Physics 215 Quantum Mechanics 1 Assignment 2

Transcription

1 Physics 15 Quntum Mechnics 1 Assignment Logn A. Morrison Jnury, 16 Problem 1 Clculte p nd p on the Gussin wve pcket α whose wve function is x α = 1 ikx x 1/4 d 1 Solution Recll tht where ψx = x ψ. Additionlly, x ˆp ψ = i ψx x Using these, let s not clculte p : x ˆp ψ = ψx x α ˆp α = dx α x x ˆp α 4 = 1 dx ikx x ik x ikx x 5 i d d d = 1 dx ik x x 6 i d d

2 Logn A. Morrison Assignment Problem 1 Pge / 9 The integrl of x x /d is zero since it is the integrl of n odd function over symmetric intervl. Thus, α ˆp α = 1 dx ik x x 7 i d d = k dx x 8 d = k π 9 1/d Now, let s clculte ˆp α ˆp α = = = = = = k 1 dx α x x ˆp α 11 dx ikx x ikx d x x 1 d dx ikx x ik x ikx x 1 d x d d [ dx ik x 1 ] x 14 d d d [ dx k ikx + x d d 1 ] x 15 4 d d Agin, the term tht goes like x x /d integrtes to zero. The other two terms yield [ α ˆp α = dx k ikx + x d d 1 ] x 4 d d [ = dx k 1 ] d + x x d 4 d = k + 1 π d 1/d 1 dxx x d 4 d = k dx x d π d 5 1/d d = k d π d 5 1/d π 1/d 1 = k + 1 d + d 5 1/1/d / = k + 1 d 1 d = k + 4 d Combining these results, we find tht p σ p = p = 5 d

3 Logn A. Morrison Assignment Problem Pge / 9 Looking t the spce wve function, we cn see tht σ x = d/. Thus, σ x σ p = 6 Hence, we hve minimum uncertinty wve function. Problem Evlute the x p uncertinty product x p for one-dimensionl prticle confined between two rigid wlls: < x < Vx = 7 otherwise Do this for both the ground stte nd excited sttes. Solution Due to the fct tht the potentil is infinite outside the intervl < x <, the spce wve function of the prticle must be zero there since it would require nd infinite mount of energy for the prticle to exist in tht region. Thus, the spcil wve function will only be non-zero in the intervl < x <. In this region, the prticle obeys the sttionry schrodinger eqution: Ĥ ψ = E ψ 8 with Ĥ = ˆp. In the position representtion, this eqution tkes the following form: m Thus, we find tht the energy eigenfunctions in the position representtion re: ψx = Eψx 9 m x ψx = A sinkx + B coskx where k = me/. Since the wve function must be zero t the endpoints, we find tht B = nd k = nπ/. Thus, the energy eigenfunctions re: ψ n x x ψ = nπx sin = 1 e inπx/ e inπx/ 1 i

4 Logn A. Morrison Assignment Problem Pge 4 / 9 The ws inserted to normlized the wvefunctions to 1. Now, let s clculte ˆp : ψ n ˆp ψ n = dx ψ n x x ˆp ψ n = 1 dx e inπx/ e inπx/ nπx sin e inπx/ e inπx/ 4 i x = inπ dx e inπx/ e inπx/ e inπx/ + e inπx/ 4 i = nπ dx e inπx/ e inπx/ 5 = nπ e inπx 1 + e inπ 1 6 inπ = 7 And now ˆp : ψ n ˆp ψ n = dx ψ n x x ˆp ψ n 8 These two results give us tht Now, let s clculte ˆx : = 1 4 n π = n π = = n π dx e inπx/ e inπx/ nπx sin e inπx/ e inπx/ 9 x dx e inπx/ e inπx/ e inπx/ e inπx/ 4 dx e inπx/ e inπx/ 41 e inπx + 1 e inπ inπ = n π 4 p = ˆp ˆp = n π To perform the integrl of x cos, we use integrtion by prts: udv = uv 4 44 ψ n ˆx ψ n = dx ψ n x x ˆx ψ n 45 = nπx dxx sin 46 = 1 nπx dx x x cos 47 = 1 nπx dxx cos 48 vdu 49

5 Logn A. Morrison Assignment Problem Pge 5 / 9 with the following substitutions: u = x; du = dx; nπx dv = cos dx; v = nπx nπ sin 5 Mking these substitutions, we find ψ n ˆx ψ n = 1 nπx dxx cos 51 = 1 x nπx nπ sin nπx sin 5 nπ = 1 + nπx 4n π cos 5 = n π = This is exctly wht we would ect since the mod squred of ech mode is symmetric bout the point x = /. Now, let s clculte ˆx 55 ψ n ˆx ψ n = dx ψ n x x ˆx ψ n 56 = nπx dxx sin 57 = 1 nπx dx x x cos 58 = 1 nπx dxx cos 59 6 Now, we use integrtion by prts gin with the following: nπx u = x ; du = xdx; dv = cos dx; v = nπx nπ sin Upon mking these substitutions, we obtin ψ n ˆx ψ n = 1 nπx dxx cos = 1 x nπx nπ sin nπ = + 1 nπx dxx sin nπ nπx dxx sin Using integrtion by prts once more, with the following nπx u = x; du = dx; dv = sin dx; v = nπx nπ cos 66

6 Logn A. Morrison Assignment Problem Pge 6 / 9 we obtin ψ n ˆx ψ n = + 1 nπ Combining these results, we obtin = + 1 nπ = + 1 nπ nπx dxx sin x nπx nπ cos + nπ nπ + nπx 4n π sin nπx cos = n π 7 x = ˆx ˆx 71 = n π 4 7 = 1 n π 7 We cn now evlute the x p uncertinty product: x p = 1 n π 74 n π = n π n π Let s check tht this obeys the uncertinty principle: x p = 1 n π 76 n π And hence, the uncertinty principle is stisfied! = n π 1 π = 4 = 4 = 1 [ ] 4 ˆx, ˆp

7 Logn A. Morrison Assignment Problem Pge 7 / 9 Problem Clculte the uncertinty p in the momentum, nd the uncertinty product p x for the following wve function ψx 1 for x < ψx = 8 otherwise Solution Note tht this wve function cn be written s the sum of two thet functions: ψx = 1 θx + θx 84 where Before we strt, let s show tht 1 x > θx = x 85 Note tht two delt functions re equl when δ 1 x f xdx = d θx = δx 86 dx δ x f xdx 87 for every function f x. Now, let f x be n rbitrry function. Since the dervivtive of the thet function is zero everywhere where x, d ɛ d dx θx f x = lim ɛ ɛ dx θx f x Using integrtion by prts, we find ɛ lim ɛ ɛ d dx θx f x = lim θx f x ɛ ɛ ɛ ɛ ɛ d θx dx f x 9 = lim f ɛ ɛ 91 = f 9 = δx f x 9

8 Logn A. Morrison Assignment Problem Pge 8 / 9 where in going from the first equlity to the second we used the fct tht the integrl of function over n intervl of length is. Now we hve shown θ x = δx. Now we re redy to determine ˆp ψ ˆp ψ = ψ x x ˆp ψ 94 = 1 i = 1 i ψ x θx + θx 95 x ψ x δx + δx 96 = 1 i ψ ψ 97 = 98 Where we used the fct tht ψ = ψ = = ψ = ψ. Now, let s clculte ˆp ψ ˆp ψ = ψ x x ˆp ψ 99 = = Now, using integrtion by prts with nd θx + θx θx + θx 1 x θx + θx δx + δx 11 x 1 u = θx + θx ; du = δx + δx dx 1 dv = δx + δx dx; v = δx + δx 14 x we find ψ ˆp ψ = θx + θx δx + δx + 15 δx + δx δx + δx dx 16 = δ + δ + δ + δ 17 = δ 18 = δ Which is infinite!

9 Logn A. Morrison Assignment Problem Pge 9 / 9 Now, let s clculte ˆx And now ˆx And hence ψ ˆx ψ = ψ ˆx ψ = = 1 = 1 = 1 ψ x x ˆx ψ dx 111 x θx + θx θx + θx 11 xdx = 115 ψ x x ˆx ψ dx 116 = 1 x θx + θx θx + θx 117 = 1 x dx 118 = = 1 Now, we hve tht x = 11 x p = δ 1 Which is formlly infinite. This infinity rises from the discontinuities of the sptil wve function. Such discontinuity results in n infinite vlue of momentum.