the time-dependent Schrödinger equation for the wave function Ψ(x

88-914 SolitonsLecture 4: Scattering TheoryIntroduction: the time-dependent Schrödinger equation for the wave function Ψ(x, t) isthe PDE∂Ψ(x, t)i = −∇ 2 Ψ(x, t) + V (x)Ψ(x, t) .∂tHere V (x) is the known potential function. Often one seeks solutions in the separatedform Ψ(x, t) = e −iEt ψ(x). This reduces the equation to the time-independent SchrödingerequationEψ(x) = −∇ 2 ψ(x) + V (x)ψ(x) .We will be considering the one-dimensional version of this−ψ ′′ + u(x)ψ = Eψ . (1)We look for solutions to the latter equation for a fixed u(x), which is assumed to tend tozero (sufficiently rapidly) at spatial infinity. We wish to find solutions ψ that are bounded,i.e. do not diverge at spatial infinity. To discuss the existence of solutions it is useful todivide into the two cases E < 0 and E > 0 1 .Case I: E < 0. Write E = −κ 2 , with κ > 0 wlog. For large positive or negative x wehave −ψ ′′ ≈ −κ 2 ψ (as u → 0) and so⎧⎨ C 1 e κx + C 2 e −κx x → ∞ψ ∼⎩ C 3 e κx + C 4 e −κx x → −∞ .Here C 1 , C 2 , C 3 , C 4 are constants. For a bounded solution we need C 1 = C 4 = 0. In generalimposing these constraints will force ψ to be zero. Only for special values of κ, known asthe (discrete) eigenvalues, will there be a nontrivial bounded solution.nontrivial bounded solution we normalize the solution by requiringAfter this normalization we will have∫ ∞−∞ψ(x) 2 dx = 1 .When there is aψ(x) ∼ ce −κxfor x → ∞for some constant c. This constant c is known as the normalization constant for the eigenvalueκ.1 We should discuss E = 0. We won’t.1

Case 2: E > 0. Write E = k 2 . For large positive or negative x we have −ψ ′′ ≈ k 2 ψ (asu → 0) and so⎧⎨ C 1 e ikx + C 2 e −ikx x → ∞ψ ∼⎩ C 3 e ikx + C 4 e −ikx x → −∞ .Here C 1 , C 2 , C 3 , C 4 are constants. We see that in the case E ≥ 0 the general solution isbounded. Typically we will work with certain special solutions of the equation. The firstspecial solutions we will consider are the Jost solutions ψ + , ψ − defined (respectively) by theconditionsψ + (x, k) ∼ e ikx , x → ∞ ,ψ − (x, k) ∼ e −ikx , x → −∞ .Existence of these solutions is guaranteed for all k, as they are each specified only by “initial”conditions at ±∞. Since the Schrödinger equation only depends on k 2 , note that ψ + (x, −k)and ψ − (x, −k) are also solutions, satisfying boundary conditionsψ + (x, −k) ∼ e −ikx , x → ∞ ,ψ − (x, −k) ∼ e +ikx , x → −∞ .The general solution can be expressed either as a linear combination of ψ + (x, k) andψ + (x, −k), or as a linear combination of ψ − (x, k) and ψ − (x, k).Physicists more commonly work with a different special solution of the Scrödinger equationknown as the scattering solution, ˆψ(x, k), determined by boundary conditions⎧⎨ eˆψ(x, −ikx + r(k)e ikx x → ∞k) ∼⎩ t(k)e −ikx x → −∞ .r(k) and t(k) are known as the reflection and transmission coefficients respectively. (For thepurpose of understanding the terminology, note that for k > 0 and E > 0 the wave functionψ(x) = e ikx of the time-independent Schrödinger equation corresponds to a wave functionΨ(x, t) = e i(kx−Et) of the time-dependent Schrödinger equation, which is a right-movingwave, while the wave function ψ(x) = e −ikx corresponds to a left-moving wave.)The existence of the scattering solution is not guaranteed, as it is specified by boundaryconditions.We now prove the following: for each real k there exists a uniquescattering solution with |r(k)| 2 + |t(k)| 2 = 1.Proof:As already mentioned, we can write the general solution of the Schrödingerequation as a linear combination of ψ − (x, k) and ψ − (x, −k). In particular, for some a(k)2

and b(k) we can always writeψ + (x, k) = a(k)ψ − (x, k) + b(k)ψ − (x, −k)ψ + (x, −k) = a(−k)ψ − (x, −k) + b(−k)ψ − (x, k)Note that if ψ 1 , ψ 2 are any two solutions of the Schrödinger equation for the same value ofk, thenψ 1 (−ψ ′′2 + u(x)ψ 2 − k 2 ψ 2 ) − ψ 2 (−ψ ′′1 + u(x)ψ 1 − k 2 ψ 1 ) = 0⇒ ψ 1 ψ ′′2 − ψ 2ψ ′′1 = 0⇒ ψ 1 ψ ′ 2 − ψ 2ψ ′ 1 = constant .(The quantity on the LHS of the last equations is known as the Wronskian of ψ 1 , ψ 2 .) Usingthe asymptotics at x = +∞ we haveψ + (x, k)ψ ′ + (x, −k)) − ψ +(x, −k)ψ ′ + (x, k) = eikx (−ike −ikx ) − e −ikx (ike ikx ) = −2ik .Using the asymptotics at x = −∞ we haveψ + (x, k)ψ ′ +(x, −k)) − ψ + (x, −k)ψ ′ +(x, k)= (a(k)e −ikx + b(k)e ikx )(ika(−k)e ikx − ikb(−k)e −ikx )(a(−k)e ikx + b(−k)e −ikx )(−ika(k)e −ikx + ikb(k)e ikx )= 2ik(a(k)a(−k) − b(k)b(−k))and thus b(k)b(−k) − a(k)a(−k) = 1.So far we have not used the assumption that k is real (as we will need to do all thistheory again later for complex k), but for real k it is quite clear that ψ + (a, −k) = ψ + (x, k) ∗ ,ψ − (a, −k) = ψ − (x, k) ∗ , a(−k) = a(k) ∗ and b(−k) = b(k) ∗ . Thus we have |b(k)| 2 − |a(k)| 2 =1, and in particular b(k) ≠ 0.Now the problem with the existence of the scattering solution is that it is specified usingboundary conditions. For any function r(k) we can uniquely specify a solution by requiringonly the condition at x = +∞. The question is can we chose an appropriate r(k) such thatthe condition at x = −∞ is also satisfied. Note the condition at +∞ impliesand henceˆψ(x, k) = ψ + (x, −k) + r(k)ψ + (x, k) ,ˆψ(x, k) = (a(−k)ψ − (x, −k) + b(−k)ψ − (x, k)) + r(k)(a(k)ψ − (x, k) + b(k)ψ − (x, −k))= (a(−k) + b(k)r(k))ψ − (x, −k) + (b(−k) + a(k)r(k))ψ − (x, k) .3

To satisfy the boundary condition at −∞ requires that r(k) = −a(−k)/b(k) and t(k) =b(−k)+a(k)r(k). Since we know that for real k, b(k) ≠ 0, we can always find an appropriater(k). For complex k we will see there is a possibility that b(k) = 0. Provided r(k) and t(k)can be defined we havesot(k) = b(−k) − a(k)a(−k)b(k)t(k)t(−k) + r(k)r(−k) =In particular for k real, |r(k)| 2 + |t(k)| 2 = 1. QED.= 1b(k) ,1b(k)b(−k) + a(−k)a(k)b(k)b(−k) = 1It is useful at this stage to work through an example. We consider the simple potentialfunctionu(x) = U 0 δ(x) .A series of straightforward calculations establishes the following: if U 0 ≥ 0 there are noeigenvalues, and if U 0 < 0 there is a single eigenvaluewith normalization constant c = √ κ.respectively,t(k) =kk + iU 0 /2κ = − U 02which clearly satisfy |r(k)| 2 + |t(k)| 2 = 1 for real k.The transmission and reflection coefficients are,r(k) = −iU 0/2k + iU 0 /2The process of taking a potential u(x) and calculating the scattering data, i.e. the eigenvaluesκ, the corresponding normalization constants c, and the transmission and reflectioncoefficients t(k) and r(k) for real k (these satisfy |r(k)| 2 + |t(k)| 2 = 1) is known as the scatteringtransform. In the next few lectures we will establish the remarkable fact that giventhe scattering data it is possible to reconstruct the potential, using what is known as theinverse scattering transform.At this stage it is worthwhile to solve the scattering problem for the following potentials:1.⎧⎨ U 0 |x| < au(x) =⎩ 0 |x| > aFor this potential you will benefit by using Maple.2. u(x) = U 0 sech 2 x. For this potential solving the Schrödinger involves associated Legendrefunctions, see (if necessary) the book of Drazin and Johnson.4

Lecture 5: The Kernel of the Jost FunctionThe inverse scattering transform proceeds through an intermediate object, known as thekernel of the Jost function (by “intermediate” I mean that we use the scattering data toconstruct the kernel of the Jost function, and then the latter to construct u(x)). This lectureis devoted to this remarkable function.Recall: the Jost solution is the solution ψ + (x, k) of −ψ ′′ + u(x)ψ = k 2 ψ satisfying thecondition ψ + (x, k) ∼ e ikx as x → ∞. Let us write ψ + (x, k) = e ikx φ(x, k). Then φ satisfiesthe differential equationφ ′′ + 2ikφ ′ − u(x)φ = 0 , φ ∼ 1 x → ∞ .It is possible to write a formal series solution for this equation in the formφ =∞∑i=0φ i (x)k i , φ 0 (x) = 1 .To verify this, just substitute the series into the equation; the coefficients φ i (x) should bedetermined by the recursionφ i+1 = i 2 (φ′′ i − u(x)φ i ) , i = 0, 1, 2, . . .and if φ 0 (x) = 1 then we will clearly have lim x→∞ φ i (x) = 0 for all i > 0, so all necessaryconditions are satisifed. A consequence of this is that for large |k| the function ψ + (x, k)−e ikxbehaves as 1 , and we can thus consider its inverse Fourier transform. That is, we can write|k|ψ + (x, k) = e ikx +∫ ∞−∞K(x, y)e iky dy (2)for some function K(x, y), known as the kernel of the Jost function. This function K(x, y)will play a central role in what follows and we will now develop its properties.Note that since ψ + (x, k) ∼ e ikx as x → ∞ we must have lim x→∞ K(x, y) = 0 (for everyy). Assuming u(x) tends to zero sufficiently fast we will even have lim x→∞ K x (x, y) = 0. Itis also the case that as a function of y K tends to zero as y → ∞, this is a consequence of thesmooth dependence of ψ + on k. Be warned, though, that while K behaves nicely at infinity,it need not even be continuous (in fact, since we have seen that its Fourier transform onlydrops off as 1/|k| for large k, we should expect discontinuities). We are going to convenientlyignore this and do things like differentiate K. Once we have the results we need we shouldgo back and do some a posteriori justification, but we will be a bit lazy about this.Substituting (2) into the Schrödinger equation and simplifying we find∫ ∞( ∫ ∞) ∫ ∞− K xx (x, y)e iky dy + u(x) e ikx + K(x, y)e iky dy = k 2 K(x, y)e iky dy .−∞−∞5−∞

Taking the Fourier inverse transform we find that K obeys the inhomogeneous wave equationK xx (x, y) − K yy (x, y) − u(x)K(x, y) = u(x)δ(x − y) .Suppose that instead of the conditions K, K x → 0 as x → ∞ we knew that K = K x = 0on the line x = M (M large). Then what we would have here is an inhomogeneous waveequation, with characteristics parallel to x = ±y, initial data on x = M and a nontrivialsource term on the line x = y. See the diagram.We deduce the following results:1. K(x, y) = 0 for x > y. (Needs verbal explanation.)2. K xx (x, y) − K yy (x, y) − u(x)K(x, y) = 0 for x < y.3. u(x) = −2(K x (x, x) + K y (x, x)) = −2 d K(x, x). (From the first result we can writedxK(x, y) = Q(x, y)H(y − x). We then haveK xx − K yy − uK = (Q xx − Q yy − uQ)H(y − x) − 2(Q x + Q y )δ(y − x)and the coefficients of δ(y − x) need to match.)Note that K is real!Note that in light of the first result we can write the relation between ψ + and K asψ + (x, k) = e ikx +∫ ∞xK(x, y)e iky dy . (3)6

Example: u(x) = U 0 δ(x). Compute the Jost solution. Findψ + (x, k) = e ikx + U 0 sin kxH(−x) .kCan find the kernel of the Jost function by Fourier transform. GetK(x, y) = U 0H(−x)2π∫ ∞−∞−iky sin kxe dk .kTo compute this integral: best to teach a “trick” way usingGet∫1 ∞e −ikq dk = δ(q) .2π −∞K(x, y) = U 0H(−x)(H(y + x) − H(y − x)) =2⎧⎨⎩U 02x < y < −x0 otherwise(Note x < y < −x implies x < 0!) You can compute u(x) from this using u(x) =−2 ddx K(x, x) and K(x, x) = − U 02H(−x). To get the latter, note that by K(x, x) we actuallymean lim y↓x K(x, y). Also remember that H(2x)H(−x) = 0.You might want to try to work out the kernel of the Jost function for the two examplesgiven as exercises in the last lecture. Be warned that in the first case there is no explicitformula for the answer, but much information can be gained by asymptotic methods.7

Lecture 6: The Inverse Scattering TransformWe start with a summary of the last two lectures: we looked at the solution of theSchrödinger equation −ψ ′′ + u(x)ψ = Eψ. For E < 0 we write E = −κ 2 . There are onlybounded solutions for certain values of κ, known as the eigenvalues.solutions by requiringWe then have∫ ∞−∞ψ 2 dx = 1 .⎧⎨ ce −κx x → ∞ψ ∼⎩ de κx x → −∞ ,We normalize thewhere the normalization constants c, d depend on κ. For E > 0 we write E = k 2 . For everyk there are bounded solutions. We can consider the Jost solutions, defined byψ ± (x, k) → e ±ikx , x → ±∞ .Either the pair ψ + (x, k), ψ + (x, −k) or the pair ψ − (x, k), ψ − (x, −k) form a basis of the spaceof solutions. We can thus writefor some functions a(k), b(k) which satisfyψ + (x, k) = a(k)ψ − (x, k) + b(k)ψ − (x, −k) (4)|b(k)| 2 − |a(k)| 2 = 1We can also consider the scattering solutions ˆψ that satisfies the boundary conditions⎧⎨ eˆψ(x, −ikx + r(k)e ikx x → ∞k) ∼⎩ t(k)e −ikx x → −∞ .We showed that this solution always exists, thatr(k) = − a(−k)b(k)and t(k) = 1b(k) ,and also that|r(k)| 2 + |t(k)| 2 = 1 .This was the content of the first lecture. In the second lecture we showed the existenceof a kernel K(x, y) such thatand with the propertiesψ + = e ikx +∫ ∞−∞K(x, y)e iky dy8

1. K(x, y) = 0 for x > y.2. K xx (x, y) − K yy (x, y) − u(x)K(x, y) = 0 for x < y.3. u(x) = −2 d K(x, x) (where K(x, x) is the limit of K(x, y) as y tends to x from above.)dxThe essence of inverse scattering is as follows: we will show how to use the scatteringdata (the κ’s, the c’s and r(k)) to find K(x, y), and from this u can be found. The finalresult can be written very simply: if we defineF (x) =∑all eigenvaluesc 2 e −κx + 1 ∫ ∞r(k)e ikx dk (5)2π −∞then K(x, y) can be determined from the Gelfand Levitan Marchenko (GLM) equationso0 = K(x, z) + F (x + z) +∫ ∞−∞F (y + z)K(x, y)dy , z > x . (6)We mention one other fact that we have already seen: for large |k|ψ + (x) = e ikx + O( ) 1|k|andψ − (x) = e −ikx + O( )( )11b(k) = 1 + O , a(k) = O ,|k||k|( )( )11t(k) = 1 + O , r(k) = O .|k||k|( ) 1|k|In particular the integral in the definition of F is well-defined. We also mention one otherfact that we have not yet seen: by computing Wronskians the coefficients a(k), b(k) can beseen to satsifya(k) = W (ψ +(x, k), ψ − (x, −k))W (ψ − (x, k), ψ − (x, −k)) = W (ψ +(x, k), ψ − (x, −k))2ik,b(k) =W (ψ +(x, k), ψ − (x, k))W (ψ − (x, −k), ψ − (x, k)) = W (ψ +(x, k), ψ − (x, k))−2ik(Reminder: W (f, g) = fg ′ − f ′ g.) Here we assume k ≠ 0.(7)The first step of proving GLM is as follows: We can represent ˆψ using the kernel of theJost function asˆψ(x, k) = ψ + (x, −k) + r(k)ψ + (x, k)= e −ikx +∫ ∞−∞e −iky K(x, y)dy + r(k)9( ∫ ∞)e ikx + e iky K(x, y)dy−∞.

Multiplying by e ikz , dividing by 2π and integrating over k gives∫1 ∞2π −∞( ˆψ(x, k) − e −ikx )e ikz dk = K(x, z) + 12π+ 1 ∫ ∞r(k)2π −∞∫ ∞−∞(∫ ∞r(k)e ik(x+z) dk−∞)e iky K(x, y)dy e ikz dk .Assuming that we can change the order in the double integral, this becomes∫1 ∞(2πˆψ(x,∫ ∞k)e ikx − 1)e ik(z−x) dk = K(x, z) + F 0 (x + z) + F 0 (y + z)K(x, y) dy , (8)−∞−∞whereF 0 (x) = 1 ∫ ∞r(k)e ikx dk .2π −∞Note the left hand side can be writtenThe logic now proceeds as follows:∫1 ∞ (ψ− (x, k)e ikx t(k) − 1 ) e ik(z−x) dk . (9)2π −∞1. We show that the Jost solutions ψ ± (x, k) can be extended to the upper half k-plane,and are analytic there (except possibly at k = 0; we omit all discussion of k = 0,which lengthens things significantly, but is merely a technical issue). It follows thatb(k) can be analytically extended to the upper half k-plane (except possible at k = 0)using the formula (7).2. We show that for large |k| (in the upper half k-plane) both ψ − (x, k)e ikx and b(k) are1 + O(1/|k|). We show that the only zeros of b(k) in the upper half k-plane are simplezeros at k = iκ, where κ are the eigenvalues of the Schrödinger equation.3. Thus for z > x the integral (9) can be computed as a sum of residues and this givesus the final form of the GLM equation.Here are the full details:1. Proof of existence and analyticity of ψ + for k in the upper half plane (ψ − is similar).Writeψ + = e ikx φ .Then φ must satisfyφ ′′ + 2ikφ ′ − uφ = 0 , φ → 1 as x → ∞ .10

By the method of “variation of constants” it can be shown that the solution ofsatisfying φ → 1 as x → ∞ isφ(x) = 1 +φ ′′ + 2ikφ ′ = f∫ ∞Thus we seek a solution φ of the integral equationφ(x) = 1 +∫ ∞xxf(y) e2ik(y−x) − 1dy .2iku(y)φ(y) e2ik(y−x) − 1dy .2ikTo construct the solution we use a standard recursive procedure (the “method ofsuccessive approximations”): let φ 0 (x) = 0 and defined a sequence φ 1 , φ 2 , . . . byφ n+1 (x) = 1 +∫ ∞xu(y)φ n (y) e2ik(y−x) − 1dy , n = 0, 1, 2, . . .2ikIf this sequence converges, its limit is the solution we seek. But does the sequenceconverge? Note that for n > 0 we haveφ n+1 (x) − φ n (x) =∫ ∞xu(y)(φ n (y) − φ n−1 (y)) e2ik(y−x) − 1dy .2ikIf k = η + iκ with κ ≥ 0 (i.e. for k in the upper half plane including the real axis),then e 2ik(y−x) is of absolute value less than 1, soClaim:|φ n+1 (x) − φ n (x)| ≤ 1|k|∫ ∞|φ n+1 (x) − φ n (x)| ≤ 1 n!x|u(y)||φ n (y) − φ n−1 (y)|dy .( ∫ 1 ∞n|u(y)|dy).|k| xFor n = 0 this is trivial as φ 1 (x) = 1, for n > 0 it is straightforward to prove byinduction. It follows that the sum∞∑n=0(φ n+1 (x) − φ n (x))converges absolutely as it is majorized by the Taylor series of( ∫ )1 ∞exp |u(y)|dy .|k| xNote all this assumes k ≠ 0. For k = 0 we have to look at the solution of the integralequationφ(x) = 1 +∫ ∞xu(y)φ(y)(y − x)dy ,which we will not do here. Analyticity of φ(x, k) follows from the standard result thatthe limit of a uniformly convergent sequence of analytic functions is itself analytic(each of the function φ n (x, k) are analytic in k).11

2. The behaviors for large |k| are obtained in the same way as in the case of real k. Asregards the location of zeros of b(k): we need to prove that all the zeros of b(k) in theupper half plane lie on the imaginary axis. and have the form k = iκ where κ is aneigenvalue of the Schrödinger equation.Suppose b(k) = 0 for k = η + iκ. We already know b(k) does not vanish on the realaxis, so we can assume κ > 0. From (7), if b(k) = 0 there must exist some constanta(k) such thatψ + (x, k) = a(k)ψ − (x, k) .[There is danger of some confusion here: for all real k the relation (4) holds. For k inthe upper half plane ψ ± (x, k) are both defined, and so is b(k), using (7), but ψ − (x, −k)need not be defined. So the relation (4) makes no sense for general k in the upper halfplane. But for those k for which b(k) = 0, we see there is such a relationship.] Now weknow that ψ + (x, k) ∼ e ikx as x → ∞ and ψ − (x, k) ∼ e −ikx as x → −∞. Since κ > 0this implies that ψ + (x, k) → 0 exponentially as x → ±∞. Consider now the complexconjugate ψ+ ∗ of ψ + . This satisfies−(ψ ∗ +) ′′ + u(x)ψ ∗ + = (k ∗ ) 2 ψ ∗ +as ψ + satisfies−ψ ′′ + + u(x)ψ + = k 2 ψ + .Multiply the first of these equation by ψ + , the second by ψ ∗ +and subtract. We obtain(ψ ∗ + ψ′ + − ψ +(ψ ∗ + )′ ) ′ == ((k ∗ ) 2 − k 2 )ψ + ψ ∗ + .Integrating, and using the fact that ψ + tends to zero at both ±∞, we deduce∫ ∞((k ∗ ) 2 − k 2 ) ψ + ψ+ ∗ dx = 0 .−∞The integral cannot vanish, so we must have k = ±k ∗ , i.e. k is either real or pureimaginary. The latter is not allowed, thus k must be pure imaginary.When b(k) = 0 with k = iκ we have⎧⎨ e −κx x → ∞ψ + →⎩ a(iκ)e κx x → −∞ .Thus at once the roots in the upper half plane of b(k) are precisely of the form iκ,where κ is an eigenvalue of the Schrödinger equation. QED.12

3. Since we now know where the poles of the integrand in (9) are, and we know that theexpression multiplying e ik(x−x) is of order 1/|k| for large k, we can now evaluate (9)using the residue theorem. We get∫ (1 ∞ ψ− (x, k)e ikx )− 1 e ik(z−x) dk = ∑ 2π −∞ b(k)κiψ − (x, iκ)e −κz res( )∣1∣∣∣∣k=iκ.b(k)Since κ is an eigenvalue we have cψ + (x, iκ) = dψ − (x, iκ) so we can write the integral∑κicrd ψ +(x, iκ)e −κz ,where r denotes res ( )∣1 ∣∣k=iκ. In terms of the Kernel of the Jost function this isb(k)∑Inserting in (8) we have∑κoricrd0 = K(x, z)+κicrd(∫ ∞)e −κ(x+z) + K(x, y)e −κ(y+z) dy−∞(∫ ∞)∫ ∞e −κ(x+z) + K(x, y)e −κ(y+z) dy = K(x, z)+F 0 (x+z)+ F 0 (y+z)K(x, y) dy ,−∞−∞(F 0 (x + z) − ∑ κThus we reach the GLM equation (6) provided) ∫ (icr∞d e−κ(x+z) + F 0 (y + z) − ∑−∞κ.icrd e−κ(y+z) )K(x, y) dy .F (x) = F 0 (x) − ∑ κicrd e−κx .From here on not done in classTo reach the final form of F (5) we need to compute the residue r. We haver = res( )∣1∣∣∣∣k=iκ= 1dbb(k)(iκ) ,dkon the assumption that the zero of b(k) at iκ is simple, which we are about to justify.Using (7) we havedbdk = W (ψ +k, ψ − ) + W (ψ + , ψ −k )+ W (ψ +, ψ − ).−2ik2ik 2When k = iκ the last term is zero, and also dψ − = cψ + . Socdbdk (iκ) = W (ψ d +k, ψ + ) + dW (ψ c −, ψ −k ).−2ik∣ k=iκ13

The right hand side of the last equation is not dependent on x, so we can evaluateit either at x = ∞ or at x = −∞. The trouble is that the first term is naturallyevaluated at +∞ and the second term at −∞. So observe thatand henceW (ψ +k , ψ + ) ′ = (ψ +k ψ ′ + − ψ′ +k ψ +) ′= ψ +k ψ ′′ + − ψ′′ +k ψ += ψ +k (u(x)ψ + − k 2 ψ + ) − (u(x)ψ + − k 2 ψ + ) k ψ += 2kψ 2 + ,∫ ∞W (ψ +k , ψ + )| x=+∞− W (ψ +k , ψ + )| x=−∞= 2k ψ+(x, 2 k)dx .−∞When k = iκ, ∫ ∞−∞ ψ2 +(x, k)dx = 1 c 2 . Putting everything together we havecdbdk (iκ) = W (ψ d +k, ψ + )| x=−∞+ d W (ψ c −, ψ −k )| x=−∞−2ik∣( ) k=iκcd W (ψ+k , ψ + )| x=+∞− 2kc +d=W (ψ 2 c −, ψ −k )| x=−∞−2ik∣ k=iκBoth the Wronskians in the last expression are zero, so we obtainandas required (see (5)).dbdk (iκ) = − icd ,F (x) = F 0 (x) + ∑ κr = icdc 2 e −κx .At this stage we can write a summary of inverse scattering transform:1. ComputeF (x) =∑all eigenvaluesc 2 e −κx + 1 ∫ ∞r(k)e ikx dk2π −∞2. Solve the linear integral equation0 = K(x, z) + F (x + z) +∫ ∞−∞to find the kernel of the Jost function K(x, y) for y > x.3. Compute u(x) as −2 ddx K(x, x). 14F (y + z)K(x, y)dy , z > x ,

Example: u(x) = U 0 δ(x) again! If U 0 > 0 then there are no discrete eigenvalues of theSchrödinger equation. We therefore haveF (x) = 1 ∫ ∞2π −∞= 1 ∫ ∞=2π⎧⎨⎩−∞r(k)e ikx dk−iU 0 /2k + iU 0 /2 eikx dk0 x > 0− U 02e xU 0/2x < 0(In the case x > 0 we close the contour in the upper half plane and there is no pole inside,since U 0 > 0; in the case x < 0 we close the contour in the lower half plane and there is asingle simple pole inside. Don’t forget the − in this case!) Now we check the GLM equation.We assume x < z. Then• K(x, z) = U 02if z < −x, zero otherwise.• F (x + z) = − U 02e (x+z)U 0/2 if z < −x, zero otherwise.• ∫ ∞−∞ F (y + z)K(x, y)dy = − U 2 04∫ −zx e (y+z)U 0/2 dy = − U 02(1 − e (x+z)U 0/2 ) if z < −x, zerootherwise.So the GLM equation holds.This was all with the assumption U 0 > 0. If U 0 < 0 we haveF (x) = − U 02 exU 0/2 + 12π= − U 02 exU 0/2 12π⎧⎨=⎩∫ ∞−∞∫ ∞−∞0 x > 0− U 02e xU 0/2x < 0r(k)e ikx dk−iU 0 /2k + iU 0 /2 eikx dk(If x < 0 we close the contour in the upper half plane and the contribution from the poleat k = −iU 0 /2 cancels the first term. If x > 0 we close in the lower half plane, there areno singularities enclosed, so the integral is zero.) Thus the cases U 0 < 0 and U 0 > 0 areactually the same.Once again, you should try to make everything work for the other two examples givenin lecture 4.15

Exercises In Scattering and Inverse ScatteringGiven a potential u(x):1. Find the eigenvalues κ and normalization constants c (and d)2. Find the Jost functions ψ ± and the functions a(k), b(k). Check |b| 2 − |a| 2 = 1.3. Check: ψ ± and b can be analytically extended to the upper half k plane (but usuallya cannot). Zeros of b in the upper half plane at k = iκ, for any eigenvalue κ.4. Check: correct large |k| behavior of ψ ± and a, b.5. Find the scattering solution ˆψ and the reflection and transmission coefficients r(k), t(k)(r = −a ∗ /b, t = 1/b). Check |r| 2 + |t| 2 = 1.6. Find (by Fourier transform) the kernel K(x, y) of the Jost function ψ + .7. Check K(x, y) = 0 for x > y. Check u(x) = −2 d K(x, x). And check that K satisfiesdxthe correct wave equation for y > x.8. Find the functionF (x) =9. Check the GLM equation∑all eigenvalues0 = K(x, z) + F (x + z) +c 2 e −κx + 1 ∫ ∞r(k)e ikx dk2π −∞∫ ∞−∞F (y + z)K(x, y)dy , z > x ,16

The case u(x) = −2sech 2 x:1. A single eigenvalue κ = 1, corresponding to the normalized eigenfunctionψ = sechx √2.For large positive x this behaves as √ 2e −x , for large negative x this behaves as √ 2e x ,so c = d = √ 2.2. Jost functions:Check |b| 2 = 1.ψ + (x, k) = e ikx (1 +ψ − (x, k) = e −ikx (1 +)2(e 2x + 1)(ik − 1)2(e −2x + 1)(ik − 1)a(k) = 0 , b(k) = ki + 1ki − 1 .)3. ψ ± and b are all analytic as functions of k except at k = −i, where they have a pole.So they are analytic in the upper half plane. In this case a can also be extended. b(k)has a single zero in the upper half plane, at k = i, corresponding to the eigenvalue 1.4. All asymptotic relations satisfied for large |k|:( ) 1ψ + (x) = e ikx + O , ψ − (x) = e −ikx + O|k|5. Scattering solutionb(k) = 1 + O( ) 1|k|ˆψ(x, k) = e −ikx (1 +Transmission and reflection coefficients:Easy to check that |t| 2 = 1., a(k) = O( ) 1|k|)2(e 2x + 1)(−ik − 1)t(k) = ki − 1ki + 1 , r(k) = 0 .6. The kernel of the Jost function:⎧⎨ −e −y sechx y > xK(x, y) =⎩ 0 y < x( ) 1|k|..,17

7. Check that−2 d (−e −x sechx ) = −2sech 2 x .dxSimilarly K xx − K yy − u(x)K = 0 for y > x.8. F (x) = 2e −x .9. We check that for z > x:∫ ∞−∞K(x, z) = −e −z sechxF (x + z) = 2e −x−zF (y + z)K(x, y)dy =∫ ∞These add to zero, so GLM is satisifed.x2e −y−z ( −e −y sechx ) dy = −e −2x−z sechx18