Pseudo-bosons, Riesz bases and coherent states 0 1 0 2 n a J F. Bagarello 7 Dipartimento di Metodi e Modelli Matematici, Facolta` di Ingegneria, ] Universit`a di Palermo, I-90128 Palermo, Italy h p e-mail: [email protected] - h t a m [ 1 v 6 Abstract 3 1 1 In a recent paper, Trifonov suggested a possible explicit model of a PT-symmetric system . 1 based on a modification of the canonical commutation relation. Although being rather 0 0 intriguing, in his treatment many mathematical aspects of the model have just been 1 neglected, making most of the results of that paper purely formal. For this reason we are : v re-considering the same model and we repeat and extend the same construction paying i X particular attention to all the subtle mathematical points. From our analysis the crucial r a role of Riesz bases clearly emerges. We also consider coherent states associated to the model. I Introduction In a recent paper, [1], Trifonov suggested a possible explicit model of a PT-symmetric system basedonamodificationofthecanonical commutationrelation(CCR). Thephysical relevance of thismodelisbasedonthefactthatitprovides aniceexampleofwhatiscalledpseudo-hermitian quantum mechanics (PHQM) in the sense discussed in [2, 3, 4] and references therein. This is an interesting approach in which the role of self-adjoint operators is replaced by operators satisfying certain rules with respect to the parity and the time reversal operators and, as a consequences, possess eigenvalues which are real or which come in conjugate pairs. However, [1] neglects many mathematical details of the model, making most of its results purely formal. Here we discuss a similar model, with the same starting point, but we focus the attention on all those results which can be rigorously proven, and to the assumptions which are needed to prove these results. The bugs in [1] will be mentioned, and our solutions will be sketched. In particular, this detailed analysis produces a somehow unexpected result, showing that Riesz bases, [5, 6], play a crucial role in our context. The paper is organized asfollows: in the next section we introduce the model andwe discuss the Fock states arising from the commutation rules considered. We will show that Riesz bases appear naturally in this context. In Section III we show how standard coherent states (CS), as well as modified CS a la Trifonov, can be introduced. In Section IV we go back to the role of Riesz bases and we discuss our final comments and future projects. II The commutation rules Let be a given Hilbert space with scalar product .,. and related norm . . In [1] two H h i k k operators a and b acting on and satisfying the following commutation rule H [a,b] = 11 (2.1) were introduced. Of course, this collapses into the CCR if b = a . It is well known that, exactly † because of this rule, a and b cannot both be bounded operators. This simple consideration was just missing in [1]. Hence we should be careful in dealing with a and b because they cannot be defined in all of . For this reason we consider the following H Assumption 1.– there exists a non-zero ϕ such that aϕ = 0 and ϕ D (b) := 0 0 0 ∞ ∈ H ∈ D(bk). k 0 ∩ ≥ 2 In other words, ϕ is annihilated by a and belongs to the domain of all the powers of b. 0 Examples of such a vector will be given below. Under this assumption we can introduce the following vectors 1 1 ϕ = bnϕ , n 0, or ϕ = bϕ , n 1, (2.2) n 0 n n 1 √n! ≥ √n − ≥ which clearly belong to for all n 0. Let us now define the unbounded operator N := ba. H ≥ Notice that N = N . It is possible to check that ϕ belongs to the domain of N, D(N), for all † n 6 n 0, and that ≥ Nϕ = nϕ , n 0. (2.3) n n ≥ Let us now take N := N = a b . The rule in (2.1) also implies that [N,b] = b, [N,a] = a, † † † − [N,a ] = a , [N,b ] = b , and moreover, that † † † † − [b ,a ] = 11, (2.4) † † which again coincides with the CCR if b = a. However, if this is not the case, this and (2.1) are † really different from the CCR. It is clear that all the commutators should be considered in the sense of the unbounded operators. To go on we need another assumption which is analogous to the previous one: Assumption 2.– there exists a non-zero Ψ such that b Ψ = 0 and Ψ D (a ) := 0 † 0 0 ∞ † ∈ H ∈ D((a )k). k 0 † ∩ ≥ Under this assumption we can define the following vectors 1 1 Ψ = (a )nΨ , n 0, or Ψ = (a )Ψ , n 1, (2.5) n † 0 n † n 1 √n! ≥ √n − ≥ which clearly belong to for all n 0, and check that they also belong to the domain of N H ≥ and that NΨ = nΨ , n 0. (2.6) n n ≥ Incidentally we notice that equation (2.6) implies that N is unbounded, as well as N. Example 1: this first example shows that the above assumptions need not to be satisfied for generic operators a and b. Let = 2(R,dν(x)), where dν(x) = dx , and let a = ip and H L 1+x2 b = x, where x and p are the quantum position and momentum operators. Then aϕ (x) = 0 0 implies that ϕ (x) is constant. Of course ϕ (x) but bϕ (x) = xϕ (x) / . Hence ϕ (x) 0 0 0 0 0 ∈ H ∈ H does not belong to D (b) and Assumption 1 is violated. ∞ 3 Example 2: the second example is that of the harmonic oscillator. In this case = H 2(R,dx), and taking a = c := 1 d +x and b = c = 1 d +x , [a,b] = [c,c ] = 11, we L √2 dx † √2 −dx † find that ϕ (x) = Ψ (x) = 1 e x2/2, which satisfies both Assumptions 1 and 2. 0 0 π1/4 − (cid:0) (cid:1) (cid:0) (cid:1) Example 3: in this third example, [1], we put = 2(R,dx) a = c + sc and b = s † s H L sc + (1 + s2)c . Hence [a ,b ] = 11 for all real s. Then a ϕ (x) = 0 implies that ϕ (x) = † s s s 0 0 N exp 1 1+s x2 , while b Ψ (x) = 0 is solved by Ψ (x) = N exp 1 1+s+s2 x2 . Here N s −2 1 s †s 0 0 s′ −2 1 s+s2 s − − and Ns′(cid:8)are s−dep(cid:9)ending normalization constants. Of course, in ordenr for both thesoe functions to be square integrable we should require that both 1+s and 1+s+s2 are positive, which is true if 1 s 1 s+s2 − − 1 < s < 1. This same condition ensures also that ϕ (x) D (b ) and that Ψ (x) D (a ): − 0 ∈ ∞ s 0 ∈ ∞ †s any polynomial multiplied for a gaussian function belongs to 2(R,dx). L A minor modification of this example is also discussed in [1]: again we have = 2(R,dx) H L and a = c+sc , but we choose b = sc+(1 s2)c . Hence [a ,b ] = 11 for all real s, ϕ (x) is s † s † s s 0 − − the same as above while Ψ (x) = N exp 1 1 s s2 x2 . The main difference with respect to 0 s′′ −2 1+−s−s2 − the previous case is in the range of s in wnhich Assumptoions 1 and 2 are satisfied: we need now to restrict s in the interval 1(1 √5), 1( 1+√5) . 2 − 2 − Example 4: in the prev(cid:0)ious example a and b are(cid:1) defined by introducing a one-dimensional deformation parameter s starting from the bosonic operators c and c . Now we generalize this † procedure, showing that also two-dimensional deformations are allowed. Let a := αc+ αc , α,µ µ † b := µα2 1c+αc , where α and µ are real constants such that α,µ = 0 and α2 = µ2(α2 1). α,µ α− † 6 6 − Hence a = b (which would trivialize the situation), and [a ,b ] = 11. The solutions of †α,µ 6 α,µ α,µ α,µ a ϕ (x) = 0 and b Ψ (x) = 0 are respectively ϕ (x) = N exp 1 µ+1 x2 , and Ψ (x) = α,µ 0 †α,µ 0 0 α,µ −2 µ 1 0 − Nα′,µexp −21 αα22+µµ((αα22−11)) x2 . Again, Nα,µ and Nα′,µ are normalizantion constaonts. For these functionsnto sat−isfy A−ssumoptions 1 and 2 it is enough to have µµ+11 > 0 and αα22+µµ((αα22−11)) > 0, which are both verified if we take α > 1 and 1 < µ < 1+ 1 . − − − α2 1 − In the above assumptions it is now easy to check that Ψ ,ϕ = δ Ψ ,ϕ for all n m n,m 0 0 h i h i n,m 0, which, if we take Ψ and ϕ such that Ψ ,ϕ = 1, becomes 0 0 0 0 ≥ h i Ψ ,ϕ = δ , n,m 0 (2.7) n m n,m h i ∀ ≥ This meansthattheΨ ’sandtheϕ ’sarebiorthogonal. Itisalso possibletoprovethefollowing n n Lemma, which will be useful in the rest of the paper 4 Lemma 1 For all n 0 we have ϕ D(a) and Ψ D(b ). Moreover n n † ≥ ∈ ∈ 0, if n = 0, aϕ = (2.8) n ( √nϕn 1, if n > 0, − and 0, if n = 0, b Ψ = (2.9) † n ( √nΨn 1, if n > 0. − Once again, the proof is simple and will not be given here. It is maybe more relevant to remark that, since the vectors in := ϕ , n 0 and in := Ψ , n 0 are not ϕ n Ψ n F { ≥ } F { ≥ } orthogonal ( ϕ ,ϕ = δ and Ψ ,Ψ = δ in general), hence equation (2.8) does not n k n,k n k n,k h i 6 h i 6 automatically imply that a ϕ = √n+1ϕ , as it would do if were an orthonormal (o.n.) † n n+1 ϕ F basis of . For the same reason bΨ = √n+1Ψ , in general. However, the sets and n n+1 ϕ Ψ H 6 F F are biorthogonal and, because of this, the vectors of each set are linearly independent. If we now call and respectively the linear span of and , and and their closures, ϕ Ψ ϕ Ψ ϕ Ψ D D F F H H by construction is complete in and is complete in . More than this, we can also ϕ ϕ Ψ Ψ F H F H prove that ∞ ∞ f = Ψ ,f ϕ , f , h = ϕ ,f Ψ , h (2.10) n n ϕ n n Ψ h i ∀ ∈ H h i ∀ ∈ H n=0 n=0 X X What is not in general ensured is that the Hilbert spaces introduced so far all coincide, i.e. that = = . With our assumptions we can only state that and . ϕ Ψ ϕ Ψ H H H H ⊆ H H ⊆ H However, in all the examples considered previously, these three Hilbert spaces really coincide and for this reason we also consider the following Assumption 3.– The above Hilbert spaces all coincide: = = . ϕ Ψ H H H We would like to mention that this problem was not considered in [1], where it was taken for granted. From (2.10) we deduce, first of all, that both and are bases in . The ϕ Ψ F F H resolution of the identity looks now ∞ ∞ ϕ >< Ψ = Ψ >< ϕ = 11, (2.11) n n n n | | | | n=0 n=0 X X where 11 is the common identity of all the Hilbert spaces and where the useful Dirac bra-ket notation has been adopted. At this stage it is natural to introduce two operators which we now write formally using again the bra-ket notation as ∞ ∞ η = ϕ >< ϕ , η = Ψ >< Ψ . (2.12) ϕ n n Ψ n n | | | | n=0 n=0 X X 5 Of course, these operators need not to be well defined: for instance the series could be not convergent, or even if they do, they could converge to some unbounded operator, so we have to be careful about domains. Again, this aspect was not considered in [1]. Let us then introduce, more rigorously, an operator η acting on a vector f in its domain ϕ D(ηϕ) as ηϕf = ∞n=0hϕn,fiϕn. We also introduce a second operator, ηΨ, acting on a vector h in its domain DP(ηΨ) as ηΨh = ∞n=0hΨn,hiΨn. Under Assumption 3, both these operators are densely defined in since D(η ) and D(η ). In particular, we find that H DϕP⊆ ϕ DΨ ⊆ Ψ η Ψ = ϕ , η ϕ = Ψ , (2.13) ϕ n n Ψ n n for all n 0, which also implies that Ψ = (η η )Ψ and ϕ = (η η )ϕ , for all n 0. Hence n Ψ ϕ n n ϕ Ψ n ≥ ≥ η η = η η = 11 η = η 1. (2.14) Ψ ϕ ϕ Ψ ⇒ Ψ ϕ− In other words, both η and η are invertible and one is the inverse of the other. Furthermore, Ψ ϕ we can also check that they are both positive defined and symmetric. One may wonder whether they are automatically bounded, then. This is not so, in general. Indeed it is not hard to construct examples of unbounded positive and symmetric operators mapping a basis of in a H biorthogonal basis. It is enough to consider a number operator Nˆ defined on an o.n. basis of , H e , n 1 , as Nˆ e = ne , n 1. Hence, calling ϕ = 1 e and Ψ = √ne , and are { n ≥ } n n ≥ n √n n n n Fϕ FΨ biorthogonal bases of . Moreover Nˆϕ = Ψ , Nˆ 1Ψ = ϕ , Nˆ > 0, but Nˆ is unbounded. A n n − n n H simple modification of example also provides an example in which the positive operator M 1 2 → mapping a basis in its biorthogonal basis and its inverse M = M 1 mapping into G1 G2 2→1 1−→2 G2 , are both unbounded. For that it is enough to define 1 G 1 e , if n is even, ne , if n is even, n2e , if n is even, ϕ = n n Ψ = n and M e = n n ( nen, if n is odd, n ( n1 en, if n is odd, 1→2 n ( n12 en, if n is odd. This is not a big surprise because, as discussed in [5], two biorthogonal bases are related by a bounded operator, with bounded inverse, if and only if they are Riesz basis. This suggests we need some more assumption to go further. In order to keep the roles of and symmetric ϕ Ψ F F we require now the following Assumption 4.– and are Bessel sequences. In other words, there exist two positive ϕ Ψ F F constants A ,A > 0 such that, for all f , ϕ Ψ ∈ H ∞ ∞ ϕ ,f 2 A f 2, Ψ ,f 2 A f 2. (2.15) n ϕ n Ψ |h i| ≤ k k |h i| ≤ k k n=0 n=0 X X 6 As a consequence, see [6], and are both Riesz bases with bounds 1 ,A and Fϕ FΨ AΨ ϕ 1 ,A . This is due to the fact that they are biorthogonal sets. In particular(cid:16)then, (cid:17) and Aϕ Ψ Fϕ (cid:16) are f(cid:17)rames. Ψ F Lemma 2 Under Assumption 4 both η and η are bounded operators. In particular we have ϕ Ψ η A , η A . (2.16) ϕ ϕ Ψ Ψ k k ≤ k k ≤ Moreover 1 1 11 η A 11, 11 η A 11. (2.17) ϕ ϕ Ψ Ψ A ≤ ≤ A ≤ ≤ Ψ ϕ Proof – We just prove that η A . Using (2.15), (2.12) and the Schwarz inequality we ϕ ϕ k k ≤ have ∞ η = sup f,η g = sup f,ϕ ϕ ,g ϕ ϕ k k k k |h i| h ih i ≤ f = g =1 f = g =1(cid:12) (cid:12) k k k k k k k k (cid:12)Xk=0 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) ∞ ∞ (cid:12) (cid:12) sup f,ϕ 2 ϕ ,g 2 A sup f g A k k ϕ ϕ ≤ v |h i| v |h i| ≤ k kk k ≤ kfk=kgk=1uuXk=0 uuXk=0 kfk=kgk=1 t t (cid:3) This Lemma implies that the domains of η and η can be taken to be all of . An ϕ Ψ H interesting consequence of our construction is the following Corollary 3 The set coincides with the dual frame ˜ of . Also, coincides with the Ψ ϕ ϕ ϕ F F F F dual frame ˜ of . Ψ Ψ F F Proof – Since Fϕ is a frame, its frame operator Sϕ, defined as Sϕf = ∞n=0hϕn,fiϕn, f ∈ H, iswell defined, boundedandinvertible. It isalso clearthatitcoincides withη . Hence, recalling P ϕ that the vectors ϕ˜ of the set ˜ are defined as ϕ˜ = S 1ϕ , we have n Fϕ n ϕ− n ϕ˜ = S 1ϕ = η 1ϕ = η ϕ = Ψ , n ϕ− n ϕ− n Ψ n n (cid:3) where we have used (2.13) and (2.14). Our second assertion can be proved similarly. 7 II.1 Connections with intertwining operators Under the assumptions we have considered so far a natural structure appears in which η Ψ and η play the role of intertwining operators between non self-adjoint operators. This looks ϕ interesting since inthe literature on the subject, [7, 8, 9, 10] and references therein, intertwining operators usually act between self-adjoint operators preserving the spectra and modifying the eigenvectors quite easily. More explicitly, suppose that h and h are two self-adjoint operators 1 2 (1) and that a third operator x satisfies the intertwining equation xh = h x. Let e be an 1 2 n (1) (1) (2) (1) eigenstate of h with eigenvalue ǫ , h e = ǫ e . Now, if e := xe = 0, then we can check 1 n 1 n n n n n 6 (2) (2) with a straight computation that h e = ǫ e . 2 n n n We begin our analysis remarking that a simple use of induction on n proves that for all n 0 the vector ϕ belongs to the domain of the operator η a η and that n ϕ † Ψ ≥ bϕ = η a η ϕ , (2.18) n ϕ † Ψ n which can also be written, recalling that ϕ = η Ψ and that η ϕ = Ψ , as bη Ψ = η a Ψ n ϕ n Ψ n n ϕ n ϕ † n or yet, as bη = η a . Hence η intertwines between b and a . Analogously we can prove that ϕ ϕ † ϕ † a η = η b, while other intertwining relations can be found just taking the adjoint of these † Ψ Ψ equations. Two interesting consequences are now the following equalities η N = Nη and N η = η N, (2.19) Ψ Ψ ϕ ϕ whose proof is straightforward. It is now trivial to check that our results are coherent with the standard technique of intertwining operators, but for the lack of self-adjointness of the operators N and N. This is suggested by the fact that N and N have the same eigenvalues. More in details, using for instance the intertwining relation N η = η N, together with the ϕ ϕ equations ϕ = η Ψ and NΨ = nΨ , we can easily check that ϕ are eigenstates of N with n ϕ n n n n eigenvalue n. The computation goes as follows: Nϕ = N η Ψ = η NΨ = η (nΨ ) = nϕ . n ϕ n ϕ n ϕ n n It isalso worth noticing that condition (2.19) isa pseudo-hermiticity conditionfor theoperators N andN. Indeedwehaveη N η 1 = N andη N η 1 = N, whichbecauseofthepropertiesof Ψ Ψ− † ϕ † ϕ− η andη , areexactlytheconditionswhichstatethatN andNarepseudo-hermitianconjugate, Ψ ϕ [2]. We recall that this was just the main motivation in [1] for considering the commutation rules in (2.1). 8 II.2 Inverting the construction What we have discussed so far shows in particular that, under Assumptions 1-4, b and a are † necessarily related by η as in (2.18) or, equivalently, as in b = η a η 1. We are now interested ϕ ϕ † ϕ− in considering the inverse construction, i.e. in considering as our starting point again two operators a and b satisfying [a,b] = 11 under the assumption that b is related to a in the way † shown above, and check what happens. Let therefore a be a given operator defined on a dense domain of a given Hilbert space, D(a) , with adjoint a densely defined. We now consider a bounded, positive, operator T, † ⊆ H with bounded inverse such that a dense subset of exists, , with T 1 : D(a ). Hence − † H E E → the operator b := T a T 1 is densely defined since D(b ). We assume that T † − T E ⊆ [a,b ] = 11 (2.20) T Remark:– If we work in the Assumptions 1-4 above, taking T as the frame operator of ϕ F we get an example of this settings. Asbefore, weneedtoextractacertainsetofconditionsifwewanttodeducesomeinteresting results. The first assumption is exactly Assumption 1 above, which we now write as Assumption I.– there exists a non-zero ϕ such that aϕ = 0 and T 1ϕ D (a ). 0 0 − 0 ∞ † ∈ H ∈ Hence (2.2) can be extended also to this settings and we have 1 1 ϕ = bn ϕ , n 0, or ϕ = b ϕ , n 1, (2.21) n √n! T 0 ≥ n √n T n−1 ≥ which are in the domain of the operator N := b a and satisfy the eigenvalue equation N ϕ = T T T n nϕn, for all n ≥ 0. As before we define NT := NT† = a†b†T. There is no need of require here the analogous of Assumption 2. Indeed, if we define Ψ0 := T−1ϕ0, it is first clear that b†TΨ0 = 0. Moreover, due to Assumption I above, Ψ D (a ). Hence the vectors 0 ∞ † ∈ 1 1 Ψ = (a )nΨ , n 0, or Ψ = (a )Ψ , n 1 n † 0 n † n 1 √n! ≥ √n − ≥ are well defined, belong to the domain of N , satisfy the eigenvalue equation N Ψ = nΨ , T T n n n 0, and the following relation holds ≥ ϕ = TΨ , Ψ = T 1ϕ , (2.22) n n n − n 9 for all n 0. Notice that these lookexactly like theequations in(2.13). Moreover, if Ψ ,ϕ = 0 0 ≥ h i 1, then Ψ ,ϕ = δ . From this biorthogonality condition two important estimates on ϕ n m n,m n h i k k and Ψ can be deduced. For instance, since for all n 0 n k k ≥ 1 = Ψ ,ϕ = T 1ϕ ,ϕ = T 1/2ϕ 2, n n − n n − n h i k k (cid:10) (cid:11) we deduce that ϕ = T1/2T 1/2ϕ T1/2 . (2.23) n − n k k k k ≤ k k Analogously we can prove that Ψ T 1/2 , n 0. Defining , , and as n − ϕ Ψ ϕ Ψ k k ≤ k k ∀ ≥ F F H H before, we consider now the following: Assumption II.– The above Hilbert spaces all coincide: = = . ϕ Ψ H H H This is exactly our previous Assumption 3. Hence, [5], since and are two (biorthogo- ϕ Ψ F F nal)basesof relatedbyaboundedoperatorT withboundedinverse, are arenecessarily ϕ Ψ H F F Riesz bases. Moreover, defining η and η as in (2.12), it is easy to check that they coincide ϕ Ψ with T and T 1, so that they are bounded operators with bounded inverse, mapping into − Ψ F and vice-versa. Moreover, and are dual frames of each other. Finally, η and η are ϕ ϕ Ψ ϕ Ψ F F F the frame operators respectively of and . Hence essentially the same general structure ϕ Ψ F F discussed previously is recovered. We will continue with this analysis in the last section, where other aspects and applications of Riesz bases in this context will be considered. III Coherent states In [1] a family of CS for the model has been introduced. Again, in our opinion some more mathematical care is required. For this reason we carry on our own analysis, focusing the attention on those points which may create some problems. Our analysis is also motivated by the work in [11], where the role of non-orthogonal bases in the description of coherent states is discussed. We work here under Assumptions 1-4 of the previous section. Hence there exist ϕ and Ψ 0 0 in such that aϕ = b Ψ = 0. Also, ϕ D (b) and Ψ D (a ). Let us introduce the 0 † 0 0 ∞ 0 ∞ † H ∈ ∈ z-dependent operators U(z) = exp zb za , V(z) = exp za zb , (3.1) † † { − } { − } 10