Quantization of the charge in Coulomb plus harmonic potential
QQuantization of the charge in Coulomb plus harmonic potential
Yoon-Seok Choun and Sang-Jin Sin ∗ Department of Physics, Hanyang University, 222 Wangshimni-ro, Sungdong-gu, Seoul, 04763, South Korea (Dated: June 12, 2020)We consider two models where the wave equation can be reduced to the effective Schr¨odingerequation whose potential contains both harmonic and the Coulomb terms, ω r − a/r . The equationreduces to the biconfluent Heun’s equation, and we find that the charge as well as the energy mustbe quantized and state dependent. We also find that two quantum numbers are necessary to countradial degrees of freedom and suggest that this is a general feature of differential equation withhigher singularity like the Heun’s equation.
1. Introduction:
Since Schr¨odinger established theequation in his name, it has been believed that for anyconfining potential, there exists discrete energy levels al-though we may not write the analytic solution explicitly.However, recent experience [1, 2] told us that it may notbe the case. When the potential has higher singularity,we need higher regularity condition. As a consequence,there is no normalizable solution unless potential itself isquantized.In this paper, we consider two models where the waveequation can be reduced to the effective Schr¨odingerequation whose potential contains both harmonic term ω r and the Coulomb term − a/r . The equation of mo-tion reduces to the biconfluent Heun’s equation, and wefind that the charge as well as the energy must be quan-tized. That is, both energy and charge must depends onthe states.We also find that due to the higher singularity, newquantum number appears. For example, in sphericallysymmetric case, apart from the radial quantum number N and two angular ones L, m , one more quantum number K appears. It turns out that only when we combine twoquantum numbers N, K , the full radial degree of free-dom can be counted. We suggest that the presence ofextra quantum numbers to count correct radial degreesof freedom is a general feature of differential equationwith higher singularity like the Heun’s equation.
2. A quark model with Coulomb and linearscalar potential
Lichtenberg et.al[3] found a semi-relativistic Hamilto-nian which leads to a Krolikowski type second order dif-ferential equation [7–9] in order to calculate meson andbaryon masses. In the center-of-mass system, the totalenergy H of two free particles of masses m , m , is H = (cid:113) (cid:126) p c + m c + (cid:113) (cid:126) p c + m c (1)Let S be the Lorentz scalar interaction and V be the in-teraction which is a time component of a 4-vector. Thenit is natural to incorporate the V and S into (1) by mak- ∗ [email protected]; [email protected] ing the replacements H → H + V, m i → m i + 12 S, i = 1 , . (2)We set m = m = 0 and introduce V = − a/r andstudy its effect for the spin-free Hamiltonian which wasproposed for the meson ( q ¯ q ) system in [10–13]. Then wehave (cid:16) E + ar (cid:17) ψ ( r ) = 4 (cid:34) c (cid:18) br (cid:19) + c (cid:18) P r + (cid:126) L ( L + 1) r (cid:19)(cid:35) ψ ( r )(3) where b is a real positive constant and we used (cid:126) p = P r + (cid:126) L ( L +1) r with P r = − (cid:126) ( ∂ ∂r + r ∂∂r ). The lin-ear scalar potential is for the confinement of the quarksbound by a QCD flux string with constant string ten-sion b . Previously, we investigated the model in the case V = 0 [1] and concluded that for the consistency of thespectrum the current quark should have zero mass. Herewe want to introduce V = − a/r and understand its effectin the presence of the confining potential.Factoring out the behavior near r = 0 by ψ ( r ) = r ˜ L f ( r ), above equation becomes d f ( r ) dr + 2( ˜ L + 1) r df ( r ) dr + (cid:18) E − b r + E a r (cid:19) f ( r ) = 0 , (4)where a = a/ (cid:126) c , b = bc/ (cid:126) , E = E/ (cid:126) c and ˜ L = − / (cid:112) ( L + 1 / − a / . If we further factor out the near- ∞ behavior by f ( r ) = exp (cid:0) − b r (cid:1) y ( r ) and introduce ρ = (cid:112) b / r , we get ρ d ydρ + (cid:0) µρ + ερ + ν (cid:1) dydρ + (Ω ρ + β ) y = 0 . (5)with µ = − ε = 0, ν = 2( ˜ L + 1), β = (cid:15)a andΩ = (cid:15) − (2 ˜ L + 3) , with (cid:15) = E / (cid:112) b . (6)This equation is a biconfluent Heun’s equation which hasa regular singularity at the origin and an irregular singu-larity of rank two at the infinity[5, 6].Substituting y ( ρ ) = (cid:80) ∞ n =0 d n ρ n into (5), we obtain therecurrence relation: d n +1 = A n d n + B n d n − for n ≥ , with (7) A n = − εn + β ( n + 1)( n + ν ) , B n = − Ω + µ ( n − n + 1)( n + ν ) . (8) a r X i v : . [ h e p - ph ] M a y For n = 0 term, only d , d appear and give d = A d .Notice that when a = 0, we have A n = β ( n + 1)( n + ν ) = (cid:15)a ( n + 1)( n + 2( ˜ L + 1)) = 0 , (9)so that the three term recurrence relation given in eq.(7) is reduced to two term recurrence relation between d n +1 and d n − and the Heun’s equation is reduced tohypergeometric one. That is, in this scaling, the Coulombparameter is precisely the term increasing the singularityorder. Similarly, if b = 0, the system can also be mappedto a hypergeometric type. The problem rises only whenboth potential terms are present.Now, unless y ( ρ ) is a polynomial, ψ ( r ) is divergent as ρ → ∞ . Therefore we need to impose regularity condi-tions by which the solution is normalizable. If we imposetwo conditions [5, 6], B N +1 = d N +1 = 0 where N ∈ N , (10)the series expansion becomes a polynomial of degree N :as one can see from eq. (7), eq. (10) is sufficient to give d N +2 = d N +3 = · · · = 0 recursively. Then the solution isa polynomial of order N , y N ( ρ ) = (cid:80) Ni =0 d i ρ i . The ques-tion whether imposing both equations in eq(10) is reallynecessary for the finite solution was studied numericallyand was concluded affirmatively in our earlier work [1].In general, d N +1 = 0 will define a N + 1-th order poly-nomial P N +1 in a , so that Eq. (10) gives (cid:15) N,L = (cid:112) N + 2 ˜ L + 3 , P N +1 ( a ) = 0 . (11)where the first comes from B N +1 = 0, and it is noth-ing but the usual energy quantization condition. Belowwe will examine the meaning of the second condition byconstructing explicitly the expressions of a few low orderpolynomial P N +1 , which are given in the appendix.One surprising fact is that for a given N, L , there aremany solutions which we can index by an integer K whichis smaller than N . Depending on whether N is even orodd, the distribution of solutions of P N +1 ( a ) = 0, isdifferent. For low lying L , the number of roots increaseswith L but not regularly. However, for L ≥ [ N/ − N/
2] + 2. Here, [ x ] isthe integer part of x >
0. The presence of extra quan-tum number is natural from the algebraic point of view.But it is rather suprising from the counting degree offreedom. We postpone the dynamics of associated K tonext section where we discuss the problem with a simplermodel.Table I shows some real roots of a ’s for each L withfixed N = 8; here, a i is the i -th root of a with given N, L . Similarly, Table II shows real roots of a ’s for each L when N = 9.For lower value of K , we can find an approximate fit-ting function. For example for K = 0 and for odd N , itis given by a ,NL ≈ .
22 tan − (cid:18) ( L + 1) . − . . N . + 0 . (cid:19) + 0 . . (12) a a a a a L=0 0 none none none noneL=1 0 2.35525 7.90698 none noneL=2 0 2.97179 11.2403 21.9815 noneL=3 0 3.43735 13.3617 28.3483 44.4635L=4 0 3.81341 14.9937 32.6448 54.7228L=5 0 4.12728 16.3243 35.9753 61.791
TABLE I. Roots of a for N = 8. a a a a a L=0 0.374151 none none none noneL=1 0.580422 4.80626 none none noneL=2 0.71935 6.26714 16.0299 24.9066 noneL=3 0.828203 7.32404 19.5432 35.397 48.5634L=4 0.917807 8.17078 22.1613 41.6303 63.7813L=5 0.993589 8.87727 24.2768 46.338 73.3714
TABLE II. Roots of a for N = 9. ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● L a (a) ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● Ν a (b) FIG. 1. (a) Fitting of a data by eq.(12), as functions of L .The lowest line is for N = 1, the top line is for N = 25. (b)Fitting of a data by eq.(12) as functions of odd N with L .the lowest line is for L = 0, the top one is for L = 25. In bothfigures (a)(b), N is odd. We calculated 338 different values of a ’s at various( N, L ) and the result is the dots inFig. 1. These datafits well by above formula. Notice also that for even N , a = 0 is always a solution for any L .By substituting eq.(12) into eq.(11), we can fit the ex-perimental data of E , which is the hadron mass. E N,L (cid:39) (cid:114) (cid:126) c · bc (cid:104) N + 2 + (cid:113) (2 L + 1) − a ,NLK (cid:105) , (13)where a = a/ (cid:126) c . What is surprising is the fact thatthe charge parameter a should be quantized as valuesapproximately given in eq.(12) if the charge is comingin the presence of the linear scalar potential which givesthe confinement. Our treatment gives the analytic resultsin the presence of the both linear potential together withCoulomb potential. However, we must also comment thatin the presence of the quark mass our method breaksdown.
3. Quantum dot with Coulomb and harmonicpotential
Here we consider Non-relativistic Schr¨odingerequation with Coulomb potential and external harmonicoscillator potential for a system of two electrons in a threedimensional Euclidean space [14–17]. The Schr¨odingerequation is given by (cid:20) − (cid:126) m (cid:18) d dr + 2 r ddr (cid:19) + V eff( r ) (cid:21) ψ ( r ) = Eψ ( r ) , (14)with V eff( r ) = ω r − ar + (cid:126) m L ( L + 1) r , (15)Introducing ρ = r (cid:16) mω (cid:126) (cid:17) / , above equation becomes ρ d ψdρ + 2 ρ dψdρ + (cid:18) (cid:15) − ρ + a ρ − L ( L + 1) ρ (cid:19) ψ = 0 . (16)where (cid:15) = Eω (cid:114) m (cid:126) , a = a √ ω (cid:18) (cid:126) m (cid:19) − / . (17)Factoring out the behavior near ρ = 0 by ψ ( ρ ) = ρ L f ( ρ ),it becomes d f ( ρ ) dρ + 2( L + 1) ρ df ( ρ ) dρ + (cid:18) (cid:15) − ρ + a ρ (cid:19) f ( ρ ) = 0 . (18)Factoring out near ∞ behavior by f ( ρ ) = e − ρ / y ( ρ ), weget the standard form eq.(5) with µ = − , ε = 0 , ν = 2( L + 1) , β = a , Ω = (cid:15) − (2 L + 3) . Similarly, if we impose eq.(10), the series expansion be-comes a polynomial of degree N . the solution becomesa polynomial y N ( ρ ) = (cid:80) Ni =0 d i ρ i . In general, d N +1 = 0will define a ( N + 1)-th order polynomial P N +1 in a , sothat Eq. (10) gives (cid:15) N,L = 2 N + 2 L + 3 , P N +1 ( a ) = 0 . (19)where the first comes from B N +1 = 0 which is the energyquantization condition. Below we will examine the mean-ing of the second equation. To do that we need explicitexpressions of a few lower order polynomial P N +1 : P ( a ) = a , P ( a ) = a − L + 1) , P ( a ) = a − L + 5) a , P ( a ) = a − L + 3) a + 144( L + 1)( L + 2) , P ( a ) = a − L + 7) a + 32(89 + 16 L (2 L + 7)) a . (20) In appendix, we gave a few low order polynomial y N ( ρ )with d = 1.We have seen that a and ω are related by eq. (17)and P N +1 ( a ) = 0 does not contain any dimensionfulparameter. This means that a/ √ ω should be a solutionof a polynomial equation, which depends on N, L . Suchextra quantization is a consequence of the Heun’s equa-tion. For the hypergeometric equations, the recurrencerelation is reduced to two term after factoring out theasymptotic behavior. There, we do not have d N +1 = 0.Hence to have a normalizable polynomial solution, we N = 4 N = 5 a a a L=0 0 24.701 115.299L=1 0 41.8531 178.147L=2 0 58.414 241.586L=3 0 74.7438 305.256L=4 0 90.9604 369.04L=5 0 107.114 432.886 a a a L=0 6.38432 64.8131 208.803L=1 10.9664 102.965 306.069L=2 15.2359 140.155 404.609L=3 19.3928 176.898 503.709L=4 23.4959 213.403 603.101L=5 27.5688 249.768 702.664
TABLE III. Roots of a only need to fine tune just one parameter, the energy,For the Heun’s equation, we have to impose two con-straints, which in turn request the charge quantizationof the system. In short, its higher singularity requestshigher regularity condition. This is the origin of thecharge quantization.Notice that a depends on the quantum numbers thatparametrize quantum states. It means that when theelectron make a transition from one state to another, thecharge parameter must be changed to a new value. Thisraises the question, how dynamics of one particle canchange the potential energy which is determined by thesurrounding system. In fact, V is not the potential butthe potential energy. The potential belongs to the sur-roundings while the potential energy contains both sur-rounding and particle information. Therefore a shouldbe written as product of particle’s charge q times thecharge Q which makes the potential φ Q , so that V = qφ Q .When one say charge is quantized, what we mean isthe quantization of q . In short, when the potential en-ergy has higher singularity, the charge as well as the en-ergy should depends on the state. At first, this conceptwas rather drastic, but this is consequence of request-ing d N +1 , whose necessity was confirmed by numericalinvestigation: without it, the shooting method did notwork.Notice that in this model, the energy (cid:15) is linear in N, L and does not depend on a quantized value of a . Table IIIshows all roots of a ’s for each L for N = 4 , a depends on threequantum number, we choose the K = 0 sector of a withgiven ( N, L ). Then, Fig. 2 shows us that a is roughlylinear in N, L for odd N . For even N , the K = 0 sectorgives a = 0.For the figure, we calculated 338 different values of a ’s at various ( N, L ). From the explicit calculation, wefind the following pattern: List N+1 a in the increasingorder such that a ,K is K -th one, K = 0 , , · · · , [ N/ − x ] means interger part of the positive real number x . Then although the total number of nodes is N , someof them are in the negative region of ρ . The polynomialwith a K has N − (cid:98) N/ (cid:99) + K nodes in the region ρ > K counts the number of nodes that crossed ρ = 0 compared with K = 0 in the positive domain.In three dimension, spinless Hydrogen atom has threequantum number: N, L, m : N for radial and the othertwo for angular momentum. However, in the pres-ence of the harmonic potential, the charge and energy ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● L a (a) ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● Ν a (b) FIG. 2. Roots of the smallest a , as function of L and N. In(a) the lowest line is for N = 1, the top line is for N = 25.the lowest line is for L = 0, the top one is for L = 25. In bothfigures (a)(b), N is odd. - - - ρ - - - ( ρ ) (a) K = 0, L = 0 - - - ρ ( ρ ) (b) K = 1, L = 0 - ρ - ( ρ ) (c) K = 2, L = 0 FIG. 3. y for various a K with K = 0 , , L = 0.For each a K , the number of positive roots is given by N −(cid:98) N/ (cid:99) + K = K + 3. have discrete values depending on four quantum numbers N, L, m, K , which shows apparent mismatch between thenumber of degrees of freedom and that of quantum num-bers. However, as we have shown above, with K = 0,only half of the nodes of radial wave function are onpositive region. This means that the radial solution forfixed K , say K = 0, can not span arbitrary shape of ra- dial function in the positive region. In fact, K counts thenumber of nodes which is moved from negative to positveregion compared with K = 0 case. This means that N together with K counts full radial degrees of freedom,and without the extra quantum number K , the solutionscan not be a basis of the radial wave functions.We expect that the presence of extra quantum numberto count correct radial degrees of freedom is a general fea-ture of differential equation with higher singularity likethe Heun’s equation.
4. Discussion : Caruso et.al[14] investigated non-relativistic 2-D radial Schr¨odinger equation which canbe related to ours just by shifting L to L + 1 / ω ,the coefficient of the harmonic potential. The quantiza-tion of ω would imply that the single particle dynamicschanges the potential’s parameter, which does not soundplausible. In our case, a is split into particle charge q andcharge Q in the potential, so that Couomb term can bewritten as V Coulomb = qφ Q ( r ). q is a property of the par-ticle, therefore dependence of the particle charge on thestate is natural although the concept is still not familiarso far. In field theory, charge depends on probe energyscale due to the renormalization. So the state dependenceof the charge can be regarded as discrete renormalizationof the charge induced by smoothing out process of the thesingularity of the potential. ACKNOWLEDGEMENTS
This work is supported by Mid-career Researcher Pro-gram through the National Research Foundation of Ko-rea grant No. NRF-2016R1A2B3007687. [1] Y. S. Choun and S. J. Sin, “Chiral symmetry and Heun’sequation,” arXiv:1909.07215 [hep-ph], Int. J. Mod. Phys.A , No.07, 2050038(2020).[2] Y. S. Choun and S. J. Sin, “Bridging the Chiral symme-try and Confinement with Singularity.,” Phys. Lett. B , No.07, 135433(2020).[3] Lichtenberg, D. B., Namgung, W., Predazzi, E. andWills,J. G., “Baryon masses in a relativistic quark-diquark model,” Phys. Lett. , 1653(1982).[4] G¨ursey, F., Comments on hardronic mass formulae, inA. Das., ed., From Symmetries to Strings: Forty Yearsof Rochester Conferences , World Scientific, Singapore,(1990).[5] Ronveaux, A.,
Heun Differential Equations , Oxford Uni-versity Press, (1995).[6] Slavyanov, S. Yu., Lay W.
Special Functions: A Uni-fied Theory Based on Singularities , Oxford MathematicalMonographs, Oxford University Press, Oxford, (2000).[7] Krolikowski, W., “Relativistic three-body equation forone Dirac and two Klein-Gordon particles,” Acta Phys.Pol. B. (5), 387–391(1980). [8] Krolikowski, W., “Solving nonperturbatively the breitequation for parapositronium,” Acta Phys. Pol. B. (9),891–895(1980).[9] Todorov, I. T., “Quasipotential Equation Correspondingto the Relativistic Eikonal Approximation,” Phys. Rev.D , 2351(1971).[10] Catto, S. and G¨ursey, F., “Algebraic treatment of effec-tive supersymmetry,” Nuovo Cim. A. (1985)201.[11] Catto, S. and G¨ursey, F., “New realizations of hadronicsupersymmetry,” Nuovo Cim. A, (1985)685.[12] Catto, S., Cheung, H. Y., Gursey, F., “Effective Hamilto-nian of the relativistic Quark model,” Mod. Phys. Lett.A , (1991)3485.[13] G¨ursey, F., Comments on hardronic mass formulae, inA. Das., ed., From Symmetries to Strings: Forty Yearsof Rochester Conferences , World Scientific, Singapore,(1990).[14] Caruso, F., Martins, J., Oguri, V., “Solving a two-electron quantum dot model in terms of polynomial solu-tions of a biconfluent Heun equation,” Ann. Phys. ,130 (2014). [15] Reimann, S.M., Manninen M., “Electronic structure ofquantum dots,” Reviews of Modern Physics, , 1283(2002). [16] Sikorski, Ch., Merkt, U., , “Spectroscopy of electronicstates in InSb quantum dots,” Physical Review Letters , 2164 (1989).[17] Merkt U., Huser, J., Wagner, M., “Energy spectra of twoelectrons in a harmonic quantum dot,” Physical ReviewB , 7320 (1991). Appendix A: P N for N = 0 , , ..., P ( a ) = a , P ( a ) = a (cid:18) (cid:113) (2 L + 1) − a (cid:19) − (cid:18) (cid:113) (2 L + 1) − a (cid:19) , P ( a ) = a (cid:18) − a (cid:18) (cid:113) (2 L + 1) − a (cid:19) + 12 + 8 (cid:113) (2 L + 1) − a (cid:19) , P ( a ) = − a + a (cid:18)
85 + 4 L ( L + 1) + 16 (cid:113) (2 L + 1) − a (cid:19) − a (cid:18)
47 + 10 L ( L + 1) + 25 (cid:113) (2 L + 1) − a (cid:19) + 144 (cid:18) L + L + 1 + (cid:113) (2 L + 1) − a (cid:19) (A1) Appendix B: y N ( ρ ) polynomials for N = 0 , , ..., We lists expressions of a few lower order polynomial y N ( ρ ): y ( ρ ) = 1 ,y ( ρ ) = 1 − a ρ L ) ,y ( ρ ) = 1 + ( a − L + 1)) ρ − a (2 L + 3) ρ L + 1)(2 L + 3) y ( ρ ) = 1 + a (36 + 28 L − a ) ρ + 6( L + 2)( a − L + 1)) ρ − L + 2)(2 L + 3) a ρ L + 1)( L + 2)(2 L + 3) y ( ρ ) = 1 + 196( L + 1)( L + 2)(2 L + 3)(2 L + 5) (cid:40) (384( L + 1)( L + 2) − L ) a + a ) ρ + 4(2 L + 5)(52 + 40 L − a ) a ρ − L + 2)(2 L + 5)(16 + 16 L − a ) ρ − L + 2)(2 L + 3)(2 L + 5) a ρ (cid:41) y ( ρ ) = 1 + 1960( L + 1)( L + 2)( L + 3)(2 L + 3)(2 L + 5) (cid:40) − a (6880 + 2384 L + 24 L (354 − a ) + a ( a − ρ + 10( L + 3)(720( L + 1)( L + 2) − L ) a + a ) ρ + 40 a ( L + 3)(2 L + 5)(68 + 52 L − a ) ρ − L + 2)( L + 3)(2 L + 5)(20 + 20 L − a ) ρ − a ( L + 2)( L + 3)(2 L + 3)(2 L + 5) ρ (cid:41)(cid:41)