Problem Set 2 Solutions

Transcription

1 Chemistry 362 Dr. Jen M. Stnr Problem Set 2 Solutions 1. Determine the outcomes of operting the following opertors on the functions liste. In these functions, is constnt.).) opertor: /x ; function: x e x x x ) e x = e x x e x b.) opertor: ; function: 3x The result is 3x No further simplifiction cn be one. c.) opertor: 2 /x 2 ; function: x sin 3x 2 x 2 x sin 3x) = sin 3x + 3x cos 3x) x = 3cos 3x + 3cos 3x 9x sin 3x = 6 cos 3x 9x sin 3x. 2. Consier two opertors, ˆ A = x n ˆ B = /x. 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 x A ˆ B ˆ f x x e x ) ) = x e x x e x ) = x e x x 2 e x. b.) B ˆ A ˆ f x) B ˆ A ˆ f x) = B ˆ A ˆ f x = x x x x ) e x x 2 e x ) ) = 2x e x x 2 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

2 2 3. Which of the following functions re eigenfunctions of /x? e ikx, cos kx, kx, e x 2. In these functions, k n re constnts. For those functions tht re eigenfunctions, give the eigenvlues..) x eikx = ike ikx The function e ikx is n eigenfunction of /x with eigenvlue ik. b.) cos kx = k sin kx x The function cos kx is not n eigenfunction of /x. c.) x kx = k The function kx is not n eigenfunction of /x..) 2 x e x = 2x e x 2 The function e x 2 is not n eigenfunction of /x. 4. Demonstrte whether or not the function f x) = e x 2 is n eigenfunction of the opertor 2 /x 2. If the function is n eigenfunction, give the eigenvlue. The first erivtive is 2 x e x = 2x e x 2. The secon erivtive is 2 2 x 2 e x = 4 2 x 2 2)e x 2. Since this result oes not equl constnt times the originl function, f x) = e x 2 the opertor 2 /x 2. is NOT n eigenfunction of

3 3 5. For prticle moving in three imensions, the clssicl expression for the kinetic energy T is given by: T = 1 2m p x 2 + p 2 2 y + p z ). Give the corresponing expression for the quntum mechnicl kinetic energy opertor in three imensions, T ˆ. 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 2 p ˆ α =! 2 2 α 2. Substituting these momentum opertors, the quntum mechnicl kinetic energy opertor ˆ T becomes ˆ T = 1 # 2 2m!2! 2 2! 2 2 & x 2 y 2 z 2, n fctoring out the! 2 terms gives T ˆ =!2 2m # 2 x y & z 2. Since ˆ T involves three vribles rther thn just one, it is more pproprite to use prtil erivtives, T ˆ =!2 2m 2 x y & z 2 ). 6. Inicte which functions re cceptble wvefunctions: x 2 - not cceptble goes to s x ) e x - not cceptble goes to s x ) e x - not cceptble goes to s x ) e x2 - cceptble

4 4 7. Normlize the function Ψx) = sin nπ x & ), where n is positive integer n is constnt. The rnge of x is to. You 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 to. Therefore, for normliztion we require Substituting Ψ norm = NΨ les to Substituting Ψx) = sin nπ x & ), we hve x) x = 1. Ψ 2 norm N 2 Ψ 2 x) x = 1. N 2 sin 2 # nπx & x = 1. To evlute the integrl, we cn look it up in tble. From the CRC, sin 2 bx x = x 2 sin2bx 4b. By mking the substitution b = nπ, we obtin sin 2 nπx & ) x = + x 2 2nπx. - sin& ), 4nπ /. Evluting the limits yiels sin 2 nπx & ) x = 2 & sin 2nπ )) & 4nπ 4nπ sin ) ). Note tht since sin2nπ = n sin =, the first n lst terms rop out, n we re left with sin 2 nπx & ) x = 2. Finlly, substituting this result into the normliztion conition, N 2 sin 2 # nπx & x = 1, we hve N 2 " = 1, or N = 2 # 2 &. Therefore, the normlize wvefunction Ψ norm is given by Ψ norm = 2 nπx sin& ).

5 5 8. Is the function in problem 7 n eigenfunction of the liner momentum opertor, ˆ p x? Since p ˆ x = i!, the eigenvlue eqution woul be x ˆ p x Ψx) = i! x Ψx) = i! nπx sin * x & ) p ˆ Ψx) = i! nπ x & * cos nπx * ) & ) Since the right sie is not equl to constnt times the originl function, the function Ψx) from problem 7 is NOT n eigenfunction of the opertor p ˆ x.