_{\varepsilon}\;\rho)^{(|l|-1)/2}
  e^{-(m\varepsilon+\lambda_{\varepsilon})\rho/2}
  M\left(
    d_{\varepsilon},1+|l|,\lambda_{\varepsilon}\;\rho
  \right),
\end{equation}
where
\begin{equation}
\lambda_{\varepsilon}=1-E\varepsilon,
\end{equation}
and
\begin{equation}
  d_{\varepsilon}=\frac{1+|l|}{2}+
  \frac{2m^2\omega \varepsilon-k_{\varepsilon}^2}
  {4 \gamma \lambda_{\varepsilon}}.
\end{equation}

The deformed spectrum are obtained by establishing as convergence
criterion the condition $d_{\varepsilon}=-n$.
In this manner, the deformed energy spectrum is given by
\begin{equation}
  \label{eq:energy_def}
  E^{2}-m^{2}=2m\omega(2n+|l|-sl)\lambda_{\varepsilon}.
\end{equation}
Thus, solving Eq. \eqref{eq:energy_def} for $E$, the deformed energy
levels are explicitly given by
\begin{equation}\label{eq:energy_def_ex}
  E=  \pm \sqrt{m^2+2m\omega
    \left(
      2n+|l|-sl
  \right)}-m\varepsilon\omega(2n+|l|-sl),
\end{equation}
and its unnormalized wave functions are of the form
\begin{equation}\label{eq:def_eigen_def}
    f_{\varepsilon}(\rho)
  =(\lambda_{\varepsilon}\;\rho)^{(|l|-1)/2}
  e^{-(m\varepsilon+\lambda_{\varepsilon})\rho/2}
  M\left(
    -n,1+|l|,\lambda_{\varepsilon}\;\rho
  \right).
\end{equation}
In deriving our results we have neglected terms of $O(\varepsilon^2)$.

\begin{figure}
  \centering
  \includegraphics*[width=\columnwidth]{fig2}
  \caption{ \label{fig:fig2}
    (Color online) The deformed energy levels,
    Eq. \eqref{eq:energy_def_ex}, for the 2D $\kappa$-Dirac oscillator
    for $n=10$, $s=1$ and for different values of $l$.
    We use units such as $m=\omega=1$.}
\end{figure}

We can observe that the particle and antiparticle energies in the 2D
$\kappa$-Dirac oscillator are different, as a consequence of charge
conjugation symmetry breaking caused by the deformation parameter in the
same manner as observed in the three-dimensional one
\cite{PLB.2014.731.327}.
Notice that, $\varepsilon=0$, exactly conducts to the results for the
energy levels and wave functions of the previous section for the
usual (undeformed) 2D Dirac oscillator, revealing the consistency of the description
here developed.
It is worthwhile to note that the infinity degeneracy present in the
usual two-dimensional Dirac oscillator is preserved by the deformation, but
affecting the separation of the energy levels.
The distance between the adjacent energy levels decreases as the
deformation parameter increases.
In Fig. \ref{fig:fig2} is depicted the undeformed and deformed energy
levels for some values of the deformation parameter for $n=10$.
In Ref. \cite{PLB.2013.719.467}, we have determined a upper bound for
the deformation parameter.
Taking into account this upper bound, the product $m\varepsilon$ should
be smaller than $0.00116$.
In this manner, using units such as $m=1$, we must consider values for
the deformation smaller than $0.00116$.

\section{Conclusions}
\label{sec:conclusion}

In this letter, we considered the dynamics of the 2D Dirac oscillator
in the context of $\kappa$-Poincar\'{e}-Hopf algebra.
Using the fact that the deformation does not break the angular symmetry,
we have derived the $\kappa$-deformed radial differential equation whose
solution has led to the deformed energy spectrum and wave functions.
We verify that the energy spectrum and wave functions are modified by
the presence of the deformation parameter $\varepsilon$.
Using values for the deformation parameter lower than the upper
bound $0.00116$, we have examined the dependence of the energy of the
oscillator with the deformation.
The deformation parameter modifies the energy spectrum and wave
functions the Dirac oscillator, preserving the infinity degeneracy, but
affecting the distance between the adjacent energy levels.
Finally, the case $\varepsilon=0$, exactly conducts for the results for
the usual 2D Dirac oscillator.

\section{Acknowledgments}
We would like to thank Rodolfo Casana for discussions on
dimensionality of the deformed Dirac equation.
This work was supported by the
Funda\c{c}\~{a}o Arauc\'{a}ria (Grant No. 205/2013 (PPP) and
No. 484/2014 (PQ)),
and the Conselho Nacional de Desenvolvimento
Cient\'{i}fico e Tecnol\'{o}gico (Grants No. 482015/2013-6 (Universal),
No. 306068/2013-3 (PQ)) and FAPEMA (Grant No. 00845/13).

\bibliographystyle{model1a-num-names}
\input{k-2d-dirac-oscillator-rev2.bbl}

\end{document}

config: arxiv


TITLE: Conjectured identity involving an elliptic integral and a Bessel function
QUESTION [6 upvotes]: Let $K$ be the complete elliptic integral of the first kind and $K_\nu$ a modified Bessel function of the second kind. I'm wondering whether the following identity is true and if so, how to prove it:
$$\int_{1}^\infty e^{-2 a u^2} u^{1/2}K\left(\frac{u-1}{2u}\right) du \overset{?}{=}\sqrt{\frac{\pi}{32a}} e^{-a}K_{1/4}(a)\tag{1}$$
for $a$ real and strictly positive. I've searched several integral tables but could not find anything that looks related.
The reason why I suspect this might be true is the following. Start from the integral
$$Z(g)=\int_{-\infty}^{\infty} \frac{dx}{\sqrt{2\pi}}\, \exp{\left(-\frac{1}{2} x^2 -\frac{g}{4!}x^4\right)} \tag{2}$$
This can be expressed in terms of a Bessel function, for example as in this question,
$$Z(g)=\sqrt{\frac{3}{2\pi g}} \exp{\left(\frac{3}{4g}\right)} K_{1/4}\left(\frac{3}{4 g}\right)$$
However, going back to $(2)$, expanding the second exponential and exchanging the order of integration and summation, one gets the asymptotic series
$$Z(g)\sim \sum_{n=0}^\infty \frac{(4n)!}{2^{2n}(2n)!n!}\left(-\frac{g}{24}\right)^n$$
Its Borel transform sums to a complete elliptic integral of the first kind, as in this question:
$$\mathcal{B}Z(g)= \sum_{n=0}^\infty \frac{(4n)!}{2^{2n}(2n)!(n!)^2}\left(-\frac{g}{24}\right)^n = \frac{2}{\pi} \frac{1}{\left(1+2g/3\right)^{1/4}}K\left(\frac{-1+\sqrt{1+2g/3}}{2\sqrt{1+2g/3}}\right)$$
The Borel transform converges for $|g| < 3/2$, however the right hand side makes sense for all $g > -3/2$. Now the statement $(1)$ is equivalent to the assertion that $Z$ is equal to its Borel sum, that is $Z(g) \overset{?}{=} \int_0^\infty e^{-t} \mathcal{B}Z(gt)$ after replacing $u^2 = 1+2gt/3$ and $a=3/(4g)$ in the integral for cleanliness.

REPLY [4 votes]: With the change of integration variables $u = \sqrt {1 + t}$, your original identity reads
$$
K_{1/4} (a) = \sqrt \pi  e^{ - a} \sqrt {2a} \int_0^{ + \infty } {e^{ - 2at} \frac{2}{{\pi (1 + t)^{1/4} }}K\!\left( {\frac{{\sqrt {1 + t}  - 1}}{{2\sqrt {1 + t} }}} \right)dt} .
$$
By entry 19.2 on page 195 of F. Oberhettinger, L. Badii, Tables of Laplace Transforms, we have
$$
K_\nu  (a) = \sqrt \pi  e^{ - a} (2a)^{\lambda  - 1/2} \int_0^{ + \infty } {e^{ - 2at} t^{\lambda  - 1} \frac{1}{{\Gamma (\lambda )}}{}_2F_1 \! \left( {\nu  + \tfrac{1}{2}, - \nu  + \tfrac{1}{2};\lambda ; - t} \right)dt}
$$
for any $\nu \in \mathbb C$, $\Re a>0$ and $\Re \lambda >0$. Here ${}_2 F_1$ stands for the Gauss hypergeometric function. Thus it is enough to show that
$$
\frac{2}{{\pi (1 + t)^{1/4} }}K\!\left( {\frac{{\sqrt {1 + t}  - 1}}{{2\sqrt {1 + t} }}} \right) = {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - t} \right),\;\;t>0.
$$
This is shown in https://math.stackexchange.com/q/4267825.
Proof using the asymptotics of the Bessel function. We start with the expansion
$$
F(a): = \sqrt {\frac{2}{\pi a}} e^a K_{1/4} (a) = \sum\limits_{n = 0}^N {( - 1)^n \frac{{(4n)!}}{{128^n n!(2n)!}}\frac{1}{{a^{n + 1} }}}  + R_{N} (a),
$$
where
$$\tag{1}
\left| {R_{N} (a)} \right| \le \frac{{(4N + 4)!}}{{128^{N + 1} (N + 1)!(2N + 2)!}}\frac{1}{{\left| a \right|^{N + 2} }}
$$
provided that $|\arg a|\leq\frac{\pi}{2}$ and $N\geq 0$, see for instance https://doi.org/10.1098/rspa.1990.0058. Now with an arbitrary $c>0$
$$
f(t): = \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} F(a)da}  = \sum\limits_{n = 1}^N {( - 1)^n \frac{{(4n)!}}{{128^n n!^2 (2n)!}}t^n }  + \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} R_{N} (a)da}
$$
for any $t>0$. Differentiating $N$ times gives
$$\tag{2}
f^{(N)} (t) = ( - 1)^N \frac{{(4N)!}}{{128^N N!(2N)!}} + \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} a^N R_{N} (a)da}
$$
where, by $(1)$,
$$\tag{3}
\left| {\frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} a^N R_{N} (a)da} } \right| \le \frac{{(4N + 4)!}}{{128^{N + 1} (N + 1)!(2N + 2)!}}\frac{{e^{ct} }}{{2c}}.
$$
Stirling's formula then implies that
$$
\frac{{\left| {f^{(N)} (t)} \right|}}{{N!}} \le C\frac{N}{{2^N }}\frac{{e^{ct} }}{c}
$$
with an absolute $C>0$. Thus, by Taylor's formula and the Cauchy–Hadamard theorem, $f(t)$ is analytic in the tubular neighbourhood $U:=\left\{ {t:\text{dist}(t,\mathbb R^ +  ) < 2} \right\}$ of the positive axis. Employing $(2)$ and $(3)$ with $t=0$ and letting $c\to +\infty$ shows that
$$
\frac{{f^{(N)} (0)}}{{N!}} = ( - 1)^N \frac{{(4N)!}}{{128^N N!^2 (2N)!}},
$$
i.e., the Maclaurin series of $f(t)$ and ${}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \frac{t}{2}} \right)$ are the same. By analytic continuation, $f(t)={}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \frac{t}{2}} \right)$ for any $t\in U$. Accordingly,
$$
F(a) = \int_0^{ + \infty } {e^{ - at} {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \tfrac{t}{2}} \right)dt}  = 2\int_0^{ + \infty } {e^{ - 2at} {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - t} \right)dt}
$$
provided $\Re a >0$.


TITLE: permutations with specified repetition counts
QUESTION [0 upvotes]: Problem: Determine the number of permutations of the characters for:
AABBBC
How can I calculate a problem like this generally, given a set of characters and a number of times each has to appear?

REPLY [0 votes]: Presumably you know six distinct letters can be rearranged in $6!$ distinct ways.   A straightforward count of permutations.
Now suppose two of those letters were not distinct.   $\rm BEA_1A_2CD$ is indistinguishable from $\rm BEA_2A_1CD$ when the numbers rub off, and so forth.   Every arrangement is one of a set of arrangement where only the identical letters are in different positions.   There are $2!$ ways to arrange two distinct letters, so what we need to do is divide by that.
There are $6!/2!$ distinct ways to rearrange $\rm AABCDE$.
Can you now count the distinct ways to rarrange $\rm AABBBC$?


text:
\DOC RING

\TYPE {RING : (term -> num) * (num -> term) * conv * term * term * term * term * term * term * term * thm * thm * (term -> thm) -> term -> thm}

\SYNOPSIS
Generic ring procedure.

\DESCRIBE
The {RING} function takes a number of arguments specifying a ring structure and
giving operations for computing and proving over it. Specifically the call is:
{
  RING(toterm,tonum,EQ_CONV,
       neg,add,sub,inv,mul,div,pow,
       INTEGRAL_TH,FIELD_TH,POLY_CONV)
}
\noindent where {toterm} is a conversion from constant terms in the structure
to rational numbers (e.g. {rat_of_term} for the reals), {tonum} is the opposite
(e.g. {term_of_rat} for the reals), {EQ_CONV} is an equality test conversion
(e.g. {REAL_RAT_EQ_CONV}), {neg} is negation, {add} is addition, {sub} is
subtraction, {inv} is multiplicative inverse, {div} is division, {pow} is
power, {INTEGRAL_TH} is an integrality theorem and {FIELD_TH} is a field
theorem (see below) and {POLY_CONV} is a polynomial normalization theorem for
the structure as returned by {SEMIRING_NORMALIZERS_CONV} (e.g. {REAL_POLY_CONV}
for the reals).

The integrality theorem essentially states that if a product is zero, so is one
of the factors (i.e. the structure is an integral domain), but this is stated
in an unnatural way to allow application to structures without negation. It is
permissible in this case to use boolean variables instead of operators such as
negation and subtraction. The precise form of the theorem (notation for natural
numbers, but this is supposed to be over the same structure):
{
  |- (!x. 0 * x = 0) /\
     (!x y z. x + y = x + z <=> y = z) /\
     (!w x y z. w * y + x * z = w * z + x * y <=> w = x \/ y = z)
}

The field theorem is of the following form. It is not logically necessary, and
if the structure is not a field you can just pass in {TRUTH} instead. However,
it is usually beneficial for performance to include it.
{
  |- !x y. ~(x = y) <=> ?z. (x - y) * z = 1
}

It returns a proof procedure that will attempt to prove a formula that, after
suitable normalization, can be considered a universally quantified Boolean
combination of equations and inequations between terms of the right type. If
that formula holds in all integral domains, it will prove it. Any ``alien''
atomic formulas that are not natural number equations will not contribute to
the proof.

\FAILURE
Fails if the theorems are malformed.

\EXAMPLE
The instantiation for the real numbers (in fact this is already available under
the name {REAL_RING}) could be coded as:
{
  let REAL_RING =
    let REAL_INTEGRAL = prove
     (`(!x. &0 * x = &0) /\
       (!x y z. (x + y = x + z) <=> (y = z)) /\
       (!w x y z. (w * y + x * z = w * z + x * y) <=> (w = x) \/ (y = z))`,
      REWRITE_TAC[MULT_CLAUSES; EQ_ADD_LCANCEL] THEN
      REWRITE_TAC[GSYM REAL_OF_NUM_EQ;
                  GSYM REAL_OF_NUM_ADD; GSYM REAL_OF_NUM_MUL] THEN
      ONCE_REWRITE_TAC[GSYM REAL_SUB_0] THEN
      REWRITE_TAC[GSYM REAL_ENTIRE] THEN REAL_ARITH_TAC)
    and REAL_INVERSE = prove
     (`!x y:real. ~(x = y) <=> ?z. (x - y) * z = &1`,
      REPEAT GEN_TAC THEN
      GEN_REWRITE_TAC (LAND_CONV o RAND_CONV) [GSYM REAL_SUB_0] THEN
      MESON_TAC[REAL_MUL_RINV; REAL_MUL_LZERO; REAL_ARITH `~(&1 = &0)`]) in
    RING(rat_of_term,term_of_rat,REAL_RAT_EQ_CONV,
         `(--):real->real`,`(+):real->real->real`,`(-):real->real->real`,
         `(inv):real->real`,`(*):real->real->real`,`(/):real->real->real`,
         `(pow):real->num->real`,
         REAL_INTEGRAL,REAL_INVERSE,REAL_POLY_CONV);;
}
\noindent after which, for example, we can verify a reduction for cubic
equations to quadratics entirely automatically:
{
  # REAL_RING
     `p = (&3 * a1 - a2 pow 2) / &3 /\
      q = (&9 * a1 * a2 - &27 * a0 - &2 * a2 pow 3) / &27 /\
      z = x - a2 / &3 /\
      x * w = w pow 2 - p / &3 /\
      ~(p = &0)
      ==> (z pow 3 + a2 * z pow 2 + a1 * z + a0 = &0 <=>
          (w pow 3) pow 2 - q * (w pow 3) - p pow 3 / &27 = &0)`;;
}

\SEEALSO
ideal_cofactors, NUM_RING, REAL_FIELD, REAL_RING, real_ideal_cofactors,
RING_AND_IDEAL_CONV.

\ENDDOC

subset_name: curated


TITLE: Does the $\sum\limits_{n = 1}^\infty {\frac{{\sin (n)}}{{\sqrt {{n^3} + {{\cos }^3}(n)} }}} $ series converges?
QUESTION [1 upvotes]: $$\sum\limits_{n = 1}^\infty  {\frac{{\sin (n)}}{{\sqrt {{n^3} + {{\cos }^3}(n)} }}} $$
I tried to check with Maplesoft and Microsft Excel and seems this series is divergent.
Is my conjecture true? How can I prove it?
Thanks in advance.

REPLY [2 votes]: For $n \geq 2$ we have
$$
\frac{|\sin n|}{\sqrt{n^{3}+\cos^{3}(n)}} \leq \frac{1}{\sqrt{n^{3} + \cos^{3}(n)}} \leq \frac{1}{\sqrt{n^{3}-1}}.
$$
We have
$$
\frac{1}{\sqrt{n^{3}-1}} \sim \frac{1}{n^{3/2}}
$$
as $n \to \infty$,
so the series $\sum_{n \geq 2}\frac{1}{\sqrt{n^{3}-1}}$ converges by limit comparison test;
hence by comparison test we conclude that the series
$$
\sum_{n \geq 1}\frac{\sin n}{\sqrt{n^{3}+\cos^{3}(n)}}
$$
converges absolutely,
and the convergence follows.


TITLE: Line diving 2 triangles in a plane in equal halves by area
QUESTION [2 upvotes]: Can you prove or disprove that for any 2 triangles in a plane, there will always be a line passing through them both which divide each of them into 2 halves of equal area? If said line exists, how can we find it?

REPLY [1 votes]: Of course one can make appeal to the ham-sandwich-theorem. But for two triangles it should be possible to obtain, at least in principle, an explicit solution.
We first have to describe the set of lines dividing a single triangle into two parts of equal area. To this end consider the triangle $T$ with vertices $(0,0)$, $(1,0)$ and $(0,1)$. A line cutting off the origin together with half the area of this triangle has an equation of the form
$${x\over u}+{y\over v}=1\ ,$$
whereby $u,v\in\>]0,1[\>$ and $uv={1\over2}$. The family of these lines can therefore be described in the form
$$2v^2x+y-v=0,\qquad0<v<1\ .$$
The envelope of this family turns out to be an arc $\gamma$ of the hyperbola $xy={1\over8}$, whereby the endpoints of this arc are the midpoints $\bigl({1\over2},{1\over4}\bigr)$ and $\bigl({1\over4},{1\over2}\bigr)$ of the medians  through the vertices $(1,0)$ and $(0,1)$. Of course these medians themselves are tangent to the arc $\gamma$ in these points, since they are halving the area of $T$. By affinity it is not necessary to do a similar computation for cutting off another vertex of $T$: In all we obtain three arcs of hyperbolas that form a "hyperbolic triangle" with cusps in the midpoints of the medians of $T$. It is possible to rotate a line $\ell$ smoothly $180^\circ$ along these three arcs so that $\ell$ touches (at least) one of the arcs at all times.

If we now are given two arbitrary triangles $T_1$, $T_2$ then  each of them has its own "hyperbolic triangle" $H_i$. The rotating $\ell$ of $H_1$ will then at a certain moment be tangent to one of the arcs of  $H_2$ (one would have to check this). In this position the line $\ell$ halves both triangle areas simultaneously.


TITLE: $\varphi (t)$ is a maximal solution of $x'=f(x) \implies \varphi (t+c)$ is a maximal solution
QUESTION [0 upvotes]: Consider a function $f:]a,b[\to\mathbb R$ satisfying the Lipschitz condition.

Let $\varphi : ]\alpha , \beta[ \to ]a,b[$ be a maximal solution, $c\in\mathbb R$. Show the function $$\varphi ^ c: ]\alpha -c,\beta -c[ \to ]a,b[$$
$$ \varphi ^c (t):= \varphi (t+c)$$
is also a maximal solution.
Let $\varphi,\psi$ be maximal solutions of $x'=f(x)$ with a common image. Prove one is a translation of the other.

For 1. I firstly proved "$\varphi$ is a solution $\implies$ $\varphi ^c$ is a solution". To prove it's also a maximal solution I tried the contrapositive:
Suppose $\varphi ^ c $ isn't a maximal solution. Then there exists an extension $\varphi _I^c$ of $\varphi ^c$ defined on an interval $I$ where $]\alpha - c,\beta -c[\subset I$. Now I need to build a function $\varphi_I$ defined on the interval $I+c$ such that $\varphi_I (t)$=$\varphi (t)$ for all $t\in ]\alpha,\beta[$ to conclude but it's not clear to me how should I do this.
For the second question, I don't understand the intuition behind it. If both solutions have a common image then there exists $t,s$ such that $\varphi(t)=\psi (s)$. Rewriting we have $\varphi(t)=\psi (t+c)$ where $c = s-t$. I don't know where to go from here. Any help is appreciated.

REPLY [0 votes]: For 1. I think it is enough to observe that, if $\varphi\colon I \to \mathbb{R}$ is any solution of $x'=f(x)$, then $\varphi^c\colon I^c := I-c \to\mathbb{R}$ is also a solution.
For 2., let $x_0 := \varphi(t_0) = \psi(s_0)$ (in your notation).
Since $f$ is Lipschitz continuous, $\varphi$ is the unique maximal solution of the Cauchy problem $x'=f(x)$, $x(t_0) = x_0$.
On the other hand, also $\psi^c$, for $c:= t_0 - s_0$, is a solution of the same Cauchy problem, hence $\psi^c = \varphi$.


TITLE: Simplify $Im \left(\frac{az+b}{cz+d}\right)$
QUESTION [2 upvotes]: Let $z \in \mathbb{H}$, where $\mathbb{H}$ denotes the half plane $\mathbb{H}=\{z \in \mathbb{C}:Im(z)>0\}$. Let
\begin{equation*}
f(z)=\frac{az+b}{cz+d}
\end{equation*}
which is called a Mobius Transformation, and let $ad-bc>0$. I want to show that $Im(f(z))>0$.

Following this solution, I should use $Im(z)=\frac{z-\overline{z}}{2i}$. Applying this formula, I get
\begin{align*}
Im (f(z))
&=Im \left(\frac{az+b}{cz+d}\right) \\
&=\frac{\frac{az+b}{cz+d}-\overline{\left(\frac{az+b}{cz+d}\right)}}{2i}
\end{align*}
but I am not sure how to show that this equals $\frac{ad-bc}{c^2+d^2}$. Is there a nice and clean way to simplify $Im(f(z))$? Thank you!

REPLY [4 votes]: There is an error in the answer you linked. We cannot expect $Im (f(z))$ not to depend on $z$; otherwise, $f$ would map $\mathbb{H}$ to an horizontal line, which contradicts its analyticity.
I think the best way is to multiply by the conjugate of the denominator:
$$Im \left( \frac{az+b}{cz+d}\right) = Im \left( \frac{(az+b)(c \overline{z}+d)}{(cz+d)(c \overline{z}+d)}\right) = \frac{1}{|cz+d|^2} Im \left( acz\overline{z}+bc\overline{z}+adz+bd\right) = \frac{(ad-bc) Im(z)}{|cz+d|^2}.$$

REPLY [1 votes]: $$\frac{az+b}{cz+d}=\frac{(az+b)(c\bar{z}+d)}{|cz+d|^2}$$
And one has
$$(az+b)(c\bar{z}+d)=acz\bar{z}+adz+bc\bar{z}+bd$$
$acz\bar{z}+bd$ is real and $\mathcal{I}(adz+bc\bar{z})=(ad-bc)\mathcal{I}(z)$
Putting the above together, one gets
$$\mathcal{I}\left(\frac{az+b}{cz+d}\right)=\frac{ad-bc}{|cz+d|^2}\mathcal{I}(z)$$


TITLE: what is the expected magnitude projection of random unit vector?
QUESTION [0 upvotes]: For a random unit vector x, write it as $\sum_i a_ie_i$ where $\sum_i a_i^2=1$ and $e_i$ is the basis unit vector. What would be the $E[|a_i|]$?
How to calculate the magnitude $a_i$ relation between different dims?

REPLY [2 votes]: We can do this integration explicitly using hyperspherical coordinates. Assume $n$ dimensions, and let the angles $\theta_i$, $1 \le i \le n-1$ be the spherical angles, and, without loss of generality, choose $\theta_1$ to be the angle between the random vector and the projected vector. Additionally, let $S_n$ be the $n-1$ dimensional volume of the unit $n$-sphere. Then the resulting integral is
$$
E[|a_i|] = \frac{1}{S_n}\int_0^\pi...\int_0^\pi \int_0^{2\pi} |\cos\theta_1|\left(\sin^{n-2}\theta_1 ...\sin\theta_{n-2}\right)d\theta_1...d\theta_{n-2}d\theta_{n-1} \\= \frac{2S_{n-1}}{S_n}\int_0^{\pi/2} \cos\theta_1\sin^{n-2}\theta_1d\theta_1 = \frac{2S_{n-1}}{(n-1)S_n}
$$
We also have $S_n = 2\pi^{n/2}/\Gamma(n/2)$, so the expected value is
$$
E[|a_i|] = \frac{2 \Gamma(n/2)}{(n-1)\sqrt{\pi}\Gamma(n/2-1/2)} = \frac{2^{n-1}\Gamma(n/2)^2}{\pi \Gamma(n)}
$$
For large $n$, this scales as $\sqrt{2/\pi} n^{-1/2}$, which matches the intuitive expectation that $E[|a_i|]\approx n^{-1/2}$.