This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chemistry 120A, Spring 2009 Problem Set 2 Solutions Due February 9, 2009 Problems 1. Another particle in a box problem. Suppose that a particle in a box (of length L ) is represented at an instant in time by a state  ψ i with wavefunction: ψ ( x ) = Nx 2 L 2 x 2 (a) Evaluate the normalization constant, N , such that ψ ( x ) is normalized (( i.e. ) h ψ  ψ i = 1) h ψ  ψ i = 1 = Z L ψ * ( x ) ψ ( x ) dx 1 = Z L Nx 2 L 2 x 2 · Nx 2 L 2 x 2 dx 1 = N 2 Z L x 4 L 2 x 2 2 dx 1 = N 2 Z L x 4 x 4 2 L 2 x 2 + L 4 dx 1 = N 2 Z L x 8 2 L 2 x 6 + L 4 x 4 dx 1 = N 2 1 9 x 9 2 7 L 2 x 7 + 1 5 x 5 L 4 L 1 = N 2 1 9 L 9 2 7 L 9 + 1 5 L 9 1 = N 2 8 315 L 9 N = r 315 8 L 9 (b) Evaluate the average energy which would be obtained by making measurements on a very large number of iden tically prepared systems using the expectation value expression, h E i = h ψ  ˆ H  ψ i , for normalized wavefunctions. h E i = h ψ  ˆ H  ψ i = Z L r 315 8 L 9 x 2 L 2 x 2 ¯ h 2 2 m d 2 dx 2 ! r 315 8 L 9 x 2 L 2 x 2 dx = ¯ h 2 2 m 315 8 L 9 Z L x 2 L 2 x 4 2 L 2 12 x 2 dx 1 = ¯ h 2 2 m 315 8 L 9 Z L 2 x 2 L 4 14 x 4 L 2 + 12 x 6 dx = ¯ h 2 2 m 315 8 L 9 2 3 L 7 14 5 L 7 + 12 7 L 7 = ¯ h 2 2 m 315 8 L 9 44 105 L 7 = 33 4 ¯ h 2 mL 2 2. A followon to ’Another particle in a box problem.’ Same particle in the same box, in the same state,  ψ i (a) A very natural basis in which to represent  ψ i = X j  φ j i c j is the basis of energy eigenfunctions,  φ n i , where n is the quantum number. Evaluate the first few expansion coefficients ( c n =1 ,c n =2 ,c n =3 ,c n =4 ). For a particle in a box, the energy eigenfunctions are  φ n i = r 2 L sin nπx L c n = h φ n  ψ i = Z L r 2 L sin nπx L r 315 8 L 9 x 2 L 2 x 2 dx = r 2 L r 315 8 L 9 Z L x 2 L 2 x 2 sin nπx L dx = √ 315 2 L 5 L n 5 π 5 n 2 π 2 n 2 + 12 L 4 n 2 π 2 π 2 n 2 + 12 x 2 L 2 + n 4 π 4 x 4 o cos nπx L L + √ 315 2 L 5 L n 5 π 5 2 Lnπx n L 2 π 2 n 2 + 12 2 n 2 π 2 x 2 o sin nπx L L ⇒ since both sin( nπ ) = 0 and sin(0) = 0 , the sine term vanishes when limits are applied = √ 315 2 L 5 L n 5 π 5 h 2 π 2 n 2 + 12 L 4 n 2 π 2 L 4 π 2 n 2 + 12 + n 4 π 4 L 4 cos( nπ ) 2 π 2 n 2 + 12 L 4 cos(0) i = √ 315 2 L 5 L 5 n 5 π 5 π 2 n 2 + 12 " 2 n 2 π 2 + n 4 π 4 π 2 n 2 + 12 ! cos( nπ ) 2 # c n = √ 315 2 n 5 π 5 π 2 n 2 + 12 " 2 n 2 π 2 + n 4 π 4 π 2 n 2 + 12 ! cos( nπ ) 2 # c n =1 = 0 . 898 c n =2 = . 429 c n =3 = 0 . 0791 c n =4 = . 0537 2 (b) Show that the average energy measured for the system  ψ i = X j  φ j i c j described in the basis of energy eigenfunctions (  φ n i corresponding to energy E n ) h E i = X n  c n  2 E n h E i = D ψ ˆ H ψ E = X n D ψ ˆ H φ n E h φ n  ψ i = X n h ψ  E n  φ n ih φ n  ψ i = X n E n h ψ  φ n ih φ n  ψ i = X n E n c * n c n = X n  c n  2 E n (c) Compare your average energy from 1(b) with the average energy you would estimate using your first 4 terms of the basis representation in 2(a), via the alternative expression derived in 2(b). Explain why this new resultof the basis representation in 2(a), via the alternative expression derived in 2(b)....
View
Full
Document
 Spring '09
 HEADGORDON
 dx, Hermitian

Click to edit the document details