Given a weighted 2 space with weights associated with an entire function, we consider pairs of weighted shift operators, whose commutators are diagonal operators, when considered as operators over a general Fock space. We establish a calculus for the algebra of these commutators and apply it to the general case of Gelfond–Leontiev derivatives. This general class of operators includes many known examples, such as classic fractional derivatives and Dunkl operators. This allows us to establish a general framework, which goes beyond the classic Weyl–Heisenberg algebra. Concrete examples for its application are provided.

It is a well-known fact that the Bargmann–Fock space arises via the Bargmann representation of the Heisenberg group. This results in a close connection between these two; see Refs. 8, 15, or 18. Moreover, this fact also explains why the classic Bargmann–Fock space is a central object in quantum physics. The importance of the Bargmann–Fock space stems from the fact that in this space the dual of the derivative operator is the multiplication operator Mz. Since these operators are Fourier (as well as Fischer) duals of each other, it provides the basis for the study of pseudo-differential operators on Fock spaces and of Toeplitz operators. Furthermore, the Lie group induced by the arising Lie algebra has some interesting consequences, such as its translation invariance.

This automatically leads to the question if we can use the same or similar algebraic methods in other settings where the duals with respect to the integral transform are different from the above. Instead of concentrating on the derivative and multiplication operators whose commutator is the identity and, therefore, immediately leads to the Heisenberg group,8 we first look at the dual pair of integration and back-shift operators. However, the commutator between the two is not the identity operator, but rather a (infinite-dimensional) diagonal operator when applied to the standard (non-normalized) basis en(z) = zn. This is an important point since in many other cases, such as fractional derivatives of Gelfond–Leontiev type, we encounter similar algebraic properties. In fact, as will be seen later, the class of Gelfond–Leontiev operators of generalized differentiation provides a general setting, which includes standard examples, such as classic fractional derivatives (such as Caputo and Riemann–Liouville derivatives, the latter by changing the ground state11), Dunkl (or difference-differential) operators,16 and the q-derivative,5,12,13,17 and, therefore, has a broad range of applications in physics. We only recall that while fractional differential operators are being applied in many areas, such as fractional mechanics, Dunkl operators appear naturally in the study of Calogero–Sutherland–Moser models for n-particle systems.16 In these cases, even the commutator of the derivative and multiplication operator Mz does not lead to the identity operator, but to a diagonal operator when applied to the standard basis. The one exception is the case of the q-derivative where we can replace the commutator by the q-commutator as an alternative. In our general case of fractional derivatives, this leads to the well-known fact that, in general, there is no finite Leibniz rule for fractional derivatives. While this seems to be a problem, it also means that the resulting Lie algebra is much richer. Therefore, we need an approach, which allows us to work with such structures where the commutator is a diagonal operator when applied to an appropriate basis. To this end, we are going to study commutator relations over generalized Fock spaces. As a principal example, we are going to look into the important case of the backward-shift operator, which is dual of the integration operator in the standard case. This allows us to consider settings that a priori are quite different, but which fit in the general analysis we are doing.

The outline of this paper is as follows. In Sec. II, we discuss the backward shift operator in the classical Fock space. In Sec. III, we study the Fock space associated with the Gelfond–Leontiev operator. A first example of the calculus on the diagonal is given in Sec. IV, where the family of spaces studied in Ref. 4 is considered. Finally, in Sec. V, we consider the general setting of pairs of weighted shift operators.

By starting with the backward-shift operator, we can calculate its dual in Fock spaces with more general weights than usually considered in the literature, but which continue to be reproducing kernel Hilbert spaces.

Let us recall the definition of the Fock space. Let $dA(z)≔1πe−|z|2dxdy,z=x+iy$, be a weighted Lebesgue measure in $C$. The classic Fock space $F1$ is the set of all entire functions $f:C↦C$ such that
$‖f‖1≔1π∫C|f(z)|2dA(z)12=∑n=0∞n!|fn|2<∞,$
where $f(z)=∑n=0∞fnzn$.
The classical Fock space $F1$ can be seen as the unique (up to a strictly positive multiplicative factor for the norm) Hilbert space of power series defined in a neighborhood of the origin for which the complex derivative and the operator Mz of multiplication by the variable z are closed operators defined on the span of polynomials and adjoint to each other there. Their commutator
$[∂,Mz]=∂Mz−Mz∂=id$
(2.1)
is equal to the identity operator id. In fact, it then follows that the elements are entire functions; see Refs. 6 and 7.
Another important operator in this context is the backward-shift operator R0 defined by
$(R0f)(z)=f(z)−f(0)z,z≠0,f′(0),z=0,$
(2.2)
for functions analytic in a neighborhood of the origin. It is easily seen (see the lemma below) that R0 is a contraction from the Fock space into itself, with its adjoint being the integration operator I defined by
$(If)(z)=∫[0,z]f(s)ds,$
(2.3)
where the functions are assumed analytic in an open convex neighborhood of the origin (and in the whole of $C$ in the case of the Fock space).

Lemma 2.1.

For the case of the classic Fock space $F1$, we have for the backward-shift operator R0 the following properties:

1. R0 is a contraction in the Fock space.

2. Its adjoint operator $R0*$ is the integration operator.

Proof.
Let $f∈F1$, that is, f is an entire function $f(z)=∑n=0∞fnzn$ with a finite $F1$-norm,
$‖f‖12=∑n=0∞|fn|2n!<∞.$
We have
$(R0f)(z)=∑n=0∞fn+1zn,$
and so
$‖R0f‖12=∑n=0∞|fn+1|2n!≤∑n=0∞|fn+1|2(n+1)!≤‖f‖12.$
Since R0 is bounded, it is enough to check that $R0*=I$ on monomials. Let $n,m∈N0$. We have
$Izn,zm1=zn+1n+1,zm1=δn+1,m(n+1)!n+1=δn+1,mn!,$
while, for m ≥ 1,
$zn,R0zm1=zn,zm−11=δn,m−1n!,$
and hence the result for m ≥ 1. For m = 0, the equality is trivial,
$⟨Izn,1⟩1=zn+1n+1,11=0,whilezn,R011=zn,01=0.$

The condition $R0*=I$, in fact, is a characterization of the Fock space (up to a strictly positive multiplicative factor for the norm). Now, the commutator of R0 and I is equal on monomials to
$[R0,I](zn)=1,n=0,−znn(n+1),n=1,2,….$
(2.4)
We will denote by D0 the (formal) diagonal operator defined by the right-hand side of (2.4), i.e.,
$D0=100⋯0⋯0−120⋯0⋯00−12⋅3⋯0⋯⋮⋮⋮ ⋮ 000⋯−1n(n+1)⋯⋮⋮⋮ ⋮ ≔diag1,−12,−12⋅3,…,−1n(n+1),…,$
(2.5)
and so
$[R0,I]=D0.$
(2.6)
It is interesting to already note that (still formally at this stage)
$TrD0=0.$
(2.7)
We remark that formula (2.4) also makes sense for such spaces different from the Fock space in which case we do not have $R0*=I$. Furthermore, (2.4) suggests that the study of the algebra generated by R0 and I also requires diagonal operators as coefficients.
We begin with the definition of generalized differentiation and integration operators with respect to a given entire function. Let the function
$φ(z)=∑n=0∞znφn$
(3.1)
be an entire function with order ρ > 0 and degree σ > 0, that is, such that $limn→∞n1ρ|φn|n=σeρ1ρ$.
Furthermore, we always assume the normalization
$φ0=1.$
(3.2)

Definition 3.1.
Assume that φn > 0 for $n∈N0$. We define the Hilbert space $H(φ)$ as the set of all entire functions endowed with the inner product,
$⟨f,g⟩φ=∑n=0∞φnfnḡn,$
(3.3)
where $f(z)=∑n=0∞fnzn$ and $g(z)=∑n=0∞gnzn$.

The inner product (3.3) induces the norm
$‖f‖φ2=∑n=0∞φn|fn|2,f∈H(φ).$
(3.4)

Proposition 3.2.
The Hilbert space $H(φ)$ is a reproducing kernel Hilbert space with the reproducing kernel
$k(z,ω)=φ(zω̄).$
(3.5)

Remark 3.3.

When the sequence φn is increasing and φ0 = 1, the space $H(φ)$ is a de Branges–Rovnyak space. See Ref. 1.

Definition 3.4.
Let φ be as in (3.1). We define the Gelfond–Leontiev (G–L) operator of generalized differentiation with respect to φ, denoted as φ, as the operator acting on function f analytic in a neighborhood of the origin, $f(z)=∑n=0∞fnzn$, by
$f(z)=∑n=0∞fnzn↦∂φf(z)=∑n=1∞fnφnφn−1zn−1.$
(3.6)

It is well known that $∂φ*=Mz$ over the space $H(φ)$. We now introduce the corresponding generalization of the integration operator I.

Proposition 3.5.
The operator R0 is densely defined and closed in $H(φ)$. It is a contraction when the sequence $(φn)n=0∞$ is non-decreasing. The adjoint of R0 is given by
$Iφzn=φnφn+1zn+1,n=0,1,2,….$
(3.7)

Proof.
The operator R0 is closed due to the fact that in a reproducing kernel Hilbert space, convergence in norm implies pointwise convergence. We consider a sequence of functions $f̃k:C↦C$ in $H(φ)$ converging in norm to $f∈H(φ)$ and such that the sequence $(R0f̃k)k=1∞$ is convergent, with limit $g∈H(φ)$. Then, for every ω ≠ 0 in the domain of analyticity of the elements of $H(φ)$,
$limk→∞f̃k(ω)=f(ω),limk→∞R0f̃k(ω)=limk→∞f̃k(ω)−f̃k(0)ω=g(ω).$
It follows that g(ω) = R0f(ω) for ω ≠ 0 and for ω = 0 by analytic continuation.
Since the sequence $(φn)n=0∞$ is non-decreasing, that is, φnφn+1 for $n∈N0$, we have
$R0f(z)=R0∑n=0fnzn=∑n=1fnzn−1,$
and hence,
$‖R0f‖φ2=∑n=1∞φn−1|fn|2≤∑n=1∞φn|fn|2≤‖f‖φ2−|f0|2.$
(3.8)
We now compute the adjoint of R0. Let $n,m∈N0$. We have
$R0zn,zmφ=zn−1,zmφ=φn−1δn−1,m,n≥1,0,n=0,$
while
$zn,Iφzmφ=zn,φmφm+1zmφ=φm+1φmφm+1δn,m+1=φmδn,m+1=φn−1δn−1,m,n≥1,0,n=0,$
so that $R0*=Iφ$.□

We note that inequality (3.8) is the structure inequality, which characterizes de Branges–Rovnyak spaces; see Ref. 2 (Theorem 3.1.2, p. 83) and Ref. 1 for further on this point.

Definition 3.6.
Let φ be as in (3.1). We define the Gelfond–Leontiev (G–L) operator of generalized integration with respect to φ, denoted as Iφ, as the operator acting on functions f analytic in a neighborhood of the origin,
$f(z)=∑n=0∞fnzn↦Iφf(z)=∑n=0∞fnφnφn+1zn+1.$
(3.9)

Example 3.7.
For φ(z) = ez, we have φn = Γ(n + 1), n = 0, 1, 2, …, so that
$∂φf(z)=∂φ∑n=0∞fnzn=∑n=1∞fnnzn−1=∂zf(z),$
(3.10)
$Iφf(z)=Iφ∑n=0∞fnzn=∑n=0∞fnn+1zn+1=If(z).$
(3.11)

Example 3.8.
When φ is the Mittag–Leffler function,
$φ(z)=E1ρ,μ(z)=∑n=0∞znΓμ+nρ,ρ>0,μ∈Cs.t.Re(μ)>0,$
(3.12)
we have that $φ(z)=E1ρ,μ(z)$ is an entire function of order ρ and type 1 [see Ref. 9 (p. 56)]. We obtain $φn=Γμ+nρ$, and operator (3.6) becomes the Dzrbashjan–Gelfond–Leontiev operator,
$∂ρ,μf(z)=∑n=1∞fnΓμ+nρΓμ+n−1ρzn−1.$
In a similar way, operator (3.9) is then
$Iρ,μf(z)=∑n=0∞fnΓμ+nρΓμ+n+1ρzn+1.$

Lemma 3.9.

The function $E1ρ,μ(zω̄)$ is positive definite if μ > 0.

This is an immediate consequence since for ρμ > 0, we have $Γ(μ+nρ)>0$ and, therefore, φn > 0 too.

Example 3.10.
A second example from a different area than the one above is the rank-one case of a Dunkl operator (also differential-difference operator) where the reflection group $G={id,σ}$ acts on $R$ by the reflection σ(x) = −x. Given a multiplicity constant $κ∈C(Re(κ)>0)$, we get the first-order rational Dunkl operator attached to $G$ and κ defined as
$Tf(x)≔f′(x)+κf(x)−f(−x)x.$
The above differential-difference operator is linked to the function φ with
$φ(z)=ezF11(κ,2κ+1;−2z).$
We obtain as coefficients (see Ref. 16)
$φ2n=(2n)!κ+12n12n and φ2n+1=(2n+1)!κ+12n+112n+1,$
(3.13)
where $(x)n≔x(x+1)⋯(x+n−1),x∈R\Z$, denotes the rising factorial. Hence, easy calculations carry
$∂φf(z)=∑n=1∞fnφnφn−1zn−1=∑n=0∞f2n+1φ2n+1φ2nz2n+∑n=1∞f2nφ2nφ2n−1z2n−1=∑n=0∞f2n+1(2n+1)!κ+12n+112n12n+1(2n)!κ+12nz2n+∑n=1∞f2n(2n)!κ+12n12n12n(2n−1)!κ+12nz2n−1=∑n=0∞f2n+1(2n+2κ+1)z2n+∑n=1∞f2n2nz2n−1=:Tf(z).$
(3.14)
As such, the rank-one operator T can be seen as a special case of the Gelfond–Leontiev operator φ. Moreover, the function $φ(zω̄)$ is positive definite since coefficients (3.13) are positive. Likewise, we obtain
$Iφf(z)=∑n=0∞fnφnφn+1zn+1=∑n=0∞f2nφ2nφ2n+1z2n+1+∑n=1∞f2n−1φ2n−1φ2nz2n=∑n=0∞f2n(2n)!κ+12n12n+112n(2n+1)!κ+12n+1z2n+1+∑n=1∞f2n−1(2n−1)!κ+12n12n12n(2n)!κ+12nz2n=∑n=0∞f2nz2n+12n+2κ+1+∑n=1∞f2n−1z2n2n.$
(3.15)

Remark 3.11.
Let q ∈ (0, 1). The case where φn is the q deformation of n!, that is,
$φn=nq!=∏j=0kjq,$
with $jq=1+q+⋯+qj−1$ for j > 0 and $0q=1$, corresponds to the case of the so-called q-calculus. We refer, in particular, to the papers5,12,13,17 for a study of the corresponding Fock spaces and commutations relations.

Let us start our discussion of properties and expressions of diagonal operators arising from commutators with the special case of the p-Fock space. These spaces have been investigated.4 For us, they are particularly interesting since they correspond to the choice of
$φ(z)=∑n=0∞zn(n!)p,$
that is, φn = (n!)p for $p∈N$ or, in other words, they represent the case of powers of the factorials.

Definition 4.1.
We define the p-Fock space $Fp,p∈N$, as the set of all entire functions $f:C↦C$ such that
$‖f‖p≔∑n=0∞|fn|2(n!)p<∞.$

Hereby, the particular case p = 1 is the classic Fock space $F1$ already mentioned in Sec. I. The inner product between two functions $f,g∈Fp$ is given by
$⟨f,g⟩=∑n=0fngn̄(n!)p, where f(z)=∑n=0∞fnzn,g(z)=∑n=0∞gnzn,$
and can also be expressed as an integral with respect to a two-dimensional positive measure; see Ref. 4 for the latter. Note that such measures are called Bargmann representations (see Ref. 5) and are of special interest when they are radial; we refer to Ref. 5 for the definition and for further information. We remark that the p-Fock space $Fp$ is a reproducing kernel Hilbert space with reproducing kernel
$kp(z,ω)=∑n=0∞znω̄n(n!)p$
since point-evaluation functionals are continuous on the corresponding Fock space.

Note that the space $F2$ was considered in Ref. 10 (Lemma 4, p. 181) and plays an important role in the theory of discrete analytic functions; see Ref. 3 for details.

In these spaces, the backward-shift operator has the following properties.

Theorem 4.2.
Let $p∈N$. Then, we have the following:
1. The backward-shift operator R0 is a contraction in $Fp$.

2. Its adjoint operator $R0*$ is given by

$R0*≔(IR0)p−1I,$
where I stands for the integration operator (2.3).

Proof.

The proof that R0 is bounded in $Fp,p=2,3,…$, follows the same lines as the proof in the classic case p = 1.

For the proof of the second statement, we first observe the action of $(IR0)p−1I$ on monomials,
$(IR0)Izn=(IR0)zn+1n+1=Iznn+1=zn+1(n+1)2,n=0,1,2,…,$
so that by induction one has
$(IR0)p−1Izn=(IR0)p−1zn+1n+1=zn+1(n+1)p.$
Hence, given $f(z)=∑n=0∞fnzn$ and $g(z)=∑n=0∞gnzn$ in $Fp$, we obtain
$⟨f,(IR0)p−1Ig⟩p=f,(IR0)p−1I∑n=0gnznp=f,∑n=1∞gn−1znnpp=∑n=1∞fngn−1(n!)pnp=∑n=0∞fn+1gn(n!)p=∑n=0∞fn+1zn,gp=⟨R0f,g⟩p.$
(4.1)
Therefore, $R0*=(IR0)p−1I$ in the Fock space $Fp$.□

Remark 4.3.

In Ref. 4, it was shown that for the multiplication operator and the derivative operator, i.e., with A = Mz and B = z, we have a similar formula for the adjoint A* = (BA)p−1B in $Fp$.

Now, we want to discuss the commutation relations in terms of diagonal operators.

By a diagonal operator, we mean a (possibly unbounded) linear operator such that
$Dzn=dnzn,n=0,1,….$
As usual, we denote such a diagonal operator by
$D=diag(d0,d1,d2,…),$
(4.2)
and we denote by $D$ the space of such operators. We remark that $D$ is a commutative ring.
For given $D∈D$, we define its forward and backward diagonal shifts as
$D(1)=diag(0,d0,d1,…),$
(4.3)
$D(−1)=diag(d1,d2,d3,…),$
(4.4)
that is to say, D(1)zn = dn−1zn (under the convention that d−1 = 0) and D(−1)zn = dn+1zn. We note that
$D(1)(−1)=D,whileD(−1)(1)=PD,∀D∈D,$
(4.5)
where P ≔ diag(0, 1, 1, 1, …).
We also recall that
$R0Izn=R0zn+1n+1=znn+1,IR0zn=0,n=0,Izn−1=znn,n=1,2,….$
(4.6)
Although neither R0 nor I belongs to $D$, we have $[R0,I]=D0∈D$, where D0 is the diagonal operator (2.5) linked to the Fock space $F1$.

This allows us to state the following lemma.

Lemma 4.4.

On the linear span of the polynomials, it holds that

1. DI = ID(−1),

2. ID = D(1)I,

3. DR0 = R0D(1), and

4. D(−1)R0 = R0D for every $D∈D$.

Proof.

Under the usual convention 0z−1 = 0, we have the following:

1. $DIzn=Dzn+1n+1=dn+1zn+1n+1=I(dn+1zn)=ID(−1)zn$,

2. $IDzn=I(dnzn)=dnzn+1n+1=D(1)(zn+1n+1)=D(1)Izn$,

3. DR0zn = Dzn−1 = dn−1zn−1 = R0(dn−1zn) = R0D(1)zn, and

4. D(−1)R0zn = D(−1)zn−1 = dnzn−1 = R0dnzn = R0Dzn for $n∈N0$.

This lemma induces us to consider diagonal operators with forward and backward shifts of order m,
$D(m)≔diag(0,…,0︸m−times,d0,d1,d2,…),D(−m)≔diag(dm,dm+1,dm+2,…),m∈N.$
This corresponds to
$D(m)zn=dn−mzn,n≥m,0,n
Since these m-shift operators, as well as R0I, IR0, are diagonal operators, they commute. Hence, in the same way as in the previous lemma, we have
$D(m)R0I=R0ID(m),D(m)IR0=IR0D(m),$
(4.7)
$D(−m)R0I=R0ID(−m),D(−m)IR0=IR0D(−m).$
(4.8)
Based on this, we now can study linear decompositions of powers of these operators where the coefficients belong to $D$. Here, we would also like to remark that the diagonal operators are closely linked to the operators Mα,q defined in Ref. 5 (Theorem 4.3).

However, we need an auxiliary lemma before stating our main results.

Lemma 4.5.
For every $D∈D$ and $n,k∈N0$, it holds that
$DnR0k=R0k(D(k))n.$
(4.9)

Proof.
For the left-hand side, we have
$DnR0kzm=Dnzm−k=(dm−k)nzm−k,m≥k,$
and zero otherwise, while for the right-hand side, we obtain
$R0k(D(k))nzm=R0k(dm−k)nzm=(dm−k)nzm−k,m≥k,$
with again zero otherwise.□

While we state the lemma in a general form of principal importance, for us will be the case of n = 1.

Theorem 4.6.
For R0 and I acting on the linear span of polynomials and D0 defined by (2.5), it holds that
$(IR0)n=∑k=1nΛk,nIkR0k,$
(4.10)
where $Λk,n∈D$ are given by the recurrence relation,
$Λk,n=Λk−1,n−1+∑l=1kD0(l)Λk,n−1,$
(4.11)
with initial values
$Λn,n=diag(1,1,1,…)=id and Λ0,n=diag(0,0,0,…),n∈N,$
(4.12)
and with
$Λk,n=diag(0,0,0,…),$
(4.13)
for remaining values k, n.

Proof.
For n = 1, we have
$(IR0)1=Λ1,1IR0,$
which is obviously true as Λ1,1 = diag(1, 1, 1, …).
Proceeding inductively, assume that at rank n,
$(IR0)n=∑k=1nΛk,nIkR0k.$
Then,
$(IR0)n+1=(IR0)nIR0=∑k=1nΛk,nIkR0kIR0.$
Using (2.6), we obtain
$IkR0kIR0=IkR0k−1(IR0+D0)R0=IkR0k−1IR02+IkR0k−1D0R0=IkR0k−1IR02+D0(1)IkR0k,$
and reiterating, we get
$IkR0kIR0=Ik+1R0k+1+∑l=1kD0(l)IkR0k,k=1,2,….$
Thus, the original formula becomes
$∑k=1nΛk,nIkR0kIR0=∑k=1nΛk,nIk+1R0k+1+∑l=1kD0(l)IkR0k=∑k=1n+1Λk−1,n+Λk,n∑l=1kD0(l)IkR0k=∑k=1n+1Λk,n+1IkR0k,$
where Λk,n+1 are defined by formula (4.11) at rank n for j, k = 1, …, n and Λn+1,n+1 is the identity diagonal. Therefore, $(IR0)n+1=∑k=1n+1Λk,n+1IkR0k$, and the result holds for all integers.□

Note that the same follows for $(R0I)n$, albeit with $D0(l)$ replaced by $D0(−l)$. To make it easier to get a clear idea, we now illustrate (4.10) and (4.11) for the cases n = 2 and n = 3.

1. Case n = 2: We have
$(IR0)2=I(IR0+D0)R0=I2R02+ID0R0=I2R02+D0(1)IR0$
so that Λ2,2 = id and $Λ1,2=D0(1)$, and (4.11) for n = 2 and k = 1 becomes
$Λ1,2=Λ0,1+D0(1)Λ1,1,$
which holds in view of (4.12)-(4.13).
2. Case n = 3: Iterating (2.6) and using the case n = 2, we now have
$(IR0)3=(IR0)2(IR0)=(I2R02+D0(1)IR0)(IR0)=I2R0(R0I)R0+D0(1)(I2R02+D0(1)IR0)=I2R0(IR0+D(1))R0+D0(1)I02R02+(D0(1))2IR0=I2(IR0+D0)R02+D0(1)I2R02+D0(1)(I02R02+D0(1)IR0)=I3R03+(D0(2)+2D0(1))I2R02+D0(1)IR0$
so that
$Λ3,3=id,Λ2,3=2D0(1)+D0(2),andΛ1,3=(D0(1))2.$
(4.14)
Equation (4.11) for n = 3 and k = 1 and k = 2, respectively, becomes
$Λ1,3=Λ0,2+D0(1)Λ1,2,Λ2,3=Λ1,2+(D0(1)+D0(2))Λ2,2,$
which are verified by (4.14) in view of (4.13)-(4.13).

We now compute the coefficients Λk,n in terms of D0.

Proposition 4.7.
For $D0(l)$ as defined above,
$Λk,n=∑|α|=n−k∏t=1k∑l=1tD0(l)αt,$
where |α| ≔ α1 + ⋯ + αk and αi are non-negative integers.

Proof.
To prove this relation, we will show that this definition fulfills the recurrence relation for Λk,n. If k = 0, there are no elements to sum, and thus, Λ0,n = 0. Now, for n ≥ 1,
$Λn,n=∑|α|=0∏t=1n∑l=1tD0(l)αt=1.$
Thus, this form satisfies the boundary conditions of our recursive relation. It is useful to introduce the notation
$Ct=∑l=1tD0(l).$
(4.15)
Thus, for $k∈N$,
$Ctn=∑l=1tD0(l)n.$
Then,
$Λk+1,n+1=Λk,n+∑l=1k+1D0(l)Λk+1,n,∑|α|=(n−1)−(k−1)∏t=1k+1Ctαt=∑α1+⋯+αk=n−k∏t=1kCtαt+Ck+1∑α1+⋯+αk+1=n−k−1∏t=1k+1Ctαt.$
Rewriting the right-hand side with explicit terms for all αi gives the following:
$∑α1+⋯+αk=n−kC1α1⋯CkαkCk+10+∑α1+⋯+αk+1=n−k−1C1α1⋯CkαkCk+1αk+1+1.$
Note that the first sum can be seen as all terms in which αk+1 = 0 for $∑t=1k+1αt=n−k$, while the second sum can be seen as all terms in which αk+1 > 0 for $∑t=1k+1αt=n−k$. Thus, the equality proposed above holds and this is a valid representation for the exact form of Λk,n.□

Proposition 4.8.
For 0 < k < n, we have the formula for the entries of the matrices Λk,n,
$(Λk,n)i,j=0 if i≠j or i=j=1,n−k+1k−1 if i=j=2,1i−1n−k∑|α|=n−k∏s=1i−2−si−1−sαs if 2k+1.$
(4.16)

Proof.
Since we have
$Λk,n=∑|α|=n−k∏t=1k∑l=1tD0(l)αt∈D,$
the non-diagonal entries will be zero. For the diagonal entries $(Λk,n)i,i$, straightforward calculations give
$∑l=1tD0(l)i,i=0 if i=1,1i−1 if 2≤i≤t+1,−t(i−1)(i−t−1) if i>t+1.$
Hence, we have $(Λk,n)1,1=0$ and
$(Λk,n)2,2=∑|α|=n−k1=n−k+1k−1.$
For the remaining entries i > 2, we obtain the following:
1. If ki − 1, then
$∏t=1k∑l=1tD0(l)αti,i=∏t=i−1k∑l=1tD0(l)αt∏t=1i−2∑l=1tD0(l)αti,i=1i−1αi−1+⋯+αk(−1)α1+⋯+αi−21(i−1)α1+⋯+αi−21α12α2⋯(i−2)αi−2(i−2)α1(i−3)α2⋯1αi−2=1i−1n−k∏s=1i−2−si−1−sαs.$
2. If k < i − 1, then
$∏t=1k∑l=1tD0(l)αti,i=(−1)α1+⋯+αk1(i−1)α1+⋯+αk1α12α2⋯kαkkα1(k−1)α2⋯1αk=1i−1n−k∏s=1k−si−1−sαs.$

For future computations, it will be convenient to introduce the shifted version of Λk,n under the map (4.4).

Proposition 4.9.
Let $Γk,n=Λk,n(−1)$. Then,
$Γk,n=∑|α|=n−k∏t=1k∑l=1tD0(l−1)αt=∑|α|=n−k∏t=1k∑l=0t−1D0(l)αt.$
(4.17)
Then,
$R0Λk,n=Γk,nR0$
(4.18)
and
$Γk,n=Γk−1,n−1+∑l=1kD0(l−1)Γk,n−1$
(4.19)
with the same boundary conditions (4.12) and (4.13) as for Λk,n.

Proof.

(4.17) follows from Proposition 4.7, while (4.18) follows from (4.4), and (4.19) is a consequence of (4.11).□

Proposition 4.10.
In terms of the entries of the matrices Γk,n (0 < k < n), we have the formula
$(Γk,n)i,j=0 if i≠j,n−k+1k−1 if i=j=1,1in−k∑|α|=n−k∏s=1i−1−si−sαs if 2≤i=j≤k,1in−k∑|α|=n−k∏s=1k−si−sαs if i=j>k.$
(4.20)

Proof.
Since we have $Γk,n=∑|α|=n−k∏t=1k∑l=0t−1D0(l)αt∈D$, the non-diagonal entries will be zero. For the diagonal entries $(Γk,n)i,i$, again straightforward calculations give
$∑l=0t−1D0(l)i,i=1i if 1≤i≤t,−ti(i−t) if i>t.$
Hence, we have again $(Λk,n)1,1=n−k+1k−1$, and for the remaining entries i ≥ 2 we obtain the following:
1. If ki, then
$∏t=1k∑l=0t−1D0(l)αti,i=∏t=ik∑l=0t−1D0(l)αt∏t=1i−1∑l=0t−1D0(l)αti,i=1iαi+⋯+αk(−1)α1+⋯+αi−1iα1+⋯+αi−11α12α2⋯(i−1)αi−1(i−1)α1(i−2)α2⋯1αi−1=1in−k∏s=1i−1−si−sαs.$
2. If k < i, then
$∏t=1k∑l=0t−1D0(l)αti,i=1iα1+⋯+αk(−1)α1(−2)α2⋯(−k)αk(i−1)α1(i−2)α2⋯(i−k)αk=1in−k∏s=1k−si−sαs.$

These considerations allow us to compute the commutator of R0 and its adjoint. Recall that D0 defined by (2.5) satisfies [R0, I] = D0. In $F1$, we have I* = R0 and so $[R0*,R0]=−D0$ in that space. We now compute this commutator in a general $Fp$ space.

Theorem 4.11.
For any $Fp$ as defined above,
$[R0,R0*]=D0p+∑k=1p(Λk+1,p+1(−1)−Λk,p)IkR0k.$

Proof.
The commutator is given by
$[R0,R0*]=R0R0*−R0*R0=R0(IR0)p−1I−(IR0)p−1IR0=R0(IR0)p−1I−(IR0)p.$
Using (4.18) and expansion (4.10) for $(IR0)n$, we have
$[R0,R0*]=R0∑k=1p−1Λk,p−1IkR0kI−∑k=1pΛk,pIkR0k=∑k=1p−1Γk,p−1R0IkR0kI−∑k=1pΛk,pIkR0k.$
Using standard commutator relations, we obtain
$R0IkR0kI=Ik+1R0k+1+(Ck+1̂+Ck̂)IkR0k+(Ck̂)2Ik−1R0k−1$
using the notation
$Ct̂=∑l=1tD(l−1).$
(4.21)
Then,
$[R0,R0*]=∑k=1p−1Γk,p−1Ik+1R0k+1+(Ck+1̂+Ck̂)IkR0k+(Ck̂)2Ik−1R0k−1−∑k=1pΛk,pIkR0k.$
Now, analyzing the first term in the first summation,
$∑k=1pΓk,p−1Ik+1R0k+1=∑k=2p+1Γk−1,p−1IkR0k=∑k=1pΓk−1,p−1IkR0k$
as Γp,p−1 = 0. Now, we compute
$Γ1,n=(Γ0,n−1+(C1̂)Γ1,n−1)=(C1̂)n−1,n=1,2,….$
The third term in the first summation is then given by
$∑k=1pΓk,p−1(Ck̂)2Ik−1R0k−1=∑k=0p−1Γk+1,p−1(Ck+1̂)2IkR0k=∑k=1p−1Γk+1,p−1(Ck+1̂)2IkR0k+Γ1,p−1(C1̂)2=∑k=1pΓk+1,p−1(Ck+1̂)2IkR0k+(C1̂)p.$
Hence, the summation becomes
$[R0,R0*]=∑k=1p((Γk−1,p−1+Γk,p−1(Ck+1̂+Ck̂)+Γk+1,p+1(Ck+1̂)2))IkR0k+(C1̂)2−Λk,p=∑k=1p((Γk,p+Γk+1,p(Ck+1̂)))IkR0k+(C1̂)2−Λk,p=∑k=1p(Γk+1,p+1−Λk,p)IkR0k+(C1̂)p.$
Therefore, since $C1̂=∑l=11D0(l−1)=D0$,
$[R0R0*]=D0p+∑k=1p(Γk+1,p+1−Λk,p)IkR0k.$

We now work in the space $H(φ)$ (see Definition 3.1), where
$⟨zn,zm⟩=φnδn,m,n=0,1,….$
(5.1)
We now consider two (possibly unbounded but closed) operators A and B on $H(φ)$,
$Azn=0,n=0,anzn−1,n≥1,$
(5.2)
$Bzn=bnzn+1,n=0,1,….$
(5.3)

Lemma 5.1.

The conclusions of Lemma 4.4 are still valid with A instead of R0 and B instead of I, namely, we have the following:

1. DB = BD(−1),

2. BD = D(1)B,

Proof.
The proofs directly follow the arguments of Lemma 4.4, and we only give the proof of the last claim. For n ≥ 1, we have
$AD(zn)=A(dnzn)=dnanzn−1=D(−1)Azn.$

The commutator $[A,B]∈D$ and will be denoted by D(a, b). We have
$(D(a,b))n=a1b0,n=0,an+1bn−anbn−1,n≥1.$
(5.4)
We will assume that A and B are densely defined and closed.

Proposition 5.2.
It holds that
$(BA)n=∑k=1nΛk,nBkAk,$
where Λk,n is defined as before, with D = D(a, b) = [A, B].

Proof.

Theorem 5.3.
The operators A and B satisfy A* = B if and only if
$an+1̄φn=bnφn+1,n=0,1,….$
(5.5)

Proof.
For m ≥ 1,
$⟨A*zn,zm⟩=⟨zn,Azm⟩=⟨zn,amzm−1⟩=am̄φnδn,m−1,$
(5.6)
while
$Bzn,zm=bnzn+1,zm=bnφn+1δn+1,m.$
For n = m + 1, we have
$an+1̄φn=bnφn+1,$
that is, (5.5).□

An immediate consequence of the above theorem is the following.

Proposition 5.4.

The set of sequences (a, b) satisfying (5.5) is a real infinite-dimensional vector space.

Remark 5.5.

We have the following:

• For A = and B = Mz, we have an = n and bn = 1 and (5.5) becomes (n + 1)φn = φn+1, that is, φn = c · n! for some c > 0, and we get the Fock space, as expected.

• For A = R0 and B = I, we have an = 1 and $bn=1n+1$ and (5.5) becomes now $φn=φn+1n+1$, which also leads to the same φn as above.

• If A = R0 and B = Mz, we have an = 1 and bn = 1 and φn = c. Here, we get the Hardy space, but we are not in the setting of entire functions anymore.

Remark 5.6.

Since we assumed the operators to be closed, A* = B is equivalent to A = B*.

Example 5.7.
A solution to (5.5) is given by
$an=φnφn−1andbn=1,$
(5.7)
i.e.,
$A=∂φandB=Mz.$
(5.8)
Another solution to (5.5) is given by
$an=1andbn=φnφn+1,$
(5.9)
i.e.,
$A=R0andB=Iφ.$
(5.10)
In the last case when bn = n!, the operator Iφ reduces to the standard integration operator.

Theorem 5.8.
Let $p∈N$, p ≥ 2. Then, the operators A and B defined satisfy A* = (BA)p−1B if and only if
$an+1̄φn=bnpan+1p−1φn+1,n=0,1,….$

Proof.
We already computed ⟨A*zn, zm⟩ in the Proof of Theorem 5.3. On the other hand, for n ≥ 1, BAzn = B(anzn−1) = anbn−1zn. Thus,
$(BA)p−1Bzn,zm=bn(BA)p−1zn+1,zm=bn(an+1bn)p−1δn+1,m.$
Thus, comparing with (5.6), we have
$am̄φnδn,m−1=bn(an+1bn)p−1δn+1,m,$
i.e., with m = n + 1,
$an+1̄φn=bnpan+1p−1φn+1,n=0,1,….$

Theorem 5.9.
Let $p∈N$, p ≥ 2. Then, the operators A and B satisfy B* = (AB)p−1A if and only if
$bn−1̄φn=anpbn−1p−1φn−1,n=1,2,….$
(5.11)

Proof.
We have
$⟨B*zn,zm⟩=⟨zn,Bzm⟩=⟨zn,bmzm+1⟩=bm̄φm+1δn,m+1.$
On the other hand, ABzn = A(bnzn+1) = an+1bnzn for $n∈N0$. Thus, for n ≥ 1,
$⟨(AB)p−1Azn,zm⟩=an⟨(AB)p−1zn−1,zm⟩=⟨an(anbn−1)p−1zn−1,zm⟩=an(anbn−1)p−1δn−1,mφn−1.$
Thus,
$bm̄φm+1δn,m+1=an(anbn−1)p−1δn−1,mφn−1.$
Setting n = m + 1, we obtain
$bn−1̄φn=anpbn−1p−1φn−1,n=1,2,….$

Thus, given such an operator A, these formulas can be used to find an associated integration operator such that the adjoint of A can be expanded in this way. Similarly, given B, one can find an associated differentiation operator for this expansion.

In this case, for A and B satisfying A* = (BA)m−1B and with D = [A, B], we have
$[A*,A]=Dm+∑k=1mΛk,m−Λk+1,m+1BkAk,$
and for those satisfying B* = (AB)m−1A and with D = [B, A], we have
$[B*,B]=Dm+∑k=1mΛk,m−Λk+1,m+1AkBk.$
An important case to consider is when D is the identity. In this case, Λk,n = Γk,n = S(n, k), and this expansion simplifies to
$[B*,B]=I+∑k=1m(k+1)S(n,k+1)AkBk,$
reproducing the results from Ref. 4 for B = Mz and A = z. The recurrence relation for these numbers can then be reformatted to yield
$(k+1)S(n,k+1)=S(n,k)−S(n+1,k+1).$
Note that if we take D to the identity operator, then Λk,n = Γk,n = S(n, k), where S(n, k) are the Stirling numbers of the second kind. The recurrence relation for these numbers can then be rewritten to yield
$(k+1)S(n,k+1)=S(n,k)−S(n+1,k+1),$
and using this in our formula above reproduces the formula from Ref. 4.
We are now looking at some concrete examples. In the case of A = R0 and B = Iφ, we have D = [R0, Iφ], which is a diagonal operator,
$D=diag(d0,d1,d2,…),$
where the diagonal entries are given by
$di=φiφi+1−φi−1φi,i≥0,$
under the convention φ−1 = 0.

This leads to the following expressions of the matrix entries for Λk,n and Γk,n.

Proposition 5.10.
In terms of the entries of both matrices Λk,n and Γk,n (0 < k < n), we have
$(Λk,n)i,j=0 if i≠j or i=j=1,n−k+1k−1φ0φ1n−k if 2=i=j,∑|α|=n−kφi−2φi−1αi−1+⋯+αk∏t=1i−2φi−2φi−1−φi−t−2φi−t−1αt if 2k+1$
(5.12)
and
$(Γk,n)i,j=0if i≠j,n−k+1k−1φ0φ1n−kif 1=i=j,∑|α|=n−kφi−1φiαi+⋯+αk∏t=1i−1φi−1φi−φi−t−1φi−tαtif 1k.$
(5.13)

Proof.
The proof follows the same lines as in Propositions 3.8 and 3.9. Again, we have both $Λk,n,Γk,n∈D$ so that all non-diagonal entries are zero. For $(Λk,n)i,i$, we have
$(Λk,n)i,i=∑|α|=n−k∏t=1k∑l=1tD(l)i,i$
so that $(Λk,n)1,1=0$.
For the remaining values, straightforward calculations give
$∑l=1tD(l)i,i=φi−2φi−1 if 2≤i≤t+1,φi−2φi−1−φi−t−2φi−t−1 if i>t+1$
so that we obtain for the Λk,n-entries the following:
1. If i = 2, then
$∏t=1k∑l=1tD(l)αti,i=φ0φ1n−k.$
2. If 2 < ik + 1, then
$∏t=1k∑l=1tD(l)αti,i=∏t=i−1k∑l=1tD(l)αt∏t=1i−2∑l=1tD(l)αti,i=φi−2φi−1αi−1+⋯+αk∏t=1i−2φi−2φi−1−φi−t−2φi−t−1αt.$
3. If i > k + 1, then
$∏t=1k∑l=1tD(l)αti,i=∏t=1kφi−2φi−1−φi−t−2φi−t−1αt.$
Therefore, the result holds true for the Λk,n-entries. In a similar way, we have for the Γk,n-entries that
$∑l=0t−1D(l)i,i=φi−1φi if 1≤i≤t,φi−1φi−φi−t−1φi−t if i>t$
so that the following holds:
1. If i = 1, then
$∏t=1k∑l=0t−1D(l)αti,i=φ0φ1n−k.$
2. If 1 < ik, then
$∏t=1k∑l=0t−1D(l)αti,i=∏t=ik∑l=0t−1D(l)αt∏t=1i−1∑l=0t−1D(l)αti,i=φi−1φiαi+⋯+αk∏t=1i−1φi−1φi−φi−t−1φi−tαt.$
3. If i > k, then
$∏t=1k∑l=0t−1D(l)αti,i=∏t=1kφi−1φi−φi−t−1φi−tαt.$

We conclude this section with an application of Proposition 5.10.

Example 5.11.
We consider the rank-one case of a Dunkl operator as in Example 3.10, which is linked to the function
$φ(z)=ezF11(κ,2κ+1;−2z)$
and with coefficients
$φ2n=(2n)!κ+12n12n and φ2n+1=(2n+1)!κ+12n+112n+1.$
As seen before [see (3.15)], we have
$φ2nφ2n+1=12n+2κ+1,φ2n−1φ2n=12n.$
Therefore, we have for the diagonal entries of Λk,n the following:
1. $(Λk,n)1,1=0$.

2. $(Λk,n)2,2=n−k+1k−111+2κn−k$.

3. If 2 < ik + 1,
• i being even, we have
$(Λk,n)i,i=∑|α|=n−kφi−2φi−1αi−1+⋯+αk∏t=1i−2φi−2φi−1−φi−t−2φi−t−1αt=∑|α|=n−k1i−1+2καi−1+⋯+αk∏l=1i/2−11i−1+2κ−1i−2lα2l−1×∏l=1i/2−11i−1+2κ−1i−1−2l+2κα2l=1i−1+2καn−k∑|α|=n−k∏l=1i/2−11−2l−2κi−2lα2l−1−2li−1−2l+2κα2l,$
• i being odd, we have

$(Λk,n)i,i=∑|α|=n−kφi−2φi−1αi−1+⋯+αk∏t=1i−2φi−2φi−1−φi−t−2φi−t−1αt=∑|α|=n−k1i−1αi−1+⋯+αk∏l=1(i−1)/21i−1−1i−2l+2κα2l−1×∏l=1(i−1)/2−11i−1−1i−1−2lα2l=1i−1αn−k∑|α|=n−k2−i+2κ1+2καi−2∏l=1(i−1)/2−11−2l+2κi−2l+2κα2l−1−2li−1−2lα2l.$
4. If i > k + 1,and
• i being even, we have for k = 2m,

$(Λk,n)i,i=∑|α|=n−k∏t=1kφi−2φi−1−φi−t−2φi−t−1αt=∑|α|=n−2m∏l=1m1i−1+2κ−1i−1−2l+2κα2l1i−1+2κ−1i−2lα2l−1=−1i−1+2κn−2m∑|α|=n−2m∏l=1m2li−1−2l+2κα2l2l+2κ−1i−2lα2l−1,$
while for k = 2m + 1, we get
$(Λk,n)i,i=∑|α|=n−k∏t=1kφi−2φi−1−φi−t−2φi−t−1αt=∑|α|=n−2m−1∏l=1m1i−1+2κ−1i−1−2l+2κα2l∏l=1m+11i−1+2κ−1i−2lα2l−1=−1i−1+2κn−2m−1∑|α|=n−2m−12m+2κ+1i−2m−2α2m+1∏l=1m2li−1−2l+2κα2l2l+2κ−1i−2lα2l−1,$
• i being odd, we have for k = 2m,

$(Λk,n)i,i=∑|α|=n−k∏t=1kφi−2φi−1−φi−t−2φi−t−1αt=∑|α|=n−2m∏l=1m1i−1−1i−1−2lα2l1i−1−1i−2l+2κα2l−1=−1i−1n−2m∑|α|=n−2m∏l=1m2li−1−2lα2l2l−2κ−1i−2l+2κα2l−1,$
while for k = 2m + 1, we get
$(Λk,n)i,i=∑|α|=n−k∏t=1kφi−2φi−1−φi−t−2φi−t−1αt=∑|α|=n−2m−1∏l=1m1i−1−1i−1−2lα2l∏l=1m+11i−1−1i−2l+2κα2l−1=−1i−1+2κn−2m−1∑|α|=n−2m−12m−2κ+3i−2m−2+2κα2m+1∏l=1m2li−1−2lα2l2l−2κ−1i−2l+2κα2l−1.$

In this paper, we provided a general framework on how to handle commutators of operators, which act differently according to the basis elements. Using our approach of working with diagonal operators, we could obtain representations of the commuting relation both as operators and action on the basis elements. This allows for a variety of avenues to expand upon these results. For our concrete formulas, the considered operators are those that shift the basis element by one power, such as multiplication and backward shift operators. Our method also allows us to consider more general shift operators.

Another avenue of extension is to study other explicit examples, such as the Hardy space. Then, $R0*=Mz$ and the commutator
$[R0,Mz]f=(R0Mz−MzR0)f=f(0)$
corresponding to
$D0=diag(1,0,0,…).$
We then leave, as observed earlier, the realm of entire functions.

In terms of applications, the obtained framework can be used not only to calculate the corresponding Lie algebras but also to extend methods and results from the case of the standard Fock space to more general situations involving Gelfond–Leontiev derivatives as considered in the paper. In particular, questions such as pseudodifferential operators or function theories based on Fischer duality can be considered.

Daniel Alpay acknowledges the Foster G. and Mary McGaw Professorship in Mathematical Sciences, which supported this research. He also thanks Professor Alain Yger for introducing him to the work of Ore.14

P. Cerejeiras and U. Kähler were supported by Portuguese funds through the CIDMA—Center for Research and Development in Mathematics and Applications, and the Portuguese Foundation for Science and Technology (“FCT–Fundação para a Ciência e a Tecnologia”) within Project Nos. UIDB/04106/2020 and UIDP/04106/2020.

Trevor Kling acknowledges the Schmid College of Science and Technology (Chapman University) for an undergraduate summer research grant.

The authors have no conflicts to disclose.

Daniel Alpay: Conceptualization (equal); Investigation (equal); Writing – original draft (equal); Writing – review & editing (equal). Paula Cerejeiras: Conceptualization (equal); Investigation (equal); Writing – original draft (equal); Writing – review & editing (equal). Uwe Kähler: Conceptualization (equal); Investigation (equal); Writing – original draft (equal); Writing – review & editing (equal). Trevor Kling: Conceptualization (equal); Investigation (equal); Writing – original draft (equal); Writing – review & editing (equal).

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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