Chemistry 460 Dr. Jen M. Stnr Homework Problem Set 1 Solutions 1. Determine the outcomes of operting the following opertors on the functions liste. In these functions, is constnt..) opertor: / ; function: x e x ( x ) e x = e x x e x b.) opertor: ; function: 4x + 9 The result is 4x + 9 -- no further simplifiction cn be one. c.) opertor: / ; function: x sin 3x ( x sin 3x) = ( sin 3x + 3x cos 3x) = 3cos 3x + 3cos 3x 9x sin 3x = 6 cos 3x 9x sin 3x.
. Consier two opertors, ˆ A = x n ˆ B = /. For the function f (x) = x e x, where is constnt, clculte the following quntities..) A ˆ B ˆ f ( x) A ˆ B ˆ f ( x) = x A ˆ B ˆ f x ( x e x ) ( ) = x e x x e x ( ) = x e x x e x. b.) B ˆ A ˆ f ( x) B ˆ A ˆ f ( x) = B ˆ A ˆ f x = x ( x ) e x ( x e x ) ( ) = x e x x e x. c.) Are the results from prts () n (b) the sme? If the results re not the sme, it is si tht the two opertors o not commute. No, in this cse, the results re not the sme. Tht is, A ˆ B ˆ f x opertors A ˆ n B ˆ o not commute. ( ) ˆ B A ˆ f x ( ). Therefore, the
3 3. The trnsltion opertor, ˆ T h, is efine by the rule: istnce of trnsltion..) Show whether or not ˆ T h is liner opertor. ˆ T h f (x) = f (x + h), where h represents the For linerity, we wnt to consier the ction of T ˆ h on the function given by f ( x) + b g( x) : ˆ T h Therefore the opertor ˆ T h is liner. [ f ( x) + b g( x) ] = f x + h ( ) + b g( x + h) = T ˆ h f ( x) + b T ˆ h g( x). b.) Evlute the expression T ˆ ( 1 3 T ˆ 1 + ) x, where h = 1, f (x) = x, n T ˆ h mens to pply the trnsltion opertor twice. ˆ T h f (x) = f (x + h), for f (x) = x n h = 1 we hve: T ˆ 1 f ( x) = f ( x +1), so T ˆ 1 x = ( x +1). Therefore, T ˆ ( 3 T ˆ 1 1 + ) x = T ˆ 1 x 3 ˆ T 1 x + x = T ˆ 1 ( x +1) 3 ( x +1) + x = ( x + ) 3 ( x +1) + x. The result cn be left in the form shown bove, or it cn be simplifie s shown below: T ˆ ( 3 T ˆ 1 1 + ) x = x + ( ) 3 ( x +1) + x ( ) + x = x + 4x + 4 3 x + x +1 = x +1.
4 4. Which of the following functions re eigenfunctions of /? e ikx, cos kx, kx, e x. In these functions, k n re constnts. For those functions tht re eigenfunctions, give the eigenvlues. () eikx = ike ikx The function e ikx is n eigenfunction of / with eigenvlue ik. (b) cos kx = k sin kx The function cos kx is not n eigenfunction of /. (c) kx = k The function kx is not n eigenfunction of /. () e x = x e x The function e x is not n eigenfunction of /.
5 5. Which of the following functions re eigenfunctions of eigenfunction.? Give the eigenvlues for ech.) e x ex = e x, the function e x is n eigenfunction of the opertor with eigenvlue 1. b.) x x =, the function x is NOT n eigenfunction of the opertor. c.) e x ( is constnt) The first erivtive is e x = x e x. The secon erivtive is e x = ( 4 x )e x. this result oes not equl constnt times the originl function, f (x) = e x is NOT n eigenfunction of the opertor /..) 3 cos x 3cos x ( ) = 3cos x = 1 3cos x ( ), the function 3 cos x is n eigenfunction of the opertor with eigenvlue 1. e.) sin x + cos x opertor ( sin x + cos x) = ( sin x + cos x), the function sin x + cos x is n eigenfunction of the with eigenvlue 1.
6 6. For prticle moving in three imensions, the clssicl expression for the kinetic energy T is given by: T = 1 m p x + p ( y + p z ). Give the corresponing expression for the quntum mechnicl kinetic energy opertor, ˆ T, in three imensions. To convert clssicl component of momentum, p α, where α = x, y,or z, into quntum mechnicl opertor ˆ p α, we use the reltion: p ˆ α = i!. α For the momentum squre (s ppers in the kinetic energy expression), we therefore hve: p ˆ α =! α. Substituting these momentum opertors, the quntum mechnicl kinetic energy opertor ˆ T becomes: ˆ T = 1 # m!!! & % $ y z (, ' n fctoring out the! terms gives: T ˆ =! m # % $ + y + & z (. ' ˆ T involves three vribles rther thn just one, it is more pproprite to use prtil erivtives: T ˆ =! m $ x + y + ' & % z ). ( 7. Inicte which functions correspon to cceptble quntum mechnicl wvefunctions on the intervl, [ ]..) x - not cceptble (goes to s x ) b.) e x - not cceptble (goes to s x ) c.) ex - not cceptble (goes to s x ).) e x ( is constnt) - cceptble
7 $ 8. Consier the wvefunction ψ(x) = sin nπ x ' & ), where n is positive integer n is constnt. % (.) Normlize the function ψ x ( ) on the intervl 0 to. [Note: you probbly will hve to look up the integrl in tble.] For normlize wvefunction Ψ norm, the integrl over ll spce must be equl to 1. Let Ψ norm = NΨ, where N is the normliztion constnt. For this problem, the rnge of the coorinte x is 0 to. Therefore, for normliztion we require: (x) = 1. Ψ norm 0 Substituting Ψ norm = NΨ les to N Ψ (x) = 1. 0 $ Substituting Ψ(x) = sin nπ x ' & ), we hve N sin # nπx & % ( = 1. To evlute the integrl, we cn look % ( 0 $ ' it up in tble. From the hnout of useful integrls, sin bx = x sinbx 4b. By mking the substitution b = nπ, we obtin: sin $ nπx ' & ) = 0 % ( + x $ nπx '. - sin& ) 0, 4nπ % (/ 0. Evluting t the limits yiels: sin $ nπx ' & ) = 0 % ( $ ' $ & sin( nπ )) & 0 % 4nπ ( % 4nπ sin ( 0 ) ' ). ( Note tht since sinnπ = 0 n sin0 = 0, the first n lst terms rop out, n we re left with sin $ nπx ' & ) = 0 % (. Finlly, substituting this result into the normliztion conition, N sin # nπx & % ( = 1, we hve 0 $ ' N " % $ ' = 1, or N = # &. Therefore, the normlize wvefunction Ψ norm is given by: Ψ norm = $ nπx ' sin& ). % (
8 8.) continue b.) Is ψ ( x) n eigenfunction of the momentum opertor, p ˆ x? p ˆ x = i!, the eigenvlue eqution woul be: ˆp x Ψ(x) = i! Ψ(x) = i! nπ x sin ˆp x Ψ(x) = i! nπ cos nπ x. the right sie is not equl to constnt times the originl function, Ψ(x) is NOT n eigenfunction of the opertor p ˆ x.