MA 342N Assignment 1 Due 24 February 2016
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1 M 342N ssignment Due 24 February 206 Id: 342N-s206-.m4,v. 206/02/5 2:25:36 john Exp john. Suppose that q, in addition to satisfying the assumptions from lecture, is an even function. Prove that η(λ = 0, i.e. that there are no mixed terms (θϕ or ϕθ terms in the spectral decomposition. Hint: Show that Solution: In general, if y satisfies then ỹ satisfies ψ + (λ, x = ψ (λ,x, ψ (λ, x = ψ + (λ,x, ϕ(λ, x = ϕ(λ,x, θ(λ, x = θ(λ,x. y +λy = qy ỹ +λỹ = qỹ where ỹ(x = y( x q(x = q( x. In particular, if q is even then q = q ỹ satisfies the same differential equation as y. So ψ + (λ, x, ψ (λ, x, ϕ(λ, x θ(λ, x all satisfy y +λy = qy. From ψ + (λ,x e i λx as x + it follows that ψ + (λ, x e i λx as x, so ψ + (λ, x = ψ (λ,x. Similarly, or as a corollary, ψ (λ, x = ψ + (λ,x. lso ϕ(λ, x ϕ(λ,x satisfy the same initial value problem so are, by the fundamental existence uniqueness theorem, the same function. Similarly, θ(λ, x θ(λ, x are the same.
2 Id: 342N-s206-.m4,v. 206/02/5 2:25:36 john Exp john 2 Substituting x for x in the equation gives (ψ (λ,x ψ + (λ,x = (ϕ(λ,x θ(λ,x (ψ + (λ,x ψ (λ,x = ( ϕ(λ,x θ(λ,x ( α(λ β(λ γ(λ δ(λ ( α(λ β(λ γ(λ δ(λ hence ( β(λ α(λ (ψ (λ,x ψ + (λ,x = (ϕ(λ,x θ(λ,x δ(λ γ(λ It follows that thus Integrating, hence ( α(λ β(λ = γ(λ δ(λ λ+iτ iτ ( β(λ α(λ δ(λ γ(λ α(λδ(λ α(λδ(λ β(λγ(λ = 2. α(µδ(µ α(µδ(µ β(µγ(µ dµ = λ 2 η(λ = Consider the set of functions y which satisfy y +λy = 0 for x 0, are continuous at x = 0, whose derivatives are continuous except for a jump discontinuity of the form lim (x lim (x = y(0. x 0 +y x 0 y These are close enough to the situation considered in lecture that we can still do scattering theory spectral theory. If we take the limit of our special solutions to y +λy = qy with { h if x < w/2, q(x = 0 if x > w/2 with y y continuous at x = ±w/2 as w 0 h in such a way that = wh is constant then we get solutions to the problem above. Formally this is often written as y +λy = qy with q(x = δ(x, where δ is the Dirac delta. Unfortunately that doesn t really make any sense, even if the equations are interpreted in the sense of distributions. The simplest way of treating the problem is to ignore both delta functions the h 0 limit just view the equation as y +λy = 0 with the jump condition given above..
3 Id: 342N-s206-.m4,v. 206/02/5 2:25:36 john Exp john 3 Find the scattering matrix for this problem, verify that it is unitary. Note: The basic existence uniqueness theorem still applies here, despite the strangeness of the equation at x = 0. That s not hard to prove, but it s also not interesting, so don t bother. You can just assume there is a unique solution for any choice of initial data at any point, except thatif wearetaking dataatx = 0thenweneed tospecify the average of the limits ( lim (x+ lim (x 2 x 0 +y x 0 y rather than the (probably non-existent y (0. Hint: The functions θ ϕ are still useful, but their definitions need to be modified as described above: θ(λ,0 =, ϕ(λ,0 = 0, 2 ( lim (λ,x+ lim (λ,x x 0 +θ x 0 θ = 0, ( lim (λ,x+ lim (λ,x =. 2 x 0 +ϕ x 0 ϕ Solution: With, as usual s = λ, these two solutions are { cos(sx sin(sx if x < 0, θ(λ,x = 2s cos(sx+ ϕ(λ,x sin(sx if x > 0, 2s = sin(sx. s The solutions defined by their behaviour as x + are given by the equations ψ + (λ,x = e isx = cos(sx+isin(sx = θ(λ,x+ χ + (λ,x = e isx = cos(sx isin(sx = θ(λ,x+ ( 2 is ( 2 +is if x > 0. The solutions defined by their behaviour as x are ψ (λ,x = e isx = cos(sx isin(sx = θ(λ,x+ ( 2 +is χ (λ,x = e isx = cos(sx+isin(sx = θ(λ,x+ ( 2 is
4 Id: 342N-s206-.m4,v. 206/02/5 2:25:36 john Exp john 4 if x < 0. Forgetting the intermediate steps, ψ + (λ,x = θ(λ,x+ ( 2 is if x > 0 χ + (λ,x = θ(λ,x+ ψ (λ,x = θ(λ,x+ χ (λ,x = θ(λ,x+ ( 2 +is ( 2 +is ( 2 is if x < 0. These equations must in fact hold everywhere, as a consequence of the existence uniqueness theorem. little algebra shows that so ψ + (λ,x+ψ (λ,x = 2θ(λ,x = χ + (λ,x+χ (λ,x ψ + (λ,x ψ (λ,x = ( 2isϕ ( λ,x χ + (λ,x χ (λ,x = (+2isϕ ( λ,x χ = 2 (χ + +χ 2 (χ + χ = 2 (ψ + +ψ +2is 2 2is (ψ + ψ = 2is ψ + 2is 2is ψ + χ + = 2 (χ + +χ + 2 (χ + χ = 2 (ψ + +ψ + +2is 2 = 2is 2is ψ + 2is ψ + The scattering matrix is therefore ( a(λ b(λ = c(λ d(λ 2is ( 2is 2is 2is (ψ + ψ.
5 Id: 342N-s206-.m4,v. 206/02/5 2:25:36 john Exp john 5 3. Find the spectral representation for the same problem. Solution: We have ( α(λ β(λ (ψ (λ,x ψ + (λ,x = (ϕ(λ,x θ(λ,x γ(λ δ(λ where so Then, integrating, α = 2 +is, β = 2 αδ βγ = +2is, αβ αδ βγ = 4 is 2, βγ αδ βγ = 2, γδ αδ βγ = +2is. is, γ =, δ =. αβ αδ βγ dλ = 4 λ i 3 λ3/2, βγ αδ βγ = λ 2, γδ αδ βγ = i λ+ 2 log ( 2i λ. Integrating from iτ to λ + iτ taking /π times the limit of the imaginary part as τ 0 +, { 0 if λ 0, ξ(λ = λ 3/2 if λ > 0, 3π η(λ = 0. ζ(λ is more complicated depends on the sign of. If is positive then { 0 if λ 0, ζ(λ = λ /2 + arctan 2 λ if λ 0. π 2π
6 Id: 342N-s206-.m4,v. 206/02/5 2:25:36 john Exp john 6 If is negative then if λ < 2 /4, 2π ζ(λ = 0 if 2 /4 < λ 0, λ /2 + arctan 2 λ if λ 0. π 2π The case = 0 was of course done in lecture, since that s just q = 0. The spectral representation is then f(x 2π + 2π 0 0 2π ϕ( 2 /4,x ϕ(λ,xϕ(λ, xf( xd xλ /2 dλ θ(λ,xθ(λ, xf( xd x 4λ/2 4λ+ dλ 2 ϕ( 2 /4, xf( xd x, with the last term present only if < 0.
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