We have applied partial algebraization technique to the cut-off Coulomb potential −Ze2/(r + β). It has been found that for a spin j representation 2j + 1 exact solutions are obtained but they belong to excited states of potentials for different values of β. Degeneracies observed in the spectrum have been compared with exact numerical results.Primijenili smo tehniku djelomične algebrizacije na odrezani Coulombov potencijal −Ze2/(r+β). Nađeno je da se za spin j dobiva 2j+1 točnih rješenja, koja pripadaju pobuđenim stanjima potencijala za različite vrijednosti β. Opažene degeneracije u spektru uspoređene su s točnim numeričkim rezultatima

Panchanan, Swati

Roychoudhury, Rajkumar

Varshni, Y. P.

English

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ISSN1330–0016CODENFIZBE7PARTIAL ALGEBRAIZATION AND CUT-OFF COULOMB POTENTIALSWATI PANCHANAN∗, RAJKUMAR ROYCHOUDHURY+andY. P. VARSHNI++∗Ananda Ashram Sarada Vidyapeeth, 104, Barrackpore Trunk Road, Calcutta - 700 035,India+Electronics Unit, Indian Statistical Institute, 203, Barrackpore Trunk Road, Calcutta -700 035, India++Department of Physics, University of Ottawa, Ottawa, Canada - IC IN 6N5Received 7 October 1992UDC 530.145Original scientific paperWe have applied partial algebraization technique to the cut-off Coulomb potential−Ze2/(r + β). It has been found that for a spin j representation 2j + 1 exactsolutions are obtained but they belong to excited states of potentials for differentvalues of β. Degeneracies observed in the spectrum have been compared with exactnumerical results.1. IntroductionRecently the partial algebraic technique has been applied to a class of problemswhich are partially solvable1−3). (For a recent exposition see Karman and Oliver4)).Relation between this technique and that of supersymmetric quantum mechanicshas also been shown by Roy and Roychoudhury5). Attempts has been made toapply partial algebraization6) to the nonpolynomial potentialV (x) =x21 + gx2.FIZIKA B 2 (1993) 1, 11–19 11panchanan et al.: partial algebraization . . .However this technique does not give any non-trivial results for the non-polynomialpotential. This motivated us to see whether another important nonlinear type po-tential viz. the cut-off Coulomb potential −Ze2/(r + β) can be treated by thistechnique. We show that a hidden SU(2) symmetry exists for this potential andthough the general formula for En for any n can be found explicitly independent ofβ, the restriction on β itself is such that for each β only one solution is obtained.However, we found that there is a Coulomb like degeneracy in those solutions byhaving recourse to numerical analysis.2. ResultsBefore we cast the cut-off Coulomb potential problem in SU(2) symmetric formwe briefly describe the method of partial algebraization. Given a Schrödinger equa-tionHψ = Eψ , (1)we perform an imaginary gauge transformation on the wave function ψ(x)7)ψ(x) → ψ(x)e−f(x) , (2)thenH = −12d2dx2+ V (x) , (3)HG = −12d2dx2+A(x)ddx+∆V , (4)where∆V = V (x) +12A′(x)−12A2(x) , (5)whilef(x) =∫A(x)dx . (6)The gauge transformed eigenvalue equation readsHGψ̃(x) = Eψ̃(x) (7)Next we consider a finite dimensional representation of the SU(2) group with spins.The generators of the group areT+ = 2jξ − ξ2ddξ12 FIZIKA B 2 (1993) 1, 11–19panchanan et al.: partial algebraization . . .T 0 = −jξ − ξ2ddξ(8)T− =ddξThe corresponding finite dimensional representation isRj = (1, ξ, ξ2, . . . , ξ2j) . (9)We choose the gauge such a way that HG can be written asHG =∑a,b±0CabTaT b +∑a,b±0CaTa + constant, (10)where Cab and Ca are numerical coefficients. Using (8), (10) can be written asHG = −12P4(ξ)d2dξ2+ P3(ξ)ddξ+ P2(ξ) , (11)where Pn(ξ) denotes at most a polynomial of degree n.To bring (11) in Schrödinger like form we putx =∫dξP−1/24 (ξ) = F (ξ), say, (12)thenHG = −12d2dx2+P ′4 + 4P34P1/24ddx+ P2 . (13)Now the basis can be chosen as{ψ̄}+ (1, ξ, ξ2, . . . , ξ2j , ψ̄2j+2, ψ̄2j+3, . . .) (14)whereξ = F−1(x) , (15)and ψ̄2j+k is an arbitrary set of functions orthogonal to (1, ξ, . . . , ξ2j) with weightexp(−2f(x)). Then H has the formH =(H1 00 H2), (16)where H1 is finite matrix and can be diagonalized.FIZIKA B 2 (1993) 1, 11–19 13panchanan et al.: partial algebraization . . .With these tools we now attempt to solve the cut-off Coulomb problem8) forwhich the Schrödinger equation is (h̄ = m = c = 1)−12φ′′(r)−12φ′(r) + VE(r)φ(r) = Eφ(r) (17)where we have written the radial part of the Schrödinger equation andVE = V (r) +l(l + 1)2r2= −Ze2r + β+l(l + 1)2r2. (18)(Henceforth we shall take Ze2 = 1). Putting ψ = r−1ϕ(r), equation (16) reducestod2dr2φ+ (2E − 2VE(r))φ(r) = 0 . (19)Now we choose the gauge f(r) as (here we take ξ = r)f(r) = ar − (l + 1) log r − log(r + β) . (20)A(r) is given by [see equation (5)]A(r) =ddrf(x) = a−l + 1r−1r + β(21)∆V (r) = VE(r) +l + 1r−12(a−l + 1r−1r + β)2+12(r + β)2= V (r) +l + 1r− a−a22+(a+l + 1β)12(r + β)2−l + 1rβ(22)and finallyHG = −12d2dr2+(a−l + 1r−1r + β)ddr+∆V (r) . (23)Here we make a slight improvisation of the partial algebraization method followingRef. 9ΩG = 2r(r + β)(E −Hg) , (24)thenΩG = (r2 + rβ)d2dr2−(2ar2 + 2arβ − 2r(l + 1)− 2r − 2β(l + 1)) ddr+r(2− 2al − 4a)− 2aβ(l + 1) + 2(l + 1) , (25a)14 FIZIKA B 2 (1993) 1, 11–19panchanan et al.: partial algebraization . . .where we have takena2 = −2E , (25b)to remove r2 term from ΩG. The eigenvalue equation now readsΩGψ̃ = 0, where ψ̃ = ψef(r) (25c)We immediately see that ΩG can be expressed in the form given by the r.h.s. of(9). Explicitly we write,ΩG = AT20 +DT−T 0 + FT−T+ +GT+ +HcT− + IT 0≡ (A− F )r2d2dr2+Drd2dr2−Gr2ddr(26)+(A− 2jA+ 2jF − 2F + I)rddr+ (Hc − jD +D)ddr+2jGr +Aj2 + 2jF − 2jGr − Ij (27)(Hc has nothing to do with H, the Hamiltonian).Comparing with (25a) and equating the coefficients of r2 d2dr2 , r2 ddr etc., sepa-rately, one gets 7 simultaneous equations in A, F , G etc. Solving them we get (forj /=0)A =2jj + 1+2(l + 1)(1− aβ)j(j + 1)−2aβ − 2(l + 2)j + 1(28)D = β (29)G = 2a =22j + l + 2(30)Hc =β2l + j + 1(31)F =j − 1j + 1+2(l + 1)(1− aβj(j + 1)−2aβ − 2(l + 2)j + 1)(32)I =2j+2j − 2j + 1+2(l + 1)(1− aβ)j(j + 1)−2(j + 2)(aβ − 1− 2)j + 1(33)FIZIKA B 2 (1993) 1, 11–19 15panchanan et al.: partial algebraization . . .From (22b) and (30),−E =12(2j + l + 2)2(34)Though E is apparently independent of β, it depends on β through l. Because, aswill be shown below, if we fix l, β is also fixed. Below we discuss several cases.1. j = 0For this case solutions (28), (32) and (33) will be inconsistent unlessaβ = 1 (35)orβ =1a= l + 2 from Eq. (30))As expected there is only one solution for a fixed l which is given byψ̃ = C0 (36)(C0 being a constant), orψ = C0rl+1(r + β)e−r/β (37)andE = −a22= −12(l + 2)2(38)2. j = 12Here β can not be determined from the set of equations (28) to (34). However fori = 1/2, ψ̃ will be at most a linear function of r and we follow the method of Ref.7 to determine ψ̃. We writeψ̃ = C1(r − r0) . (39)To find r0 we put (39) in (25c) and equate powers of r2, r etc. to zero separately.After some elementary algebra we geta =1l + 3(40)r0 = (a− 1/β)−1 (41)and(l + 2)a2β2 − 3aβ(l + 2) + (2l + 3) = 0 . (42)16 FIZIKA B 2 (1993) 1, 11–19panchanan et al.: partial algebraization . . .From (42)aβ =3(l + 2)± (9(l + 2)2 − 4(2l + 3)(l + 2))1/22(l + 2). (43)Hence for same l we get two different values of β. However, since r0 will have twodifferent signs for these different values of β, the eigenvalue will be of ground stateTable 1.l βE0 (groundstate)E∗10 2.0 −.1230000 −.05447411 3.0 −.0555556 −.03105462 4.0 −.0312500 −.01994773 5.0 −.0200000 −.01387104 6.0 −.0138881 −.01019695 7.0 −.0102041 −.00780936 8.0 −.0078125 −.00617127 9.0 −.0061728 −.0049992∗Numerical results.Table 2.l β r0 E∗1 E1 E∗20 1.902 5.1762 −.05551(7.098) (−5.1962)1 2.945 11.16515 −.056013 −.03125 −.01996(9.0551) (−7.1652)2 3.964 19.1421 −.03125 −.02000 −.013873(11.0355) (−9.1421)3 4.9751 29.1246 −.02003 −.01388 −.01020(13.0241) (−11.1246)4 5.9815 41.11048 −.01390 −.0102 −.00781(15.0185) (−13.11088)5 6.9857 55.0999 −.01021 −.00781 −.00617(17.01447) (−15.0997)∗Numerical results.FIZIKA B 2 (1993) 1, 11–19 17panchanan et al.: partial algebraization . . .for a particular potential and will be the excited state for another potential. HenceSU(2) symmetry does not give two states for the same potential contrary to theclaim made in the original papers on partial algebraization1−3). Table 2 gives theeigenvalues obtained from the present work together with the numerical values forE0 and E2 states (the numbers within brackets are for negative values of r0). Itmay be noted that our exact solutions for j = 0 and j = 1/2 completely agreewith those obtained from supersymmetric quantum mechanics technique19) whichagain supports the result of Ref. 4 for N = 0 and N = 1, respectively. However forN > 1 our technique can be carried on to higher values of j. We give below theresults for j = 1.3. j = 1HereE = −1(l + 4)2We take ψ(r) = (r − r0)(r − r1).We apply the method of Ref. 9 to determine r0 and r1. Elimination of r0 andr1 gives a cubic equation in β. Though in principle the equation could be solvedanalytically, we present below only a set of numerical solutions. The signs of r0 andr1 show that the energy eigenvalues are those of the first excited states.In Table 3 energy eigenvalues for l = 0 to l = 3 are presented for the case j = 1for a particular set of β values. Similar analysis can produce exact results for thecase j = 32 , 2, etc., but then β has to be solved numerically.Table 3.l β r0 r1 Energy0 6.6108 8.9069 −4.7372 −0.6251 8.7617 16.8375 −6.8875 −.042 10.8364 26.7687 −8.9492 −.027773 12.8745 38.7024 −10.9820 −.020408References1) A. V. Turbiner and A. G. Ushveridze, Phys. Lett. A 126 (1987) 181;2) A. V. Turbiner, Communication Math. Phys. 118 (1988) 46,7;3) M. Shifman and A. V. Turbiner, ITEP 174 Preprint Moscow 1988;4) N. Karman and P. Olver, J. Math. Anal. Appl. 145 (1990) 342;5) P. Roy and R. Roychoudhury, Phys. Lett. A 139 (1989) 427;6) P. Roy and R. Roychoudhury, J. Phys. A 23 (1990) 1657;18 FIZIKA B 2 (1993) 1, 11–19panchanan et al.: partial algebraization . . .7) M. Shifman, Int. Journ. Mod. Phys. A 4 (1989) 2897;8) See for example: C. H. Mehta and S. H. Patil, Phys. Rev. A 17 (1975) 43; P. P. Royand K. Mahata, J. Phys. A 22 (1989) 3161; R. Dutta, U. Mukherjee and Y. P. Varshni,Phys. Rev. A 34 (1986) 7,77;9) M. A. Shifman, Int. J. Mod. Phys. A 4, 13 (1989) 3311;10) A. A. Sinha and R. Roychoudhury, J. Phys. A 23 (1990) 386)7.DJELOMIČNA ALGEBRIZACIJA I ODREZANI COULOMBOV POTENCIJALSWATI PANCHANAN∗, RAJKUMAR ROYCHOUDHURY+ iY. P. VARSHNI++∗Ananda Ashram Sarada Vidyapeeth, 104, Barrackpore Trunk Road, Calcutta - 700 035,India+Electronics Unit, Indian Statistical Institute, 203, Barrackpore Trunk Road, Calcutta -700 035, India++Department of Physics University of Ottawa, Ottawa, Canada - IC IN 6N5Received 7 October 1992UDK 530.145Original scientific paperPrimijenili smo tehniku djelomične algebrizacije na odrezani Coulombov potencijal−Ze2/(r+β). Nad–eno je da se za spin j dobiva 2j+1 točnih rješenja, koja pripadajupobud–enim stanjima potencijala za različite vrijednosti β. Opažene degeneracije uspektru uspored–ene su s točnim numeričkim rezultatima.FIZIKA B 2 (1993) 1, 11–19 19

Djelomična algebrizacija i odrezani Coulombov potencijal

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