Non-uniqueness in the one-dimensional inverse scattering problem

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1 Inverse Problems 1 (1985) Printed in Great Britain Non-uniqueness in the one-dimensional inverse scattering problem Tuncay Aktosun and Roger G Newton Department of Physics, Indiana University, Bloomington, IN 47405, USA Received 3 June 1985 Abstract. The Schrodinger equation in one dimension is considered for the case when at least one reflection amplitude at zero energy is unity. It is shown that if there exists a corresponding potential that causes no negative-energy bound states then there is a oneparameter family of potentials that causes the same scattering at all energies. Two explicit examples are given. 1. Introduction If the unitary S matrix of the Schrodinger equation in one dimension is continuous at k = 0 (as it is when IX~~VEL~(IR)) then it must be real there. For the transmission amplitude T and the reflection amplitudes RI and R, from the left and right, in terms of which there are then the following possibilities (taking into account that det S= T/F, where is the complex conjugate of T, S(-k)=S(k), and listing only the simplest cases, when the leading terms go as integral powers of k): (1) (2) T(0) # 0, det S(0) = 1 T(k) = ick + o(k), c f 0, det S(0) = - 1 (a)r,=r,=-l (b) R, =R, = 1 (3) T(k)=ck2 +o(k2),cf0,dets(o)=1 RI=-R,= 1 (4) ~(k) = o(k3). If 1x1 VEL', case (20) is generic, while case (1) is exceptional. In case (1) one says that there is a 'half-bound' state, because it implies the existence of a bounded solution of the Schrodinger equation at k=o. (Generically both solutions grow linearly in at least one direction.) None of the other possibilities can occur if (1x1 + Ix12)V~L1. As is well known, the inverse scattering problem for such potentials has a unique solution (Faddeev 1964, Deift and Trubowitz 1979). On the other hand, if 1x1 VE L' then the other listed possibilities may occur, and explicit examples of (2b) (Abraham et al 1981, Brownstein 1982), (3) (Moses 1983) and (4) /85/ The Institute of Physics 29 1

2 292 TA ktosun and R G Newton (Brownstein 1982) are known. They are also all known to be associated with more than one potential. In this paper we shall study cases (2b) and (3), and we shall show that if a potential leads to an S matrix in class (2b) or class (3) then there is always a one-parameter family of potentials that is associated with the same S matrix. In 5 4 we show that the previously known examples of ambiguity in the inverse problem are special cases of such families. Our results are closely related to those of Sabatier (1984), who came to the same conclusion in case (2b) by a different procedure. In the matrix inversion method (Newton 1980, 1983) the unique potential for S(k) with R,(O)=R,(O)=-I and the family of potentials for are both constructed together, as we shall see in 5 2. Section 3 deals with case (3). 2. The general ambiguity in case (2b) This can be split into three parts. (a) The solution of the inverse scattering problem given by Newton (1980, 1983) is based on the solution of the following Riemann-Hilbert problem. Problem Hj(S). Let a continuous 2 x 2 matrix-valued function S(k) on R with the following properties be given: where 3; is the transpose of S; (iii) SSt =II where St is the adjoint of S; (iv) (S(k)-n)EL2(R); (4 k?],lk) E L'([R); (vi) S(0) = - Q; (vii) G(O)-G(CO)=-~Z where o(k) = - 1 i In det S( k). Find a 2 x 2 matrix-valued function F(k)=E(-k) that is the boundary value of an analytic function holomorphic + in C, such that lim F(k)=A A= (' Ikl-+m 1-1 I), (F(k)-A) E L2(lR) and for real k F(- k) = QS(-k)F(k)Z.

3 Non-uniqueness in 1D inverse scattering 293 Note that (vi) and (vii) imply that we are restricting ourselves to the case that is generic when 1x1 VE L', with no bound or half-bound states. (Our example in 8 4 will show that for potentials outside that class (vii) need not guarantee that.) Comparison with equation (3.30) of Newton (1983) shows that the Jost matrix J corresponding to S is related to F by J=AF-'. It is worth noting explicitly that equation (1) is such that each column of F satisfies an autonomous equation. If f(l) and f(2) are the first and second columns of F, respectively, then as lkl+ c;o,f(1)+(~),f(2)+(-~) and f("(-k)= QS(-k)f("(k) (2) f("(- k) = - QS(- k)f(*)( k). (3) Ifwe defir~ef(~):=if(~), the~~f(~)-+(;) as lki+ CO and f(3)(- k)= QZS(-k)Zf(3)(k). (4) Th~sf(~) corresponds to S'=ZSZ asf") does to S. If we denote by F, the solution of H;(S,), where S,( k) = exp(ikzx)s(k) exp(-iklx), then the first column of F, is the scattering solution Y(l) of the Schrodinger equation with a potential V, whose S matrix is S(k) (see Newton (1983)) and the second column is ZY(2), where Y(2) is the solution of the Schrodinger equation with a potential V, whose S matrix is ZSZ, if it exists. If the first corresponds to T, R, and R,, then the second corresponds to T, -Rr, -RP Condition (v) ensures that the solution of H;(S,) is 'miraculous' and that VI and V, exist if F, does (Newton 1984). Taking determinants of both sides of (1) and defining ((-k) :=-4 det F(k) we get ((-k) = det S(-k)((k). (5) The unique solution of this equation which is the boundary value of a function holomorphic in C + and approaches 1 as /k/ + CO is given by 2 ((k) = exp(io( k) + ;.c P J : dk' E) (here P denotes Cauchy's principal value). As k+o this function vanishes linearly. In the usual case, when T vanishes linearly at k = 0 and is holomorphic in + C, c(k) = T(k). If we write near k= 0 F(k) =M+ O( 1) then we obtain from (1) and (vi) that M(d +I) = 0 and hence where a and b are real, so that F(-k)=F(k). Now, if Ha@) has a solution F then -4 det F must satisfy (5) and hence is given by (6). Thus det F cannot vanish in 6' and F-' is holomorphic too. Lemma I. If F, and F2 are two solutions of Hj(S) with the same ratio of a/b, then F, = F2.

4 294 TAktosun andr G Newton ProoJ It follows from (1) that Fl(-k)ZFl(k)-l =F,(-k)lF,(k)-' and therefore Fz(-k)'-lFl(-k)=ZFz( k)-'f,(k)z. The left-hand side is holomorphic in C- and the righthand side in C +. Near the origin where b,a, - a2b, = 0. Therefore FF 'F, is bounded at k = 0, and FT'F, is in Lz(iR). Thus, since F, EA, F, EA at infinity, by Liouville's theorem, F2 =Fl. QED In other words, the specification of the ratio a/b makes the solution of Hj(S) unique if it exists. We shall see below that this ratio can be given arbitrarily, so that we arrive at a one-parameter family of solutions of Hj(S). However, this lack of uniqueness of Hj(S,) affects the second column of F, only. The first column is uniquely determined and leads, as is known, to a unique potential V, (if S satisfies the Faddeev-Deift-Trubowitz conditions) whose S matrix is S. The lack of uniqueness of the solution of Hj(S,.) then leads to a oneparameter family of potentials V, whose S matrix is ZSZ, because the second column of F depends on the ratio of a/b. (b) In order to solve Hj(S) we proceed to remove the pole of F-' at k=o by the same method by which the bound-state poles were removed by Newton (1980, 1983). Let P be the orthogonal projection, Pz = P = Pt, such that PM= 0: Define and Then the eqn&!tior! 1 b2 -ab p=- a2+b2 (-ab a')' II( k) := (k + i)p/k + 2-P F '(k) := II(k)F(k). F '(-k) = QS'(-k)F'(k)Z for F is equivalent to (l), where S '( k) = QII(k)QS(k)II(- k)- '. Since k+i det SI= (det S) - k-i' (vii) implies that o'(0) - or( CO) = 0 if or( k) = -4i In det S'(k). The Riemann-Hilbert problem Hj(S') for F is of the normal kind, with det F'(O)# 0. If it has a solution then it is unique. Equation (8) then leads to a one-parameter family of solutions of Hj(S). It should be noted, however, that although Sr(k)-' = QS'(-k)Q, S' does not satisfy (ii) and (iii).

5 Non-uniqueness in 1 D inverse scattering 295 (c) Let us now convert Hj(S) into an integral equation by Fourier transformation. Defining f(a)=(1/2n) j" dk(f(-k)-a) -W exp(ika) we find from (1) for a > 0 G(a)=(1/27r) Jw dk Q(S(-k)-sl) exp(ika) -m because for a < 0, r(a) = 0 owing to the analyticity of F(k) in + C. The operator 9 whose kernel is G(a + p), a, p E ir +, is known to be compact, self-adjoint, and its spectrum lies in [-1, 11. The first column q(') of r satisfies the equation for a > 0, where 7 =(:). A solution of (2) is connected to a solution of (14), and a solution of (3) to one of (15). It is known (Newton 1985) that, though a solution of (14) (or (15)) does not necessarily lead to a solution of (2) (or (3)), if one does then so do all. But (2) has a unique solution (if the Faddeev-Deift-Trubowitz conditions are satisfied). Therefore (14) implies that 9 does not have the eigenvalue 1. A sufficient condition for a solution of (14) (or (15)) to yield a solution of (2) (or (3)) is that g# does not have the eigenvalue -1 (or + l), where g# is the operator whose kernel is G(-a-P), a, PE R+ (Newton 1980, 1983). But 9# is related to s as 9 is to S. Therefore the absence of 1 in the spectrum of 9 ensures that any solution of leads to a solution of <(2)(=) = - G(- - 1'" d,/j c;- a-/?jt(?)(,/jj a>o (i5a) JO g(2)(- k) = - QS(k)g(')( k). (34 However, (I5a) always has a solution (Newton 1985); hence so does (3a). Consequently, the equation g(')(-k)= QS(k)g")(k) (24 cannot have a solution without poles because (2a) and (3a) together give F'(- k) = QS(k)F'(k)l (14 if F' is formed out of the columns ofg'') and g('). (The Levinson theorem for s implies that no solution of (la) is possible without poles. That the poles occur in (2a) and not in (3a) is

6 296 TAktosun and R G Newton in agreement with the fact that (5) implies that near k=o the second column of F-' is bounded.) We thus conclude that 9 must have the eigenvalue -1, which prevents the solution of ((')(cy)=g(-cy)t + dp G(-cY-/~)(")(/~) CY>O ( 14~) im from leading to a solution of (2a). The presence of -1 in the spectrum of makes the solution of (15) non-unique and accounts for the non-uniqueness of the solution of (3) if it exists. However, we do not have any assurance that (3) has a solution (with the appropriate properties). 3. Class(3) When T(k) vanishes quadratically at k=o then det S(O)= 1. If there are no negative- energy bound states then the Levinson theorem says, instead of (vii), that cr(0)- o(co)= -E. We must then be looking for a solution of a modified problem Hj'(S) such that det F(k)= ak2 + o(k2), where a+ 0. Assuming that S is continuous and F(k)= MO + km, + O(k2) we find from (1) that MO = ZMoZ if S(0) = QZ. It follows that either F(k)=(O ibk c+idk iak )+O(k2) or c+idk ibk F(k)=jiok ) +O(k2) where a, b, c and d are real. We shall assume the first form; the arguments are similar if it is the second. If S(0) = - QI then either or F(k)= ( iak c+idk ibk ) + O(k2) ibk c+idk F(k)=io iak ) +0(k2) and again the arguments are similar. Assuming the form of (1 6) for F, we have c+idk -iak F-'( k) = - 0 Lemma 2. If F, and F, are two solutions of the same form (16) or (17) of Hj'(S) with S(O)= QZ with the same ratio of a/c, then they are identical. Pro@ The proof proceeds like that of lemma 1 and depends on the boundedness of FT'F,. But i(a,c2-a2c1) 0 1 O)+o(l)' FT'F, = a2b2k (0

7 Non-uniqueness in ID inuerse scattering 29 7 Therefore alc2 - azci = 0 implies the required boundedness. Similarly if both are of form (1 7). If S(0) = - QI then an analogue of lemma 2 holds if both F, and F, are of form (18) or both of form (1 9). Thus the result is that in each of the two cases of class (3) we must specify the form of F, either (1 6) or (1 7 ) in the first case and either (1 8) or (1 9) in the second, and in all cases the ratio alc must be given in order for Hj (S) to have at most one solution. To remove the singularity from F- we define II(k) := - where A= 1 -ik and F and Sr are as in (8) and (10). Then and ~ (0) - ai( CO) = 0. (1 -ik)(2-ik) det S =(det S) (1 + ik)(2 + ik) 4. Two examples As our first example we take 1 which is in class (2a). Its unique potential is given by V(x)= 26(x). The associated is in class (2b); this S matrix was considered by Abraham et a1 (1981) and by Brownstein (1982). The family of potentials corresponding to S is given by 80(x) 80(-~) V(a, x) = - 26(x) + + (1 +a+2x), (1 + b-2~)~ where a > 0 and b = l/a; 0(x) is the Heaviside function. Note that V(a, x)= V(b, -x). The two scattering solutions are k 2i/k Y(I)(a, k, x)=o(x) exp(ikx) - k+i (I+ 1 +c+2xj and Y( (a, k, x)= Y( )(b, k, -x). (We interpret 6(x)@(x)= fd(x).) For a+o and a-+ CO we obtain the two potentials of Abraham et a1 and for a= 1 that of Brownstein. The k= 0 limit of Y is given by Y( )(Q, 0, X) 2 2b l+b-2x 1+c+2x

8 298 TAktosun and R G Newton which is in L2(W). Therefore, unless a=o or a= CO there is a bound state at k=o. (Brownstein noticed this in his example.) For a=o, k=o two linearly independent solutions are Y1=O(x)(l +2x)- +O(-x) Y2 = O(x) 4x3 + 6x2 + 3x 1 +2x and those for a+ CO, k=o are obtained by x+-x. Thus in the cases of Abraham et a1 there is a half-bound state, i.e. one solution is bounded but not square integrable (Brownstein noticed this also). It is remarkable that in this example the existence of a bound state or half-bound state cannot be recognised from the S matrix but depends on a free parameter. It thus violates Levinson s theorem. (Note, however, that this simple example is not in the class discussed generally in this paper.) As a second example we take S= which is in class (3). This example was considered by Moses (1983). The family of potentials associated with this S matrix is V(C, x ) = 4O(x) (fix + l)(a(x) - 2c + 1) 1 (4x) + Cl2 + 4O(-x) where c > 1, l/b + l/c = 1 and a(x) = (2fi/3)x3 + 2x2 + fix. solutions are Y ( )(c, k, x)= O(-x) (b - fix)2 The two scattering and ( L ) [ ( : k k, x) = 0(-x) T exp(-i kx) PI + O(x) exp(-i kx) 1 -- P, - 7 P3 where +A?, i 1 \1 cay \dxj 1 +-.?2-7.?3 \ k k IJ k2 1 T= R =-R - k2 + ifik- 1 -kz+ifik-l P,(c,x)=- fi b-fix

9 Non-uniqueness in ID inverse scattering 299 The k+o limit of Y (I) is given by k-0 and a linearly independent solution is l)o(-x)p, (3c- 1)2 a(x) + 4c a(x) + c 5b3-3b2-9b-3 1) ( b-fix The first of these is square integrable; thus there is a zero-energy bound state. The two potentials given by Moses are obtained in the limits as c+ CO and c+ 1 (though there is an error in the first): V(CO, x)=40(-~)(1 -fix)-' For c+ CO we find and a linearly independent solution Y=33(x)+@(-x) (-+(I-fix)2j. 2 On the other hand, for c+ 1 and a!inex!y igdependext sdii:ioii i 4 i Y u(x) ). a(x) + 1 For c+l one solution is bounded but not square integrable, while for c-co neither solution is bounded. Thus there is a k= 0 bound state for all values of c > 1, a half-bound state for c+ 1 and neither for c+ CO. We note that in both of these examples of a continuous ambiguity there are zeroenergy bound states. The half-bound states are associated with a discrete two-fold non-uniqueness. Acknowledgment This work was supported in part by the US National Science Foundation.

10 300 TAktosun and R G Newton References Abraham P B, DeFacio B and Moses H E 1981 Phys. Rev. Lett Brownstein K R 1982 Phys. Rev. D Deift P and Trubowitz E 1979 Commun. Pure Appl. Math Faddeev L D 1964 Trudi Mat. Inst. Steklov (Engl. trans]. Am. Math. Soc ) Moses H E 1983 Phys. Rev. A Newton R G 1980 J. Math. Phys Con/ on Inverse Scattering: Theory andapplication ed. J B Bednar et al(philade1phia: SIAM) p I J. Math. Phys Inverse Problems Sabatier P C 1984 Inverse Problems of Acoustic and Elastic Waves ed. E Santosa et a1 (Philadelphia: SIAM) P 82