We study a harmonic molecule confined to a one-dimensional box with impenetrable walls. We explicitly consider the symmetry of the problem for the cases of different and equal masses. We propose suitable variational functions and compare the approximate energies given by the variation method and perturbation theory with accurate numerical ones for a wide range of values of the box length. We analyze the limits of small and large box size

Amore, P.

Fernández, F. M.

English

P. Amore

F. M. Fernández

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Acta Polytechnica

Two-particle Harmonic Oscillator in a One-dimensional Box

CTU Open Journal Systems (Czech Technical University, Prague / České vysoké učení technické v Praze)

Acta Polytechnica Vol. 50 No. 5/2010Two-particle Harmonic Oscillator in a One-dimensional BoxP. Amore, F. M. Ferna´ndezAbstractWe study a harmonic molecule conﬁned to a one-dimensional box with impenetrable walls. We explicitly consider thesymmetry of the problem for the cases of diﬀerent and equal masses. We propose suitable variational functions andcompare the approximate energies given by the variation method and perturbation theory with accurate numerical onesfor a wide range of values of the box length. We analyze the limits of small and large box size.Keywords: harmonic oscillator, diatomic molecule, conﬁned system, one-dimensional box, point symmetry, avoidedcrossings, perturbation theory, variational method.1 IntroductionDuring the last decades, there has been great interest in the model of a harmonic oscillator conﬁned to boxesof diﬀerent shapes and sizes [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]. Sucha model has been suitable for the study of several physical problems, ranging from dynamical friction in starclusters [4] to magnetic properties of solids [6] and impurities in quantum dots [23].One of the most widely studied models is given by a particle conﬁned to a box with impenetrable walls at−L/2 and L/2 bound by a linear force that produces a parabolic potential–energy function V (x) = k(x−x0)2/2,where |x0| < L/2. When x0 = 0 the problem is symmetric and the eigenfunctions are either even or odd; suchsymmetry is broken when x0 = 0. Although interesting in itself, this model is rather artiﬁcial because the causeof the force is not speciﬁed. It may, for example, arise from an inﬁnitely heavy particle clamped at x0. In sucha case we think that it is more interesting to consider that the other particle also moves within the box.The purpose of this paper is to discuss the model of two particles conﬁned to a one-dimensional box withimpenetrable walls. For simplicity we assume that the force between them is linear. In Sec. 2 we introducethe model and discuss some of its general mathematical properties. In Sec. 3 we discuss the solutions of theSchro¨dinger equation for small box lengths by means of perturbation theory. In Sec. 4 we consider the regime oflarge boxes and propose suitable variational functions. In Sec. 5 we compare the approximate energies providedby perturbation theory and the variational method with accurate numerical methods. Finally, in Sec. 6 wesummarize the main results and draw additional conclusions.2 The ModelAs mentioned above, we are interested in a system of two particles of masses m1 and m2 conﬁned to a one-dimensional box with impenetrable walls located at x = −L/2 and x = L/2. If we assume a linear force betweenthe particles then the Hamiltonian operator readsHˆ = − h¯22(1m1∂2∂x21+1m2∂2∂x22)+k2(x1 − x2)2 (1)and the boundary conditions are ψ = 0 when xi = ±L/2. It is convenient to convert it to a dimensionless formby means of the variable transformation qi = xi/L that leads to:Hˆd =m1L2h¯2Hˆ = −12(∂2∂q21+ β∂2∂q22)+λ2(q1 − q2)2 (2)where β = m1/m2, λ = km1L4/h¯2 and the boundary conditions become ψ = 0 if qi = ±1/2. Without loss ofgenerality we assume that 0 < β ≤ 1.The free problem (−∞ < xi <∞) is separable in terms of relative and center-of-mass variablesx = x1 − x2X =1M(m1x1 +m2x2) , M = m1 +m2 (3)17Acta Polytechnica Vol. 50 No. 5/2010respectively, that lead toHˆ = − h¯22(1M∂2∂X2+1m∂2∂x2)+ V (x), m =m1m2M. (4)In this case we can factor the eigenfunctions asψKv(x1, x2) = eiKXφv(x)−∞ < K <∞, v = 0, 1, . . . (5)where φv(x) are the well–known eigenfunctions of the harmonic oscillator, and the eigenvalues readEKv =h¯2K22M+ h¯√km(v +12). (6)However, because of the boundary conditions, any eigenfunction is of the form ψ(x1, x2) = (L2/4 − x21)(L2/4 − x22)Φ(x1, x2), where Φ(x1, x2) does not vanish at the walls. We clearly appreciate that the separa-tion just outlined is not possible in the conﬁned model.When β < 1 the transformations that leave the Hamiltonian operator (including the boundary conditions)invariant are: identity Eˆ : (q1, q2) → (q1, q2) and inversion ıˆ : (q1, q2) → (−q1,−q2). Therefore, the eigenfunc-tions of Hˆd are the basis for the irreducible representations Ag and Au of the point group S2 [24] (also calledCi by other authors).On the other hand, when β = 1 (equal masses) the problem exhibits the highest possible symmetry. Thetransformations that leave the Hamiltonian operator (including the boundary conditions) invariant are: identityEˆ : (q1, q2)→ (q1, q2), rotation by π C2 : (q1, q2)→ (q2, q1), inversion ıˆ : (q1, q2)→ (−q1,−q2), and reﬂection ina plane perpendicular to the rotation axis σh : (q1, q2)→ (−q2,−q1). In this case, the states are basis functionsfor the irreducible representations Ag, Au, Bg, and Bu of the point group C2h [24].3 Small boxWhen λ 1 we can apply perturbation theory choosing the unperturbed or reference Hamiltonian operator tobe Hˆ0d = Hˆd(λ = 0). Its eigenfunctions and eigenvalues are given byϕ(0)n1,n2(q1, q2) =⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩2 cos[(2i− 1)πq1] cos[(2j − 1)πq2] Ag2 sin(2iπq1) sin(2jπq2) Ag2 cos[(2i− 1)πq1] sin(2jπq2) Au2 sin(2iπq1) cos[(2j − 1)πq2] Au, i, j = 1, 2, . . .(0)n1,n2 =π22(n21 + βn22), n1, n2 = 1, 2, . . . (7)There is no degeneracy when β < 1, except for the accidental one that takes place for particular values of βwhich we will not discuss in this paper. The energies corrected to ﬁrst order read[1]n1,n2 =π22(n21 + βn22)+ λπ2n21n22 − 3(n21 + n22)12π2n21n22. (8)When β = 1 the zeroth–order states ϕ(0)n1,n2 and ϕ(0)n2,n1 (n1 = n2) are degenerate, but it is not necessary toresort to perturbation theory for degenerate states in order to obtain the ﬁrst-order energies. We simply takeinto account that the eigenfunctions of Hˆ0d adapted to the symmetry of the problem are18Acta Polytechnica Vol. 50 No. 5/2010ϕ(0)n1,n2(q1, q2) =⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩√2− δij {cos[(2i− 1)πq1] cos[(2j − 1)πq2]+ cos[(2j − 1)πq1] cos[(2i− 1)πq2]} Ag√2 {cos[(2i− 1)πq1] cos[(2j − 1)πq2]− cos[(2j − 1)πq1] cos[(2i− 1)πq2]} Bg√2 {cos[(2i− 1)πq1] sin[2jπq2] + sin[2jπq1] cos[(2i− 1)πq2]} Au√2 {cos[(2i− 1)πq1] sin[2jπq2]− sin[2jπq1] cos[(2i− 1)πq2]} Bu√2− δij {sin[2iπq1] sin[2jπq2] + sin[2jπq1] sin[2iπq2]} Ag√2 {sin[2iπq1] sin[2jπq2]− sin[2jπq1] sin[2iπq2]} Bg(9)where i, j = 1, 2, . . . They give us the energies corrected to ﬁrst order as [1]n (S) =〈ϕ(0)ij (S)∣∣∣ Hˆd ∣∣∣ϕ(0)ij (S)〉 whereS denotes the irreducible representation. Since some of these analytical expressions are rather cumbersome forarbitrary quantum numbers, we simply show the ﬁrst six energy levels for future reference:[1]1 (Ag) = π2 +λ(π2 − 6)12π2[1]1 (Au) =5π22+λ(108π4 − 405π2 − 4 096)1 296π4[1]1 (Bu) =5π22+λ(108π4 − 405π2 + 4 096)1 296π4[1]2 (Ag) = 4π2 +λ(2π2 − 3)24π2[1]3 (Ag) = [1]1 (Bg) = 5π2 +λ(3π2 − 10)36π2. (10)The degeneracy of the approximate energies denoted [1]3 (Ag) and [1]1 (Bg) is broken at higher perturbationorders as shown by the numerical results in Sec. 5.4 Large boxWhen L → ∞ the energy eigenvalues tend to those of the free system (6). More precisely, we expect that thestates with ﬁnite quantum numbers n1, n2 at L = 0 correlate with those with K = 0 when L→∞:limλ→∞(β, λ)√λ=√1 + β(v +12), v = 0, 1, . . . (11)Besides, we should take into account that the symmetry of a given state is conserved as L increases from 0 to∞.When β < 1 we expect that the states approachψKv(x,X) =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩cos(KX)φ2v(x) Agsin(KX)φ2v+1(x) Agsin(KX)φ2v(x) Aucos(KX)φ2v+1(x) Au(12)as L→∞.19Acta Polytechnica Vol. 50 No. 5/2010For the more symmetric case β = 1 the states should beψKv(x,X) =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩cos(KX)φ2v(x) Agsin(KX)φ2v(x) Aucos(KX)φ2v+1(x) Busin(KX)φ2v+1(x) Bg. (13)Obviously, perturbation expressions (8) or (10) are unsuitable for this analysis and we have to resort to otherapproaches.In order to obtain accurate eigenvalues and eigenfunctions for the present model we may resort to theRayleigh-Ritz variational method and the basis set of eigenfunctions of Hˆ0d given in equations (7) and (9).Alternatively, we can also make use of the collocation method with the so-called little sinc functions (LSF)that proved useful for the treatment of coupled anharmonic oscillators [25]. In this paper we choose the latterapproach.Another way of obtaining approximate eigenvalues and eigenfunctions is provided by a straightforwardvariational method proposed some time ago [26]. The trial functions suitable for the present model are of theformϕ(q1, q2) =(14− q21)(14− q22)f(c, q1, q2)e−a(q1−q2)2(14)where c = {c1, c2, . . . , cN} are linear variational parameters, which would give rise to the well known Rayleigh-Ritz secular equations, and a is a nonlinear variational parameter. Even the simplest and crudest variationalfunctions provide reasonable results for all values of λ, as shown in Sec. 5.The simplest trial function for the ground state of the model with β < 1 isϕ(q1, q2) =(14− q21)(14− q22)e−a(q1−q2)2. (15)Note that this function is the basis for the irreducible representation Ag. We calculate w(a, λ) =〈ϕ∣∣∣Hˆd∣∣∣ϕ〉 /〈ϕ|ϕ〉 and obtain λ(a) from the variational condition ∂w/∂a = 0 so that [w(a, λ(a)), λ(a)] is a suitable parametricrepresentation of the approximate energy. In this way we avoid the tedious numerical calculation of a for eachgiven value of λ and obtain an analytical parametric expression for the energy, which we do not show herebecause it is rather cumbersome. We just mention that the parametric expression is valid for a > a0 where a0is the greatest positive root of λ(a) = 0.When β = 1 we choose the following trial functions for the lowest states of each symmetry typeϕAg (q1, q2) =(14− q21)(14− q22)e−a(q1−q2)2ϕAu(q1, q2) =(14− q21)(14− q22)(q1 + q2)e−a(q1−q2)2ϕBu(q1, q2) =(14− q21)(14− q22)(q1 − q2)e−a(q1−q2)2ϕBg (q1, q2) =(14− q21)(14− q22)(q21 − q22)e−a(q1−q2)2. (16)5 ResultsFig. 1 shows the ground-state energy for β = 1/2 calculated by means of perturbation theory, the LSF methodand the variational function (15) for small and moderate values of λ. Fig. 2 shows the results of the lattertwo approaches for a wider range of values of λ. We appreciate the accuracy of the energy provided by thesimple variational function (15) for all values of λ. The reader will ﬁnd all the necessary details about theLSF collocation method elsewhere [25]. Here we just mention that a grid with N = 60 was suﬃcient forthe calculations carried out in this paper. Fig. 3 shows that (λ)/√λ calculated by the same two methodsfor β = 1/2 approaches√3/8 as suggested by equation (11). Fig. 4 shows the ﬁrst six eigenvalues (λ) forβ = 1/2 calculated by means of the LSF collocation method. The level order to the left of the crossings20Acta Polytechnica Vol. 50 No. 5/20100 50 100 150 200λ8101214εFig. 1: Perturbation theory (dotted line), LSF (points) and variational (solid line) calculation of the ground-state energy(λ) for β = 1/20 500 1000 1500 2000λ1015202530εFig. 2: Variational (line) and LSF (points) calculation of the ground-state (λ) for β = 1/210-1 100 101 102 103 104 105 106λ012345678910ε/λ1/2Fig. 3: Variational (line) and LSF (points) calculation of the ground-state (λ)/√λ for β = 1/2. The horizontal linemarks the limit√3/821Acta Polytechnica Vol. 50 No. 5/20100 200 400 600 800 1000λ01020304050607080εFig. 4: First six eigenvalues for β = 1/2 calculated by means of the LSF method0 50 100 150 200λ010203040506070εFig. 5: First six eigenvalues for β = 1. Circles, dotted line and solid line correspond to the LSF collocation approach,perturbation theory and variation method, respectively. The level order is Ag < Au < Bu < 2Ag < 3Ag < Bg100 101 102 103 104 105 106λ110100ε / λ1/2A gAuBuBgFig. 6: Variational (solid line) and LSF (symbols) calculation of (λ)/√λ for β = 1. The horizontal lines mark the limits1/√2 and 3/√222Acta Polytechnica Vol. 50 No. 5/2010is 1(Ag) < 1(Au) < 2(Au) < 2(Ag) < 3(Ag) < 3(Au). Note the crossings between states of diﬀerentsymmetry and the avoided crossing between the states 2Au and 3Au.Fig. 5 shows the ﬁrst six eigenvalues for β = 1 calculated by means of perturbation theory, the LSF methodand the variational functions (16) for small values of λ. The energy order is 1(Ag) < 1(Au) < 1(Bu) <2(Ag) < 3(Ag) < 1(Bg) and we appreciate the splitting of the energy levels 3(Ag) and 1(Bg) that doesnot take place at the ﬁrst order of perturbation theory, as discussed in Sec. 3. Finally, Fig. 6 shows (λ)/√λfor suﬃciently large values of λ. We appreciate that the four simple variational functions (16) are remarkablyaccurate and that (λ)/√λ → 1/√2 for the ﬁrst two states of symmetry Ag and Au and (λ)/√λ → 3/√2 forthe next two ones of symmetry Bu and Bg. These results are consistent with equation (13), which suggests thatthe energies of the states with symmetry A and B approach√2(2v + 1/2) and√2(2v + 3/2), respectively. Infact, Fig. 6 shows four particular examples with v = 0.6 ConclusionsThe model discussed in this paper is diﬀerent from those considered before [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15, 16, 17, 18, 19, 20, 21, 22, 23], because in the present case the linear force is due to the interaction betweentwo particles. Although the interaction potential depends on the distance between the particles the problemis not separable and should be treated as a two-dimensional eigenvalue equation. It is almost separable for asuﬃciently small box because the interaction potential is negligible in such a limit, and also for a suﬃcientlylarge box where the boundary conditions have no eﬀect. It is convenient to take into account the symmetry ofthe problem and classify the states in terms of the irreducible representations because it facilitates the discussionof the connection between both regimes.The model may be suitable for investigating the eﬀect of pressure on the vibrational spectrum of a diatomicmolecule, and in principle one can calculate the spectral lines by means of the Rayleigh-Ritz or the LSFcollocation method [25].The simple variational functions developed some time ago [26] and adapted to present problem in Sec. 4provide remarkably accurate energies for all box size values, and are, for that reason, most useful for showingthe connection between the two regimes and for verifying the accuracy of more elaborate numerical calculations.References[1] Auluck, F. 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Phys., 80 (4), 1984, p. 1 569–1 575.Dr. Paolo AmoreE-mail: paolo.amore@gmail.comFacultad de Ciencias, CUICBASUniversidad de ColimaBernal Dı´az del Castillo 340, Colima, Colima, MexicoDr. Francisco M. Ferna´ndezE-mail: fernande@quimica.unlp.edu.arINIFTA (Conicet, UNLP)Division Quimica TeoricaDiagonal 113 y 64 S/NSucursal 4, Casilla de correo 16, 1900 La Plata, Argentina24

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Two-particle Harmonic Oscillator in a One-dimensional Box

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