5.61 Fall 2013 Problem Set #3

Transcription

1 5.61 Fall 013 Problem Set #3 1. A. McQuarrie, page 10, #3-3. B. McQuarrie, page 10, #3-4. C. McQuarrie, page 18, # McQuarrie, pages 11-1, # A. McQuarrie, page 13, #3-17. B. McQuarrie, page 13, #3-18. C. McQuarrie, page 17, # Solve for the eergy levels of the particle cofied to a rig as a crude model for the electroic structure of bezee. The two dimesioal Schrödiger Equatio, i polar coordiates, is 1 µ r r r r + 1 r φ + U ( r,φ) ψ = Eψ. For this problem, U(r,φ) = for r a, but whe r = a, U(a,φ) = 0. A. This implies that ψ(r,φ) = 0 for r a. Why? ψ B. If ψ(r,φ) = 0 for r a, the r = 0. What is the simplified form of the Schrödiger Equatio that applies whe the particle is cofied to the rig? C. Apply the boudary coditio that ψ(a,φ) = ψ(a,φ + π) to obtai the E eergy levels. D. The C C bod legth i bezee is Å. Thus a circle which goes through all 6 carbo atoms has a radius r = Å. Use this to estimate the = = 1 trasitio for bezee treated as a particle o a rig. The logest wavelegth allowed electroic trasitio for real bezee is at 66 Å. Explai why the agreemet is ot perfect. Problem Set #3 Fall, 013 Page 1

2 5. This problem illustrates tuelig. It requires either iterative or graphical solutio (see McQuarrie problems 4-49 ad 4-54 for ispiratio). This is oe of my favorite guided magical mystery tours. Cosider the potetial eergy fuctio V(x) = x a/ Regios I (x < 0) ad V (x > 0) V(x) = 0 a/10 x < a/ Regios II (x < 0) ad IV (x > 0) 3h V(x) = x < a/10 Regio III a/10 < x < a/10 A. Draw V(x) ad label the regios. Some Backgroud Two boudary coditios are ψ(±a/) = 0. The potetial, V(x), has left-right symmetry about x = 0. Thus all eigefuctios of H must be either symmetric or atisymmetric, thus ψ (x < 0) = ( 1) +1 ψ (x > 0) for = 1,,. So all you will evetually eed to do is fid either ψ(x < 0) or ψ(x > 0) ad joi them at x = 0 so that both ψ(x) ad are cotiuous at x = 0. The best way to dx begi the costructio of ψ is to write ψ,trial i Regios II ad IV as ψ,trial IV (x > 0) = Acos [k(x a/) + π/] ψ,trial II(x < 0) = Acos [k( x a/) + π/], These trial fuctios satisfy the x = a/ ad x = a/ boudary coditios for all values of k. Note that the argumets of the cosies are set up so that ψ,trial II iitially icreases from 0 as x becomes larger tha a/ ad ψ,trial iv iitially icreases from 0 as x becomes smaller tha +a/. Thus ψ,trial II ad ψ,trial IV are mirror images of each other ad ψ,trial is ecessarily cotiuous at x = 0. B. What is k, expressed i terms of E,, ad m? C. The barrier causes each of the eergy levels to be shifted to higher eergy tha the correspodig levels of the simple V(x) = 0, x a/ box E barrier E ormal > 0. Before doig ay calculatios, aswer this qualitative questio: Do you expect the level shifts of the eve- levels (ati-symmetric, ode at x = 0) Problem Set #3 Fall, 013 Page

3 to be systematically larger or smaller tha those for the odd- levels (symmetric, ati-ode at x = 0)? Why? Choose either path 1 or path Path 1: D. The wavefuctios i Regio III have the form 1/ κ = 3 me ψ III = Be κx + Ce κx π a 3h for E < ad 1/ III ikx ikx me 3π ψ = De + Ee k = a 3h for E >. You ca solve for (B,C) or (D,E) at the regio II, III boudary (x < 0) by eforcig cotiuity of ψ ad dx at x = a/10. The same solutio is valid at the regio III, IV boudary (x > 0). Show that ψ III left propagated i from the II, III boudary is equal at x = 0 to ψ III right propagated i from the III, IV boudary. Note that, at the value of E you have chose, dx is probably discotiuous at x = 0. E. To fid each of the E barrier eigeeergies, you eed to icrease the trial value of E barrier relative to E ormal util dx is cotiuous at x = 0. A graphical method, where you plot III left III right at x = 0 vs. E trial, will dx dx permit you to fid the eigeeergies. Fid E barrier for = 1,, 3, 4, 5, ad 6, ad plot them ext to E ormal =. [There will be two classes of eigefuctios for the barrier-box. The odd- fuctios will have = 0 at x = 0 ad the eve- fuctios dx x=0 will have ψ (x) = 0 at x = 0. You ca fid the eige-eergies, E barrier, by either adjustig E util is cotiuous at x = 0 or, for odd-, dx = 0, ad for eve-, ψ(0) = 0.] dx x=0 Path : D. The wavefuctios i Regio III have the form h Problem Set #3 Fall, 013 Page 3

4 ψ I 3h for E < ad B κ 1/ II 3 x = π e + Ce κx κ = me a 1/ III E 3 ψ = De ikx + Ee ikx m k π = a 3h for E >. Fid the geeral form of the wavefuctio i regios I ad II as well. Use symmetry argumets to simplify these expressios. Note that you have to use differet symmetry argumets for the odd-symmetry ad evesymmetrry eigefuctios. E. Use the fact that the wavefuctio ad its first derivative are cotiuous at the boudaries betwee regios to determie the values of the coefficiets ad therefore the eergy eigevalues for the eve-symmetry ad oddsymmetry cases that you described i part D. Do this for eergy eigevalues for =1,,3,4,5,6, ad the plot them agaist the stadard h ifiite well case where the eergy eigevalues are E =. Both Paths do F ad G F. Discuss the eergy level patter. Make whatever geeralizatios you ca about the effect of a cetral barrier o the patter of eergy levels. Suggest (without full calculatio) what would happe to this patter if the barrier were made twice as high or twice as wide. Which effect is stroger, makig the barrier higher or wider? G. Plot ψ 1 + ψ ad ψ 1 ψ ad commet o the left, right localizatio of the resultat o-eigestate probability amplitude distributios. Problem Set #3 Fall, 013 Page 4

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