Finite Volcano Potentials Admitting a Rational Eigenfunction

Transcription

1 Finite Volcano Potentials Admitting a Rational Eigenfunction Spiros Konstantogiannis spiroskonstantogiannis@gmail.com March 18 Abstract Starting with a general rational wave function, we search for potentials admitting it as a bound energy eigenfunction. We thus derive singular and regular potentials asymptotically decaying as the inverse of x squared, with the latter being simple or multiple volcanoes having a finite number of bound eigenstates. We present specific examples and examine the transition from singular to volcano potentials. Keywords: volcano potentials, bound states, rational wave functions, nodes, poles 1 March 18

2 Contents Finite Volcano Potentials Admitting a Rational Eigenfunction... 1 Contents Introduction Rational eigenfunctions The potential.... Distributing the zeros of the wave function as typical nodes or simple poles in the potential Examples of volcano potentials...1 The case m, n1...1 Transition from a singular to a regular potential, in a m, n1 case...17 A double finite volcano with four bound eigenstates, in a m6, n3 case...19 A triple finite volcano with five bound eigenstates, in a m6, n case...3 The case n...9 The case m, n References... March 18

3 1. Introduction Volcano potentials frequently appear in braneworld scenarios [1-], while they ve also been used in non-relativistic quantum mechanics [5]. A typical volcano potential includes a well, usually at the origin, having finite bottom and height on both sides, while at infinity, it can be unbounded from below minus infinity or it can tend to a constant, finite value including zero [6]. Herein, we examine volcano potentials admitting a rational bound energy eigenfunction. These volcano potentials, which include a simple or multiple well and decay asymptotically as 1 x, have a finite number of bound eigenstates, with the highest of them having fixed energy that can be set to zero. Since they are finite at infinity, we ll refer to them as finite volcano potentials.. Rational eigenfunctions We consider the function æ xö Pn ç x y x A è ø æ xö Qm ç è x ø 1 where Pn, Qm are polynomials of degrees n, m, respectively, with m ³ n + [7], and x is a positive length scale. The polynomial Qm has no zeros1, i.e. it has constant sign in, so that y x has no singularities. Then, m must be even. Notes 1. By zero, we mean real zero.. Every odd-degree polynomial has at least one zero, since it is continuous and its values at minus and plus infinity have different signs. A is the normalization constant with dimensions of wave function, i.e. [ A] L-1. Since x x is dimensionless, all coefficients of Pn have the same dimensions and, likewise, all coefficients of Qm have also the same dimensions. 3 March 18

4 Then, since the normalization constant has dimensions of wave function, the dimensions of the coefficients of the two polynomials are the same and they are eliminated. Thus, without loss of generality, we assume that the coefficients of the two polynomials are dimensionless. Besides, if pn and qm are, respectively, the leading coefficients of Pn and Qm, then æ xö æ xö Pn ç P%n ç è x ø pn è x ø æ x ö qm % æ x ö Qm ç Qm ç è x ø è x ø where the polynomials P%n and Q% m are monic in x x. Incorporating the dimensionless factor pn qm into the normalization constant, we make both polynomials monic, and thus, without loss of generality, we assume that both Pn and Qm are monic. Since Qm has no zeros, y x has no singularities. Moreover, as we see from 1, y x is infinitely many times differentiable on, i.e. it is C R. Also, since deg Qm ³ deg Pn +, y x tends to zero faster than 1 x, as x, and thus it is square integrable. Therefore, y x, as given by 1, is eligible to describe a bound energy eigenstate of some potential. We ll see that, although the wave function y x is C R, the potential can have singularities, particularly simple poles. 3. The potential Assuming that the wave function 1 is an eigenfunction of energy Er, of some potential V x, it will satisfy the time-independent Schrödinger equation, i.e. y x + m Er - V x y x h March 18

5 We denote the mass by m, so that we don t confuse it with the degree m of Qm. Solving the previous equation for the potential yields V x h y x + Er m y x Since the potential depends on the ratio y x y x, the normalization constant A is eliminated and we can omit it in the calculation of the potential. Introducing the dimensionless variable x% x x we have d 1 d dx x dx% and, in terms of x%, the potential is written as V x% h y x% + Er m x y x% 3 where now the prime denotes differentiation with respect to x%. In terms of x%, 1 is written as, omitting the normalization constant, y x% Pn x% Qm x% Differentiating with respect to x%, we obtain y x% Pn x% Pn x% Qm x% Qm x% Qm x% Differentiating once again yields y x% Pn x% Pn x% Qm x% Pn x% Qm x% + Pn x% Qm x% Pn x% Qm x% + Qm x% Qm x% Qm x% Qm3 x% Pn x% Pn x% Qm x% + Pn x% Qm x% Pn x% Qm x% + Qm x% Qm x% Qm3 x% 5 March 18

6 Q Q 1 x% 1 x% P x% - Q x% Q x% P x% + Q x% - Q x% Q x% P x% Q x% Qm x% Pn x% - Qm x% Pn x% Qm x% + Pn x% Qm x% + Pn x% Qm x% m 3 m n m m n m m m n 3 m æ Q x% P x% - Q x% Q x% P x% ö P x% n m m n ç m + Qm x% - Qm x% Qm x% n 3 ç Qm x% Pn x% è ø æ Q x% Q x% P x% - Q x% P x% ö m n m n ç m y x% ç + Qm x% - Qm x% Qm x% Pn x% ç Qm x% è ø where in the last equality, we used. Thus y x% Qm x% - Qm x% Qm x% Qm x% Pn x% - Qm x% Pn x% + y x% Qm x% Qm x% Pn x% Substituting into 3 yields V x% h æ Qm x% - Qm x% Qm x% Qm x% Pn x% - Qm x% Pn x% ö ç + Er 5 + m x ç Qm x% Qm x% Pn x% è ø In 5, the term Qm x% - Qm x% Qm x% on Pn x%, while the term Qm x% is regular and does not depend Q x% P x% - Q x% P x% m n m n Q x% P x% m n has singularities at the zeros of Pn x% if any. We ll consider only removable singularities which are practically no singularities and simple poles in the potential 5. If at the zeros of Pn x%, the potential 5 has removable singularities, it can be written as a regular function, which is actually C R. In this case, all zeros of Pn x% are simple [8], i.e. of multiplicity 1. Next, we ll derive the expression of the potential at long distances. To this end, we observe that deg Qm x% m March 18

7 deg Qm x% Qm x% m + m - m - 1 deg Qm x% m Since Qm is monic, the leading term of the polynomial Qm x% - Qm x% Qm x% is m x% m -1 - m m - 1 x% m -1 m - m m - 1 x% m -1 m + m x% m -1. The leading term of the polynomial Qm x% is x% m. Thus, at long distances, m -1 Qm x% - Qm x% Qm x% m + m x% m + m : Qm x% x% m x% In the same way, we have }* deg Qm x% Pn x% m + n - m + n - * If n,1, the polynomial Qm x% Pn x% vanishes. Also ** } % % deg Qm x Pn x m n - 1 m + n - ** If n, the polynomial Qm x% Pn x% vanishes Thus ì m + n -, n ¹ deg Qm x% Pn x% - Qm x% Pn x% í n î, Also deg Qm x% Pn x% m + n 7 March 18

8 Since the polynomials Pn and Qm are monic, the leading term of the polynomial Qm x% Pn x% - Qm x% Pn x% is n n mn x% m+ n -, and vanishes if n, as it should, since then the previous polynomial is zero. The leading term of the polynomial Qm x% Pn x% is x% m+ n, and thus, at long distances, m + n- Qm x% Pn x% - Qm x% Pn x% n n mn x% n n mn : m+n Qm x% Pn x% x% x% We observe that both terms of the potential 5 decay as 1 x% and thus, at long distances, V x% h m + m + n n mn + Er m x x% Since m + m + n n mn m - mn + n + m - n m - n + m - n m - n m - n + 1 we end up to V x% m - n m - n + 1 h m x 1 + Er x% 6 We see that, at long distances, the potential 5 is symmetric and, since m - n m - n + 1 > *, the term m - n m - n + 1 h m x x% is repulsive. We also see that, at long distances, the potential depends only on the difference m - n of the degrees of the two polynomials Qm x% and Pn x%. It does not depend on the coefficients of the two polynomials, thus it does not depend on the number of the zeros of Pn x% either. * Actually, since m ³ n +, m - n m - n + 1 ³ 6. Therefore, all potentials 5 regular or singular with the same difference m - n have the same symmetric asymptotic form 8 m - n m - n + 1 h m x x%, plus a March 18

9 constant Er, which can be set to zero choosing the infinity as reference point and V ±. In any case, since V ± Er, the energy Er is the highest bound energy of the potential 5. Proof Indeed, assuming that there exists a bound energy Er > Er, then if j x is the respective eigenfunction j x + m Er - V x j x h 7 From 5, we see that if the polynomial Pn x% has no zeros, the potential is { } continuous everywhere, while if it has, then if xmax max xi Pn xi, the potential is continuous for x > xmax. In any case, the potential is continuous at long distances, and then V x ; V ± Er, and thus from 7 we obtain j x + m Er - Er j x h æ ç è Then, since Er - Er >, j x : exp ç ± i ö m E - Er x. r h ø Thus, the probability density j x is either a non-zero constant or it oscillates. In either case, the probability density does not tend to zero, and thus the eigenstate j is not bound. Therefore, the wave function describes the highest bound eigenstate and the energy Er is the highest bound energy of the potential 5.. Distributing the zeros of the wave function as typical nodes or simple poles in the potential From 3, we see that at each simple zero of the wave function y x% which is not a zero of its second derivative, the potential has a simple pole, or equivalently, at each 9 March 18

10 simple zero of the polynomial Pn x% which is not a zero of the polynomial Qm x% Pn x% - Qm x% Pn x%, the potential 5 has a simple pole. If at a zero of y x% equivalently of Pn x% the potential 5 has a simple pole, that zero is considered as a special zero, not as a typical node [8]. If y x% equivalently Pn x% has no zeros, then y x% is the ground-state wave function [8, 9] of the potential 5, which is then regular and has only one bound state, of energy Er. The potential in this case is a simple or multiple finite volcano. If y x% equivalently Pn x% has one zero which is a simple pole of the potential 5, then y x% is the ground-state wave function, while if the zero is not pole of the potential 5, y x% is the first-excited-state wave function of the potential 5. In the first case, the potential is singular, as it has a simple pole, and it has only one bound state, of energy Er, while in the second case, the potential is regular and has two bound eigenstates, the ground state and the first-excited state, with the energy of the second being Er, and the potential in this case is a simple or multiple finite volcano. In the general case, where y x% equivalently Pn x% has r zeros, with r,1,..., n, some of them may be typical nodes in the wave function and the others may be simple poles in the potential. 5. Examples of volcano potentials The case m, n1 We ll examine the case where m and n 1. We remind that m ³ n + and even. For simplicity, we ll assume that Q x% is symmetric, i.e. Q x% x% + q x% + q 8 Since n 1, P1 x% x% + p 9 The polynomial 8 must have no zeros, and since it is monic, it must be positive for every x% Î. The non-normalized wave function is then written as 1 March 18

11 y x% x% + p x% + q x% + q 1 Assuming that the potential vanishes at infinity, i.e. V ±, yields Er. The wave function has a zero, at x% - p, which can be a typical node in the wave function or a simple pole in the potential. For - p to be a typical node, it must be a zero of the polynomial Q x% P1 x% - Q x% P1 x%. Since P1 x% 1 and P1 x%, the previous polynomial equals -Q x%, and thus - p must be a zero of Q x%, i.e. Q - p. Using that Q x% x% 3 + q x% x% x% + q, we obtain p p + q and thus p or q - p Every even-degree polynomial Rn x% with positive leading coefficient has a global minimum, i.e. R n x% ³ R, for every x% Î. This happens because Rn x% is continuous and finite for every finite x%, and it tends to plus infinity as x% tends to minus or plus infinity. Then, if e >, the polynomial R n x% - R + e, which differs from Rn x% only in the constant term, is positive for every x% Î. Thus, if the constant term of an even-degree polynomial with positive leading coefficient is big enough, the polynomial is everywhere positive, no matter what its intermediate coefficients are. Therefore, Q x% can be positive even if q is negative. For p, the wave function 1 becomes y x% x% x% + q x% + q 11 March 18

12 For q - p, the wave function 1 becomes y x% x% + p x% - p x% + q Both wave functions describe the first-excited state of a respective finite volcano potential having only two bound eigenstates, the highest of which has zero energy. Observe that if p and q, the two previous cases are merged. In any other case, i.e. if p ¹ and q ¹ - p, the wave function 1 describes the ground state of a singular potential with only one bound state having zero energy. Next, we ll examine the case where p and q. The wave function 1 then becomes y x% x% x% + q 11 with q >, so that the denominator does not vanish. For p, P1 x% x% and thus P1 x% 1, P1 x%. For q, Q x% x% + q and thus Q x% x% 3, Q x% 1 x%. Plugging into 5, we obtain, for Er, 3 æ ö h ç x% - x% + q 1 x% -8 x% 3 V x% + m x ç x% + q x% % + q x è ø 3 æ ö æ ö 6 6 h ç x% - x% + q 6 x% x% 3 h ç 16 x% - 6 x% - 6q x% - x% m x ç x% + q x% + q x% m x ç % + q % + q x x è ø è ø æ ö æ ö h ç 1 x% 6-6q x% x% h ç 5 x% 6-3q x% x% - m x ç x% + q x% + q m x ç x% + q x% + q è ø è ø 6 h 5 x% - 3q x% - x% x% + q h 3x% 6-5q x% m x m x x% + q x% + q That is V x% h 3x% 6-5q x% m x x% + q 1 1 March 18

13 The potential 1 is symmetric as a result of the wave function 11 having definite odd parity. If an eigenfunction has definite parity, its second derivative has the same parity, then the ratio y y is an even-parity function, and thus the resulting potential is also of even parity, i.e. symmetric. At long distances, the potential 1 becomes V x% 6h m x x%, in agreement with 6 for m - n 3 and Er. The potential 1 vanishes at 3 x% 6-5q x% Thus x% double zero, x% ± 5q 3 The derivative of the potential 1 is omitting the factor h m x which does not affect the sign of the derivative V x% x% x% x% - 18 x% 5-1q x% x% + q 9 x% 3 9 x% 3-3x% + q + q x% x% + q 3 + q x% - 6-5q x% x% 3 x% + q 3 æ 5 1 x% 9 - q x% 5 ö % % 9 x 5 q x ç % x + q % x + q è ø - 5q x% x% + q - 1 x% 9 + q x% 5 + 9q x% 5-5q x% 5-5q x% - 1 x% 9 + q x% 5 9 3x% 8 + q x% 5-5q x% x% + q 3 x% -3x% 8 + q x% - 5q - q x% + 5q That is V x% - x% x% + q 3 3x% 8 - q x% + 5q Setting y x%, the trinomial 3 x% 8 - q x% + 5q is written as 13 March 18

14 3y q y + 5q Its discriminant is 576q 6q 516q >, thus it has two zeros, at q ± 516q q ±.7q y1, q ± 3.79q.1q, 7.79q 6 6 Since both zeros are positive, x y x y x ± y, y.1q, 7.79q 1, 1, 1, 1, The trinomial 3y q y + 5q is then positive for y < y1 or y > y, and negative for y1 < y < y. Thus, the trinomial 3x q x + 5q is positive for 8 x < y x < y x < y y < x < y or x > y x > y x > y x < y or x > y and it is negative for y < x < y or y < x < y 1 1 Then, we have, V x > For x < y For y < x < y 1 For y < x < and the potential 1 is strictly increasing., V x <, V x > 1 For < x < y 1, V x < For y < x < y For x > y, V x > 1, V x < Finally, at y 1, and the potential 1 is strictly decreasing. and the potential 1 is strictly increasing. and the potential 1 is strictly decreasing. and the potential 1 is strictly increasing. and the potential 1 is strictly decreasing.. ±, V x Therefore, at y,, y the potential has local maxima, while at y, y it has local minima March 18

15 Since the potential is symmetric, its values at the two symmetric minima are equal, as they are at the two non-zero symmetric maxima too. The potential 1 is then a symmetric double finite volcano that, at long distances, decays as 1 x. It has only two bound eigenstates, a ground state and a first-excited state, with the energy of the second being zero. In Figure 1, the potential 1 is plotted for q.1 red line, q 1 blue line, and q 1 green line, in units ħ m x 1. Figure 1 We see that, as q increases, the minimum value of the potential increases, approaching zero from below. The value of the potential at the two minima is V y ± y1 + q ± y 1 5q ± y1 3 y1 5q y1 ħ ħ ± + 1 m x m x y1 q 3 5 ħ 3y y 5q y ħ y q y m x Thus y1 + q mx y1 + q 15 March 18

16 V y ħ y1 q y1 1 mx y1 q ± 13 Substituting y1.1q, we obtain V.1q 1.1q ħ.63q 5q.1q ħ.37q.6 q.75ħ ± m x m x 1.6q m x q The minimum value of the potential is then.75ħ m x q, and it increases as 1 q, as q increases. Since the ground-state energy must exceed the minimum value of the potential [1], it will also approach zero from below, as q increases. We expect that, as q, the ground-state energy will approach zero as 1 q or faster, otherwise the minimum value of the potential will exceed the ground-state energy. On the other hand, the first-excited-state energy does not depend on q, it is always zero. Then, as q increases, the difference between the two bound energies decreases, the ground-state energy comes closer to the first-excited-state energy. Similarly to 13, the value of the potential at the two local maxima ± is V y ħ y q y mx y q ± Substituting y 7.79q, we obtain V ± 7.79q ħ 18.37q.79 q 1.33ħ m x 77.6q m x q y Since 1.33ħ m x q >, 1.33ħ m x q is the global maximum of the potential, and thus it is the height of the two volcanoes. The depth of each of the two volcanoes is then 16 March 18

17 æ.75h 1.33h çm x q çè m x q ö.8h mx q ø and it goes to zero as 1 q, when q. The width of each volcano is infinity as 7.79q see Figure 1, thus it increases and goes to q, when q. As q increases, the two volcanoes become shallower and wider and as q, the height of the two volcanoes vanishes, while their width becomes infinite, i.e. the volcanoes decay. Transition from a singular to a regular potential, in a m, n1 case We showed that if p ¹ and q ¹ - p, the resulting potential has a simple pole at p. For p ¹ and q, the two previous conditions are satisfied and we have the case of a singular potential, with a simple pole at p. In this case, we have P1 x% x% + p Q x% x% + q with q >, so that Q x% ¹. For simplicity, we ll further assume that q 1. Then P1 x% 1, P1 x% and Q x% x% 3, Q x% 1 x% Plugging into 5, we obtain, for Er, 17 March 18

18 V x 3 + x x ħ 8 + mx x 1 x p ħ 3x 1x 1x 8x ħ x 1x 8x + mx p m x p ħ 1x 6x x ħ 5x 3x x m x ħ 5x 3x x + p x x + 1 5x + 5p x 3x 3p x x x m x p m x p ħ p m x p ħ 3x + 5 p x 5x 3p x m x x x p That is, in units ħ m x 1, the potential is V x 3x + 5p x 5x 3p x p 1 In Figure, the potential 1 is plotted for p 1 red line, p blue line, and p 1 green line. 18 March 18

19 Figure When the parameter p takes the critical value p, the two open edges of the potential pole, one at plus/minus infinity and the other at minus/plus, are glued together forming a volcano which is added to the existing volcano, resulting in a symmetric double volcano, and at the same time, one more bound eigenstate is created, with zero energy. A similar formation of a volcano takes place at every value of n m -, with m ³ and even, when the coefficients of the polynomial Qm x% Pn x% - Qm x% Pn x% take such values that one of the zeros of the previous polynomial becomes equal to one of the zeros of the polynomial Pn x%, but the correspondence between the zeros of Pn x% and the finally formed volcanoes is not one-to-one, as we ll show below. If the polynomial Pn x% has n zeros and the coefficients of the polynomial Qm x% Pn x% - Qm x% Pn x% can take such values that there exists an m - - degree polynomial Rm - x% such that Qm x% Pn x% - Qm x% Pn x% Rm- x% Pn x% 15 with Qm x% ¹, then we can transform the n simple poles of the initial potential, which has only one bound eigenstate, with zero energy, into a multiple finite volcano with n bound eigenstates, with the highest of them, i.e. the n th-excited state, having zero energy. By means of 15, the potential 5 is written as, for Er, h æ Qm x% - Qm x% Qm x% Rm - x% ö ç V x% + m x ç Qm x% Qm x% è ø 16 with Qm x% ¹. A double finite volcano with four bound eigenstates, in a m6, n3 case We ll consider a case where P3 x% is of odd parity and Q6 x% is of even parity, i.e. P3 x% x% 3 - x% 19 March 18

20 6 Q x x + q x + q x + q 6 with q >, otherwise Q 6 x has zeros. Then, P is of odd parity, thus Q P is also of odd parity, while P x of even parity and Q is of odd parity, thus Q P x and finally, Q x P x Q x P x is of odd parity. Then, since P3 is of odd parity, R has the form R x r x + r x + r Plugging into 15, we obtain 3 is is again of odd parity, x in 15 must be of even parity, and thus it 6 q x qx q 6x 6x qx qx 3x 1 r x r x r x x x + 6q x + 6q x + 6qx 18x 6x + 1qx qx + 6q x qx r x r x + r x r x + r x r x x + 6q x + 6q x + 6q x 36x + 1x q x + 8q x 1q x + q x r x r x + r x r x + r x r x x 18q x 6q x + 6q x + 1x + 8q x + q x r x + r r x + r r x r x x + 6 3q x + q 3q x + 3q + q x r x + r r x + r r x r x Comparing the same-degree terms in both sides of the last equation, we obtain r 3 17 r r 6 3q 18 r r q 3q 19 r 3q + q Substituting 17 into 18 yields r r q 1 18q 18q 18 r q 1 18 Substituting q into 19 yields March 18

21 r r r r r 1 3q 3q 8 6q r 5r 5r r 6q 8 r + r 8 + r q Substituting q into yields 5r r 8 5r r 16 1r r r 3q + + 3q + 6q r r 16 1r r 8 5r r 6q r + + q If the intermediate coefficients q and constant term q is positive, Q For q to be non-negative, 6 x has no zeros. q of Q 6 x are non-negative, and the r r r 18 For q to be non-negative, 5 r r r 5 r 8 5 r + + r For q to be positive, 8 5 r r r 8 5 r 16 1 r > < r < We observe that if r is enough negative and r 18, all three inequalities are satisfied. In the limiting case, where r 18 and 5 18 r , 9 the third inequality holds, since < < it holds 9 Then 1 March 18

22 q and q q Thus 6 Q x x and R x 3x 18x 18 Substituting into 16 yields V x 5 6 x x x x ħ mx x x 15x + 9x x x x + x ħ 6 m x ħ m x ħ 36x 15x 5x 15x 5x 9x 7x 9x 7 m x ħ 6x 9x 9x 7x 9x 7 m x ħ m x That is 6x 9x 9x 9x 7x V x ħ x 3x 3x 3x 9x 9 m x At long distances, the potential 1 takes the form V m x x m x x 1 3ħ x 6ħ 1 1 March 18

23 in agreement with the general expression 6 for m - n 3 and Er. In Figure 3, the potential 1 is plotted in units 3h m x 1. It is a symmetric double finite volcano potential with four bound eigenstates, the highest of which, i.e. the third-excited state, has zero energy and is described by the wave function y x% A x% x% - 1 x% 6 + 3, where A is the normalization constant. We see that, although the polynomial P3 x% x% 3 - x% has three zeros, the potential is still a double and not a triple volcano. Figure 3 A triple finite volcano with five bound eigenstates, in a m6, n case We ll consider a case where P x% x% - 1 x% - x% - 3 x% + Q6 x% x% 6 + q x% + q x% + q with q >, otherwise Q6 x% has zeros. 3 March 18

24 Both polynomials are of even parity, thus Q6 x% P x% - Q6 x% P x% is now of even parity, and since P x% is also of even parity, R x% must be of even parity too, and thus it has the form R x% r x% + r x% + r If Qm x% is of even parity and Pn x% has definite parity, i.e. it is of even or odd parity, then Pn x% has the same parity as Pn x%, and thus Qm x% Pn x% has also the same parity as Pn x%, while Pn x% has different parity from that of Pn x%, and since Qm x% is of odd parity, Qm x% Pn x% has the same parity as Pn x%, and thus the polynomial Qm x% Pn x% - Qm x% Pn x% has the same parity as Pn x%, and then from 15 we derive that Rm - x% must be of even parity. That is, if Qm x% is of even parity and Pn x% has definite parity, Rm - x% is of even parity. Plugging into 15, we obtain x% + q x% + q x% + q 1 x% x% r x% + r x% + r x% - 3x% + Þ q x% 3 + q x% x% 3-6 x% Þ 1 x% 8-6 x% 6 + 1q x% 6-6q x% + 1q x% - 6q x% + 1q x% - 6q - - x% 8-36 x% q x% 6 - q x% + 8q x% - 1q x% r x% 8-3r x% 6 + r x% + r x% 6-3r x% + r x% + + r x% - 3r x% + r Þ 1 x% 8-6 x% 6 + 1q x% 6-6q x% + 1q x% - 6q x% + 1q x% - 6q - -8 x% x% 6-3q x% 6 + 8q x% - 16q x% + q x% r x% 8 + r - 3r x% 6 + r - 3r + r x% + + r - 3r x% + r Þ Þ -36 x% x% 6 - q x% 6 + q x% - q x% + 18q x% + 1q x% - 6q r x% 8 + r - 3r x% 6 + r - 3r + r x% + r - 3r x% + r Þ Þ -36 x% q x% 6 + 1q - q x% + 6 3q + q x% - 6q r x% 8 + r - 3r x% 6 + r - 3r + r x% + r - 3r x% + r Comparing the same-degree terms in both sides of the last equation, we obtain r -36 March 18

25 r 3r 33 1q 3 r 3r + r 1q q r 3r 6 3q + q 5 r 6q 6 Substituting into 3 yields r q r q q r Thus 1 r 1 q 7 Substituting and 7 into yields 1 r 1* 1 1r 1* 1 1r 7 3r + r 1 q q q r 81 9r q r r + r Thus 81 9r r q + 8 Substituting 8 into 5 yields 81 9r r 3 7r 3r r 3r q q 3*3 3*7r 9r + + 1q 1 3* 3 3* 7r 9r 79 1r 3r 1q + + r 3r Thus 79 1r 3r 1 q + 9 Substituting 9 into 6 yields 5 March 18

26 79 1r 3r 79 1r 3r 1r 79 11r 1r 79 r r r 11r Thus 79 1r r + 3 Substituting 3 into 9 yields 79 1r r r r q r 1r r 79 1r + 1 1*11 * That is 79 1r q 31 Substituting 3 into 8 yields 79 1r r r 79 1r r 1r q *11 * r 81 9r That is 81 9r 11 q + 3 Since q must be positive, from 31 we obtain 79 1r 1r > < r < To simplify things, we consider the case where r vanishes. Then, 31, 3, and 7 give, respectively, 79 q > March 18

27 81 q 11 1 q 1 Thus, the polynomial Q 6 x is Q6 x x x The polynomial Q i.e. it has no zeros. 6 x is plotted in Figure and, as we see, it is everywhere positive, For r, 3 gives Figure r Then, since R 36x 55 r eq., the polynomial R x is 7 March 18

28 Substituting into 16, we obtain, in units ħ m x, x x x x x x + 3x x V x x x x x x x The previous potential is plotted in Figure 5. It is a symmetric triple finite volcano potential consisting of two deep, symmetric-around-zero volcanoes and one shallow volcano which is also symmetric around zero. The potential has five bound eigenstates, the highest of which, i.e. the fourth-excited state, is described by the wave function ψ 1 A x, where A is the normalization constant, x x x and it has zero energy. Figure 5 8 March 18

29 The case n Since P x% 1, the wave function is, including the normalization constant A, y x% A 33 Qm x% with m ³ and even. Then, the potential 5 becomes h Qm x% - Qm x% Qm x% V x% + Er m x Qm x% Since Qm x% has no zeros, the potential has no singularities. Actually it is C R. The wave function 33 has no zeros, thus it describes the ground state of the previous regular potential [8, 9], which has only one bound state, of energy Er. Using that n, from 6 we derive that, at long distances, the previous potential takes the form V x% m m + 1 h 1 + Er m x x% 3 Assuming that the potential vanishes at infinity, Er, and then it is written as V x% h Qm x% - Qm x% Qm x% m x Qm x% 35 As an example, we ll examine the case where Qm x% x% m + q, with m 1,,..., and q >, so that Qm x% ¹. In this case, the wave function has definite even parity. We have Qm x% mx% m -1, Qm x% m m - 1 x% m- Plugging into 35 yields 9 March 18

30 V x + m + m 1 m m ħ mx x q m m 1 x mx x q 8m x m m 1 x m m 1 q x m m m ħ mx x q m + 8 m 1 1 m + m 1 m + 1 m + m ħ m m m x m m qx mx x q m m m m ħ m + m x m m 1 qx ħ m m + 1 x m 1 q x m x m x ħ m x + q x + q m m x m q x m x x q That is V x ħ + 1 m 1 m m x m q x m m + mx x q 36 The potential 36 is symmetric, as a result of the wave function having definite parity. At long distances, 36 becomes V m + ħ + + ħ m m 1 x m m 1 m m 1 ħ 1 m x x m x x m x x m in agreement with 3 for E r and m m. The potential 36 vanishes at m 1 q > 1 m m m m m m m + 1 x m 1 q x q x q m + 1 m m 1 m x ± q m + 1 and at if m > 1. Since and the denominator of 36 is positive, we have m x V x > for x 1 m 1 m > q m March 18

31 1 æ m - 1 ö m V x% < for x% < ç q and x% ¹ if m > 1 è m + 1 ø Thus, if m > 1, the potential 36 is a symmetric double finite volcano, while if m 1, it is a symmetric simple finite volcano, since then it does not vanish at. In both cases, m 1 and m > 1, the potential 36 has only one bound state, its ground state, with zero energy, which is described by the wave function y x% A x% m + q. In Figure 6, the potential 36 is plotted in units h m x 1, for q 1 and m 1 red line, m blue line, and m 3 green line. Figure 6 The case m, n For simplicity, we ll assume that both P x% and Q x% are symmetric. Then, since both are also monic, they have the form P x% x% + p Q x% x% + q x% + q 31 March 18

32 Q x >. with As explained, for every value of q, there exists a q such that the global minimum of Q x is positive. Also, p, so that the zeros of P x if any are simple zeros. The wave function is, including the normalization constant A, x + p ψ A x q x q The wave function 37 has definite parity even, thus ψ has also definite parity even, and then the potential 3 is of even parity symmetric. Assuming, as usual, that the potential vanishes at infinity, E r, i.e. the energy of the eigenstate described by the wave function 37 is zero. I. If p >, P x > and the wave function 37 has no zeros, and thus it describes the ground state of the potential 5 with E r, which is regular. In this case, the potential is a symmetric double finite volcano. II. If p <, ± p. P x has two zeros, at Then, the polynomial Q x P x Q x P x is x + q x + q x + q x x 1x 6q x + q That is Q x P x Q x P x x q x q The previous polynomial is symmetric, thus its zeros if any will come in opposite pairs. Then, since P x has two opposite zeros, either none or both of them will be zeros of Q x P x Q x P x. IIa. Substituting the two zeros of P into the expression of Q x P x Q x P x, we obtain 3 March 18

33 7 ± - p + 3q ± - p - q 7 p - 3q p - q The discriminant of the previous trinomial is 9q + 8q. But q >, otherwise Q x% has zeros. Thus, the discriminant is positive and the trinomial has two zeros, the product of which is - q 7 <, thus one zero is positive and the other is negative. The negative zero is p 3q - 9q + 8q 1. Therefore, for every pair of the parameters q, q, with q >, there exists a p < such that the polynomial Q x% P x% - Q x% P x% vanishes at ± - p, and the resulting potential is regular. In this case, the potential is again a symmetric double finite volcano, but now it has three bound eigenstates, with the highest of them, i.e. the second-excited state, being described by the wave function 37, which takes the form x% + 3q - 9q + 8q y x% A x% + q x% + q 38 where we ve incorporated into the normalization constant A the factor 1 1 appearing in the expression of p. The wave function 38 has two zeros, thus it is the second-excited-state wave function of the resulting potential, with energy Er. Next, we ll calculate the potential for the case where q. In this case, p - P x% x% - q, and then 7 q, P x% x%, P x% 7 Also Q x% x% + q, Q x% x% 3, Q x% 1 x% Then, we have 33 March 18

34 + + Q x Q x Q x 3x 1x x + q x 1q x Q x x q x q and 6 6 Q x P x Q x P 16 x x + q x 1x + q Q x P x q q + q x + q x 7 7 q q q q 1 x 1 x x + 1 x q q + q x + q x x + q x 7 7 Substituting into 5, with E r, we obtain the potential q q 1 x + 7 x ħ x 1q 7 7 x 1x 6q x V x ħ mx x q x + q m x x q + x + q x 7q ħ x q x 1x 6q x 7x + 7q x + q + ħ m x + q + q m x + q 1x 6q x 7x 7q x 7q x 7q q 3x 7q x 13q x 7q q m x ħ ħ + q m x + q That is V x 3x 7q x 13q x 7q q 6 ħ mx x q + 39 At long distances, the potential 39 takes the form V x 3ħ m x x agreement with 6, for m n and E. r, in In units ħ mx 1, the potential 39 becomes V x 3x 7q x 13q x 7q q 6 + q 3 March 18

35 For q 1 7 and q 7 the previous potential is written as, respectively, x x x 3x x x 6 7 *3 7 7* x x x V x 1 1 x + 7x + 1 7x x 9x 91x 7 7x + 1 That is V x And 6 17x 9x 91x 7 7x x x 13 x 3x x x 6 7 *3 7 * 7*5 7* x x x V x 1 x + 7x + 7x x 98x 36x 56 7x + That is V x 6 17x 98x 36x 56 7x + Both potentials are plotted in Figure 7 red and blue line, respectively. Each of them has three bound eigenstates, with the highest of them, i.e. the second-excited state, being of zero energy. We see that, as q increases, the bottom of the symmetric double volcano approaches zero. Then, since the ground-state energy must exceed the minimum value of the potential [1], both the ground-state and the first-excited-state energies approach zero, as q increases. In other words, as q increases, the two lower bound energies of each volcano potential get closer to the third, which is fixed to zero. 35 March 18

36 Figure 7 IIb. If p ¹ 3q - 9q + 8q 1, the polynomial Q x% P x% - Q x% P x% does not vanish at the two simple zeros of P x% and then the potential is singular, it has two simple poles, at ± - p. In this case, the two zeros of the wave function 37 are not typical nodes, they are special zeros, and the wave function is the ground-state wave function of the potential 5 with Er, which is a symmetric singular potential with only one bound state, its ground state, having zero energy. As the coefficient p reaches the critical value 3q - 9q + 8q 1, the two open edges at each of the two simple poles of the potential are glued together forming a volcano and, at the same time, two more bound eigenstates are created, with the highest of them, i.e. the second-excited state, having zero energy. For the case where q, we saw that the critical value of p is - q 7. We ll examine what happens if p e q 7, where e > -1, so that p <. For e, we obtain the critical value of p. 36 March 18

37 The polynomials P and Q +, P x x ε 1 q 7 x are, respectively, Q x x + q Q x Q x Q x Q x The regular term of the potential 5, i.e. the term is the same as before, i.e. Q x Q x Q x x 1q x 6 + Q x x q Q x P x Q x P x Q x P x while the singular term is, q 1 x Q x P x Q x P x 7 Q P q + q x 1+ ε 7 Substituting the two previous terms into 5, we obtain, assuming again that the potential vanishes at infinity, i.e. E r, V x ħ m x 1 x q 6 x 1qx 7 x q + q + q x 1+ ε 7 To simplify the calculation, we further assume that q 7. Then, in units ħ mx 1, the previous potential is written as x + 7 ε x x 1+ ε ε V x But ε x 1* 7x 1 1 1x 6* 7x 7 1 x + 7 x 1+ x + 7 x March 18

38 1 1 ε ε 1 ε x 7 1x ε x 6 x + 1+ ε x 7x 8 x x x x + x x + x + x x + + x ε ε x 1 1 x 8x 1 x 9 Thus, we end up to the potential V x ε ε 8 6 3x 1 1+ x 8x + 1+ x ε For ε, the potential has two simple poles, at ± 1+ ε with ε > 1. For ε, a volcano is formed at each pole and the potential then becomes a double volcano, with three bound eigenstates, the highest of which, i.e. the second-excited state, has zero energy. In Figures 8-1, the potential is plotted for ε.1, ε, and ε.1, respectively. Figure 8 38 March 18

39 Figure 9 Figure 1 39 March 18

40 6. References [1] Martin Gremm, Phys.Lett. B [] W. T. Cruz et al 11 EPL [3] Lisa Randall, Raman Sundrum, Phys.Rev.Lett.83:69-693, [] Michael Martin Nieto, Phys.Lett. B [5] Sayan kar and Rajesh R. Parwani 7 EPL 8 3. [6] Ratna Koley, Sayan Kar, Phys.Lett.A363: , 7. [7] Zafar Ahmed, Eur. J. Phys [8] Spiros Konstantogiannis, On the Ground State of Potentials with, at Most, Finite Discontinuities and Simple Poles, ntials_with_at_most_finite_discontinuities_and_simple_poles. [9] M. Moriconi, Am. J. Phys. 75, [1] L. D. Landau and E. M. Lifshitz, Quantum Mechanics Pergamon Press, Second Edition, March 18