A Dirac Delta Operator

Mathematics and Statistics 9(2): 179-187, 2021 http://www.hrpub.org DOI: 10.13189/ms.2021.090213

Juan Carlos Ferrando

Centro de Investigacion Operativa, Universidad Miguel Hernandez, E-03202 Elche, Spain

Received February 1, 2021; Revised March 18, 2021; Accepted April 9, 2021 Cite This Paper in the following Citation Styles (a): [1] Juan Carlos Ferrando, ”A Dirac Delta Operator,” Mathematics and Statistics, Vol.9, No.2, pp. 179-187, 2021. DOI: 10.13189/ms.2021.090213. (b): Juan Carlos Ferrando, (2021). A Dirac Delta Operator. Mathematics and Statistics, 9(2), 179-187. DOI: 10.13189/ms.2021.090213.

Abstract If T is a (densely deﬁned) self-adjoint opera- operator λ 7→ δ (λI − T ) at T by tor acting on a complex Hilbert space H and I stands for Z +∞ the identity operator, we introduce the delta function opera- f (T ) = f (λ) δ (λI − T ) dλ (1.1) tor λ 7→ δ (λI − T ) at T . When T is a bounded operator, −∞ then δ (λI − T ) is an operator-valued distribution. If T is unbounded, δ (λI − T ) is a more general object that still retains for each f ∈ C (R), i. e., for each real-valued continuous func- some properties of distributions. We provide an explicit repre- tion f (λ). Here dλ is the Lebesgue measure of R, but the sentation of δ (λI − T ) in some particular cases, derive vari- right-hand side of (1.1) is not a true integral. If T is a bounded ous operative formulas involving δ (λI − T ) and give several operator, we shall see at once that δ (λI − T ) must be regarded applications of its usage in Spectral Theory as well as in Quan- as a vector-valued distribution, i. e., as a continuous linear map tum Mechanics. from the space D (R) into the locally convex space L (H) of Keywords Hilbert Space, Self-adjoint Operator, Vector- the bounded linear operators (endomorphisms) on H equipped valued Distribution, Spectral Measure with the strong operator topology [10, 11], whose action on f ∈ D (R) we denote as an integral. If T is unbounded we shall see that δ (λI − T ) still retains some useful distributional-like properties. The previous equation means

Z +∞ δ (λI − T ) hy, f (T ) xi = f (λ) hy, δ (λI − T ) xi dλ (1.2) 1 The delta function −∞

The scalar delta ‘function’ λ 7→ δ (λ − a) along with its for each (x, y) ∈ D (f (T )) × H, where D (f (T )) stands for derivatives were introduced by Paul Dirac in [1], and later in the domain of the self-adjoint operator f (T ). [2, Section 15], although its definition can be traced back to Let us recall that if T is a (densely defined) self-adjoint op- Heaviside. The rigorous treatment of this object in the context erator, there is a unique spectral family {Eλ : λ ∈ R} of self- of distribution theory is due to Laurent Schwartz [6, 12]. In this adjoint operators defined on the whole of H that satisfy (i) paper we extend the definition of δ (λ − a) from real numbers Eλ ≤ Eµ and EλEµ = Eλ for λ ≤ µ, (ii) lim→0+ Eλ+x = to self-adjoint operators on a Hilbert space H. We denote by Eλx, and (iii) limλ→−∞ Eλx = 0 and limλ→∞ Eλx = x in H x ∈ H D (T ) T D (R) = lim−→ D ([−n, n]) the linear space of infinitely differen- for all . The domain of consists of those tiable complex-valued functions of compact support, equipped x ∈ H such that with the inductive limit topology. As usual in physics we shall Z +∞ 2 2 assume that the scalar product in H is anti-linear for the first |λ| d kEλxk < ∞. variable. −∞ If T is a densely defined self-adjoint operator1 on H and I In this case, the spectral theorem (cf. [8, Section 107]) and stands for the identity operator, we define the delta function the Borel-measurable functional calculus provide a self-adjoint operator f (T ) defined by 1In what follows σ (T ) will denote the spectrum of T . Recall that the residual spectrum of a self-adjoint operator T is empty, so that σ (T ) = Z +∞ σp (T ) ∪ σc (T ), where σp (T ) denotes the point spectrum (the eigenvalues) f (T ) = f (λ) dE (1.3) and σ (T ) the continuous spectrum of T . λ c −∞ 180 A Dirac Delta Operator

for each Borel-measurable function f (λ), whose domain If λ 7→ Y (λ − µ) denotes the unit step function at µ ∈ R, given by Y (λ − µ) = 0 if λ < µ and Y (λ − µ) = 1 if λ ≥ µ, Z +∞ 2 2 since Eλ = Y (λI − T ) for each λ ∈ R, formally D (f (T )) = x ∈ H : |f (λ)| d kEλxk < ∞ −∞ 0 dEλ/dλ = Y (λI − T ) . (1.7) is dense in H. Observe that if T is bounded, f (T ) need not So, from (1.6) and (1.7) we get Y 0 (λI − T ) = δ (λI − T ). be bounded. Moreover, since λ 7→ Eλ is constant on the set R \ σ (T ) of T , an open set in R, equation (1.3) tell us that Proposition 1. If T is a bounded self-adjoint operator on H f (λ) need not be defined on \ σ (T ). 1 R and f ∈ C (R), then Thanks to (1.3) the definition of δ (λI − T ) may be extended Z +∞ to Borel-measurable functions by declaring that the equation 0 0 (1.1) holds for (x, y) ∈ D (f (T )) × H and each Borel func- f (λ) δ (λI − T ) dλ = −f (T ) . −∞ tion f. But, by reasons that will become clear later, we shall restrict ourselves to those Borel functions which are continuous The same equality holds if T is unbounded but f ∈ D (R). at each point of σp (T ). Moreover, working with the real and complex parts, no difficulty arises if the function f involved If T is a self-adjoint operator and f ∈ Bp (R), then in the equation (1.1) is complex-valued (except that f (T ) is Z + ∞ Z + ∞ 2 2 no longer a self-adjoint operator whenever Imf 6= 0). Thus, |f (λ)| δ (λI − T ) dλ = |f (λ)| dEλ. unless otherwise stated, we shall assume that both in (1.1) and −∞ −∞ (1.3) the function f is complex-valued. Note that the complex where the latter equality is the definition of |f (T )|2. So, we Stieltjes measure d hEλx, yi need not be dλ-continuous. In have the following result. what follows we shall denote by Bp (R) the linear space over C consisting of all complex-valued Borel-measurable functions Proposition 2. If T is self-adjoint and f ∈ Bp (R), then of one real variable which are continuous on σp (T ). ∞ + ∞ If fn → f in D (R), the sequence {fn}n=1 is uniformly Z hf (T ) y, f (T ) xi = |f (λ)|2 hy, δ (λI − T ) xi dλ bounded and fn (x) → f (x) at each x ∈ . So, if T is R −∞ bounded on H (equivalently, self-adjoint on the whole of H) it turns out that fn (T ) → f (T ) in the strong operator topol- for every x, y ∈ D (f (T )). ogy [3, 10.2.8 Corollary]. Therefore, in this case δ (λI − T ) is an L (H)-valued distribution. Proof. We adapt a classic argument. Indeed, for every x, y ∈ As all integrals considered so far are over σ (T ), we have D (f (T )) we have Z + ∞ δ (λI − T ) = 0 ∀λ∈ / σ (T ) . (1.4) hf (T ) y, f (T ) xi = f (λ) d hf (T ) y, Eλxi . −∞ Also δ (−λI + T ) = δ (λI − T ) for all λ ∈ . On the other R E E = E µ ≤ λ hE y, xi hand, if µ ∈ σ (T ) and y is an eigenvector corresponding to Since µ λ µ whenever , and λ does not p µ the eigenvalue µ, clearly depend on , by splitting the integral we get Z + ∞ Z λ δ (λI − T ) y = δ (λ − µ) y (1.5) f (µ) d hEµy, Eλxi = f (µ) d hy, Eµxi , (1.8) −∞ −∞ for every λ ∈ R. In the particular case when Ta is the linear where clearly the first integral is hf (T ) y, Eλxi. Plugging operator defined on H by Tax = ax for a fixed a ∈ R, then Ta d hf (T ) y, Eλxi into (1.8), we are done. is a self-adjoint linear operator with σ (Ta) = σp (Ta) = {a}. In this case δ (λI − Ta) x = δ (λ − a) x for every x ∈ H, i. e., Corollary 3. Under the same conditions of the previous theo- δ (λI − Ta) = δ (λ − a) I. rem, the equality D E Since equality f (T )† y, x = hy, f (T ) xi holds for all Z + ∞ 2 2 x, y ∈ D (f (T )) and each f ∈ Bp (R), we may infer that kf (T ) xk = |f (λ)| hx, δ (λI − T ) xi dλ (1.9) −∞ hy, δ (λI − T ) xi = hδ (λI − T ) y, xi holds for every x ∈ D (f (T )). ∞ holds (in a ‘distributional’ sense) for all x, y ∈ D (T ). This Proposition 4. If T is self-adjoint and {fn}n=1 is a uniformly suggests that in certain sense δ (λI − T ) may be regarded (pos- bounded sequence in Bp (R) such that fn → f pointwise on sibly for almost all λ ∈ R) as a Hermitian operator on D (T ). R with f ∈ Bp (R), then fn (T ) x → f (T ) x for every x ∈ Let us also point out that as equation (1.1) holds for all f ∈ D (T ). D ( ), in a distributional sense we have R Proof. This is a straightforward consequence of preceding d corollary and the Lebesgue dominated convergence theo- hy, E xi = hy, δ (λI − T ) xi (1.6) dλ λ rem. Mathematics and Statistics 9(2): 179-187, 2021 181

This proposition holds in particular if fn → f in D (R). If we denote by L (H) the linear space of all linear endo- Hence, even in the unbounded case, δ (λI − T ) behaves as a morphisms on H, the next theorem summarize some previous vector-valued distribution-like object. results. Proposition 5. Let (λ, µ) 7→ g (λ, µ) be a function deﬁned on Theorem 7. If T is a densely deﬁned self-adjoint operator on 2 R such that g (λ, ·) ∈ L1 (R) for every λ ∈ R and g (·, µ) ∈ a Hilbert space H, there is an L (H)-valued linear map δT on Bp (R) for every µ ∈ R. If the parametric integral Bp (R), whose action on f ∈ Bp (R) we denote by Z + ∞ Z + ∞ f (λ) = g (λ, µ) dµ hδT , fi = f (λ) δ (λI − T ) dλ, −∞ −∞ is continuous on R and makes sense if we replace λ by a self- such that hδT , fi = f (T ). If {fn} ⊆ Bp (R) is uniformly adjoint operator T , the value of the integral bounded and fn (t) → f (t), with f ∈ Bp (R), for all t ∈ R Z +∞ Z +∞ then hδT , fni x → hδT , fi x for all x ∈ H. If T is bounded, g (λ, µ) δ (λI − T ) dµ dλ δT is an L (H)-valued distribution, so hδT , fi is a bounded −∞ −∞ operator on H. In addition δ (λI − T ) = 0 if λ∈ / σ (T ) and does not depend on the integration ordering. hy, δ (λI − T ) xi = hδ (λI − T ) y, xi for x, y ∈ D (T ).

Proof. Since g (·, µ) ∈ Bp (R) for every µ ∈ R., one has Z +∞ 2 Explicit form of δ (λI − T ) g (T, µ) = g (λ, µ) δ (λI − T ) dλ, −∞ If Q is a vector-valued distribution, the Fourier transform which implies of Q is deﬁned as the vector valued distribution FQ on S (R) hFQ, fi = hQ, Ffi F −1 Z +∞ Z +∞ such that . As usual, we denote by f (T ) = g (λ, µ) δ (λI − T ) dλ dµ. the inverse Fourier transform. −∞ −∞ Theorem 8. If T is a self-adjoint operator, the identity On the other hand, by the deﬁnition of δ (λI − T ) we have 1 Z + ∞ Z +∞ Z +∞ δ (λI − T ) = eit(λI−T )dt (2.1) f (T ) = g (λ, µ) dµ δ (λI − T ) dλ, 2π −∞ −∞ −∞ holds for every λ ∈ R, and the action f (T ) of δ (λI − T ) on for (x, y) ∈ D (T ) × H. So, the proposition follows. f ∈ S (R) is given by Theorem 6. If T is a self-adjoint operator on H, then Z +∞ Z +∞ f (λ) it(λI−T ) +∞ f (T ) = e dλ dt. Z −∞ −∞ 2π f (λ) δ λI − T 2 dλ = (1.10) 0 Proof. Setting δT (λ) = δ (λI − T ) observe that Z +∞ 1 n √ √ o √ δ λI − T − δ λI + T f (λ) dλ 1 −itT (FδT )(t) = √ e . 0 2 λ 2π if λ > 0 and f ∈ B ( ), both members acting on D T 2. p R Indeed, if f ∈ S (R) we have Proof. First note that T 2 ≥ 0. Hence σ T 2 ⊆ [0, + ∞), Z + ∞ which implies that δ λI − T 2 = 0 if λ < 0. Since T 2 is a hFδT , fi = hδT , Ffi = (Ff)(λ) δ (λI − T ) dλ −∞ f ∈ B ( ) self-adjoint operator, for p R we have Z + ∞ 1 −itT Z +∞ = (Ff)(T ) = √ f (t) e dt. f (λ) δ λI − T 2 dλ = f T 2 2π −∞ 0 Consequently On the other hand, it is clear that −1 1 −itT Z +∞ f (λ) √ Z +∞ δT = F √ e . (2.2) √ δ λI − T dλ = f µ2 δ (µI − T ) dµ 2π 0 2 λ 0 Functionally, the action of δT on f ∈ S (R) by means of equa- whereas, using that δ (−µI + T ) = δ (µI − T ), we have tion (2.2) becomes Z +∞ √ Z 0 f (λ) 2 1 −itT −1 − √ δ λI + T dλ = f µ δ (µI − T ) dµ hδT , fi = √ e , F f (t) (2.3) 0 2 λ −∞ 2π So, the right-hand side of (1.10) coincides with Consequently, we have Z +∞ Z +∞ Z +∞ 2 2 f (µ) it(µI−T ) f µ δ (µI − T ) dµ = f T hδT , fi = e dµ dt −∞ −∞ −∞ 2π since µ 7→ f µ2 is a Borel function. with the order of the integration as stated. 182 A Dirac Delta Operator

Corollary 9. If e−itT x = x (t), for x ∈ D (T ) one has Example 11. The spectral family of the (up to a sign) one- dimensional Quantum Mechanics momentum operator of the 1 Z +∞ free particle P = iD, where Dϕ = ϕ0, acting on the Hilbert δ (λI − T ) x = eiλtx (t) dt 2π −∞ space H = L2 (R) is given by Z +∞ iλ(s−x) and if x ∈ D (f (T )) and f ∈ S (R), then 1 1 e (Eλϕ)(x) = ϕ (x) + p.v. ϕ (s) ds 2 2πi −∞ s − x Z +∞ Z +∞ 1 iλt f (T ) x = f (λ) e dλ x (t) dt. for every regular compactly supported ϕ ∈ D (P ). 2π −∞ −∞ Proof. As is well-known P is a self-adjoint operator with Remark 10. Consider the one-parameter unitary group 2,1 D (P ) = H (R) and σc (P ) = R. Since {U (t): t ∈ R} generated by the self-adjoint operator T , that is, U (t) = exp (−itT ) for every t ∈ R. If F denotes the e−itP ϕ (x) = etDϕ (x) = ϕ (x + t) Fourier transform, equation (2.2) can be written as for a regular enough ϕ ∈ D (P ), by Corollary 9 we have 1 δ (λI − T ) = √ F −1 (U)(λ) . (2.4) 2π 1 Z +∞ {δ (µI − P ) ϕ} (x) = eiµtϕ (x + t) dt. 2π So, equation (2.3) reads as −∞ Note that the integral of the right-hand side does exist because 1 Z +∞ f (T ) = √ F −1f (t) U (t) dt. (2.5) ϕ has compact support. 2π −∞ According to the definition of Eλ for the continuous spectrum and keeping in mind the order of integration as indicated In what follows we shall compute the spectral family in Corollary 9, one has {Eλ : λ ∈ R} for some useful self-adjoint operators of Quan- Z + ∞ Z + ∞ tum Mechanics by means of the delta δ (λI − T ). Nonetheless, 1 iµt {Eλϕ} (x) = Y (λ − µ) e ϕ (x + t) dµ dt. although Eλ = Y (λI − T ), the identification 2π −∞ −∞ Z + ∞ So, since Y (λI − T ) = Y (λ − µ) δ (µI − T ) dµ −∞ 1 Z + ∞ √ Y (λ − µ) eiµtdµ = F (Y )(t) · eiλt, might be not well-defined because µ 7→ Y (λ − µ) has a jump 2π −∞ µ = λ λ ∈ σ (T ) x discontinuity at . Indeed, if p and is an bearing in mind the distributional relation eigenvector corresponding to λ, then rπ 1 1 Z + ∞ (Z λ ) F (Y )(t) = δ (t) + p.v. , (2.8) Y (λ − µ) δ (µI − T ) x dµ = δ (µ − λ) dµ x 2 iπ t −∞ −∞ we get and the right-hand integral makes no sense (see [4] for a useful 1 1 Z +∞ eiλ(s−x) discussion). If λ∈ / σp (T ) we define {Eλϕ} (x) = ϕ (x) + ϕ (s) ds 2 2πi −∞ s − x Z + ∞ where the last integral must be understood in Cauchy’s princi- Eλ = Y (λ − µ) δ (µI − T ) dµ (2.6) −∞ pal value sense.

If λ belongs to σp (T ), then (µ 7→ Y (λ − µ)) ∈/ Bp (R). In Example 12. The spectral family of the one-dimensional order to define Eλ we enlarge a little the interval of integration Quantum Mechanics kinetic energy term of the free particle, by considering the integral corresponding to the Laplace operator T = −D2 on H = 2 00 L2 (R), where D ϕ = ϕ , is given by Z λ+ δ (µ − λ) dµ 1 Z +∞ cos (λ (s − x)) − 1 −∞ (Eλϕ)(x) = p.v. ϕ (s) ds iπ −∞ s − x for small > 0. So, if λ ∈ σp (T ) we define for λ > 0 and Eλ = 0 whenever λ < 0, where ϕ is a regular Z + ∞ function with compact support belonging to D (T ). Eλ = lim Y (λ + − µ) δ (µI − T ) dµ. (2.7) →0+ −∞ Proof. In this case T is a self-adjoint operator with σ (T ) = [0, +∞). Since T = (iD)2, according to (1.10) we have The limit is well-defined since lim→0+ Eλ+ = Eλ pointwise on H. In the particular case when λ belongs to σd (T ), the 1 n √ √ o δ (λI − T ) = √ δ λI − iD − δ λI + iD discrete part of σp (T ), λ is isolated in σp (T ). 2 λ Mathematics and Statistics 9(2): 179-187, 2021 183

−1/2 −inx regarded as a functional on S (R) through dλ-integration over Z}, with ϕn (x) = (2π) e , are the solutions of the [0, +∞). Plugging eigenvalue problem iϕ0 = λϕ with ϕ (−π) = ϕ (π). So, for L2 P +∞ ϕ ∈ D (S) we have ϕ = cnϕn with 1 Z n∈Z (δ (µI ∓ iD) ϕ)(x) = eiµtϕ (x ± t) dt 2π −∞ Z π 1 inx cn = hϕ, ϕni = ϕ (x) e dx into the previous expression and keeping in mind the correct 2π −π order of integration, we see that for every n ∈ . Since σ (S) = σd (S), recalling the definition ∞ Z Z of the operator E for λ ∈ σ (S), clearly we have f (λ)(δ (λI − T ) ϕ)(x) dλ = λ d 0 √ Z + ∞ Z ∞ Z +∞ i λt 1 e (Eλϕ)(x) = lim Y (λ + − µ)(δ (µI − S) ϕ)(x) dµ f (λ) √ [ ϕ (x + t) − ϕ (x − t)] dλ dt + →0 −∞ 4π 0 −∞ λ for every λ ∈ . So, the fact that E is a bounded operator for every f ∈ S (R). By the definition of Eλ if λ > 0 and the R λ fact that δ (µI − T ) = 0 whenever µ < 0, we have yields X E ϕ = c E ϕ Z + ∞ λ n λ n n∈ (Eλϕ)(x) = Y (λ − µ)(δ (µI − T ) ϕ)(x) dµ. Z 0 Using that δ (µI − S) e−inx = δ (µ − n) e−inx and that Working out the penultimate integral with µ instead of λ and Y (λ + 0 − n) = Y (λ − n), we get f (µ) = Y (λ − µ), we obtain √ −inx + ∞ +∞ i µt X cn −inx X e Z Z e (Eλϕ)(x) = √ Y (λ − n) e = √ . Y (λ − µ) √ [ ϕ (x + t) − ϕ (x − t)] dµ dt 2π 2π n∈ n∈ , n≤[λ] 0 −∞ µ Z Z ( √ ) Z +∞ Z λ ei µt = √ dµ [ ϕ (x + t) − ϕ (x − t)] dt Remark 14. Since in the previous example S is bounded on −∞ 0 µ H = L2 [−π, π], the delta operator δ (λI − S) should be re- √ garded as a continuous endomorphism as well. In this case for λ > 0. So, by setting u = µ we get X Z +∞ Z λ δ (λI − S) ϕ = cn δ (λ − n) ϕn. dt iut (Eλϕ)(x) = [ ϕ (x + t) − ϕ (x − t)] e du. n∈Z −∞ 2π 0 Example 15. The one-dimensional Quantum Mechanics po- Now we have sition operator on L2 (R). This operator is defined on H = λ 1 Z L2 (R) by (Qϕ)(x) = xϕ (x) for every x ∈ R. Clearly √ eiutdu = 1 − eiλt F −1 (Y )(t) , σc (Q) = and ϕ ∈ D (Q) if (x 7→ x ϕ (x)) ∈ L2 ( ). More- 2π 0 R R over, it is clear that so, using that F −1 (Y (v)) = F (1 − Y (v)) (t) as well as equation (2.8), we get {exp (it (λI − Q)) ϕ} (x) = ei(λ−x)tϕ (x) .

1 Z λ rπ 1 1 √ eiutdu = 1 − eiλt δ (t) − p.v. So we have 2π 0 2 iπ t (δ (λI − Q) ϕ)(x) = δ (λ − x) ϕ (x) . which implies Hence, in this case we can write (Eλϕ)(x) = 1 Z +∞ ϕ (s) 1 Z +∞ cos (λ (s − x)) Z +∞ − ds + ϕ (s) ds {Eλϕ} (x) = Y (λ − µ) δ (µ − x) ϕ (x) dµ πi −∞ s − x πi −∞ s − x −∞ where the integrals are understood in Cauchy’s principal value Therefore, if λ 6= x we get sense. Example 13. Spectral family of the (up to a sign) one- {Eλϕ} (x) = Y (λ − x) ϕ (x) . dimensional Quantum Mechanics momentum operator S for Example 16. Explicit form of δ (λI − M) for the Hermitian a bounded particle on H = L2 [−π, π] with domain matrix of H = C3 0 {ϕ ∈ L2 [−π, π]: ϕ ∈ L2 [−π, π] , ϕ (−π) = ϕ (π)}  0 1 1  As is well-known this is a self-adjoint operator with discrete M =  1 0 1  . spectrum σ (S) = Z whose eigenfunction system {ϕn : n ∈ 1 1 0 184 A Dirac Delta Operator

−1 Proof. In this case M = PJM P with σ (M) = {−1, 2} If T is an unbounded self-adjoint operator then D (T ) 6= H and and D (T n) becomes smaller as n grows. So, the following result, makes sense only if the operator T is bounded.  −1 0 0   1 1 1  JM =  0 −1 0  ,P =  −1 0 1  Theorem 18. In general, if T is a bounded self-adjoint opera- 0 0 2 0 −1 1 tor, one has

∞ (n) Using (2.1) we get X n δ (λ) δ (λI − T ) = (−1) T n (2.9) n!  δ (λ + 1) 0 0  n=0 δ (λI − M) = P 0 δ (λ + 1) 0 P −1   which is the Taylor series of δ (λI − T ) at λI. 0 0 δ (λ − 2) Proof. Developing the operator function exp (itT ), which is Let us compute the spectral family and the projection operator well-defined by the spectral theorem, we get onto the eigenspace ker (M + I). Clearly ∞ n   1 Z +∞ X (it) Y (λ + 1) 0 0 δ (λI − T ) = e−iλt T ndt, −1 2π n! Eλ = P  0 Y (λ + 1) 0  P −∞ n=0 0 0 Y (λ − 2) so that, formally interchanging the sum and the integral, we for every λ ∈ R. If λ1 = −1, the orthogonal projection Pλ1 may write onto ker (I + M) is ∞ n 1 X F{(it) } (λ) n  1 0 0   2 −1 −1  δ (λI − T ) = √ T . 1 1 2π n! P = P 0 1 0 P −1 = −1 2 −1 n=0 λ1 3   3   0 0 0 −1 −1 2 Using the fact that √ P = E − E = E n n (n) since λ1 λ1 λ1−0 λ1 . F{(it) } (λ) = (−1) 2π δ (λ) Example 17. Consider a compact self-adjoint operator K act- for every n ∈ , we obtain (2.9). ing on a separable Hilbert space H which does not admit the N eigenvalue zero. Let {ui : i ∈ N} be a Hilbert basis of H with its corresponding sequence of real eigenvalues {λi : i ∈ N}, 3 The resolvent operator and where |λi+1| ≤ |λi| for every i ∈ N. Let us compute the action of the operator (λI − K)−1 on any x ∈ H and the operator δ (λI − T ) δ (λI − T ). Recall that the spectrum σ (T ) of a (densely defined) self- P∞ H Proof. If x ∈ H, we can write x = i=1 hx, uii ui. Since adjoint operator on a complex Hilbert space is a closed (λI − K)−1 is a bounded operator whenever λ∈ / σ (K), we subset of C contained in R (see for instance [9, 3.2]). If have z ∈ C \ σ (T ), i. e., if z is a regular point of T , and ∞ Z + ∞ −1 −1 X 1 R (z, T ) = (zI − T ) (λI − K) x = hx, uii δ (µI − K) ui dµ −∞ λ − µ i=1 denotes the resolvent operator of T at z (see [7, Definition −1 so we obtain the classic series 8.2]), the function λ 7→ (z − λ) is continuous on σ (T ). The resolvent is well-defined over H, so it is a bounded normal ∞ −1 X 1 z ∈ \ σ (T ) R (z, T ) (λI − K) x = hx, u i u . operator. If R then is even self-adjoint. λ − µ i i From (1.1) it follows that i=1 i Z +∞ 1 For the solution of the equation (I − zK) x = y with z ∈ C R (z, T ) = δ (λI − T ) dλ we get the Schmidt series −∞ z − λ

∞ −1 X 1 which is the integral form of the resolvent of T . So, by con- x = (I − zK) y = hy, uii ui sidering the complex-valued function f (λ) = (z − λ)−1 with 1 − zµi i=1 z ∈ C \ σ (T ) and using the fact that whenever z−1 ∈/ σ (T ). On the other hand, since δ (λI − K) 1 √ acts on H as a continuous endomorphism, equation F −1 (t) = 2π ieiztY (t) λ − z ∞ X δ (λI − K) x = hx, uii δ (λ − µi) ui. then, according to (2.5), for Imz > 0 we have i=1 Z ∞ −1 izt holds for every x ∈ H. (zI − T ) = −i e U (t) dt. 0 Mathematics and Statistics 9(2): 179-187, 2021 185

From here, it follows that Since in the sense of distributions −1 1 1 1 R (z, iT ) = iR (iz, −T ) = LU (z) − → δ (λ − µ) 2πi λ − i − µ λ + i − µ if Imz > 0, where L is the Laplace transform. This is the Hille-Yosida theorem which relates the resolvent with the one- as → 0+, we have parameter group of unitary transformations {U (t): t ∈ R} 1 Z Z T f (λ) 2 dλ → f (λ) δ (λ − µ) dλ generated by the self-adjoint operator . π (λ − µ) + 2 If T is a bounded self-adjoint operator, γ is a closed Jordan R R + contour that encloses σ (T ) and f (z) is holomorphic inside as → 0 . Hence, if gn is defined by the left-hand side the connected region surrounded by the path γ, the Dunford µ-parametric integral with = 1/n, then gn → f point- integral formula asserts that wise on R. So, if f ∈ D (R) and T is bounded (hence with σ (T ) compact), as can be easily checked {g }∞ is 1 Z n n=1 f (z) R (z, T ) dz = f (T ) . a uniformly bounded sequence of continuous functions, with 2πi γ sup kg k ≤ kfk , that converges pointwise on to f. n∈N n ∞ ∞ R In [13] is pointed out that (2πi)−1 R (z, T ) can be considered Thus, by [3, 10.2.8 Corollary] one has gn (T ) → f (T ) in the as the indicatrix of a vector-valued distribution with values in strong operator topology, that is ( ) L (H). Dunford integral formula is easily obtained by using 1 Z +∞ Z +∞ the δ (λI − T ) operator since, if we apply the Proposition 5 f (λ) 2 dλ δ (µI − T ) dµ −1 π (λ − µ) + 2 with g (λ, µ) = f (z (µ)) (z (µ) − λ) , where z (µ) = γ (µ) −∞ −∞ and 0 ≤ µ ≤ 1, then Z +∞ → f (µ) δ (µI − T ) dµ Z Z +∞ Z f (z) −∞ f (z) R (z, T ) dz = dz δ (λI − T ) dλ + γ −∞ γ z − λ as → 0 in the strong operator topology of L (H). Therefore, Z +∞ if T is bounded and f ∈ D (T ) then = 2πi f (λ) δ (λI − T ) dλ = 2πif (T ) . Z +∞ f (λ) n o −∞ ((λ − i) I − T )−1 − ((λ + i) I − T )−1 dλ Example 19. Derivation of the orthogonal projection operator −∞ 2πi onto ker (M + I) of the Hermitian matrix M of the Example goes to f (T ) as → 0+. This proves that for bounded T 16 by the resolvent technique. We must compute 1 −1 −1 1 Z lim ((λ − i) I − T ) − ((λ + i) I − T ) P = R (z, M) dz. →0+ 2πi λ1 2πi |z+1|=1 coincides with δ (λI − T ) as an L (H)-valued distribution. Clearly, we have  z − 1 1 1  1 5 Unitary equivalence of δ (λI − T ) R (z, M) = 1 z − 1 1 . z2 − z − 2   1 1 z − 1 Theorem 20. If T is a self-adjoint operator defined on the Using that whole of H, there exist a finite measure µ on the Borel sets of the compact space σ (T ) and a linear isometry U from Z {1, z − 1} 2πi 4πi dz = − , L2 (σ (T ) , µ) onto H such that |z+1|=1 (z + 1) (z − 2) 3 3 U −1δ (λI − T ) U = δ (λI − Q) we reproduce the result we got earlier. where (Qϕ)(x) = x ϕ (x) is the position operator. 4 The δ (λI − T ) operator as a limit Proof. According to [5] there exist a finite measure µ on the Borel sets of the compact space σ (T ) and a linear isometry U −1 from L (σ (T ) , µ) onto H such that As µ 7→ (λ ± i − µ) is continuous, for self-adjoint T 2 U −1TU ϕ = Qϕ ((λ − i) I − T )−1 − ((λ + i) I − T )−1 −1 Z +∞ 1 1 for every ϕ ∈ L2 (σ (T ) , µ). So, since U TU is a self- = − δ (µI − T ) dµ. adjoint operator on L2 (σ (T ) , µ), we have −∞ λ − i − µ λ + i − µ U −1δ (λI − T ) U = δ λI − U −1TU = δ (λI − Q) If f ∈ D (R), Proposition 5 yields as stated. Z +∞ f (λ) n o ((λ − i) I − T )−1 − ((λ + i) I − T )−1 dλ −∞ 2πi Remark 21. For such linear isometry U the equation ( ) 1 Z +∞ Z +∞ U −1δ (λI − T ) U ϕ (x) = δ (λ − x) ϕ (x) = f (λ) dλ δ (µI − T ) dµ. 2 2 π −∞ −∞ (λ − µ) + holds for every ϕ ∈ L2 (σ (T ) , µ). 186 A Dirac Delta Operator

6 Commutation relations and integrating by parts, it follows that Z +∞ Let S and T be two self-adjoint operators deﬁned on the λ [δ (λI − S) , δ (µI − T )] dλ whole of H for which equations [S, [S, T ]] = [T, [S, T ]] = 0 −∞ hold. In this case i Z +∞ = − [S, T ] seis(µI−T )ds. 2π [−itS, [−itS, −isT ]] = it2s [S, [S, T ]] = 0 −∞ Observe that a second application of equation (6.1) and a sec- and the Baker-Campbell-Hausdorff formula yields ond integration by parts yield δ (λI − S) δ (µI − T ) = Z +∞ λµ [δ (λI − S) , δ (µI − T )] dλ dµ = 1 ZZ ei(tλ+sµ)e−itSe−isT dt ds = −∞ 2 +∞ +∞ (2π) 2 i Z Z R − [S, T ] se−isT µeisµdµ ds = 1 ZZ −st 2π i(tλ+sµ) 2 [S,T ] −i(tS+sT ) −∞ −∞ 2 e e e dt ds. (2π) 2 Z +∞ R [S, T ] δ (s) {1 − isT } e−isT ds = [S, T ] Likewise, since [T, [T,S]] = [S, [T,S]] = 0 one has −∞ as expected. δ (µI − T ) δ (λI − S) =

1 ZZ −st i(tλ+sµ) 2 [T,S] −i(tT +sS) 2 e e e dt ds 7 A remark on the Stone formula 2 (2π) R 1 ZZ st i(tλ+sµ) 2 [S,T ] −i(tS+sT ) Let T be a self-adjoint operator densely defined on a Hilbert = 2 e e e dt ds. 2 (2π) R space H. If A is a Borel set in σ (T ), defining So, using that Z +∞ E (A) := χA (λ) dEλ (7.1) −∞ ist ist ist exp i [S, T ] −exp − i [S, T ] = 2i sin [S, T ] 2 2 2 where χA stands for the characteristic function of A (which is a bounded Borel function), then E is an L (H)-valued finitely we have additive and pointwise countably additive measure (i.e., count- able additivity under the strong operator topology of L (H)) on [δ (λI − S) , δ (µI − T )] = the σ-algebra A of Borel subsets of σ (T ). So, if the charac- i ZZ ist teristic function χA of A with respect to R is continuous on sin [S, T ] ei(tλ+sµ)e−i(tS+sT ) dt ds. 2 σ (T ) then 2π 2 2 p R Z +∞ For position Q and momentum P of a one-dimensional par- E (A) = χA (λ) δ (λI − T ) dλ. ticle, one has H = L2 (R) and [Q, P ] = i~I. Therefore −∞ [Q, [Q, P ]] = [P, [Q, P ]] = 0 and For −∞ < a < b < ∞ and > 0, we have

[δ (λI − Q) , δ (µI − P )] = Z b Z +∞ 1 1 i ZZ st − δ (µI − T ) dµ dλ i(tλ+sµ) −i(tQ+sP ) a −∞ λ − i − µ λ + i − µ 2 sin − ~ e e dt ds. 2π 2 2 ( ) R Z +∞ Z b 2i dλ = δ (µI − T ) dµ = 2 2 According to Theorem 8, if [δ (λI − S) , δ (µI − T )] acts on −∞ a (λ − µ) + f (λ) = λ, formally we have Z b − µ a − µ 2i arg tan − arg tan δ (µI − T ) dµ. Z +∞ R λ [δ (λI − S) , δ (µI − T )] dλ = + −∞ If the limit as → 0 the bracketed function is equal to 0 if i ZZ ist µ ∈ R \ [a, b], equal to π if a < µ < b and equal to π/2 if sin [S, T ] 2 µ ∈ {a, b}. So, if a, b∈ / σp (T ) so that χ and χ both 2π 2 2 (a,b) [a,b] R B ( ) Z +∞ belong to p R , setting λeitλ dλ eisµe−i(tS+sT ) dt ds. Z b −1 −∞ 1 2in dλ gn (µ) := π (λ − µ)2 + n−2 So, using the distributional equality a for each n ∈ and Z +∞ N itλ 2π 0 λe dλ = δ (t) (6.1) f (µ) := χ (µ) + χ (µ) , −∞ i (a,b) [a,b] Mathematics and Statistics 9(2): 179-187, 2021 187 then g (µ) → f (µ) for every µ ∈ and sup kg k ≤ 1 DOI: 10.1119/1.19283 n R n∈N n ∞ which, according to Proposition 4, implies that gn (T ) x → f (T ) x for every x ∈ D (T ). In other words [5] Halmos, P., What does the spectral theorem say? Amer. Math. Monthly 70, 241–247, (1963). Z b 1 −1 −1 lim (λ − i − T ) − (λ + i − T ) dλ DOI: 10.1080/00029890.1963.11990075 →0+ 2πi a Z +∞ 1 = χ(a,b) + χ[a,b] δ (µI − T ) dµ, [6] Horvath,´ J., Topological Vector Spaces and Distributions, 2 −∞ Vol. I, Addison-Wesley, Reading, Massachusetts, 1966. holds pointwise on the domain D (T ) of T . Hence, by virtue of (7.1) we get [7] Moretti, V., Spectral Theory and Quantum Mechanics, 2nd Edition, Unitext 110, Springer, 2013. 1 Z b lim (λ − i − T )−1 − (λ + i − T )−1 dλ = + [8] Riesz, F., Nagy, B. Sz.-, Functional Analysis, Dover Pub- →0 2πi a 1 1 1 1 lications, Mineola, 1990. E ((a, b)) + E ([a, b]) = E (a, b) + E (a) + E (b) 2 2 2 2 [9] Schmudgen,¨ K., Unbounded Self-adjoint Operators on which is Stone’s formula. Hilbert Space, Springer, Dordrecht, 2012.

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