TITLE: Is there a "square root" of Klein–Gordon equation which is different from Dirac equation?
QUESTION [7 upvotes]: Legend has it that Dirac arrives at the Dirac equation (all equations are in Planck units $c= \hbar = 1$ in this post):
$$
i\gamma^\mu\partial_\mu\psi - m\psi =0
$$
by taking the square root of the operator
$$
\partial^\mu\partial_\mu
$$
of the Klein–Gordon equation which is the relativistic version of the Schrodinger equation:
$$
(\partial^\mu\partial_\mu + m^2)\psi =0
$$
However, is Dirac equation the only "square root" of Klein–Gordon equation? Is there any other "square root" of Klein–Gordon equation?
It turns out there is a "square root" of Klein–Gordon equation which is different from Dirac equation:
$$
i\gamma^\mu\partial_\mu\psi - me^{\theta i\gamma_5} \psi = 0
$$
where the "complex" fermion mass term is endowed with both CP even scalar part and CP odd pseudoscalar $i\gamma_5$ part:
$$
m  e^{\theta i\gamma_5}   = m\cos\theta   + m i\gamma_5 \sin\theta  .
$$
Can we verify that the above "modified" Dirac equation is indeed the "square root" of Klein-Gordon equation?
Let's get cracking on the nitty-gritty:
$$
m^2\psi \\
= (me^{-\theta i\gamma_5}) (me^{\theta i\gamma_5})\psi \\
= (me^{-\theta i\gamma_5}) (i\gamma^\mu\partial_\mu)\psi \\
= (i\gamma^\mu\partial_\mu) (me^{\theta i\gamma_5}) \psi \\
=(i\gamma^\mu\partial_\mu)(i\gamma^\nu\partial_\nu) \psi \\
= -\partial^\mu\partial_\mu\psi.
$$
Voila! We indeed recover the classic Klein-Gordon equation, without any funny "complex" factor. Note that in the 4th line we leveraged the crucial anti-commuting property between $\gamma_5$ and $\gamma^\mu$.
Can anyone please verify the above derivation is correct or wrong? Or is it actually the same as the original Dirac equation?

Added note
The fun fact is that after an axial rotation of the fermion field
$$
\psi \rightarrow e^{-\frac{1}{2}\theta i\gamma_5} \psi.
$$
the "complex" mass term in the Lagrangian format can be transformed into a scalar mass term (while leaving the kinetic term intact):
$$
m\bar{\psi} e^{\theta i\gamma_5} \psi \rightarrow m\bar{\psi} \psi.
$$
In other words, via redefining the fermion field, the CP odd part of the mass $m i\gamma_5 \sin\theta$ can be effectively rotated away. That being said, we know that the fermion mass is generated via the Higgs mechanism. Rotating to the "real" mass is achievable if only there is one Higgs boson. If you fancy a fancy-schmancy beyond standard model involving multiple Higgs bosons, the said rotation can only make one Higgs boson CP even, while leaving the other Higgs boson with a CP breaking pseudoscalar $i\gamma_5$ phase.
Of course these multi-Higgs boson scenarios are not usually considered in entry-level QFT books. Thus the "modified" Dirac equation is usually not mentioned at all. But do you think the usual QFT text books should mention it just for fun?
If you treat the right-handed and left-handed spinors separately (such as Weyl spinors), you would have more freedom in axial rotations without explicitly invoking the pseudoscalar. This is exactly what is going on in the CKM-related rotations. But there is still not enough freedom in case of three generations to cancel out all the pseudoscalars in the family-mixing parameters, therefore you are left with a residual CP violating phase in the electroweak sector.
In a nutshell, the most general "square root" counterpart of the Klein–Gordon equation is the Dirac equation with a "complex" mass term $m e^{\theta i\gamma_5} \psi$. The "real" mass Dirac equation is merely a special case of $\theta = 0$.

More added note:
There is a separate issue of the scalar mass term
$$
m\bar\psi \psi
$$
being imaginary if you plug in the Grassmann number components from $\psi$ and $\bar\psi$. See details here.

REPLY [1 votes]: Uniqueness is relatively broader reasearch topic: we need to check what is applicable in which context: take a look at this: The pseudodifferential operator square root of the Klein–Gordon equation  - Claus Lammerzahla, 1993, ZARM. Also see: https://www.zarm.uni-bremen.de


TITLE: About 3-fold log canonical singularity
QUESTION [2 upvotes]: As far as I know, log canonical surface singularities were classified. How about higher dimensional case?
I especially want to know whether given 3-fold singularity is log canonical or not.
Let $f$ be a holomorphic function near $0 \in \mathbb C^4$ and $D=\{ f=0\}$.  Then how can I check that whether it is log canonical or not?
For example,
Let $f=x_2x_4+(x_3+x_4)^2+(x_1x_4+x_2x_3)(x_1+x_2+x_3+x_4)+(x_1x_4+x_2x_3)^2$.
Is it log canonical? Can you give any answers or references?

REPLY [6 votes]: Yes, it is log canonical.
Some overarching comments
A quick way is simply to check whether it is F-pure after reduction to characteristic p for some p.  (Actually, before [this paper][1], one would have to check infinitely many primes, but now we know that fpt($f$ mod $p$) $\leq$ lct($f$) for all p).
Fortunately, one can check this very easily in a computer (I bet blowups would also be do-able here, although certainly a lot more work...).
In characteristic $p = 7, 13$, I checked that $\langle f^{p-1} \rangle^{[1/p]} = \langle 1 \rangle = \mathbb{Z}/p[x_1,x_1,x_3,x_4]$.  This proves that the pair $(\mathbb{A}^4, f^1)$ is $F$-pure and hence log canonical back in characteristic zero.  For some additional background, you could see these survey articles:  Smith-Zhang, Schwede-Tucker, Blickle-Schwede, Patakfalvi-Schwede-Tucker.  The original result that the FPT is $\leq$ the LCT is for $p \gg 0$ is essentially due to Hara-Watanabe (except they didn't define the FPT, that was Takagi-Watanabe and Mustata-Takagi-Watanabe, see the references contained in the above articles).
The actual computation
Ok, I did this in Macaulay2, and I used [PosChar.m2][6] (a Macaulay2 package, some documentation is [HERE][7]).  I then entered the following commands:
loadPackage "PosChar"
p = 7
R=ZZ/p[x1,x2,x3,x4]
f = x2*x4+(x3+x4)^2+(x1*x4+x2*x3)*(x1+x2+x3+x4)+(x1*x4+x2*x3)^2
ethRoot(f^(p-1), 1)

Macaulay2 output
ideal 1
Ideal of R

which proves that it is $F$-pure and hence log canonical back in characteristic zero.  Anyway, the key computation which I am sure you could also do by hand in this case is called "Fedder's Criterion" (see the above articles).
A caveat
The package PosChar.m2 also has a function called "IsSharplyFPurePoly" which could also work here.  However, the implementation is slower than what I did above (I need to go fix that, I will try to update this answer when I do that).
Update:
PosChar.m2 has been depricated.  The right package is the TestIdeals.m2 package, which should be built into Macaulay2.


TITLE: In how many ways can we form a garland using distinct red, yellow, and blue flowers if flowers of the same colour must be together?
QUESTION [0 upvotes]: In how many ways can we form a garland using $3$ different red, $5$ different yellow and $4$ different blue flowers, if flowers of the same colour must be together?
My approach:
I made $3$ groups of $3$ different flower colours and applied circular permutations clockwise, then I found internal permutations and multiplied, but this approach isn’t giving me the correct answer.

REPLY [1 votes]: First, arrange the colours.  If we proceed clockwise from red, there are two ways to do so, $rby$ and $ryb$.  Now, arrange the groups of flowers of each colour internally.

 For invariance under rotation, we obtain $$2 \cdot 3!5!4!$$  For invariance under rotations and reflection, we divide this result by $2$ to equate clockwise and anti-clockwise rotations.


text:
\begin{document}

\thispagestyle{empty}

\vspace*{2cm}

\begin{center}
{\Large \bf Finite-dimensional  pseudo-bosons: a non-Hermitian version of the truncated harmonic oscillator}   \vspace{2cm}\\

{\large F. Bagarello}\\
 DEIM -Dipartimento di Energia, ingegneria dell' Informazione e modelli Matematici,
\\ Scuola Politecnica, Universit\`a di Palermo, I-90128  Palermo, Italy\\
and\\ INFN, Sezione di Napoli.\\
e-mail: fabio.bagarello@unipa.it

\end{center}

\vspace*{2cm}

\begin{abstract}
\noindent We propose a deformed version of the commutation rule introduced in 1967 by Buchdahl to describe a particular model of the truncated harmonic oscillator. The rule we consider is defined on a $N$-dimensional Hilbert space $\Hil_N$, and produces two biorhogonal bases of $\Hil_N$ which are eigenstates of the Hamiltonians $h=\frac{1}{2}(q^2+p^2)$, and of its adjoint $h^\dagger$. Here $q$ and $p$ are non-Hermitian operators obeying $[q,p]=i(\1-Nk)$, where $k$ is a suitable orthogonal projection operator. These eigenstates are connected by  ladder operators constructed out of $q$, $p$, $q^\dagger$ and $p^\dagger$. Some examples are discussed.

\end{abstract}

\vspace{2cm}

{\bf Keywords:--} pseudo-bosons; PT-quantum mechanics; truncated Harmonic oscillator

{\bf MSC classification:--} 46N50; 81Q12

\vfill

\newpage

\section{Introduction}\label{sectI}

Quantum mechanics is often thought to be naturally associated to self-adjoint (or Hermitian\footnote{We will use these two words as synonymous here.}) operators. In particular, the dynamics is deduced out of a self-adjoint Hamiltonian, and the {\em observables} of the system are almost always assumed to be self-adjoint as well.

In recent years, mainly since the seminal work by Bender and Boettcher, \cite{ben1}, it was understood that self-adjointness is not an essential requirement, since other operators exist, not self-adjoint, having purely real (and discrete) spectra. We refer to \cite{ben,mosta,bagbook} for some reviews on this alternative approach. What is interesting, from a mathematical point of view, is that orthonormal (o.n.) bases of eigenstates are replaced by biorthonormal sets which can be, or not, bases of the Hilbert space where the physical system lives. Also, different scalar products can play a role, and this different products produce different adjoints of the same operators. Moreover, the role of pseudospectra in connection with unbounded operators becomes relevant, \cite{petr}. Then, in a sense, loosing self-adjointness makes the mathematical structure reacher. Not only that: from a physical point of view the situation is also rather interesting since, for instance, some so-called PT-symmetric Hamiltonians can be naturally used to describe quantum systems with gain and loss phenomena, see \cite{gain1,gain2} and references therein.

In recent years, in connection with this kind of operators, we have developed a rather general formalism based on some suitable deformations of the canonical commutation and anti-commutation relations (CCR and CAR). These deformations produce what we have called $\D$-pseudo bosons and  pseudo-fermions. A rather complete review on both these topics can be found in \cite{baginbagbook}, to which we refer for several details and for some physical applications. Later on a similar framework was proposed for quons and for generalized Heisenberg algebra, \cite{bagquons,bagcurgaz}.

Here we consider a deformation of a different commutation rule, originally considered in \cite{buc}, and later analyzed in \cite{bagchi}, in connection with a truncated version of the harmonic oscillator. The operator $c$ considered in these papers obeys the following rule
\be
[c,c^\dagger]=\1-N\,K,
\label{11}\en
in which $N=2,3,4,\ldots$ is a fixed natural number, while $K$ is a self-adjoint projection operator, $K=K^2=K^\dagger$, satisfying the equality $Kc=0$. The presence of the term $N\,K$ in (\ref{11}) makes it possible to find a representation of $K$ and $c$ in terms of $N\times N$ matrices. In fact,  in absence of this term we would recover the CCR, which does not admit any finite-dimensional representation. Here, on the other hand, $K$, $c$ and $c^\dagger$ act on a $N$-th dimensional Hilbert space, which we call $\Hil_N$. In \cite{buc} it was shown that the matrices for $c$ and $c^\dagger$ are essentially the truncated versions of the analogous, infinite-dimensional, matrices for the bosonic annihilation and creation operators. In \cite{buc} it was also discussed how to construct an orthonormal (o.n.) basis of eigenvectors of the self-adjoint operator $H_0=\frac{1}{2}(Q_0^2+P_0^2)$, where $Q_0=\frac{c+c^\dagger}{\sqrt{2}}$ and $P_0=\frac{c-c^\dagger}{\sqrt{2}\,i}$ are the {\em truncated } position and momentum operators. These vectors turn out to be eigenvectors of both $H_0$ and $K$, and their explicit construction is strongly based on the fact that $H_0$ is a positive operator, other than being self-adjoint. This automatically imposes a lower bound on the possible eigenvalues of $H_0$, bound which was used in \cite{buc} to construct the set of eigenvectors. We will see that, in our extended case, positivity is apparently lost, so that we cannot adopt the same construction as in \cite{buc} for the  eigenvectors of our new Hamiltonian $h$, constructed in analogy with $H_0$. Moreover, since $h\neq h^\dagger$, it is natural to analyze also what happens for $h^\dagger$, and this will produce a biorthogonal set of eigenvectors of $h^\dagger$, see Section \ref{sectIII}, which is a basis for $\Hil_N$.

The article is organized as follows: in the next section we discuss our deformed version of the commutation rule (\ref{11}), and we construct a set of eigenvectors for the related truncated non self-adjoint harmonic oscillator, with Hamiltonian $h$, see above. We call the operators $a$ and $b$ appearing in this deformation {\em finite-dimensional pseudo-bosons} (FDPBs), since they can be seen as a truncated version of the $\D$-PBs considered in \cite{baginbagbook}. We show explicitly how our construction works for some fixed values of $N$, and then we generalize the procedure to generic $N$. Incidentally we will find that the procedure proposed here is much direct than that considered in \cite{buc}.  In Section \ref{sectIII} the biorthogonal set of eigenvectors of $h^\dagger$ is deduced. We also show how these FDPBs are related to the operators $c$ and $K$ in (\ref{11}). In Section \ref{sectIV} we discuss two examples, while our conclusions are given in Section \ref{sectV}.

\section{Deformed commutation rules}\label{sectII}

The main object of our research is the following deformed version of the commutation rule (\ref{11}):
\be
[a,b]=\1-Nk.
\label{21}\en
Here $N$ can be any fixed integer larger than 1, and $k$ is an orthogonal projector: $k=k^2=k^\dagger$. Extending what is done in \cite{buc} we also require that $ka=bk=0$. Moreover, $a$ and $b$ are not, in general, one the adjoint of the other: $b\neq a^\dagger$. This is, in a sense, close to what was done in \cite{pb1} first, and in \cite{pf1} later, for CCR and CAR, and, in fact, what we will show here, is that we recover the same {\em global} functional structure (raising and lowering relations, biorthogonal sets, non-Hermitian number-like operators,....) as in the cited papers, even if we work here in finite-dimensional Hilbert spaces of dimension not necessarily equal to 2, as we did in \cite{pf1}.

The first remark is that operators obeying the commutation rule in (\ref{21}) can also be represented as matrices acting on a $N$-dimensional Hilbert space $\Hil_N$. This can be easily seen as follows: let $S_0$ be an $(N-1)\times(N-1)$ invertible matrix, and let $s$ be a non-zero complex number. Then, if $S$ is the diagonal block matrix with blocks $S_0$ and $s$, $S^{-1}$ exists (but, in general, $S^{-1}\neq S^\dagger$) and, since (\ref{11}) is implemented in $\Hil_N$, we can easily define three new  matrices $a=ScS^{-1}$, $b=Sc^\dagger S^{-1}$ and $k=SKS^{-1}$. These operators, since $K$ commutes with $S^\dagger S$,satisfy (\ref{21}), as well as the equalities  $k=k^2=k^\dagger$ and $ka=bk=0$. So we see that, at least in this situation, (\ref{21}) can be represented in $\Hil_N$. Of course, other (higher-dimensional) representations could also exist. However, from now on, $a$, $b$ and $k$ will be considered as operators on $\Hil_N$.

We start our analysis by introducing two (non-Hermitian) position and momentum-like operators:
$$
q=\frac{a+b}{\sqrt{2}},\qquad p=\frac{a-b}{\sqrt{2}\,i},
$$
so that $a=\frac{q+ip}{\sqrt{2}}$ and $b=\frac{q-ip}{\sqrt{2}}$. As in \cite{buc}, we introduce the operator $h=\frac{1}{2}(p^2+q^2)$. Despite of its expression, and of what happens in \cite{buc}, $h$ is not Hermitian ($h\neq h^\dagger$). Moreover, it is  not even manifestly  positive ($h\ngeq0$), due to the fact that both $q$ and $p$ are not Hermitian. Nevertheless, we will show later in this section that the eigenvalues of $h$ are indeed strictly positive for all possible choices of $N$. After few computations it is easy to deduce the following equalities:
\be
[a,h]=a-\frac{1}{2}\,Nak,\qquad
[b,h]=-b+\frac{1}{2}\,Nkb,
\label{22}\en
as well as
\be
\left\{
    \begin{array}{ll}
h=ba+\frac{1}{2}(\1-Nk)=ab-\frac{1}{2}(\1-Nk),\\
\{a,b\}=2h,\\
kh=hk=-\frac{1}{2}(\1-N)k,
\end{array}
        \right.
\label{23}\en
which in particular imply that $[h,k]=0$.  Then we can look for common eigenstates of $h$ and $k$, which we call $\varphi_{h',k'}$:
\be
\left\{
    \begin{array}{ll}
h\varphi_{h',k'}=h'\varphi_{h',k'},\\
k\varphi_{h',k'}=k'\varphi_{h',k'}.
\end{array}
        \right.
\label{23a}\en
Of course, since $k=k^2$, $k'$ can only be 0 and 1. In particular, in analogy with what happens in \cite{buc}, the only (possibly) non zero vector $\varphi_{h',k'}$, when $k'=1$, is the vector with $h'=\frac{1}{2}(N-1)$, $\varphi_{\frac{1}{2}(N-1),1}$; all the other vectors, $\varphi_{h',1}$, if $h'\neq \frac{1}{2}(N-1)$, turn out to be zero. In general, the vectors $\varphi_{h',k'}$ are not mutually orthogonal in $h'$, since $h\neq h^\dagger$, while they are orthogonal in $k'$, since $k=k^\dagger$:
\be
\left<\varphi_{h',k'},\varphi_{h'',k''}\right>=\left<\varphi_{h',k'},\varphi_{h'',k'}\right>\,\delta_{k',k''}.
\label{24}\en

It is now possible to prove that, if $a\,\varphi_{h',k'}\neq0$, then this vector must be proportional to $\varphi_{h'-1+\frac{1}{2}Nk',0}$. This follows from the following facts: first, since $ka=0$, $k(a\varphi_{h',k'})=0$. Secondly, using (\ref{22}), we have
$$
h\left(a\varphi_{h',k'}\right)=\left([h,a]+ah\right)\varphi_{h',k'}=\left(h'-1+\frac{1}{2}Nk'\right)(a\varphi_{h',k'}).
$$
Hence our claim follows. In particular we have
\be
a\varphi_{h',0}=0 \quad \Leftrightarrow \quad h'=\frac{1}{2}.
\label{25}\en
In fact, let us assume that $\varphi_{h',0}\neq0$ but $a\varphi_{h',0}=0$. Then, using (\ref{23}), we have
$$
0=b\left(a\varphi_{h',0}\right)=\left(h-\frac{1}{2}(\1-Nk)\right)\varphi_{h',0}=\left(h'-\frac{1}{2}\right)\varphi_{h',0},
$$
so that $h'=\frac{1}{2}$. The proof of the converse implication, i.e. that $a\varphi_{\frac{1}{2},0}=0$, needs to be postponed but it is essentially based on the fact that $\Hil_N$ has dimension $N$. In fact, we will see that acting with $a$ and $b$ on vectors of the form $\varphi_{h',k'}$ we can produce $N$ linearly independent (l.i.) vectors, including $\varphi_{\frac{1}{2},0}$. Their linear independence is due to the fact that they correspond to different, strictly positive, values of $h'$ (so, even if they are not orthogonal, they are still l.i.), or to different values of $k'$ (so they are orthogonal and, therefore, l.i., too). Then, if $a\varphi_{\frac{1}{2},0}$ is different from zero, it would be proportional to $\varphi_{-\frac{1}{2},0}$. This vector, being eigenstate of $h$ with eigenvalue $h'=-\frac{1}{2}$ different from the other ones (see below), would be the $N+1$-th l.i. vector in a space with dimension $N$. This is clearly impossible. Hence (\ref{25}) follows. Notice that, in particular, this also implies that $h$ admits only strictly positive eigenvalues, even in absence of an manifest positivity, which was used in \cite{buc} to deduce the analogous of (\ref{25}).

After showing that $a$ annihilates $\varphi_{\frac{1}{2},0}$, we need now to show that $b$ annihilates the vector $\varphi_{\frac{1}{2}(N-1),1}$:
\be
b\varphi_{\frac{1}{2}(N-1),1}=0.
\label{26}\en
 To check this, we start observing that, if $kb\varphi_{h',k'}\neq0$, then it must be proportional to $\varphi_{h'+1-\frac{1}{2}N,1}$. First of all, since $k^2=k$, it is clear that in this case $kb\varphi_{h',k'}$ must
an eigenstate of $k$ with eigenvalue 1. Now, using (\ref{22}) we find first that
$$
hb\varphi_{h',k'}=\left(bh+b-\frac{1}{2}\,Nkb\right)\varphi_{h',k'}=\left(h'+1-\frac{1}{2}\,Nk\right)b\varphi_{h',k'},
$$
which, when left-multiplied by $k$, produces
$$
h\left(kb\varphi_{h',k'}\right)=\left(h'+1-\frac{1}{2}\,N\right)\left(kb\varphi_{h',k'}\right).
$$
Then, as stated,  if $kb\varphi_{h',k'}\neq0$, it must be proportional to $\varphi_{h'+1-\frac{1}{2}N,1}$. It is now possible to see that the only possibility for having $kb\varphi_{h',k'}\neq0$ is that $h'=N-\frac{3}{2}$. In other words: if $h'\neq N-\frac{3}{2}$ then $kb\varphi_{h',k'}=0$, independently of $k'$.

To prove this claim we use (\ref{22}) and the equality $kh=-\frac{1}{2}(\1-N)k$ in (\ref{23}). Few algebraic manipulations produce now the equality $kbh=\left(N-\frac{3}{2}\right)kb$. Therefore $\left(h'-\left(N-\frac{3}{2}\right)\right)kb\varphi_{h',k'}=0$, which is only possible, if $kb\varphi_{h',k'}\neq0$, when $h'=N-\frac{3}{2}$. Then, if $h'\neq N-\frac{3}{2}$, $kb\varphi_{h',k'}$ must be zero. If we now compute  $kb\varphi_{N-\frac{3}{2},k'}$ we find, because of what deduced before, that this is proportional to $\varphi_{\frac{1}{2}(N-1),1}$. Notice that, in principle, $kb\varphi_{N-\frac{3}{2},k'}$ could still be zero. However, inspired by the results in \cite{buc}, we will assume that this is not so, and check this assumption in explicit examples.

Formula (\ref{26}) is now a simple consequence of the fact that $bk=0$. Indeed we have, for what we have deduced so far, that $b\varphi_{\frac{1}{2}(N-1),1}$ must be proportional to $b\left(kb\varphi_{N-\frac{3}{2},k'}\right)$, which is the zero vector since $bk=0$. Summarizing we have two different vectors, $\varphi_{\frac{1}{2},0}$ and $\varphi_{\frac{1}{2}(N-1),1}$, which are annihilated respectively by $a$ and $b$. This is very close to what happens in \cite{buc}, where we have also two vectors annihilated by the operator $c$ in (\ref{11}), and by its hermitian conjugate $c^\dagger$. A similar feature is observed in ordinary CAR, where the lowering operator annihilates the vacuum and its adjoint annihilates the upper lever. Moreover, in \cite{buc}, it is shown that $c$ behaves as a sort of lowering operator, while $c^\dagger$ behaves as a raising operator. We expect that a similar behavior can be deduced here for $a$ and $b$, and this is in-fact what we will see now.

\subsection{Two preliminary examples}

Before discussing the general case (i.e. generic $N>2$), we briefly discuss   how the construction works when $N=2$ and when $N=3$. It is worth stressing that our construction is significantly different from the one proposed in \cite{buc}, because of the many properties related to the particular structure arising from (\ref{11}), which are lost here.

\vspace{2mm}

If $N=2$ the commutation rule in (\ref{21}) become $[a,b]=\1-2k$. The Hilbert space $\Hil_2$ is two-dimensional, and the two vectors in (\ref{25}) and (\ref{26}), $\varphi_{\frac{1}{2},0}$ and $\varphi_{\frac{1}{2},1}$, turn out to be orthogonal: $\left<\varphi_{\frac{1}{2},0},\varphi_{\frac{1}{2},1}\right>=0$, since they correspond to different eigenvalues of the Hermitian operator $k$. Then $\F_\varphi^{(2)}=\{\varphi_{\frac{1}{2},0},\varphi_{\frac{1}{2},1}\}$ is an o.n. basis for $\Hil_2$ (assuming a {\em good} normalization). Of course, the eigenvalue $\frac{1}{2}$ of the Hamiltonian $h$ is degenerate.
As discussed before, $a\varphi_{\frac{1}{2},0}$ must be zero. Otherwise, it would be proportional to $\varphi_{-\frac{1}{2},0}$, so that the vectors $\varphi_{\frac{1}{2},0}$, $\varphi_{\frac{1}{2},1}$ and $a\varphi_{\frac{1}{2},0}$ would be l.i., in contrast with the fact that $\dim(\Hil_2)=2$. Hence $a\varphi_{\frac{1}{2},0}=0$.

We have shown in (\ref{26}) that $b\varphi_{\frac{1}{2},1}=0$. So, in order to fully understand the situation, in this simple case we still need to compute $a\varphi_{\frac{1}{2},1}$ and $b\varphi_{\frac{1}{2},0}$.
We will now prove that a non zero constant $\nu_{\frac{1}{2},0}$ exists such that
\be
a\varphi_{\frac{1}{2},1}=\nu_{\frac{1}{2},1}\,\varphi_{\frac{1}{2},0}, \qquad \mbox{and}\qquad b\varphi_{\frac{1}{2},0}=\nu_{\frac{1}{2},1}^{-1}\varphi_{\frac{1}{2},1}.
\label{27}\en
In fact, since $ha=ah-a+ak$, we see first that $ha\varphi_{\frac{1}{2},1}=(ah-a+ak)\varphi_{\frac{1}{2},1}=\frac{1}{2}a\varphi_{\frac{1}{2},1}$.
Moreover $k\left(a\varphi_{\frac{1}{2},1}\right)=0$. Then, a non zero constant  $\nu_{\frac{1}{2},1}$ should exist such that  $a\varphi_{\frac{1}{2},1}=\nu_{\frac{1}{2},1}\varphi_{\frac{1}{2},0}$. As for the second relation in (\ref{27}), this can be deduced by applying now the operator $b$ from the left to the first one and using the equality $[a,b]=\1-2k$.

\vspace{2mm}

The case $N=3$ is surely more interesting because, as we will see, the vectors $\varphi_{h',k'}$ do not form an o.n. basis for $\Hil_3$.

In this case the commutation rule in (\ref{21}) becomes $[a,b]=\1-3k$ and, since $\frac{1}{2}(N-1)=1$, our {\em extreme} vectors in (\ref{25}) and (\ref{26}) are $\varphi_{\frac{1}{2},0}$ and $\varphi_{1,1}$. Since they correspond to  different eigenvalues of the Hermitian operator $k$, they are orthogonal: $\left<\varphi_{\frac{1}{2},0},\varphi_{1,1}\right>=0$. But they are just two vectors in a three-dimensional space. Therefore, they cannot be a basis for $\Hil_3$. A third vector, l.i. with respect to the these  two, can be easily constructed. To do this, let us consider the vector $a\varphi_{1,1}$. Using (\ref{22}) and the fact that $ka=0$ we deduce that
$$
h\left(a\varphi_{1,1}\right)=\frac{3}{2}\left(a\varphi_{1,1}\right), \qquad k\left(a\varphi_{1,1}\right)=0.
$$
These imply that, if $a\,\varphi_{1,1}\neq0$, then it is an eigenvector of $h$, with eigenvalue $\frac{3}{2}$, and of $k$, with eigenvalue $0$. Hence we can introduce a vector, $\varphi_{\frac{3}{2},0}$, and a non-zero (complex) number $\nu_{1,1}$, such that $a\,\varphi_{1,1}=\nu_{1,1}\varphi_{\frac{3}{2},0}$. Let then consider the set $\F_\varphi^{(3)}=\{\varphi_{\frac{1}{2},0},\varphi_{\frac{3}{2},0},\varphi_{1,1}\}$. Its vectors are l.i., since they are orthogonal or they correspond to different eigenvalues of $h$. Hence $\F_\varphi^{(3)}$ is a basis for $\Hil_3$. Furthermore, using the commutation rule between $a$ and $b$, it follows that $b\,\varphi_{\frac{3}{2},0}=2\nu_{1,1}^{-1}\varphi_{1,1}$. In a similar way we can also prove that
$$
a\,\varphi_{\frac{3}{2},0}=\nu_{\frac{3}{2},0}\,\varphi_{\frac{1}{2},0},\qquad \mbox{ and } \qquad b\,\varphi_{\frac{1}{2},0}=\nu_{\frac{3}{2},0}^{-1}\,\varphi_{\frac{3}{2},0},
$$
for some non zero $\nu_{\frac{3}{2},0}$. This is because  $a\,\varphi_{\frac{3}{2},0}$ turns out to be an eigenstate of $h$ and $k$, with eigenvalues $\frac{1}{2}$ and $0$ respectively. Hence the first equality above follows. The second is a consequence of this first and of the commutator $[a,b]=\1-3k$. We see that, with a slight abuse of language, $a$ behaves as a lowering operator while $b$ behaves as a raising operator.

\subsection{Larger $N$}

\vspace{3mm}

The situation is just a little more complicated if we take now $N\geq4$:
 we have two vectors $\varphi_{\frac{1}{2},0}$ and $\varphi_{\frac{1}{2}(N-1),1}$ which are mutually orthogonal in $\Hil_N$. A third l.i. vector can be defined via the action of $a$ on $\varphi_{\frac{1}{2}(N-1),1}$. In fact $a\varphi_{\frac{1}{2}(N-1),1}$, if it is non-zero, is an eigenstate of $h$ and $k$ with eigenvalues $N-\frac{3}{2}$ and $0$ respectively:
$$
h\left(a\varphi_{\frac{1}{2}(N-1),1}\right)=\left(N-\frac{3}{2}\right)\left(a\varphi_{\frac{1}{2}(N-1),1}\right), \qquad\mbox{  and }\qquad k\left(a\varphi_{\frac{1}{2}(N-1),1}\right)=0.
$$
Then, a non zero constant $\nu_{\frac{1}{2}(N-1),1}$ exists such that $a\varphi_{\frac{1}{2}(N-1),1}=\nu_{\frac{1}{2}(N-1),1}\varphi_{N-\frac{3}{2},0}$. Now, since $N\geq4$, $N-\frac{3}{2}\neq \frac{1}{2}$. Therefore $\{\varphi_{\frac{1}{2},0},\varphi_{N-\frac{3}{2},0},\varphi_{\frac{1}{2}(N-1),1}\}$ are l.i., for the usual reasons, but they cannot form a basis for $\Hil_N$. Acting with $b$ on  $\varphi_{N-\frac{3}{2},0}$ we go back to $\varphi_{\frac{1}{2}(N-1),1}$. More in details, $b\varphi_{N-\frac{3}{2},0}=(N-1)\nu_{\frac{1}{2}(N-1),1}^{-1}\varphi_{\frac{1}{2}(N-1),1}$. If we go further, considering now $a\varphi_{N-\frac{3}{2},0}$, we find that this vector satisfies the following eigenvalue equations:
$$
h\left(a\varphi_{N-\frac{3}{2},0}\right)=\left(N-\frac{5}{2}\right)\left(a\varphi_{N-\frac{3}{2},0}\right), \qquad\mbox{  and }\qquad k\left(a\varphi_{N-\frac{3}{2},0}\right)=0.
$$
Therefore, if $a\varphi_{N-\frac{3}{2},0}\neq0$, a non zero  $\nu_{N-\frac{3}{2},0}$ exists such that $a\varphi_{N-\frac{3}{2},0}=\nu_{N-\frac{3}{2},0}\varphi_{N-\frac{5}{2},0}$. Also, $b\varphi_{N-\frac{5}{2},0}=(N-2)\nu_{N-\frac{3}{2},0}^{-1}\varphi_{N-\frac{3}{2},0}$. Of course, if $N>4$, $N-\frac{5}{2}\neq \frac{1}{2}$. Therefore $\{\varphi_{\frac{1}{2},0},\varphi_{N-\frac{5}{2},0},\varphi_{N-\frac{3}{2},0},\varphi_{\frac{1}{2}(N-1),1}\}$ are still l.i., but, again, they cannot be a basis for $\Hil_N$. So we consider $a\varphi_{N-\frac{5}{2},0}$, and the usual arguments show that this is proportional to $\varphi_{N-\frac{7}{2},0}$. Hence we have two possibilities: either $N-\frac{9}{2}=\frac{1}{2}$, which means that $N=5$, or, when  $N-\frac{9}{2}>\frac{1}{2}$, $N>5$. In the first case $\F_\varphi^{(5)}=\{\varphi_{\frac{1}{2},0},\varphi_{N-\frac{7}{2},0},\varphi_{N-\frac{5}{2},0},
\varphi_{N-\frac{3}{2},0},\varphi_{\frac{1}{2}(N-1),1}\}$ are five l.i. vectors in $\Hil_5$, so they do form a basis. In fact, acting again with $a$ on $\varphi_{N-\frac{7}{2},0}$ would produce a vector proportional to $\varphi_{\frac{1}{2},0}$, which is clearly not l.i. with respect to the ones in $\F_\varphi^{(5)}$. On the other hand, if $N>5$, we can continue our lowering procedure, acting with $a$ on $\varphi_{N-\frac{7}{2},0}$ and getting, this time, a vector which is surely not proportional to $\varphi_{\frac{1}{2},0}$. Of course, while $a$ acts as a lowering operator, $b$ behaves as a raising operator. However, due to the fact that $\Hil_N$ is finite-dimensional, $b$ annihilates $\varphi_{\frac{1}{2}(N-1),1}$, as we have seen. The situation is shown in Figure \ref{fig3}.

\begin{figure}
\begin{center}

\begin{picture}(450,145)

\put(5,127){\line(1,0){65}} \put(7,127){\vector(-1,0){2}}\put(40,145){\makebox(0,0){$b$}}
\put(5,135){\line(1,0){65}} \put(68,135){\vector(1,0){2}}\put(40,120){\makebox(0,0){$a$}}
\multiput(-10,120)(0,-5){10}{\line(0,-1){2}}\put(-15,95){\makebox(0,0){$a$}}\put(-10,70){\vector(0,-1){2}}
\put(-10,60){\makebox(0,0){$0$}}

\put(100,127){\line(1,0){65}} \put(103,127){\vector(-1,0){2}}\put(130,145){\makebox(0,0){$b$}}
\put(100,135){\line(1,0){65}} \put(163,135){\vector(1,0){2}}\put(130,120){\makebox(0,0){$a$}}

\put(310,127){\line(1,0){65}} \put(313,127){\vector(-1,0){2}}\put(340,145){\makebox(0,0){$b$}}
\put(310,135){\line(1,0){65}} \put(373,135){\vector(1,0){2}}\put(340,120){\makebox(0,0){$a$}}

\multiput(200,130)(3,0){20}{\line(2,0){1}}

\multiput(395,120)(0,-5){10}{\line(0,-1){2}}\put(400,95){\makebox(0,0){$b$}}\put(395,70){\vector(0,-1){2}}
\put(395,60){\makebox(0,0){$0$}}

\put(-10,130){\makebox(0,0){$\varphi_{\frac{1}{2},0}$}}
\put(85,130){\makebox(0,0){$\varphi_{\frac{3}{2},0}$}}
\put(180,130){\makebox(0,0){$\varphi_{\frac{5}{2},0}$}}
\put(285,130){\makebox(0,0){$\varphi_{\frac{N}{2}-3,0}$}}
\put(405,130){\makebox(0,0){$\varphi_{\frac{1}{2}(N-1),1}$}}

\end{picture}

\end{center}

\vspace{-1.5cm}

\caption{{\protect\footnotesize The ladder operators and their action in $\Hil_N$.}}
\label{fig3}
\end{figure}

\vspace*{2mm}

{\bf Remarks:--} (1) We first observe that the values of the $\nu_{h',k'}$ in our construction are unfixed. The reason is that they will be only (partially) fixed by the biorthogonality condition discussed in the next section.

(2) The vectors $\varphi_{h',k'}$ are eigenstates not only of $h$ and $k$, but also of the operators $\hat M=ab$ and $\hat N=ba$. In fact we find:
\be
\left\{
    \begin{array}{ll}
\hat N\varphi_{\frac{1}{2}(N-1),1}=(N-1)\varphi_{\frac{1}{2}(N-1),1},\\
\hat N\varphi_{N-\frac{3}{2},0}=(N-2)\varphi_{N-\frac{3}{2},0},\\
\hat N\varphi_{N-\frac{5}{2},0}=(N-3)\varphi_{N-\frac{5}{2},0},\\
\ldots\\
\ldots\\
\hat N\varphi_{\frac{1}{2},0}=0,
\end{array}
        \right.
\label{28}\en
while
\be
\left\{
    \begin{array}{ll}
\hat M\varphi_{\frac{1}{2}(N-1),1}=0,\\
\hat M\varphi_{N-\frac{3}{2},0}=(N-1)\varphi_{N-\frac{3}{2},0},\\
\hat M\varphi_{N-\frac{5}{2},0}=(N-2)\varphi_{N-\frac{5}{2},0},\\
\ldots\\
\ldots\\
\hat M\varphi_{\frac{1}{2},0}=\varphi_{\frac{1}{2},0}.
\end{array}
        \right.
\label{29}\en
This is not surprising since one can check that $[\hat N,h]=[\hat M,h]=[\hat N,k]=[\hat M,k]=0$. In fact, recalling that $ka=bk=0$, we see that $[k,\hat M]=[k,ab]=0$. Moreover, since $ba=ab+Nk-\1$, we deduce that $[k,\hat N]=[k,ba]=[k,ab+Nk-\1]=0$. The fact that we also have $[\hat N,h]=[\hat M,h]=0$ follows now from the first line in (\ref{23}) which implies that $\hat M=h+\frac{1}{2}(\1-Nk)$ and $\hat N=h-\frac{1}{2}(\1-Nk)$, recalling further that $[h,k]=0$.

\section{The biorthogonal set}\label{sectIII}

In the literature on non self-adjoint Hamiltonians it is widely discussed how biorthogonal sets play an essential role in the description of the physical system $\Sc$: the eigenstates of the Hamiltonian $H_\Sc$ of $\Sc$ are not orthogonal, while they are biorthogonal to the elements of the set of eigenstates of $H_\Sc^\dagger$. For this reason, recalling that the operator $h=\frac{1}{2}(p^2+q^2)$ introduced in Section \ref{sectII} is not Hermitian, $h\neq h^\dagger$, it is natural to consider the problem of the diagonalization of $h^\dagger$. For that we take the adjoint of (\ref{21}) and we observe that, introducing $A=b^\dagger$ and $B=a^\dagger$, and recalling that $k=k^\dagger$, we get
\be
[A,B]=\1-Nk.
\label{31}\en
Then the pair $(A,B)$ satisfies the same commutation rule as the original pair $(a,b)$. Now, calling $Q=\frac{A+B}{\sqrt{2}}$, $P=\frac{A-B}{\sqrt{2}\,i}$ and $H=\frac{1}{2}(P^2+Q^2)$, it follows that $Q=q^\dagger$, $P=p^\dagger$ and $H=h^\dagger$. The analogous of formulas (\ref{22}) and (\ref{23}) can be deduced for these operators. For instance $\{A,B\}=2H$, $BA=H-\frac{1}{2}(\1-Nk)$, $AB=H+\frac{1}{2}(\1-Nk)$, $[H,k]=0$ and so on. Moreover, we also have $Bk=kA=0$. We call now $\psi_{h',k'}$ the common eigenstates of $H$ and $k$:
\be
\left\{
    \begin{array}{ll}
H\psi_{h',k'}=h'\psi_{h',k'},\\
k\psi_{h',k'}=k'\psi_{h',k'}.
\end{array}
        \right.
\label{32}\en
Then, standard arguments show that the sets $\F_\varphi=\{\varphi_{h',k'}\}$ and $\F_\psi=\{\psi_{h',k'}\}$ are biorthogonal and, if we choose properly the normalization of the vectors, they are also biorthonormal:
\be
\left<\psi_{h',k'},\varphi_{h'',k''}\right>=\delta_{h',h''}\delta_{k',k''}.
\label{33}\en

The construction of the set $\F_\psi^{(N)}$ reflects that of $\F_\varphi^{(N)}$ for the different values of $N$. For instance, if $N=2$, we only have two vectors $\F_\psi^{(2)}=\{\psi_{\frac{1}{2},0},\psi_{\frac{1}{2},1}\}$ and they satisfy the following equations:
$$
A\psi_{\frac{1}{2},0}=B\psi_{\frac{1}{2},1}=0, \qquad \mbox{and }\quad A\psi_{\frac{1}{2},1}=\mu_{\frac{1}{2},1}\psi_{\frac{1}{2},0},\quad B\psi_{\frac{1}{2},0}=\mu_{\frac{1}{2},1}^{-1}\psi_{\frac{1}{2},1},
$$
for some non-zero $\mu_{\frac{1}{2},1}$. This has to be related to $\nu_{\frac{1}{2},1}$ if we want $\F_\varphi^{(2)}$ and $\F_\psi^{(2)}$ to be biorthonormal. In fact, while $\left<\psi_{\frac{1}{2},1},\varphi_{\frac{1}{2},0}\right>=
\left<\psi_{\frac{1}{2},0},\varphi_{\frac{1}{2},1}\right>=0,$
automatically, if we further choose the normalization of $\varphi_{\frac{1}{2},0}$ and $\psi_{\frac{1}{2},0}$ by requiring $
\left<\psi_{\frac{1}{2},0},\varphi_{\frac{1}{2},0}\right>=1,
$
then we also get
$
\left<\psi_{\frac{1}{2},1},\varphi_{\frac{1}{2},1}\right>=1
$
at least if \be \overline{\nu_{\frac{1}{2},1}}\,\mu_{\frac{1}{2},1}=1.\label{add1}\en A similar situation is recovered for larger values of $N$. For instance, if $N=3$, the set $\F_\psi^{(3)}=\{\psi_{\frac{1}{2},0},\psi_{\frac{3}{2},0},\psi_{1,1}\}$ obeys the following ladder rules, see Figure \ref{fig2}:
\be
 A\psi_{1,1}=\mu_{1,1}\psi_{\frac{3}{2},0},\quad A\psi_{\frac{3}{2},0}=\mu_{\frac{3}{2},0}\psi_{\frac{1}{2},0},\quad B\psi_{\frac{3}{2},0}=2\mu_{1,1}^{-1}\psi_{1,1}, \quad B\psi_{\frac{1}{2},0}=\mu_{\frac{3}{2},0}^{-1}\psi_{\frac{3}{2},0},
\label{add2}\en
for some suitable and non zero $\mu_{1,1}$ and $\mu_{\frac{3}{2},0}$, as well as $A\psi_{\frac{1}{2},0}=B\psi_{1,1}=0$. Using now (\ref{21}) and assuming that $\left<\psi_{\frac{1}{2},0},\varphi_{\frac{1}{2},0}\right>=1$ we recover biorthonormality of the sets $\F_\psi^{(3)}$ and $\F_\varphi^{(3)}$  if we require
\be\overline{\nu_{\frac{3}{2},0}}\,\mu_{\frac{3}{2},0}=\frac{1}{2}\,\overline{\nu_{1,1}}\,\mu_{1,1}=1.\label{34}\en

It is straightforward to extend these results to larger $N$. In all cases, the following resolution of the identity in $\Hil_N$ is satisfied:
$$
\sum_{(h',k')\in\G_N}|\psi_{h',k'}\left>\right<\varphi_{h',k'}|=\sum_{(h',k')\in\G_N}|\varphi_{h',k'}\left>\right<\psi_{h',k'}|=\1,
$$
where we have used the Dirac bra-ket notation and where the sum is extended, for each fixed $N$, to all the possible pairs of $(h',k')$, see Section \ref{sectII}. We have called $\G_N$ this set.

\vspace{3mm}

The two biorthonormal sets $\F_\varphi^{(N)}$ and $\F_\psi^{(N)}$ can be used, together, to represent the operators $a$, $b$, and their adjoints, as a sum of rank-one operators. For instance, if $N=2$, we have
\be
a=\nu_{\frac{1}{2},1}\,|\varphi_{\frac{1}{2},0}\left>\right<\psi_{\frac{1}{2},1}|, \qquad b=\nu_{\frac{1}{2},1}^{-1}\,|\varphi_{\frac{1}{2},1}\left>\right<\psi_{\frac{1}{2},0}|,
\label{39}\en
whose adjoints are $a^\dagger=\overline{\nu_{\frac{1}{2},1}}\,|\psi_{\frac{1}{2},1}\left>\right<\varphi_{\frac{1}{2},0}|$ and $b^\dagger=\overline{\nu_{\frac{1}{2},1}^{-1}}\,|\psi_{\frac{1}{2},0}\left>\right<\varphi_{\frac{1}{2},1}|$. It is interesting to observe that the operators $a^\dagger$ and $b^\dagger$ given here coincide with $B$ and $A$ if condition (\ref{add1}) is satisfied.

In a similar way, if $N=3$, $a$ and $b$ can be written as
\be
a=\nu_{1,1}\,|\varphi_{\frac{3}{2},0}\left>\right<\psi_{1,1}|+\nu_{\frac{3}{2},0}\,
|\varphi_{\frac{1}{2},0}\left>\right<\psi_{\frac{3}{2},0}|, \qquad b=2\nu_{1,1}^{-1}\,|\varphi_{1,1}\left>\right<\psi_{\frac{3}{2},0}|+
\nu_{\frac{3}{2},0}^{-1}\,|\varphi_{\frac{3}{2},0}\left>\right<\psi_{\frac{1}{2},0}|,
\label{310}\en
and the adjoint $a^\dagger$ and $b^\dagger$ which we deduce out of these behaves as the operators $B$ and $A$ in (\ref{add2}) if conditions in (\ref{34}) are satisfied. The same is true for larger values of $N$.

It turns out that  $a^\dagger$ and $b^\dagger$ are respectively equal to the ladder operators $B$ and $A$ for $\F_\psi^{(4)}$. This is a general characteristic of the construction: the same constraints on $\mu_{h',k'}$ and $\nu_{h',k'}$ which make of $\F_\varphi^{(N)}$ and $\F_\psi^{(N)}$ biorthonormal bases ensure that the adjoint of the representations of $a$ and $b$ coincide exactly with $B$ and $A$.

\subsection{Relation with (\ref{11})}\label{sectIII.1}

At the beginning of Section \ref{sectII} we have already discussed how (\ref{21}) can be obtained from (\ref{11}), by means of a similarity map. In this section we discuss the inverse construction, i.e. we show how, starting from (\ref{21}), it is possible to construct two operators, $c$ and $K$, which obey the commutation rule in (\ref{11}) and such that $Kc=0$. Our construction is similar, but not identical, to that proposed for pseudo-fermions, \cite{pf1}.

We start introducing the operators
\be
S_\varphi=\sum_{(h',k')\in\G_N}|\varphi_{h',k'}\left>\right<\varphi_{h',k'}|, \qquad S_\psi=\sum_{(h',k')\in\G_N}|\psi_{h',k'}\left>\right<\psi_{h',k'}|.
\label{36}\en
These are bounded, invertible, Hermitian and positive. Moreover, they are one the inverse of the other, $S_\varphi S_\psi=S_\psi S_\varphi=\1$, and satisfy the following:
\be
S_\varphi \psi_{h',k'}=\varphi_{h',k'}, \qquad S_\psi \varphi_{h',k'}=\psi_{h',k'},
\label{37}\en
for all $(h',k')\in\G_N$. $S_\psi$ admits an unique positive square root, which is also invertible. Hence we can define
\be
e_{h',k'}=S_\psi^{1/2}\varphi_{h',k'}, \qquad c=S_\psi^{1/2}aS_\psi^{-1/2},
\label{38}\en
for $(h',k')\in\G_N$.  Now, it is a simple computation to prove that the set $\F_e=\{e_{h',k'}\}$ is an o.n. basis for $\Hil_N$. It is also possible to check that $c^\dagger$, other than being equal to $S_\psi^{-1/2}a^\dagger S_\psi^{1/2}$, can also be written as $c^\dagger=S_\psi^{1/2}bS_\psi^{-1/2}$. In fact $S_\psi^{-1/2}a^\dagger S_\psi^{1/2}=S_\psi^{1/2}bS_\psi^{-1/2}$ if and only if $S_\psi b=a^\dagger S_\psi$, i.e. if $S_\psi$ intertwines between $b$ and $a^\dagger$. This equality can be easily deduced by considering the action of $S_\psi b$ and $a^\dagger S_\psi$ on the vectors of $\F_\varphi$. In fact, it turns out that $S_\psi b\varphi_{h',k'}=a^\dagger S_\psi\varphi_{h',k'}$ for all $(h',k')\in\G_N$, at least if we assume that $\mu_{h',k'}=\nu_{h',k'}$ for all $(h',k')\in\G_N$. This further constraint simplifies the conditions we have found before to guarantee the biorthonormality of the sets $\F_\varphi^{(N)}$ and $\F_\psi^{(N)}$ and in the analysis of the representation of the operators $a$, $a^\dagger$, $b$ and $b^\dagger$ in terms of these vectors.

Now, since $c=S_\psi^{1/2}aS_\psi^{-1/2}$ and $c^\dagger=S_\psi^{1/2}bS_\psi^{-1/2}$, it turns out that $[c,c^\dagger]=\1-NK$, where $K$ is defined as $K=S_\psi^{1/2}kS_\psi^{-1/2}$. Of course $K^2=K$ and $Kc=0$. Moreover $K=K^\dagger$ if and only if $k$ commutes with $S_\psi$, which is true:
$
S_\psi k\varphi_{h',k'}=k'S_\psi \varphi_{h',k'}=k'\psi_{h',k'}=k\psi_{h',k'}=kS_\psi\varphi_{h',k'}.
$
Hence, since $\Hil_N$ has finite dimension, $[k,S_\psi]=0$.

\vspace{2mm}

Summarizing we can say that it is possible to deform (\ref{11}) to get (\ref{21}), but we can also go the other way around, i.e. we can consider (\ref{21}) as our starting point, and use the eigenvectors of $h$, $h^\dagger$ and $k$ constructed out of $a$ and $b$ to define new operators satisfying (\ref{11}).

\section{Examples}\label{sectIV}

In this section we consider a pair of examples. The first one is more mathematical, while in the second we connect our general settings with a truncated version of the Swanson model, \cite{swan}, which is very well known among the PT-quantum mechanical community, being one non trivial example of manifestly non self-adjoint Hamiltonian which is isospectral to the standard (i.e., self-adjoint) harmonic oscillator.

\subsection{An example with $N=4$}

Let $a$ and $b$ the following four-by-four matrices:
$$
a=\frac{1}{1+\alpha^3}\left(
                        \begin{array}{cccc}
                          (1-\sqrt{2})\alpha^2 & 1+\sqrt{2}\alpha^3 & (\sqrt{2}-1)\alpha & 0 \\
                          -\sqrt{2}\alpha & \sqrt{2}\alpha^2 & \sqrt{2} & \sqrt{3}\alpha(1+\alpha^3) \\
                          \alpha^3 & \alpha & -\alpha^2 & \sqrt{3}(1+\alpha^3) \\
                          0 & 0 & 0 & 0 \\
                        \end{array}
                      \right)
$$
and
$$
b=\frac{1}{1+\alpha^3}\left(
                        \begin{array}{cccc}
                          \alpha & -\alpha^2 & \alpha^3 & 0 \\
                          1+\sqrt{2}\alpha^3 & (\sqrt{2}-1)\alpha & (1-\sqrt{2})\alpha^2 & 0 \\
                          \sqrt{2}\alpha^2 & \sqrt{2} & -\sqrt{2}\alpha & 0 \\
                          -\sqrt{3}\alpha & \sqrt{3}\alpha^2 & \sqrt{3} & 0 \\
                        \end{array}
                      \right),
$$
where $\alpha$ is a real constant different from $-1$. These operators satisfy the commutation rule $[a,b]=\1-4k$, where $k$ is the diagonal matrix on $\Hil_4$ with three zeros and a single one in the main diagonal: $k=diag(0,0,0,1)$. Hence $ka=bk=0$. The hamiltonian $h=ba+\frac{1}{2}(\1-4k)$ looks like
$$
h=\frac{1}{1+\alpha^3}\left(
                        \begin{array}{cccc}
                          \frac{1}{2}(1+3\alpha^2) & \alpha & -\alpha^2 & 0 \\
                          -\alpha^2 & \frac{1}{2}(3+5\alpha^3) & \alpha & 0 \\
                          -2\alpha & 2\alpha^2 & \frac{1}{2}(5+\alpha^3) & 0 \\
                          0 & 0 & 0 & \frac{3}{2} \\
                        \end{array}
                      \right),
$$
which is manifestly not self-adjoint if $\alpha\neq0$: $h\neq h^\dagger$. We will show now how the procedure proposed in this paper can be applied and produces two biorthogonal bases of eigenvectors of $h$ and $h^\dagger$, which are also eigenvectors of the operator $k$. We start by looking at the vector $\varphi_{\frac{3}{2},1}$ which is annihilated by $b$: $b\varphi_{\frac{3}{2},1}=0$. This forces $\varphi_{\frac{3}{2},1}^T$, the transpose of $\varphi_{\frac{3}{2},1}$, to be of the following form:
$
\varphi_{\frac{3}{2},1}^T=\left(
                          \begin{array}{cccc}
                            0 &0 &0 &
                            \gamma_4 \\
                          \end{array}
                        \right),
$
where $\gamma_4$ could be any non zero complex number, (almost) fixed later by the normalization.

Now, since $a\varphi_{\frac{3}{2},1}=\nu_{\frac{3}{2},1}\varphi_{\frac{5}{2},0}$, we easily check that
$
\varphi_{\frac{5}{2},0}^T=\frac{\sqrt{3}\,\gamma_4}{\nu_{\frac{3}{2},1}}\left(
\begin{array}{cccc}
0 &\alpha &1 &0 \\
                          \end{array}
                        \right).
$
Still, since $a\varphi_{\frac{5}{2},0}=\nu_{\frac{5}{2},0}\varphi_{\frac{3}{2},0}$, we deduce that
$
\varphi_{\frac{3}{2},0}^T=\frac{\sqrt{6}\,\gamma_4}{\nu_{\frac{3}{2},1}\,\nu_{\frac{5}{2},0}}\left(
                          \begin{array}{cccc}
                          0 &1 &0 &0 \\
                          \end{array}
                        \right),
$
while, using the equality $a\varphi_{\frac{3}{2},0}=\nu_{\frac{3}{2},0}\varphi_{\frac{1}{2},0}$, we get
$
\varphi_{\frac{1}{2},0}^T=\frac{\sqrt{6}\,\gamma_4}{\nu_{\frac{3}{2},1}\,\nu_{\frac{5}{2},0}\,\nu_{\frac{3}{2},0}}\left(
                          \begin{array}{cccc}
                          1 &0 &\alpha &0 \\
                          \end{array}
                        \right).
$
An explicit check shows that $a\varphi_{\frac{1}{2},0}=0$, and that $b\varphi_{\frac{5}{2},0}=3\nu_{\frac{3}{2},1}^{-1}\varphi_{\frac{3}{2},1}$, $b\varphi_{\frac{3}{2},0}=2\nu_{\frac{5}{2},0}^{-1}\varphi_{\frac{5}{2},0}$ and $b\varphi_{\frac{1}{2},0}=\nu_{\frac{3}{2},0}^{-1}\varphi_{\frac{3}{2},0}$, as they should.

\vspace{2mm}

The biorthogonal set $\F_\psi^{(4)}$ can be constructed in a similar way: we start looking for a four dimensional vector $\psi_{\frac{3}{2},1}$ which is annihilated by $B=a^\dagger$. This vector is
$
\psi_{\frac{3}{2},1}^T=\left(
                          \begin{array}{cccc}
                          0 &0 &0 &
                          \gamma_4' \\
                          \end{array}
                        \right).
$
Then, acting with $A$ several times on $\psi_{\frac{3}{2},1}$, we find the other vectors of $\F_\psi^{(4)}$. In particular, since $A\psi_{\frac{3}{2},1}=\mu_{\frac{3}{2},1}\psi_{\frac{5}{2},0}$, we find
$
\psi_{\frac{5}{2},0}^T=\frac{\sqrt{3}\,\gamma_4'}{(1+\alpha^3)\mu_{\frac{3}{2},1}}\left(
                          \begin{array}{cccc}
                         -\alpha &\alpha^2 &1 &
                          0 \\
                          \end{array}
                       \right),
$
while $\psi_{\frac{3}{2},0}$ and $\psi_{\frac{1}{2},0}$ can be deduced by acting one or two times with $A$ on $\psi_{\frac{5}{2},0}$. We get
$$
\psi_{\frac{3}{2},0}=\frac{\sqrt{6}\,\gamma_4'}{(1+\alpha^3)\mu_{\frac{3}{2},1}\mu_{\frac{5}{2},0}}\left(
                          \begin{array}{c}
                            \alpha^2 \\
                            1 \\
                            -\alpha \\
                            0 \\
                          \end{array}
                        \right),\qquad \psi_{\frac{1}{2},0}=\frac{\sqrt{6}\,\gamma_4'}{(1+\alpha^3)
                        \mu_{\frac{3}{2},1}\mu_{\frac{5}{2},0}\mu_{\frac{3}{2},0}}\left(
                          \begin{array}{c}
                            1 \\
                            -\alpha \\
                            \alpha^2 \\
                            0 \\
                          \end{array}
                        \right).
$$
A direct check shows that $\left<\varphi_{h',k'},\psi_{h'',k''}\right>=0$ if $(h',k')\neq(h'',k'')$. As for the normalization, the situation is the following: $\left<\varphi_{\frac{3}{2},1},\psi_{\frac{3}{2},1}\right>=1$ if $\gamma_4\gamma_4'=1$; then, $\left<\varphi_{\frac{5}{2},0},\psi_{\frac{5}{2},0}\right>=1$ if $\nu_{\frac{3}{2},1}\mu_{\frac{3}{2},1}=3$; now, $\left<\varphi_{\frac{3}{2},0},\psi_{\frac{3}{2},0}\right>=1$ if $\nu_{\frac{5}{2},0}\mu_{\frac{5}{2},0}=2$. If all these equalities hold, then $\left<\varphi_{\frac{1}{2},0},\psi_{\frac{1}{2},0}\right>=1$ if $\nu_{\frac{3}{2},0}\mu_{\frac{3}{2},0}=1$.  The sets $\F_\varphi^{(4)}$ and $\F_\psi^{(4)}$ obtained in this way are eigenstates of $h$, $h^\dagger$ and of $k$, with the {\em right eigenvalues}.

If we now, for instance, fix $\gamma_4=1$, $\nu_{\frac{3}{2},1}=\mu_{\frac{3}{2},1}=\sqrt{3}$, $\nu_{\frac{5}{2},0}=\mu_{\frac{5}{2},0}=\sqrt{2}$ and $\nu_{\frac{3}{2},0}=\mu_{\frac{3}{2},0}=1$, we obtain the following families of vectors:
$$
\varphi_{\frac{1}{2},0}=\left(
                          \begin{array}{c}
                            1 \\
                            0 \\
                            \alpha \\
                            0 \\
                          \end{array}
                        \right),\quad \varphi_{\frac{3}{2},0}=\left(
                          \begin{array}{c}
                            \alpha \\
                            1 \\
                            0 \\
                            0 \\
                          \end{array}
                        \right),\quad \varphi_{\frac{5}{2},0}=\left(
                          \begin{array}{c}
                            0 \\
                            \alpha \\
                            1 \\
                            0 \\
                          \end{array}
                        \right),\quad \varphi_{\frac{3}{2},1}=\left(
                          \begin{array}{c}
                            0 \\
                            0 \\
                            0 \\
                            1 \\
                          \end{array}
                        \right),
$$
while
$$
\psi_{\frac{1}{2},0}=\frac{1}{(1+\alpha^3)}\left(
                          \begin{array}{c}
                            1 \\
                            -\alpha \\
                            \alpha^2 \\
                            0 \\
                          \end{array}
                        \right),\, \psi_{\frac{3}{2},0}=\frac{1}{(1+\alpha^3)}\left(
                          \begin{array}{c}
                            \alpha^2 \\
                            1 \\
                            -\alpha \\
                            0 \\
                          \end{array}
                        \right), \, \psi_{\frac{5}{2},0}=\frac{1}{(1+\alpha^3)}\left(
                          \begin{array}{c}
                            -\alpha \\
                            \alpha^2 \\
                            1 \\
                            0 \\
                          \end{array}
                        \right), \, \psi_{\frac{3}{2},1}=\left(
                          \begin{array}{c}
                            0 \\
                            0 \\
                            0 \\
                            1 \\
                          \end{array}
                        \right).
$$
If we fix, for concreteness, $\alpha={1}{2}$, we find in particular that the operators $S_\varphi$ and $S_\psi$ are
$$
S_\varphi=\frac{1}{4}\left(
\begin{array}{cccc}
 5 & 2 & 2 & 0 \\
 2 & 5 & 2 & 0 \\
 2 & 2 & 5 & 0 \\
 0 & 0 & 0 & 4 \\
\end{array}
\right),
\qquad S_\psi=\frac{1}{27}\left(
\begin{array}{cccc}
 28 & -8 & -8 & 0 \\
 -8 & 28 & -8 & 0 \\
 -8 & -8 & 28 & 0 \\
 0 & 0 & 0 & 27 \\
\end{array}
\right).
$$
Following Section \ref{sectIII.1}, and (\ref{38}) in particular, we get the following {\em self-adjoint version} of the system:

$$
c=\left(
\begin{array}{cccc}
 \frac{1}{9} \left(1+\sqrt{2}+\sqrt{3}-\sqrt{6}\right) & \frac{1}{9} \left(4+\sqrt{2}+2 \sqrt{3}\right) & \frac{1}{9} \left(-2+\sqrt{2}+\sqrt{6}\right) & \frac{1}{\sqrt{3}}-1 \\
 \frac{1}{9} \left(1-2 \sqrt{2}-\sqrt{3}\right) & \frac{1}{9} \left(-2+\sqrt{2}+\sqrt{6}\right) & \frac{2}{9} \left(2+2 \sqrt{2}-\sqrt{3}+\sqrt{6}\right) & \frac{1}{\sqrt{3}} \\
 \frac{1}{9} \left(1+4 \sqrt{2}-2 \sqrt{6}\right) & \frac{1}{9} \left(1+\sqrt{2}+\sqrt{3}-\sqrt{6}\right) & \frac{1}{9} \left(1-2 \sqrt{2}-\sqrt{3}\right) & 1+\frac{1}{\sqrt{3}} \\
 0 & 0 & 0 & 0 \\
\end{array}
\right),
$$
and the following o.n. basis:
$$
e_{\frac{1}{2},0}=\left(
\begin{array}{c}
 \frac{1}{3}+\frac{1}{\sqrt{3}} \\
 \frac{1}{3}-\frac{1}{\sqrt{3}} \\
 \frac{1}{3} \\
 0 \\
\end{array}
\right),\quad e_{\frac{3}{2},0}=\left(
\begin{array}{c}
 \frac{1}{3} \\
 \frac{1}{3}+\frac{1}{\sqrt{3}} \\
 \frac{1}{3}-\frac{1}{\sqrt{3}} \\
 0 \\
\end{array}
\right), \quad e_{\frac{5}{2},0}=\left(
\begin{array}{c}
 \frac{1}{3}-\frac{1}{\sqrt{3}} \\
 \frac{1}{3} \\
 \frac{1}{3}+\frac{1}{\sqrt{3}} \\
 0 \\
\end{array}
\right), \quad e_{\frac{3}{2},1}=\left(
\begin{array}{c}
 0 \\
 0 \\
 0 \\
 1 \\
\end{array}
\right).
$$
It is easy to check that $c$, together with its adjoint $c^\dagger$, satisfies (\ref{11}), and that the vectors $e_{h',k'}$ are eigenstates of $H_0=c^\dagger c+\frac{1}{2}(\1-4K)$ and $K$, with the {\em right} eigenvalues. So our original system can be mapped into a system as those described in \cite{buc}. However, in order to perform such a mapping, it should be stressed that the eigenvectors $\varphi_{h',k'}$ (or the $\psi_{h',k'}$) should be found first, to construct $S_\varphi$ and its inverse, $S_\psi$, and then their square roots. These are in fact the essential ingredients of formula (\ref{38}).

\subsection{A truncated Swanson model}

In \cite{buc} the commutation rule (\ref{11}) have been used to consider a truncated version of the harmonic oscillator, living in the finite dimensional Hilbert space $\Hil_N$, and then considering its limit for diverging $N$. In \cite{bgv} it has been discussed that the self-adjoint Hamiltonian of the oscillator produces, using similarity transformations, several non self-adjoint quadratic Hamiltonians with known spectra and eigenstates which may, or may not, form bases for the infinite-dimensional Hilbert space where the model is defined. Among these Hamiltonians, one can recover the ones for the shifted harmonic oscillator and for the Swanson model. All these systems can be described in terms of pseudo-bosonic operators. Hence it might be interesting to consider the truncated versions of this models. For instance, following \cite{baginbagbook}, we introduce
$$
H_\theta=\frac{1}{2}\left(p^2+x^2\right)-\frac{i}{2}\,\tan(2\theta)\left(p^2-x^2\right), $$ where $\theta$ is a real parameter taking value in
$I=\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\setminus\{0\}$.  We assume here that $[x,p]=i(\1-Nk)$. Hence $H_\theta$ can be rewritten as
$ H_\theta=\omega_\theta\left(B_\theta\,A_\theta+\frac{1}{2}(\1-Nk)\right), $ where $\omega_\theta=\frac{1}{\cos(2\theta)}$ is well
defined because $\cos(2\theta)\neq0$ for all $\theta\in I$, and where the operators $A_\theta$ and $B_\theta$ are defined as in \cite{baginbagbook}:
$$\left\{
\begin{array}{ll}A_\theta=\frac{1}{\sqrt{2}}\left(e^{i\theta}x+ie^{-i\theta}p\right),\\
B_\theta=\frac{1}{\sqrt{2}}\left(e^{i\theta}x-ie^{-i\theta}p\right),\end{array} \right. $$
which satisfy the commutation rule $[A_\theta,B_\theta]=\1-Nk$. Of course, if $\theta\neq0$, $H_\theta\neq H_\theta^\dagger$. The $N$ eigenstates of $H_\theta$ and  those of $H_\theta^\dagger$ can be constructed as shown previously. In particular they are also eigenstates of the projection operator $k$, and correspond to the eigenvalues $\frac{\omega_\theta}{2}$, $\frac{3\,\omega_\theta}{2}$, $\frac{5\,\omega_\theta}{2}$ and so on, depending on the explicit value of $N$. Of course, as in the infinite dimensional Swanson model, since $H_\theta$ and $H_\theta^\dagger$ have the same eigenvalues, we can imagine that they satisfy a suitable intertwining relations, and in fact we could explicitly deduce that
$$
H_\theta S_\varphi=S_\varphi H_\theta^\dagger.
$$
Notice that this is much better than we get for the infinite dimensional case, \cite{baginbagbook}. In that case, in fact, we are only able to prove that the analogous of the intertwining equation in above holds on each eigenstate of $H_\theta^\dagger$, but the set of these vectors is not a basis for the Hilbert space: it is only a $\D$-quasi basis. Here this problem does not exist since in the present situation the set of eigenstates of $H_\theta$ is surely a basis for the finite-dimensional space $\Hil_N$. So, in a sense, the truncated Swanson model is {\em better} than the original one.

Of course, it is clear that this {\em truncation} works also for many other quantum mechanical systems which are similar to the harmonic oscillator, as those listed in \cite{bgv}. In particular, for instance, it can be applied to the shifted harmonic oscillator whose Hamiltonian, $H_\beta=\frac{\beta}{2}\left(p^2+x^2\right)+i\sqrt{2}\,p$,  $\beta>0$, is manifestly non self-adjoint. In principle, then, we can expect that this kind of truncation can be relevant is applications, at least for some particular systems. For instance, a similar framework was used in connection with a biological system in \cite{bagentropy}. What we like in the operators used here is that they obey interesting, and simple, commutation relations so that they can produce more relevant mathematical and physical results.

\section{Conclusions}\label{sectV}

We have proposed a deformed version of the truncated harmonic oscillator which, in our opinion, is particularly interesting in connection with $PT$-quantum mechanics and with its relatives. We have shown how two biorthogonal bases can be constructed, using raising and lowering operators which are not necessarily related by an adjoint operation, and that these bases are eigenstates of two Hamiltonians, one the adjoint of the other, connected by suitable intertwining relations. In particular, these results extend, and improve, those found in \cite{abg}, where two biorthogonal bases were used to define two different pairs of ladder operators but where no closed commutation rule was deduced.

We have also discussed in details an example in $N=4$, and a possible application to the Swanson model, and in particular we have deduced that the basis property for its truncated, non-self-adjoint, version is satisfied.

\section*{Acknowledgements}

FB gratefully thanks B. Bagchi for useful discussions and for suggesting the reading of \cite{buc} and \cite{bagchi}. He also acknowledges support from the GNFM of Indam and from the University of Palermo.

config: arxiv


text:
\label{sec:applications}
We now move to the core of the paper and analyze the set of fixed point equations \eqref{main:fixed_point_equations_gordon}, or equivalently \eqref{main:fixed_point_equations_replicas}, leading to the generalization performances given by \eqref{main:generalization_errors},
 for common classifiers on our synthetic binary classification task. As already stressed, even though the results are valid for a wide range of separable convex loss and regularizers, we focus on estimators based on ERM with $\rL_2$ regularization $r(\vec{w}) = \lambda \|\vec{w}\|_2^2/2$, and with square loss (ridge regression) $l^{\rm square}(y,z)=\frac{1}{2}(y-z)^2$, logistic loss (logistic regression) $l^{\rm logistic}(y, z) = \log(1+\exp(-y z))$ or hinge loss (SVM) $l^{\rm hinge}(y, z) = \max\(0,1-yz\)$. In particular, we study the influence of the hyper-parameter $\lambda$ on the generalization performances and the different large $\alpha$ behavior generalization rates in the high-dimensional regime, and compare with the Bayes results.
 We show the solutions of the set of fixed point equations eqs.~\eqref{main:fixed_point_equations_replicas} in Figs.~\ref{fig:gen_error_ridge},~\ref{fig:gen_error_hinge},~\ref{fig:gen_error_logistic} respectively for ridge, hinge and logistic $\rL_2$ regressions.
  Let us mention that the analytic solution of the set of equations \eqref{main:fixed_point_equations_gordon} is provided only in the ridge case, for which its quadratic loss allows to derive and fully solve the equations (see SM.~\ref{appendix:applications:ridge}), and also for logistic and hinge losses in the regime of vanishing $\lambda \to 0$ (see SM.~\ref{appendix:applications:hinge}).
 Unfortunately, in general the set of equations has no analytical closed form expression and needs therefore to be evaluated numerically. It is in particular the case for logistic and hinge for finite $\lambda$, whose Moreau-Yosida regularization eq.~\eqref{eq:MY_reg} is, yet, analytical.
 However, note that \eqref{main:fixed_point_equations_gordon} are fixed point equations on scalar variables so that their numerical resolution posed no problem. The non-trivial part of the rigorous analysis, performed in Thm.~\ref{main:thm:gordon_fixed_points:classification}, is the reduction of the high-dimensional problem to the scalar fixed point equations.

 First, to highlight the accuracy of the theoretical predictions, we compare in Figs.~\ref{fig:gen_error_ridge}-\ref{fig:gen_error_logistic} the ERM asymptotic ($\ndim\to \infty$) generalization error with the performances of numerical simulations ($\ndim = 10^{3}$, averaged over $n_s=20$ samples) of ERM of the training loss eq.~\eqref{main:training_loss}. Presented for a wide range of number of samples $\alpha$ and of regularization strength $\lambda$, we observe a perfect match between theoretical predictions and numerical simulations so that the error bars are barely visible and have been therefore removed. This shows that the asymptotic predictions are valid even with very moderate sizes. As an information theoretical baseline, we also show the Bayes-optimal performances (black) given by the solution of eq.~\eqref{main:fixed_point_equations_bayes}.

 \paragraph{Ridge estimation---}  As we might expect the square loss gives the worst performances. For low values of the generalization, it leads to an interpolation-peak at $\alpha=1$. The limit of vanishing regularization $\lambda \to 0$ leads to the  \emph{least-norm} or \emph{pseudo-inverse} estimator $\hat{\vec{w}}_{\rm pseudo} = \(\mat{X}^\intercal\mat{X}\)^{-1}\mat{X}^\intercal \vec{y}$. The corresponding generalization error presents the largest interpolation-peak and achieves a maximal generalization error $e_{\rm g}=0.5$. These are well known observations, discussed  as early as in \cite{Kinzel96, Opper1990, krogh1992simple}, that are object of a renewal of interest under the name \emph{double descent}, following a recent series of papers \cite{Hastie19,Geiger19,geiger2020scaling,Belkin19,Mitra2019,Mei19,gerace2020generalisation,d2020double}. This double descent behavior for the pseudo-inverse is shown in Fig.~\ref{fig:gen_error_ridge} with a yellow line. On the contrary, larger regularization strengths do not suffer this peak at $\alpha=1$, but their generalization error performance is significantly worse than the Bayes-optimal baseline for larger values of $\alpha$. Indeed, as we might expect, for a large number of samples, a large regularization biases wrongly the training. However, even with optimized regularizations, performances of the ridge estimator remains far away from the Bayes-optimal performance.

 \paragraph{Hinge and logistic estimation---}  Both these losses, which are the classical ones used in classification problems, improve drastically the generalization error. First of all, let us notice that they do not display a double-descent behavior. This is due to the fact that our results are illustrated in the noiseless case and that our synthetic dataset is always linearly separable. Optimizing the regularization, our results in Fig.~\ref{fig:gen_error_hinge}-\ref{fig:gen_error_logistic} show both hinge and logistic ERM-based classification approach very closely the Bayes error. This might be an interesting message for practitioners, even though showing it in more realistic settings would be preferable.
 To offset these results, note that performances of logistic regression on non-linearly separable data are however very poor, as illustrated by our analysis of a \emph{rectangle door} teacher (see SM.~\ref{appendix:applications:rectangle}).

 \paragraph{Max-margin estimation---} As discussed in \cite{Rosset04}, both the logistic and hinge  estimator converge, for vanishing regularization $\lambda\to0$, to the \emph{max-margin} solution.  Taking the $\lambda \to 0$ limit in our equations, we thus obtain the \emph{max-margin} estimator performances. While this is not what gives the best generalization error (as can be seen in Fig.\ref{fig:gen_error_logistic} the logistic with an optimized $\lambda$ has a lower error), the max-margin estimator gives very good results, and gets very close to the Bayes-error.

\paragraph{Optimal regularization---} Defining the regularization value that optimizes the generalization as
\begin{align}
	\lambda^{\rm opt}\(\alpha\) &= \argmin_{\lambda} e_{\rm g}^{\rm erm}\(\alpha, \lambda\)\,.
\end{align}
we show in Figs.~\ref{fig:gen_error_hinge}-\ref{fig:gen_error_logistic} that both optimal values $\lambda^{\rm opt}\(\alpha\)$ (dashed-dotted orange) for logistic and hinge regression decrease to $0$ as $\alpha$ grows and more data are given. Somehow surprisingly, we observe in particular that  the generalization performances of logistic regression with optimal regularization are \emph{extremely close} to the Bayes performances. The difference with the optimized logistic generalization error is barely visible by eye, so that we explicitly plotted the difference, which is roughly of order $10^{-5}$.

Ridge regression Fig.~\ref{fig:gen_error_ridge} shows a singular behavior: there exists an  optimal value (purple) which is moreover independent of $\alpha$ achieved for $\lambda^{\rm opt} \simeq 0.5708$. This value was first found numerically and confirmed afterwards semi-analytically in SM.~\ref{appendix:applications:ridge}.

\begin{figure}
    \centering
    \begin{subfigure}[b]{\textwidth}
		\centering
		\includegraphics[scale=0.44]{fig/Comparison_Bayes_ridge_scikit_lambd_lin.pdf}
		\hfill
		\includegraphics[scale=0.44]{fig/Comparison_Bayes_ridge_lambd_log.pdf}
		\caption{Ridge regression: square loss with $\rL_2$ regularization. Interpolation-peak, at $\alpha=1$, is maximal  for the pseudo-inverse estimator $\lambda=0$ (yellow line) that reaches $e_{\rm g}=0.5$.
		}
		\label{fig:gen_error_ridge}
    \end{subfigure}
    \vskip\baselineskip
    \begin{subfigure}[b]{\textwidth}
		\centering
	\includegraphics[scale=0.44]{fig/Comparison_Bayes_hinge_lambd_opt_lin.pdf}
	\hfill
	\includegraphics[scale=0.44]{fig/Comparison_Bayes_hinge_lambd_log.pdf}
		\caption{Hinge regression: hinge loss with $\rL_2$ regularization. For clarity the rescaled value of $\lambda^{\rm opt}/10$ (dotted-dashed orange) is shown as well as its generalization error $e_{\rm g}^{\rm opt}$ (dotted orange) that is slightly below and almost indistinguishable of the max-margin performances (dashed black).}
		\label{fig:gen_error_hinge}
	\end{subfigure}
\end{figure}
\begin{figure}[p]
\ContinuedFloat
    \begin{subfigure}[b]{\textwidth}
		\centering
		\includegraphics[width=0.44\linewidth]{fig/Difference_Bayes_optimal_lambda_model=logistic_inset_lin.pdf}
		\hfill
		\includegraphics[scale=0.44]{fig/Comparison_Bayes_logistic_lambd_log.pdf}
	 \caption{Logistic regression: logistic loss with $\rL_2$ regularization - The value of $\lambda^{\rm opt}$ (dotted-dashed orange) is shown as well as its generalization error $e_{\rm g}^{\rm opt}$ (dotted orange). Visually indistinguishable from the Bayes-optimal line, their difference $e_{\rm g}^{\rm opt}-e_{\rm g}^{\rm bayes}$ is shown as an inset (dashed orange).}
	 \label{fig:gen_error_logistic}
	 \end{subfigure}
     \caption{Asymptotic generalization error for $\rL_2$ regularization ($\ndim \to \infty$) as a function of $\alpha$ for different regularizations strengths $\lambda$, compared to numerical simulation (points) of ridge regression for $\ndim=10^{3}$ and averaged over $n_s=20$ samples. Numerics has been performed with the default methods \textit{Ridge}, \textit{LinearSVC},  \textit{LogisticRegression} of \texttt{scikit-learn} package \cite{scikit-learn}. Bayes optimal performances are shown with a black line and goes as $\Theta\(\alpha^{-1}\)$, while the Rademacher complexity \cite{Abbara2019} (dashed green) decreases as $\Theta\(\alpha^{-1/2}\)$. Both hinge and logistic converge to max-margin estimator (limit $\lambda=0$) which is shown in dashed black and deceases as $\Theta(\alpha^{-1})$, while Ridge decreases as $\Theta(\alpha^{-1/2})$.
     }
     \label{fig:gen_error_global}
\end{figure}

\paragraph{Generalization rates at large $\alpha$---} Finally, we turn to the very instructive behavior at large values of $\alpha$ when a large amount of data is available. First, we notice that the Bayes-optimal generalization error, whose large $\alpha$ analysis is performed in  SM.~\ref{appendix:applications:bayes}, decreases as $e_{\rm g}^{\rm bayes} \underset{\alpha \to \infty}{\sim} 0.4417 \alpha^{-1}$. Compared to this optimal value, ridge regression gives poor performances in this regime. For any value of the regularization $\lambda$ --- and in particular for both the pseudo-inverse case at $\lambda=0$ and the optimal estimator $\lambda^{\rm opt}$ --- its generalization performances decrease much slower than the Bayes rate, and goes only as $e_{\rm g}^{\rm ridge} \!\underset{\alpha \to \infty}{\sim} \!  0.2405 \alpha^{-1/2}$ (see SM.~\ref{appendix:applications:ridge} for the derivation).

Hinge and logistic regressions present a radically different, and more favorable, behavior. Fig.~\ref{fig:gen_error_hinge}-\ref{fig:gen_error_logistic} show that keeping $\lambda$ finite when $\alpha$ goes to $\infty$, does not yield the Bayes-optimal rates. However the max-margin solution (that corresponds to the $\lambda \to 0$ limit of these estimators) gives extremely good performances $e_{\rm g}^{\rm logistic,hinge} \underset{\lambda \to 0}{\sim} e_{\rm g}^{\rm max-margin} \!\underset{\alpha \to \infty}{\sim} \! 0.500 \alpha^{-1}$ see derivation in SM.~\ref{appendix:applications:hinge}).
This is the same rate as the Bayes one, only that the constant is slightly higher.
However, we do not know whether there is a general criteria that would distinguish when the decay is $\Theta (\alpha^{-1})$ or $\Theta (\alpha^{-1/2})$. Providing such a generic criteria is definitely a line of research we would like to investigate in the future.
{\color{black}
Moreover, let us point out the work \cite{sridharan2009fast} that discusses fast convergence rates $\Theta(\nsamples^{-1})$ for the hinge loss, whose analysis for only very large $\alpha$ and Lipshitz functions does not directly apply to our setting.
}

\paragraph{Comparison with VC and Rademacher statistical bounds---} Given the fact that both the max-margin estimator and the optimized logistic achieve optimal generalization rates going as $\Theta\(\alpha^{-1}\)$, it is of interest to compare those rates to the prediction of statistical learning theory bounds. Statistical learning analysis (see e.g.~\cite{vapnik2006estimation,Bartlett98,shalev2014understanding}) relies to a large extent on the \emph{Vapnik-Chervonenkis} dimension (VC) analysis and on the so-called \emph{Rademacher complexity}. The uniform convergence result states that if the Rademacher complexity or the Vapnik-Chervonenkis dimension $d_{\rm VC}$ is finite, then for a large enough number of samples the generalization gap will vanish uniformly over all possible values of parameters. Informally, uniform convergence tells us that with high probability, for any value of the weights $\vec{w}$, the generalization gap satisfies $ {\mR}_{\rm population}(\vec{w}) - {\cal R}_{\rm empirical}^{\nsamples} (\vec{w}) = \Theta\(\sqrt{d_{\rm VC} / \nsamples}\)$ where $d_{\rm VC} = \ndim-1$ for our GLM hypothesis class. Therefore, given that the empirical risk can go to \emph{zero} (since our data are separable), this provides a generalization error upper-bound $ e_{\rm g} \!\leq\!  \Theta(\alpha^{-1/2})$.

This is much worse that what we observe in practice, where we reach the Bayes rate $e_{\rm g} = \Theta(\alpha^{-1})$. Tighter bounds can be obtained using the Rademacher complexity, and this was studied recently (using the aforementioned \emph{replica method}) in \cite{Abbara2019} for the very same problem. To bring to light interesting conclusions, we reproduced their results and plotted the Rademacher complexity generalization bound in Fig.\ref{fig:gen_error_global} (dashed-green) that decreases as $\Theta\(\alpha^{-1/2}\)$ for the binary classification task eq.~\eqref{main:teacher_sign}.

One may wonder if this could be somehow improved. Another statistical-physics heuristic computation, however, suggests that, unfortunately, uniform bound are plagued to a slow rate $\Theta\(\alpha^{-1/2}\)$. Indeed, the authors of \cite{engel1993statistical} showed with a replica method-style computation that \emph{there exists} some set of weights, in the binary classification task eq.~\eqref{main:teacher_sign}, that leads to $\Theta\(\alpha^{-1/2}\)$ rates: the uniform bound is thus tight. The gap observed between the uniform bound and the almost Bayes-optimal results observed in practice in this case is therefore not a paradox, but an illustration that the price to pay for uniform convergence is the inability to describe the optimal rates one can sometimes get in practice. Therefore, we believe, that the fact this phenomena can be observed in a such simple problem sheds an interesting light on the current debate in understanding generalization in deep learning \cite{zhang2016understanding}.

Remarking our synthetic dataset is linearly separable, we may try to take this fact into consideration to improve the generalization rate. In particular, it can be done using the max-margin  based generalization error for separable data:
\begin{theorem}[Hard-margin generalization bound \cite{vapnik2006estimation,Bartlett98,shalev2014understanding}]
	Given $S=\{\vec{x}_1,\cdots, \vec{x}_\nsamples\}$ such that $\forall \mu \in [1:\nsamples], \|\vec{x}_\mu\| \leq r$. Let $\hat{\vec{w}}$ the hard-margin SVM estimator on $S$ drawn with distribution $D$. With probability $1-\delta$, the generalization error is bounded by
	\begin{align}
		e_{\rm g}(\alpha) \underset{\alpha \to \infty}{\leq} \(4 r \|\hat{\vec{w}}\|  + \sqrt{\log\(4/\delta\) \log_2 \|\hat{\vec{w}}\|} \)/\sqrt{\nsamples}\,.
	\end{align}
\end{theorem}

In our case one has $r^2 \simeq \frac{1}{\ndim} \EE_{\vec{x}} \| \vec{x} \|_2^2 = \frac{1}{\ndim}\sum_{i=1}^\ndim \EE x_i^2 = 1$. On the other hand, in the large size limit, the  norm of the estimator $\|\hat{\vec{w}}\|_2 / \sqrt{\ndim}\simeq \sqrt{q}$, that yields $e_{\rm g}(\alpha)\leq 4 \sqrt{\frac{q}{\alpha}}$. We now need to plug the values of the norm $q$ obtained by our max-margin solution to finally obtain the results. Unfortunately, this bound turns out to be even worse than the previous one. Indeed the norm of the hard margin estimator $q$ is found to grow with $\alpha$ in the solution of the fixed point equation, and therefore the margin decay rather fast, rendering the bound vacuous. For small values of $\alpha$, one finds that $q\sim \alpha$ that provides a vacuous constant generalisation bound $e_{\rm g}\leq \Theta\(1\)$, while for large $\alpha$, $q\sim \alpha^2$ that yields an even worse bound $e_{\rm g}\leq \Theta\(\sqrt{\alpha}\)$. Clearly, max-margin based bounds do not perform well in this high-dimensional example.