Pseudo-bosons: mathematics and physical examples

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Organization of the . . . What are pseudo- . . . A class of examples . . . The extended . . . The Swanson . . . A 2D system: . . . Damped harmonic . . . Comments, future . . .

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Pseudo-bosons: mathematics and physical examples

Fabio Bagarello

Dipartimento di Metodi e Modelli Matematici

Benasque July 2010

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I. Organization of the talk

1. What are pseudo-bosons? A mathematical intro- duction

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I. Organization of the talk

2. The role of Riesz bases

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I. Organization of the talk

3. Where do pseudo-bosons appear?

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I. Organization of the talk

4. The extended quantum harmonic oscillator

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I. Organization of the talk

4. The extended quantum harmonic oscillator 5. The Swanson model

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I. Organization of the talk

6. Generalized Landau levels

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I. Organization of the talk

6. Generalized Landau levels

7. The quantum damped harmonic oscillator

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II. What are pseudo-bosons? [Some math- ematics]

Let H be a given Hilbert space with scalar product h:; :i and norm k:k. Let a and b be two operators acting on H and satisfying (Trifonov, 2009)

[a; b] = 11; (1)

If b = a^y then we recover CCR. Recall that a and b cannot both be bounded operators: they cannot be dened in all of H. For this reason we consider the following

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II. What are pseudo-bosons? [Some math- ematics]

[a; b] = 11; (1)

Assumption 1. there exists a non-zero '⁰ 2 H such that a'⁰ = 0 and '⁰ 2 D¹(b) := \k0D(b^k).

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II. What are pseudo-bosons? [Some math- ematics]

[a; b] = 11; (1)

Assumption 1. there exists a non-zero '⁰ 2 H such that a'⁰ = 0 and '⁰ 2 D¹(b) := \k0D(b^k).

Then

'n = 1

pn!bⁿ'⁰; n 0; (2) belongs to H for all n 0.

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Let N := ba. Then 'n 2 D(N), for all n 0, and N'n = n'n; n 0: (3) Let us now take N := N^y = a^yb^y 6= N. We require that the following holds:

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Assumption 2. there exists a non-zero Ψ⁰ 2 H such that b^yΨ0 = 0 and Ψ⁰ 2 D¹(a^y) := \k0D((a^y)^k).

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Assumption 2. there exists a non-zero Ψ⁰ 2 H such that b^yΨ0 = 0 and Ψ⁰ 2 D¹(a^y) := \k0D((a^y)^k). Under this assumption the following vectors

Ψ_n = 1

pn!(a^y)ⁿΨ0; n 0; (4) belong to H for all n 0, and to D(N ). Moreover

N Ψ nΨ ; n 0: (5)

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Example 1: the above natural assumptions are not al- ways satised: let H = L²(R; d(x)), d(x) = ₁₊^dx_x², a = ip, b = x. Then a'⁰(x) = 0 implies that '⁰(x) is constant. Of course '⁰(x) 2 H but b'⁰(x) = x'⁰(x) =2 H. Hence '⁰(x) does not belong to D¹(b) and Assumption 1 is violated.

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Example 1: the above natural assumptions are not al- ways satised: let H = L²(R; d(x)), d(x) = ₁₊^dx_x², a = ip, b = x. Then a'⁰(x) = 0 implies that '⁰(x) is constant. Of course '⁰(x) 2 H but b'⁰(x) = x'⁰(x) =2 H. Hence '⁰(x) does not belong to D¹(b) and Assumption 1 is violated.

Example 2: the trivial case: harmonic oscillator. In this case H = L²(R; dx), and taking a = c :=

p1 2

dxd +x and b = c^y = ^p¹

2

dxd +x, [a; b] = [c; c^y] = 11, we nd that '⁰(x) = Ψ⁰(x) = ¹¹=4e ^x²⁼², which satises both Assumptions 1 and 2.

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Example 3: [Trifonov] H = L²(R; dx), as = c + sc^y and bs = sc + (1 + s²)c^y. Hence [as; bs] = 11 for all real s. as'⁰(x) = 0 ) '⁰(x) = Ns expn

1 2

1+s

1 s x²o, while bs^yΨ0(x) = 0 ) Ψ⁰(x) = N_s⁰ expn

1 2

1+s+s² 1 s+s² x²o.

Both these functions are square integrable if 1 <

s < 1. This same condition ensures also that '⁰(x) 2 D¹(bs) and that Ψ⁰(x) 2 D¹(as^y): any polynomial multiplied for a gaussian function belongs to L²(R; dx).

Minor modication of this example [Trifonov]: H = L²(R; dx), as = c + sc^y, and bs = sc + (1 s²)c^y.

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Example 4: (two-dimensional deformation of c and c^y) Let a; := c + c^y, b; := ²¹c + c^y, where and are real constants such that ; 6=

0 and ² 6= ²(² 1). Hence a;^y 6= b; and [a;; b;] = 11.

a;'⁰(x) = 0 and b;^y Ψ₀(x) = 0 produce '⁰(x) = N;exp

1 2

+ 1

1x²

; and

Ψ0(x) = N_;⁰ exp

1 2

²+(² 1) ² (² 1) x²

: These functions satisfy Assumptions 1 and 2 if > 1 and 1 < < 1 + ¹ .

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Under Assumptions 1 and 2 we have hΨn; 'mi =

n;mhΨ⁰; '⁰i for all n; m 0. Then, if hΨ⁰; '⁰i = 1, hΨn; 'mi = n;m; 8n; m 0 (6)

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n;mhΨ⁰; '⁰i for all n; m 0. Then, if hΨ⁰; '⁰i = 1, hΨn; 'mi = n;m; 8n; m 0 (6) Moreover, 8n 0 we have 'n 2 D(a) and Ψn 2 D(b^y), anda'n = p

n 'n 1;as well asb^yΨ_n = p

n Ψn 1.

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n Ψn 1. Let F' := f'n; n 0g and F^Ψ := fΨn; n 0g.

Since h'n; 'ki 6= n;k, a^y'n = p

n + 1 'n+1 is false, in general. For the same reason b Ψn 6= p

n + 1 Ψn+1.

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n Ψn 1. Let F' := f'n; n 0g and F^Ψ := fΨn; n 0g.

Since h'n; 'ki 6= n;k, a^y'n = p

n + 1 'n+1 is false, in general. For the same reason b Ψn 6= p

n + 1 Ψn+1. However, the sets F' and F^Ψ are biorthogonal and, because of this, the vectors of each set are linearly independent.

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Let D' be the linear span of F' and D^Ψ the linear span of F^Ψ, and H' and H^Ψ their closures: Hence, by construction:

f =

1

X

n=0

hΨn; f i 'n; 8f 2 H'; and

h =

1

X

n=0

h'n; f i Ψn; 8h 2 H^Ψ

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Let D' be the linear span of F' and D^Ψ the linear span of F^Ψ, and H' and H^Ψ their closures: Hence, by construction:

f =

1

X

n=0

hΨn; f i 'n; 8f 2 H'; and

h =

1

X

n=0

h'n; f i Ψn; 8h 2 H^Ψ

What is not in general ensured is that H' = H^Ψ = H.

At this stage we can only say that H' H and H^Ψ H. It is convenient (and natural) to consider Assumption 3. The above Hilbert spaces all coin-

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Then, F' and F^Ψ are bases in H. The resolution of the identity looks now

1

X

n=0

j'n >< Ψnj =

1

X

n=0

jΨn >< 'nj = 11; (7) where 11 is the identity operator on H.

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Then, F' and F^Ψ are bases in H. The resolution of the identity looks now

1

X

n=0

j'n >< Ψnj =

1

X

n=0

jΨn >< 'nj = 11; (7)

where 11 is the identity operator on H.

Let further

' =

1

X

n=0

j'n >< 'nj; ^Ψ =

1

X

n=0

jΨn >< Ψnj:

(8) These operators need not to be well dened: for in- stance the series could be not convergent, or even if they do, they could converge to some unbounded

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More rigorously, we introduce an operator ' acting on a vector f 2 D(') as 'f = P¹_n=0h'n; f i 'n, and ^Ψ, acting on a vector h 2 D(^Ψ) as ^Ψh = P1

n=0hΨn; hi Ψn. Under Assumption 3, both these operators are densely dened in H. In particular:

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'Ψ_n = 'n; ^Ψ'n = Ψ_n;

for all n 0. Then Ψn = (^Ψ')Ψ_n and 'n = ('^Ψ)'n, for all n 0. Hence

^Ψ' = '^Ψ = 11 ) ^Ψ = _'¹: (9)

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'Ψ_n = 'n; ^Ψ'n = Ψ_n;

for all n 0. Then Ψn = (^Ψ')Ψ_n and 'n = ('^Ψ)'n, for all n 0. Hence

^Ψ' = '^Ψ = 11 ) ^Ψ = _'¹: (9) Furthermore, we can also check that they are both positive dened and symmetric. In general, however, they are unbounded.

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This is not a big surprise: two biorthogonal bases are related by a bounded operator, with bounded inverse, if and only if they are Riesz bases. Then:

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Assumption 4. F' and F^Ψ are Riesz bases: there exist an o.n. basis G = fgn; n 0g and two bounded operators X and Y , with bounded inverses, such that

'n = X gn; and Ψn = Y gn; for all n 0.

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Assumption 4. F' and F^Ψ are Riesz bases: there exist an o.n. basis G = fgn; n 0g and two bounded operators X and Y , with bounded inverses, such that

'n = X gn; and Ψn = Y gn; for all n 0.

In thus case we call our pseudo-bosons regular, and both ' and ^Ψ are bounded operators with k'k A' and k^Ψk A^Ψ and D(') = D(^Ψ) = H.

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A physical interlude: connections with intertwining operators

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^Ψ and ' are intertwining operators between non self-adjoint operators:

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rst we prove (induction on n) that, 8 n 0, 'n

belongs to D('a^y^Ψ) and

b 'Ψ_n = 'a^yΨ_n: (10) Then b ' = 'a^y, so that ' intertwines between b and a^y. Analogously we can prove that a^y^Ψ = ^Ψb, and

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^ΨN = N ^Ψ and N ' = 'N : (11)

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^ΨN = N ^Ψ and N ' = 'N : (11) These results are in agrement with the standard theory on intertwining operators, but for the lack of self- adjointness of the operators N and N :

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N'n = N 'Ψ_n = 'N Ψn = '(n Ψn) = n 'n:

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N'n = N 'Ψ_n = 'N Ψn = '(n Ψn) = n 'n: It is also worth noticing that condition (11) is a pseudo- hermiticity condition for the operators N and N . In- deed we have ^ΨN _Ψ¹ = N^y and 'N^y_'¹ = N, which because of the properties of Ψ and ', are exactly the conditions which state that N and N are pseudo-hermitian conjugate, [Mostafazadeh].

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III. A class of examples from Riesz bases

Let F' := f'n; n 0g be a Riesz basis of H with bounds A and B, 0 < A B < 1. The associated frame operator S := P¹_n=0 j'n >< 'nj is bounded, positive and admits a bounded inverse.

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III. A class of examples from Riesz bases

The set F'^ˆ := f ˆ'n := S ¹⁼²'n; n 0g is an o.n.

basis of H. Hence we can dene a lowering and a raising operators a' and a^y_' on F'^ˆ as usual:

a''ˆn = p

n ˆ'n 1; a_'^y'ˆn = p

n + 1 ˆ'n+1; n 0. Hence [a'; a^y'] = 11.

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III. A class of examples from Riesz bases

The set F'^ˆ := f ˆ'n := S ¹⁼²'n; n 0g is an o.n.

basis of H. Hence we can dene a lowering and a raising operators a' and a^y_' on F'^ˆ as usual:

a''ˆn = p

n ˆ'n 1; a_'^y'ˆn = p

n + 1 ˆ'n+1; n 0. Hence [a'; a^y'] = 11.

If we now dene a := S¹⁼²a'S ¹⁼², then a'n = pn 'n 1. Hence a is also a lowering operator. How- ever, since F is not an o.n. basis in general, a^y is

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We further dene b := S¹⁼²a^y'S ¹⁼², (b 6= a^y). b acts on 'n as a raising operator: b 'n = p

n + 1 'n+1, for all n 0, and we also have [a; b] = 11.

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So we have constructed two operators satisfying (1) and which are not related by a simple conjugation.

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So we have constructed two operators satisfying (1) and which are not related by a simple conjugation.

Moreover:

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1. Assumption 1 is veried since a'⁰ = 0 and since '⁰ 2 D¹(b). In particular we nd that bⁿ'⁰ = pn! 'n, 8n 0.

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2. As for Assumption 2, it is enough to dene Ψ⁰ = S ¹'⁰. Then b^yΨ0 = 0 and Ψ⁰ belongs to D¹(a^y). In particular (a^y)ⁿ Ψ₀ = p

n! S ¹⁼²'ˆn.

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n! S ¹⁼²'ˆn. 3. Since F' is a Riesz basis of H by assumption,

then H' = H. Notice now that the vector Ψn in (4) can be written as Ψn = S ¹'n, for all n 0.

Hence F^Ψ is in duality with F' and therefore is a Riesz basis of H as well. Hence H^Ψ = H. This prove Assumption 3.

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n! S ¹⁼²'ˆn. 3. Since F' is a Riesz basis of H by assumption,

then H' = H. Notice now that the vector Ψn in (4) can be written as Ψn = S ¹'n, for all n 0.

Hence F^Ψ is in duality with F' and therefore is a Riesz basis of H as well. Hence H^Ψ = H. This prove Assumption 3.

4. As for Assumption 4, this is just our working assumption: our pseudo-bosons are regular.

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Remark:Our results show the relation between the frame operator for a certain Riesz basis, for which some perturbative expansions can be found in the lit- erature, with the so-called the metric operator, which is used both to dene a new scalar product in the Hilbert space of the theory and the so-called pseudo- hermitian conjugate of an operator, [Mostafazadeh, ...].

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IV. The extended quantum harmonic os- cillator

[J. da Providˆencia et al., Non hermitian operators with real spectrum in quantum mechanics, arXiv: quant- ph 0909.3054

H =

2 p² +x² + ip 2p;

> 0 and [x; p] = i.

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IV. The extended quantum harmonic os- cillator

[J. da Providˆencia et al., Non hermitian operators with real spectrum in quantum mechanics, arXiv: quant- ph 0909.3054

H =

2 p² +x² + ip 2p;

> 0 and [x; p] = i.

Using a = ^p¹₂ x + _dx^d

, a^y = ^p¹

2 x _dx^d

, [a; a^y] = 11, and N = a^ya, we can write H = N+(a a^y)+₂ 11 which, putting

A =a 1

; B =a^y+ 1

; )

H = (BA + 11);

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Assumption 1: nd a non zero vector '⁽₀⁾ 2 H such that A'⁽₀⁾ = 0 and '₀⁽⁾ 2 D¹(B).

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A'⁽0⁾ = 0 ) a '⁽0⁾ = ¹ '⁽0⁾ ) '⁽0⁾ is a standard coherent state with parameter ¹:

'⁽0⁾ = U( ¹)'⁰ = e ¹⁼²²

1

X

k=0

^k

pk! 'k; (1)

where a'⁰ = 0, and U( ¹) = e¹⁽^a^y ^a) is the unitary (displacement) operator: k'⁽0⁾k = k'⁰k = 1.

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A'⁽0⁾ = 0 ) a '⁽0⁾ = ¹ '⁽0⁾ ) '⁽0⁾ is a standard coherent state with parameter ¹:

'⁽0⁾ = U( ¹)'⁰ = e ¹⁼²²

1

X

k=0

^k

pk! 'k; (1)

where a'⁰ = 0, and U( ¹) = e¹⁽^a^y ^a) is the unitary (displacement) operator: k'⁽0⁾k = k'⁰k = 1.

Since kB^k'⁽0⁾k k! e²⁼, k 0, '⁽0⁾ belongs to the domain of all the powers of B. As a consequence

'⁽n⁾ = 1

pn!Bⁿ'⁽₀⁾; (2) is well dened for all n 0.

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Assumption 2: B^yΨ⁽₀⁾ = 0 ) Ψ0⁽⁾ = '⁽0 ⁾ = U( ¹)'⁰ = U ¹( ¹)'⁰ and k(A^y)^kΨ⁽₀⁾k k! e²⁼, k 0. Hence

Ψ⁽_n⁾ = 1

pn!(A^y)ⁿΨ⁽₀⁾; (3) is also well dened for all n 0.