A quasi-exactly solvable model: two charges in a magnetic field, subject to a non-Coulomb mutual interaction
aa r X i v : . [ qu a n t - ph ] S e p A quasi-exactly solvable model: two charges in a magnetic field,sub ject to a non-Coulomb mutual interaction
Michael Kreshchuk
School of Physics and Astronomy, University of MinnesotaMinneapolis MN 55455 USAe-mail: kreshchu @ physics.umn.edu
Abstract
We extend the class of QM problems which permit for quasi-exact solutions. Specifically, we considerplanar motion of two interacting charges in a constant uniform magnetic field. While Turbiner andEscobar-Ruiz (2013) addressed the case of the Coulomb interaction between the particles, we explore threeother potentials. We do this by reducing the appropriate Hamiltonians to the second-order polynomials inthe generators of the representation of SL (2 , C ) group in the differential form. This allows us to performpartial diagonalisation of the Hamiltonian, and to reduce the search for the first few energies and thecorresponding wave functions to an algebraic procedure. In quantum mechanics, quasi-exactly solvable (QES) problems are those which permit analytic derivationof a finite part of the (generally, infinite) spectrum. They allow for partial reducing of the Hamiltonianto a block-diagonal form. In these cases, one can find nonperturbative solutions for Hamiltonians of somespecial types. These special types are obtained from Hamiltonians of broader (generally, unsolvable) classesby constraining the entering parameters.Exploration of these systems was pioneered by Turbiner [1]. Later, this kind of systems was independentlyrediscovered by Kamran and Olver [2]. A different approach (often termed analytic ) was developed byUshveridze [3]. Various approaches to QES problems and some advances along this line of research arepresented in [4]. Among the recent advances in this field, notable are the discovery of ‘semiclassically quasi-exactly solvable potentials’ [5], as well as the results on studying elliptic A n and BC n models [6, 7]. In thecurrent paper, we employ a so-called algebraic approach, a detailed description whereof can be found in [8].All exactly and quasi-exactly solvable one-dimensional Hamiltonians known so far can be reduced tosecond-order polynomials in the generators of a representation of the SL (2 , C ) group:ˆ h = a αβ ˆ J αn ˆ J βn + b α ˆ J αn , where ˆ J αn are implemented as differential operators. Such Hamiltonians’ eigenvectors and eigenvalues arefound through an algebraic procedure. Our goal is to broaden the class of problems which can be cast intothe said form and which can, therefore, be treated by algebraic methods. A list of QES Hamiltonians of theform ˆ H = − d dx + V ( x ) (1)1as provided in [1]. When one considers the radial part of the equation for the relative motion in the problemof two particles in two or three dimensions, other QES Hamiltonians emerge. In distinction from the type (1),they contain the first derivative. In [9], several particular analytic solutions were described for a system of twoelectrons (interacting with Coulomb potentials) in a homogeneous magnetic field and an external oscillatorpotential. In [10], analytic solutions were found for two particles of opposite charges in a homogenous magneticfield. Both sets of solutions were obtained my means of a recurrence relation on the coefficients a n in an anzats (cid:0) P n a n ρ n (cid:1) exp( . . . ) for the wave function. In [11], the algebraic structure of the Hamiltonian of a system oftwo electrons in an external oscillator potential was revealed.In this paper, we consider planar motion of two interacting charged spinless nonrelativistic particles subjectto a constant uniform magnetic field. While Turbiner and Escobar-Ruiz [12, 13, 14] addressed the case of aCoulomb interaction between the particles, we here explore other potentials that can permit for the afore-described reduction of the Hamiltonian. Of a special interest are the potentials resembling effective interactionbetween charges in realistic physical settings [14].In Section 2, we briefly discuss the underlying idea of the method, an idea that ensues from the basicproperties of representations of Lie groups. In Section 3, we provide the statement of the problem and carryout some prelusory calculations. In the subsequent sections we derive and explore three reducible potentialsnever presented in the literature hitherto. Let ˆ J n be the generators of an irreducible finite-dimensional representation of the SL (2 , C ) group in somevector space: [ ˆ J n , ˆ J + n ] = ˆ J + n , [ ˆ J n , ˆ J − n ] = − ˆ J − n , [ ˆ J + n , ˆ J − n ] = 2 ˆ J n , (2)where n = 2 j + 1 is the dimension of the space, and j is the spin of the representation. These relations aresatisfied by the operators defined as ˆ J + n = 2 jρ − ρ ∂ ρ , ˆ J − n = ∂ ρ , ˆ J n = − j + ρ∂ ρ . (3)For each n , we find the corresponding vector space to be V = span { , ρ, . . . ρ j } . (4)Indeed, these monomials are eigenvectors of ˆ J n , while ˆ J + n and ˆ J − n are the raising and lowering operators. Noneof these operators can take us outside the representation space V . As a result of this, V stays invariant underthe action of any polynomial function ˆ O of the operators ˆ J n :ˆ O V ⊆ V . (5)Solving the associated spectral problem ˆ
O f ( ρ ) = − νf ( ρ ) (6)over the vector space (4) is then equivalent to diagonalisation of an n × n matrix. This becomes obviousif one implements the basis vectors (4) and the differential operators (3) with column vectors and matrices,correspondingly. 2n what follows, we shall be interested in operators of the typeˆ h = a αβ ˆ J αn ˆ J βn + b α ˆ J αn , (7)because all exactly and quasi-exactly solvable one-dimensionsl Hamiltonians known hitherto can be reducedto this form. In this way, the spectral problem for the operator ˆ h , h a αβ ˆ J αn ˆ J βn + b α ˆ J αn i f ( ρ ) = − ν f ( ρ ) , (8)is solvable by purely algebraic means. We consider a system of two non-relativistic charged spinless particles, ( e , m ) and ( e , m ), moving on aplane orthogonal to a magnetic field B = B ˆ z . We limit ourselves to a constant and uniform magnetic field, arestriction which permits us to use the symmetric gauge A , = B × ρ , . The latter circumstance simplifiesthe problem greatly and allows us to write the Hamiltonian in the form ofˆ H = ( ˆ p − e A ) m + ( ˆ p − e A ) m + V ( | ρ − ρ | ) , (9)where ~ = c = 1, while ρ , and ˆ p , = − i ∇ are the coordinates and momenta of the particles.It is natural to switch to the centre of mass reference frame (CMF): R = µ ρ + µ ρ , ρ = ρ − ρ . (10)Canonically conjugated to them are the CMF and relative momenta:ˆ P = ˆ p + ˆ p , ˆ p = µ ˆ p − µ ˆ p , (11)where M = m + m , µ i = m i M . (12)Using these formulae, one can prove thatˆ P = − i ∇ R , ˆ p = − i ∇ ρ . (13)The above new coordinates and momenta enable us to define the CMF and relative angular momenta:ˆ L = R × ˆ P , ˆ ℓ = ρ × ˆ p . (14)By direct calculation, one can check that the total pseudomomentum ˆ K and the total angular momentum ˆ L T are the integrals of the system:ˆ K = ˆ p + e A + ˆ p + e A = ˆ P + q A R + e c A ρ , (15)ˆ L T = ρ × ˆ p + ρ × ˆ p = ˆ L + ˆ ℓ , (16)[ ˆ K , ˆ H ] = [ ˆ L T , ˆ H ] = 0 , (17)where q = e + e , e c = µ e − µ e = m r (cid:18) e m − e m (cid:19) , (18)3 c being a coupling charge and m r being the reduced mass of the system: m r = m m M . (19)After a unitary transformation U = e − ie c A ρ · R , (20)the Hamiltonian becomesˆ H ′ = U − ˆ H U = ( ˆ P − q A R − e c A ρ ) M + ( ˆ p − q W A ρ ) m r + V ( | ρ − ρ | ) , (21)where q W = e µ + e µ . (22)The eigenfunctions of ˆ H and ˆ H ′ are related viaΨ ′ = Ψ e ie c A ρ · R . (23)The two cases of interest are • e c = 0 , • q = 0 and ˆ P = 0 .Both lead to similar equations; so here we are going to focus mainly on the first case.For e c = 0 , the eigenfunctions can be factorised:Ψ( R , ρ ) = χ ( R ) ψ ( ρ ) , (24)ˆ H Ψ = ( E R + E ρ )Ψ . (25)The equation for the CMF variable χ renders Landau levels (e.g., [12]). The equation for the relative motionacquires the form of " − ∇ ρ m r − ω c ˆ ℓ z + m r ω c ρ V ( | ρ | ) − E ρ ψ ( ρ ) = 0 , (26)where ω c = qBM . (27)The vanishing of the commutator [ ˆ H , ˆ ℓ z ] = 0 permits further factorisation: ψ ( ρ ) = ζ ( ρ )Φ( ϕ ) , (28)where Φ( ϕ ) is the eigenfunction of the relative angular momentum operator ˆ ℓ z :Φ( ϕ ) = e isϕ , (29)ˆ ℓ z Φ = s Φ , s = 0 , ± , ± . . . (30)The ensuing equation for ζ is: " − m r ∂ ρ − m r ρ ∂ ρ + s ρ − s ω c + m r ω c ρ V ( ρ ) − E ρ ζ ( ρ ) = 0 . (31)4 Various potentials
In [12], the case of the Coulomb interaction V ( ρ ) = e e ρ between two charges was considered. Here weexplore other potentials which permit exact solutions. We begin with V ( ρ ) = e e ρ + ϑρ + k ρ + k ρ . (32)Then the substitution ζ s ( ρ ) = e − τρ − ηρ ρ ξ p s ( ρ ) (33)furnishes the following equation for p s : " − ρ∂ ρ + (cid:0) τ ρ + 2 ηρ − − ξ (cid:1) ∂ ρ + 14 ρ (cid:0) ω c m r + 8 k m r − τ (cid:1) + ρ (2 k m r − ητ ) ++ ρ (cid:0) τ { ξ } − m r { sω c + 2 E ρ } − η (cid:1) + 8 ϑm r − ξ + 4 s ρ p s = − (cid:2) ǫ + η (1 + 2 ξ ) (cid:3) p s , (34)where ǫ = 2 m r e e . (35)If the coefficients τ , η and ξ are chosen to be τ = 14 p m r ω c + 8 k m r , (36) η = k m r τ = 2 k q ω c + k m r , (37) ξ = p s + 2 ϑm r . (38)the equation (34) becomes: " − ρ∂ ρ + (cid:0) τ ρ + 2 ηρ − − ξ (cid:1) ∂ ρ + ρ (cid:0) τ { ξ } − m r { sω c + 2 E ρ } − η (cid:1) + p s = − (cid:2) ǫ + η (1 + 2 ξ ) (cid:3) p s . (39)Our next goal is to cast the equation (39) into the form (8) or, to be more exact, into the form h − ˆ J n ˆ J − n + ˆ J + n − α ˆ J − n + β ˆ J n i p s = − ν p s , (40)which is the special case of (8). We remind that ˆ J − n , ˆ J + n , ˆ J n are generators of an irreducible n − dimensionalrepresentation of the SL (2 , C ) group.To that end, we insert into (40) the operators in the differential form, and change the variable as cρ → ρ : " − ρ∂ ρ + (cid:18) c ρ + 12 ( n − α ) + βcρ (cid:19) ∂ ρ − nc ρ p s = − (cid:18) ν − βn (cid:19) cp s . (41)5quating of the coefficients in (39) to those in (41) gives us: α = 1 + 2 ξ + n , (42) β = 2 ηc = 4 k m r ( m r ω c + 8 k m r ) / , (43) c = 2 √ τ = p m r ω c + 8 k m r . (44)The energy of the relative motion for a fixed n is: E ρ = 4 τ ( ξ + 1) − sω c m r + c n − η m r = ω c s k ω c m r ( n + 1 + p s + 2 ϑm r ) − s − k (cid:18) ω c + 8 k m r (cid:19) ω c m r . (45)We see that the infinite degeneracy of the energy for | s | >
0, which was the case in [12], does not take placefor nonzero ϑ and k .By comparing RHS of (39) and (41), we find the condition ǫ + η (1 + 2 ξ ) = c (cid:18) ν − βn (cid:19) . (46)Inserting therein the expressions (37), (43), and (44) for η , β and c from, we arrive at ǫ + 2 k q ω c + 8 k m r (1 + 2 p s + 2 ϑm r ) = p m r ω c + 8 k m r ν − n k m r ( m r ω c + 8 k m r ) / ! . (47)This equation serves as a constraint upon the permissible values of the parameters entering our problem.(These include the magnetic field, along with the constants in the expression (32) for the potential.) Permis-sible is a set of values of these parameters, for which the equation (40), or its equivalent (41), has a solution.The values of all these parameters, except one, can be chosen arbitrarily. The remaining parameter will thenassume a discrete set of values, to satisfy the quantisation condition (47). With the potential fixed, this maybe interpreted as ‘quantisation’ of the magnetic field. This interpretation will work also for the next potentialconsidered.In case k , k and ϑ are equal to zero, the above equation reduces to ν = ǫ ω c m r , (48)which coincides with the equation (25) in [12].At this point, two questions emerge. First, what if the expression under the square root in (38) is negative?Second, how to describe the fall on the centre due to the presence of the inverse-square term in the potential?The two issues turn out to be closely connected, and can thus be resolved simultaneously.For ρ → ζ ( ρ ) assumes the form of (cid:18) ∂ ρ + 1 ρ ∂ ρ − Aρ (cid:19) ζ ( ρ ) = 0 , (49) Be mindful that vanishing of k entails the vanishing of β and η . A = s + 2 ϑm r . (50)As is explained in greater detail in [15], no fall towards the centre takes place for A > s + 2 ϑm r > Consider the potential of the form V ( ρ ) = ϑρ + k ρ + k ρ + k ρ . (52)A substitution helpful in this case is ζ s ( ρ ) = e − τρ − ηρ ρ ξ p s ( ρ ) . (53)Insertion thereof into the formula (31) entails the following equation for p s : " − ρ∂ ρ + (cid:0) τ ρ + 4 ηρ − − ξ (cid:1) ∂ ρ + 14 ρ (8 k m r − ητ ) + 14 ρ (cid:0) k m r − τ (cid:1) ++ 14 ρ (cid:16) k m r + ω c m r − (cid:0) η − ξ + 2) τ (cid:1)(cid:17) + ρ (cid:0) η ( ξ + 1) − m r (2 E + sω c ) (cid:1) + 8 ϑm r − ξ + 4 s ρ p s = 0 , (54)wherefrom we find: τ = √ k m r , (55) η = k m r τ = √ m r k √ k , (56) ξ = p s + 2 ϑm r . (57)Then the equation takes form " − ρ∂ ρ + (cid:0) τ ρ + 4 ηρ − − ξ (cid:1) ∂ ρ + 14 ρ (cid:16) k m r + ω c m r − (cid:0) η − ξ + 2) τ (cid:1)(cid:17) ++ ρ (cid:0) η ( ξ + 1) − m r (2 E + sω c ) (cid:1) p s = 0 , (58)We wish to cast the above equation (58) into the standard form (40). To accomplish this, we substitute theoperators in the differential form (the variable in (40) now being called ˜ ρ ), and change the variable as c ˜ ρ → ˜ ρ : " − ˜ ρ∂ ρ + (cid:18) c ˜ ρ + cβ ˜ ρ − α + n (cid:19) ∂ ˜ ρ − c n ˜ ρ p s = − c (cid:18) ν − βn (cid:19) p s . (59)7fter we change the variable one more time, ˜ ρ → ρ , the equation gets simplified to h − ρ∂ ρ + (cid:0) c ρ + 2 βcρ − α + n + 1 (cid:1) ∂ ρ − c nρ + 2 c ( nβ − ν ) ρ i p s = 0 . (60)It acquires the needed form if we choose the parameters as: α = 1 + ξ + n , (61) β = η √ τ = k (cid:18) m r k (cid:19) / , (62) c = 2 √ τ = (2 k m r ) / . (63)The energy of the relative motion will be: E ρ = c ( nβ − ν ) m r + 2 ηm r (1 + ξ ) − sω c , (64)while the quantisation condition will read as14 ω c m r = 4 η − c n − τ ( ξ + 2) − k m r , (65)14 ω c m r = m r k k − p k m r (cid:16) n + 2 p s + 2 ϑm r + 2 (cid:17) − k m r . (66) Finally, let us explore the option V ( ρ ) = l ρ + l ρ + l ρ + l ρ − k ρ . (67)This time the needed substitution and the equation for p s are: ζ s ( ρ ) = e − τρ − ηρ ρ ξ p s ( ρ ) , (68) " − ρ ∂ ρ + (cid:0) ηρ − (2 ξ + 1) ρ − τ (cid:1) ∂ ρ + 14 ρ (cid:0) ω c m r − k m r (cid:1) − ρ (cid:0) Em r + η + sω c m r (cid:1) ++ ρ (2 ηξ + 2 l m r + η ) + 2 l m r − ξτ + τρ + 2 l m r − τ ρ p s = (cid:0) ξ − ητ − s − l m r (cid:1) p s . (69)We compare its coefficients with h − ˆ J − ˆ J + n − α ˆ J − + β ˆ J n + 2 γ ˆ J + n i p s = − νp s . (70)By substituting operators in the differential form, we get h − ρ ∂ ρ + (cid:0) γρ + ρ ( β + n − − α (cid:1) ∂ ρ − γnρ i p s = − (cid:18) ν − βn n (cid:19) p s . (71)8rom where we find: α = τ = p l m r , (72) γ = η = − l m r τl m r + τ ( n + 1) , (73) ξ = 12 + l m r τ , (74) β = − l m r τ − n . (75)The energy of the relative motion is E ρ = − sω c + η m r ! = − sω c + l √ l m r l m r + √ l m r { n + 1 } ) ! . (76)The first constraint on the parameters of the potential has the form of2 l m r − ητ − ξ + s = ν + n (cid:18) − β (cid:19) or, after we substitute the constants:2 l m r − l l m r l m r + √ l m r ( n + 1) − (cid:18)
12 + l m r √ l m r (cid:19) + s = ν + n (cid:18) l m r √ l m r + n (cid:19) . (77)In this case, the quantisation of parameters of the potential cannot be reduced to the quantisation of themagnetic field. Instead, we have one more constraint: ω c m r = 8 k . (78) Let us now discuss the case of e = − e ≡ e and show that under certain circumstances it also rendersexact solutions. Although in the general case the variables in the unitary-transformed Hamiltonianˆ H ′ = ( ˆ P − e B × ρ ) M + ( ˆ p − e ( µ − µ ) A ρ ) m r + V ( | ρ − ρ | ) (79)cannot be separated, we nevertheless can consider R -independent solutions corresponding to a compositeparticle at rest: Ψ ′ ( R , ρ ) = ψ ( ρ ) . (80)The equation for the relative motion takes the form " − ∇ ρ m r − ω q ˆ ℓ z + m r Ω q ρ V ( | ρ | ) − E ρ ψ ( ρ ) = 0 , (81) It is worth noting that in some situations, where the variables in the Hamiltonian cannot be separated, the wave functioncan nevertheless be factorised. q = eB m r (82)and ω q = eB | µ − µ | m r . (83)Further steps are analagous to the case of e c = 0 . The factorisation (28) leads to the following equationfor ζ ( ρ ) : " − m r ∂ ρ − m r ρ ∂ ρ + s ρ − s ω q + m r Ω q ρ V ( ρ ) − E ρ ζ ( ρ ) = 0 , (84)which is almost identical to (31). Below we present an inventory of changes in the solutions for the threeafore-considered potentials. In what follows, ω q always replaces ω c . For the potential of the form (32), we again use the substitute (33), changing this time (36) to τ = 14 q m r Ω q + 8 k m r . (85)The equations (36) – (44) remain unchanged. The energy of the relative motion and the quantisation conditionwill look as E ρ = 4 τ ( ξ + 1) − s Ω q m r + c n − η m r = 12 r q + 8 k m r ( n + 1 + p s + 2 ϑm r ) − sω q − k ω q (cid:16) q + k m r (cid:17) m r , (86) ǫ + 2 k q q + k m r (1 + 2 p s + 2 ϑm r ) = q m r Ω q + 8 k m r ν − n k m r (4 m r Ω q + 8 k m r ) / ! . (87) In this case, we still employ the equations (53) – (63) and arrive at the same expresssion for the energy: E ρ = c ( nβ − ν ) m r + 2 ηm r (1 + ξ ) − sω q , (88)The quantisation condition assumes the form ofΩ q m r = m r k k − p k m r (cid:16) n + 2 p s + 2 ϑm r + 2 (cid:17) − k m r . (89) In this situation, the equations (68) – (75) stay unchanged, and so do the expression for the energy andthe first constraint: E ρ = − sω q + η m r ! = − " sω q + (cid:18) l τ l m r + τ { n + 1 } ) (cid:19) , (90)10 l m r − l l m r l m r + √ l m r ( n + 1) − (cid:18)
12 + l m r √ l m r (cid:19) + s = ν + n (cid:18) l m r √ l m r + n (cid:19) . (91)The second constrain, however, is now subject to change:Ω q m r = 2 k . (92) In the presented paper, we extended the class of quasi-exactly solvable problems. We considered a problemof two charged particles placed in an external magnetic field and interacting with each other through a(generally, non-Coulomb) potential. The first of addresseed potentials was a generalizstion of the potentialstudied in [12]. All three considered Hamiltonians resemble those explored in [1]. They share a commonfeature: if all of the Hamiltonian’s parameters, except one, are fixed, then the exact solutions will exist foran infinite discrete set of values of the remaining parameter. For the first two Hamiltonians, this observationmay be written down as ‘quantisation’ of the magnetic field, with the potential’s parameters fixed. In thethird Hamiltonian, however, the sole permissible value of the magnetic field is unambiguously linked to oneof the fixed parameters; so this time it is one of the parameters of the potential that becomes subject to‘quantisation’.In all three cases, the energy depends on the parameters of the potential and magnetic field, as well as onthe magnetic quantum number s and the dimension n of the representation of the generators ˆ J n . Diagolalisinga finite n × n block of the Hamiltonian, one finds n energy levels and the corresponding polynomial wavefuntions. The author is grateful to Alexander Turbiner for suggesting this problem and for helpful advise. Theauthor also would like to thank Mikhail Shifman for meticulous reading of the manuscript and very judiciouscomments that were of great help.
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