5.61 FIRST HOUR EXAM ANSWERS Fall, 2013

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1 5.61 FIRST HOUR EXAM ANSWERS Fall, 013 I. A. Sketch ψ 5(x)ψ 5 (x) vs. x, where ψ 5 (x) is the n = 5 wavefunction of a particle in a box. Describe, in a few words, each of the essential qualitative features of your sketch. ψ5(x)ψ 5 (x) E /a 0 E 5 = h 8ma (5) 0 x a This is a probability density for n = 5 of a PIB. There are n 1 = 4 nodes. The no [ des ( are )] equally spaced. Each lobe between consecutive nodes is a half-cycle of sin 5πx a with maximum height of (/a).

2 5.61 Fall 013 First Hour Exam ANSWERS Page B. Sketch ψ5(x)ψ 5 (x) vs. x, where ψ 5 (x) isthev = 5 wavefunction of a harmonic oscillator. Describe, in a few words, each of the essential qualitative features of this sketch. ψ5(x)ψ 5 (x) 0 E 5 = ħω(5.5) 0 x This is a probability density for v = 5 of a H.O. There are v = 5 nodes. One node is at x = 0. The, 3, 4 nodes are close together because the classical p(x) function is largest near x = 0 and λ = h/p. The 1 and 5 nodes are closer to the turning points at x ± = [(ħω5.5)/k] 1/ than to the and 4 nodes. The outer lobes have the largest maximum height and area, but cannot finish at ψ 1/ 5ψ 5 =0atx ± [11ħω/k], thus have exponentially decreasing tails in the classically forbidden E<V(x) regions.

3 5.61 Fall 013 First Hour Exam ANSWERS Page 3 C. (i) Sketch ψ 1 (x) and ψ (x) for a particle in a box where there is a small and thin barrier in the middle of the box, as shown on this V (x): The wavefunctions look like this: V (x) ( h 8ma ) 10 a/100 0 a/ a The barrier has a negligible effect on the n = energy and wavefunction. However the n = 1 wave function tries to go to zero near x = a/, but it is not allowed to actually cross zero, because that would generate an extra node. In order for ψ 1 (x) to approach 0 at x = a/, the E 1 energy increases until it lies just barely below E. (ii) Make a very approximate estimate of E E 1 for this PIB with a thin barrier in the middle. Specify whether E E 1 is smaller than or larger than 3 h 8ma which is the energy level spacing between the n = and n = 1 energy levels of a PIB without a barrier in the middle. h 0 <E (0) (0) E 1 3= E ma E 8 1. D. Consider the half harmonic oscillator, which has V (x) = 1 kx for x<0 and V (x) = for x 0. The energy levels of a full harmonic oscillator are E(v) =ħω(v +1/) where ω =[k/μ] 1/. Sketch the v = 0 and v =1ψ v (x) of the half harmonic oscillator and say as much as you can about a general energy level formula for the half harmonic oscillator. A little speculation might be a good idea.

4 5.61 Fall 013 First Hour Exam ANSWERS Page 4 V (x) v =1 ψ 1 (x) v =0 ψ 0 (x) 0 x For the half-harmonic oscillator, the v = 0 wave function is the left half of the v =1 wave function of the full harmonic oscillator. The v = wave function is the left half of the v = 3 wave function of the full HO. Half HO Full HO E(v =0)= 3 ħω E(v =0)= 1 ħω E(v =1)= 7 ħ ħω ω E(v =1)= 3 ħω 1ħω E. Give exact energy level formulas (expressed in terms of k and μ) for a harmonic oscillator with reduced mass, μ, where (i) V (x) = 1 kx + V 0 E(v) =V 0 + ħω(v + 1 ) (ii) V (x) = 1 k(x x 0 ) 1 E(v) = ħω(v + ) (iii) V (x) = 1 k x where k E(v) =ħ E(v) =ħ [ 1/ =4k [ k ] 1/ (v + 1 ) μ k =4k ] k (v + 1 )=ħω(v +1/) μ

5 5.61 Fall 013 First Hour Exam ANSWERS Page 5 II. PROMISE KEPT: FREE PARTICLE ħ Ĥ = + V 0 m x ψ(x) =ae ikx + be ikx A. Is ψ(x) an eigenfunction of H? If so, what is the eigenvalue of Ĥ, expressed in terms of ħ, m, V 0, and k? Hψ ħ = [(ik) ae ikx +( ik) be ikx ] m ħ k = [ae ikx + be ikx ] m ψ is an eigenfunction of H with eigenvalue ħ k. B. Is ψ(x) an eigenfunction of m pˆ? Your answer must include an evaluation of pψ ˆ (x). ψ is not an eigenfunction of pˆ. pψ ˆ = iħ[(ik)ae ikx +( ik)be ikx ] = ħk[ae ikx be ikx ] C. Write a complete expression for the expectation value of pˆ, without evaluating any of the integrals present in pˆ. see answer in part D. D. Taking advantage of the fact that dxe icx =0 compute pˆ, the expectation value of pˆ. dxψ pψ ˆ pˆ = dxψ ψ dxħk[a e ikx + b e ikx ][ae ikx be ikx ] = dx[a e ikx + b e ikx ][ae ikx + be ikx ] ħk dx[ a b ] pˆ = dx[ a + b ] a = ħk b a + b

6 5.61 Fall 013 First Hour Exam ANSWERS Page 6 E. Suppose you perform a click-click experiment on this ψ(x) where a = 0.63 and b = One particle detector is located at x =+ and another is located at x =. Let s say you do 100 experiments. What would be the fraction of detection events at the x = + detector? The x =+ sees a particle at a 0.40 f + = = = 40% of the time a + b F. What is the expectation value of Ĥ? In part A. we found that Hψ ħ k (x) = ψ(x) m dxψ Hψ ħ k H dxψ ψ = = m ħ k =. dxψ ψ dxψ ψ m If ψ is an eigenstate of the measurement operator, every measurement yields the eigenvalue of ψ. The expectation value is the eigenvalue!

7 5.61 Fall 013 First Hour Exam ANSWERS Page 7 III. â AND â FOR HARMONIC OSCILLATOR âψ v = v 1/ ψ v 1 â ψ v =(v +1) 1/ ψ v+1 Nψ v = vψ v where N = â â A. Show that [â, â] by applying this commutator to ψ v. [â, â] =â â ââ [â, â]ψ v = â âψ v ââ ψ v = â v 1/ ψ v 1 â(v +1) 1/ ψ v+1 1/ 1/ = v v ψ v (v +1) 1/ (v +1) 1/ ψ v =[v (v + 1)]ψ v =( 1)ψ v [â, â] = 1 B. Evaluate the following expressions (it is not necessary to explicitly multiply out all of the factors of v). (i) (â ) (â) 5 ψ 3 (â ) (â) 5 5 ψ3 = 0 because â ψ 3 = â ψ 0 (3 1) 1/ but âψ 0 =0. (ii) (â) 5 (â ) ψ 3 (â) 5 (â ) has selection rule Δv = 3 (â) 5 (â ) ψ 3 =( ) 1/ (4 5) 1/ ψ 0. (iii) dxψ 3(â ) 3 ψ 0 dxψ 3 (â ) 3 ψ 0 =(1 3) 1/. (iv) What is the selection rule for non-zero integrals of the following operator product (â ) (â) 5 (â ) 4? Δv =+4 5 = +1. (v) (â+â ) = â +ââ +â â+â. Simplify using â +â to yield an expression containing â + â + terms that involve N = â â and a constant. N (â + â ) = â + â + ââ + â â ââ =[â, â ]+â â =1+N (â + â ) = â + â +N +1

8 5.61 Fall 013 First Hour Exam ANSWERS Page 8 IV. TIME-DEPENDENT WAVE EQUATION AND PIB SUPERPOSITION For the harmonic oscillator ( ) 1/ ħ xˆ = (â + â) μω ( ) 1/ ħμω pˆ = i(â â) N = â â δ ij = dxψi (x)ψ j, which means orthonormal {ψ n } Hψ n(x) =E n ψ n which means eigenvalues {E n } ( ) 1 E n = ħω n + Consider the time-dependent state A. Sketch Ψ(x, 0) and Ψ(x, t = π ω ) Ψ(x, t) = [ 1/ e ie0t/ħ ψ 0 (x)+e ie 1 ] t/ħ ψ 1 (x) [ ] = 1/ e ie0t/ħ ψ 0 (x)+e iħωt/ħ ψ 1 Ψ(x, 0) Ψ ( x, π ω ) localized on left localized on right Ψ(x, 0) = ψ 0 (x)+ψ 1 (x) Ψ(x, π ω )=ψ 0(x) ψ 1 (x)

9 5.61 Fall 013 First Hour Exam ANSWERS Page 9 B. Compute dxψ (x, t)n Ψ(x, t). dxψ 1 (x, t)n Ψ(x, t) = dx(ψ0 + e iωt ψ1)n (ψ 0 + e iωħ ψ 1 ) 1 = dx(ψ 0 + e iωt ψ1)(0 + e iωt ψ 1 ) 1 because Nψ 0 =0ψ 0 = C. Compute H = dxψ (x, t)ĥψ(x, t) and comment on the relationship of N to H. ( 1 Ĥ = ħω N + [ H = ħω N + ) ] = because Ψ is normalized to 1. H = ħω [ ] = ħω. This is not surprising because the average E in Ψ(x, t) is and E is conserved. E v=0 + E v=1 = ħω

10 5.61 Fall 013 First Hour Exam ANSWERS Page 10 D. Compute xˆ = dxψ (x, t)ˆxψ(x, t). xˆ ( ) 1/ ħ xˆ = (â + â) μω ( ) 1/ ħ xˆψ(x, t) = 1/ e ie 0t/ħ (â + â)(ψ 0 + e iωt ψ 1 ) μω ( ) 1 / ħ = 1/ e iωt/ (ψ + 1/ e iωt 1 ψ + e iωt ψ 0 ) μω ( ) 1/ 1 ħ Ψ xˆψ = (ψ0 + e iωt ψ1 )( e iωt ψ 0 + ψ 1 + 1/ e iωt ψ ) μω ( ) 1/ dxψ 1 ħ xˆψ = (e iωt + e iωt +0) μω ( 1/ 1 ħ ) = cos ωt. μω This reveals a phase ambiguity. The picture of Ψ (x, 0)Ψ(x, 0) in Part IV.A. suggests that x t starts negative and oscillates cosinusoidally. But the calculation using â, â shows that x t starts positive at t = 0. This means that the implicit phase convention for ψ v (x) is outermost lobe positive not innermost lobe positive as assumed in part A. E. Compute xˆ. The selection rule for xˆ is Δv = ±, 0. Since Ψ(x, t) contains only ψ 0 and ψ 1, there will be no Δv = ± integrals. We only need the Δv = 0 part of xˆ. ( ) ħ xˆ = (â + â ) μω ( ) ħ = (â + â +N +1) μω ħ So we want N +1. μω N +1 = H /ħω ( ) ħ ħ xˆ = () =. μω μω

11 5.61 Fall 013 First Hour Exam ANSWERS Page 11 F. Based on your answer to part E, evaluate V (x). 1 V = k xˆ 1 ħ 1 = k = ħω μω because ω =(k/μ) 1/. G. Based on your answer to parts C and F, evaluate T. From part C. H = ħω. From part F. V = ħω/. T = Ĥ V T = H ħω ħω V = ħω =. Why are T V and independent of t? Because Ψ(x, t) contains only ψ 0 and ψ 1 and the motion of xˆ or pˆ requires ψ v in Ψ(x, t) differing in v by ±.

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