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4 Scattering Theory 4 Scattering theory Exercise 4.1: Delta-function potential Consider a one-dimensional particle of mass m subject to the potential V (x)= ~uδ(x). − (a) Show that the general solutions of the time-independent Schr¨odinger equation at energy ε = ~2k2/2m can be cast in the form 1 eikx + r(k)e−ikx x< 0, ψk(x) = √2π × t(k)eikx x> 0, 1 e−ikx + r′(k)eikx x> 0, ψ′ (x) = k √2π × t′(k)e−ikx x< 0, and find explicit expressions for the amplitudes r(k), t(k), t′(k), and r′(k). What is the interpretation of the amplitudes r(k), t(k), t′(k), and r′(k)? (b) Show that one may, alternatively, cast the general solution to the stationary Schr¨odinger equation into the form 1 ψS,k(x) = cos(k x + δS), √π~v | | 1 ψA,k(x) = sign(x)sin(k x + δA), √π~v | | with v = ~k/m. Calculate the “phase shifts” δS and δA. Can you explain your result for δA? Exercise 4.2: Adiabatic switching on of interaction The construction of the stationary retarded scattering state Ψ relies on a procedure in which the interaction Vˆ is adiabatically switched on, Vˆ (t) = |Veˆ ηt/~, where η is a positive infinitesimal with the dimension of energy. The stationary retarded scattering state obeys the Lippmann-Schwinger equation Ψ(t) = Φ(t) + Gˆ Veˆ ηt/~ Ψ(t) , (1) | | 0 | 1 where Gˆ =(E Hˆ +iη)−1 is the Green function or resolvent for the system with Hamiltonian 0 − 0 Hˆ0, i.e., in the absence of the interaction Vˆ , and Φ is the eigenstate of Hˆ0 that describes the initial state. This equation was derived under the| assumption that Ψ is a true stationary state, with a time dependence e−iEt/~. Clearly, this assumption can| not be strictly valid no matter how small η is chosen,∝ because the state Ψ(t) that solves Eq. (1) evolves from the initial state Φ(t) for t to the retarded| scattering state for times t 0, which is different from |Φ(t). The→ purpose −∞ of this exercise is to show, for a specific exampl∼ e, that | nevertheless the solution to Eq. (1) is the valid solution of the time-dependent Schr¨odinger equation with interaction Vˆ (t) = Veˆ ηt/~ in the limit η 0, and you will systematically include corrections to this solution for small but finite η. ↓ Hereto, we consider a particle of mass m in one dimension, subject to the potential V (x)eηt/~ = ~uδ(x)eηt/~. The wavefunction φ(x,t) of the initial state is taken to be − ~ 1 φ(x,t)= φ(x)e−iEt/ , φ(x)= eik0x, √2π −1 where k0 = ~ √2mE. (a) Show that the Lippmann-Schwinger equation (1) gives a closed equation for the wave- function ψ(0,t) at x = 0. Solve that equation and show that the full solution of Eq. (1) reads ηt/~ 1 −iEt/~ ik0x mue ikη|x| ψ(x,t)= e e ηt/~ e , (2) √2π − i~kη + mue 1 where kη = ~ 2m(E + iη). Hint: The Green function Gˆ0 in position representation reads ′ m ′ ˆ ′ ikη|x−x | G0(x,x ; kη) = x G0 x = e 2 . | | i~ kη (b) Show that in the limit η 0 (at fixed x and t) your solution agrees with the retarded scattering state ψ(x,t) that↓ is obtained from a direct solution of the Schr¨odinger equa- tion with a time-independent potential. (c) The limit η 0 at fixed x and t obscures the switching on of the potential. In order to see the evolution↓ of the wavefunction while the potential is switched on, one should take the limit η 0 while keeping the product ηt fixed. Apply this limiting procedure to the solution to↓ the Lippmann-Schwinger equation (1) given in part (a) and interpret your answer. 2 The Lippmann-Schwinger equation (1) was derived from the time-dependent equation t i ˆ ′ ~ ′ −iH0(t−t ) ˆ ηt/ ′ ψ(t) = φ(t) ~ dt e Ve ψ(t ) (3) | | − −∞ | ′ under the assumption that ψ(t) is a stationary state, ψ(t′) = ψ(t) eiE(t−t )/~. This assumption will not be made| in the remainder of this exercise| . | (d) Specialize to the case of the delta-function potential, and show that Eq. (3) implies a closed (integral) equation for the full function ψ(0,t) of the wavefunction of the retarded scattering state at x = 0. For a solution for the full wavefunction ψ(x,t) it is sufficient to find a solution for the function ψ(0,t). This can be done using the ansatz ψ(0,t)= e−iEt/~f(ηt; η), (4) where we anticipate that the function f, which describes to what extent the time dependence of ψ(0,t) deviates from that of a stationary function, varies with time t on the slow scale ~/η. One then solves for the function f as a power series in η, f(ηt; η)= f (0)(ηt)+ ηf (1)(ηt)+ .... (5) (e) Show that truncating the expansion at zero order reproduces the solution for ψ(0,t) you obtained from the Lippmann-Schwinger equation (1). Hint: Taylor expand f(ηt′; η) around t′ = t in order to transform the integral equation for f you obtained in (d) into an ordinary equation. (f) Obtain an equation for the first-order correction f (1), and solve that equation. Is the correction to f (0) significant in the limit η 0? ↓ Exercise 4.3: Lippmann-Schwinger equation for advanced scattering states R The stationary retarded scattering state Ψi (or simply Ψi ) is related to the stationary (free) initial state Φ through the Lippmann-Schwinger| equation| | i −1 R ˆ ˆ R Ψi = Φi + Ei H0 + iη V Ψi | | − | −1 = 1+ Ei Hˆ + iη Vˆ Φi , − | 3 R where η is a positive infinitesimal. Show that the stationary advanced scattering state Ψf is related to the stationary (free) final state Φ through the related equation | | f −1 A ˆ A Ψf = Φf + Ef H0 iη V Ψf | | − − | −1 = 1+ Ef Hˆ iη Vˆ Φf . − − | Exercise 4.4: Optical theorem The generalized optical theorem reads T T ∗ = 2πi δ(E E )T T ∗ ba − ab − − i bi ai i = 2πi δ(E E )T T ∗ with E = E = E . − − f fa fb a b f Here the labels a and b specify initial and final states. (a) Show that the generalized optical theorem follows from the relation T = Φ Vˆ ΨR ba b| | a R between the T matrix, the free state Φb , the scattering state Ψa and the interaction V , and the Lippmann-Schwinger equation.| | (b) Show that the optical theorem, Im T = π δ(E E ) T 2, ii − i − f | fi| f follows from the generalized optical theorem. (c) Use the generalized optical theorem to show that the scattering matrix, S = δ 2πiT δ(E E ), ba ba − ba b − a is unitary. 4 Exercise 4.5: Analicity and Causality In complex analysis you learn that the integral of a function f(z) over a contour γ in the complex plane can be deformed in those regions where f(z) is analytic. Specifically, you learn that (according to the residue theorem) the contour integral over a closed path γ of a function f(z), which is analytic up to some isolated points z ,...,z where f(z) has poles, { 0 N } can be calculated from the residues of f(z) at those points which are enclosed by γ, γ dz f(z)=2πi( 1) Resz f, − i γ i=0,...,n where ( 1)γ = 1( 1) if γ encircles the enclosed poles z ,...,z counter-clockwise (clock- − − { 0 n} wise). The residue Reszi f is defined through the Laurent expansion of f(z) near z = zi, ∞ f(z)= a (z )(z z )n, Res f = a (z ). n i − i zi −1 i n=−k(zi) If the coefficient a = 0, the integer k(z ) is referred to as the “order” of the pole at −k(zi) i z = zi. (a) Calculate the integral ∞ 1 I(z0,z1)= dx , (x z )(x z ) −∞ − 0 − 1 where z0 and z1 are complex numbers with Im z0 > 0 and Im z1 < 0. (b) Repeat the calculation of I(z0,z1) for complex numbers z0 and z1 with Im z0 > 0 and Im z1 > 0. (c) Calculate the integral ∞ −iE(t−t′)/~ ± ′ dE e F (Ek,t t )= , − 2π E E iη −∞ − k ± distinguishing the cases t>t′ and t<t′. (d) In the lecture you solved the Lippmann-Schwinger equation for the advanced and −1 retarded scattering states with help of the propagators Gˆ(Ei)=(Ei Hˆ iη) . Use your answer to (c) to explain why these are called the “retarded”− and± “advanced” propagators ? 5 (e) Show that F +(E ,t t′)=[F −(E ,t′ t)]∗. k − k − Exercise 4.6: One-dimensional free particle propagator In the lecture you learned that the retarded, free particle propagator in a one dimensional system in the position-represenatation is ik|x−x′| ′ −1 ′ e G0(x,x ; k)= x (Ek Hˆ0 + iη) x = , | − | ivk~ 2 with k = 2mEk/~ and vk = ~k/m. (a) Show that the propagator satisfies the equation ~2 E + ∂2 G (x,x′; k)= δ(x x′). k 2m x 0 − −1 (b) Show that the retarded propagator Gˆ0 =(Ek Hˆ0 + iη) is diagonal in the k repre- sentation, − δ(k k ) k Gˆ (k) k = G (k , k ; k)= 1 − 2 . 1| 0 | 2 0 1 2 E E + iη k − k1 Since Gˆ is diagonal in the k representation, we write G (k′; k) G (k′, k′; k).
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