#12.3.7 - Quantum Mechanics - 1: HW 6 Solutions Leo...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Quantum Mechanics - 1: HW 6 Solutions Leo Radzihovsky Paul Martens November 12, 2007 1 Problem 1 Show that H ~ E 1 ~ E 2 commutes with ~ L , where ~ E 1 , 2 are vector operators. The definition of a vector operator. h E (1 or 2) i ,L j i = ~ ijk E k (1) (transforms like a vector under rotations) h E (1) i E (2) i ,L j i = E (1) i h E (2) i ,L j i + h E (1) i ,L j i E (2) i (2) = ~ ijk E (1) i E (2) k + E (1) k E (2) i (3) = 0 (4) Because ~ E 1 ~ E 2 commute with ~ L all functions of ~ E 1 ~ E 2 commute with ~ L h H ~ E 1 ~ E 2 , ~ L i = 0 (5) 2 Problem 2 Show that L z =- ~ ( x y- y x ) =- ~ relating coordinates x = r sin cos (6) y = r sin sin (7) z = r cos (8) and derivatives y = r y r + y + y (9) x = r x r + x + x (10) 1 r = p x 2 + y 2 + z 2 (11) z r = cos (12) calculating derivatives r x = x r = sin cos (13) r y = y r = sin sin (14) derivatives can be taken easily by making use of the chain rule and equation (12) x z r =- z r 2 r x =- sin x =- sin x (15) solving for x . x = 1 r cos cos (16) the calculation for y is almost identical y = 1 r cos sin (17) finding derivatives y x = tan (18) taking derivatives of both sides- tan x = sec 2 x (19) solving for x . x =- sin r sin (20) a similar argument can be used for y y = cos r sin (21) Putting all of it together L z- ~ = x y- y x = r sin cos ( y r r + y + y )- r sin sin ( x r r + x + x ) (22) L z =- ~ (23) 2 3 Problem 3 Suppose we have three identical bosonic particles constrained to move on a circle in an equilateral configuration and want to find the wavefunction and the energy spectrum. The Hamiltonian of this system is H = L 2 z 2 I (24) where I = 3 mr 2 (25) The eigenstates of this Hamiltonian are the eigenstates of the L z operator. L z | ` z i = ~ ` z | ` z i (26) ` z ( ) = e ` z (27) The position of each of the particles are 1 = (28) 2 = + 2 3 (29) 3 = + 4 3 (30) Because these are bosons, we must require that the full wavefunction be symmetric under permutations. For the rigid structure above this amounts to m ( + 2 3 ) = m ( ). Therefore, ` z = 3 n , where n Z . n ( ) = e 3 n (31) E n = 9 ~ 2 n 2 2 I (32) 4 Problem 4 Express Y m ` s in Cartesian coords 4.1 Y Y = 1 4 (33) 4.2 Y 1 1 Y 1 1 = r 3 8 sin e | {z } sin cos sin sin (34) = r 3 8 x y p x 2 + y 2 + z 2 !...
View Full Document

This note was uploaded on 04/09/2008 for the course PHYS 502 taught by Professor Hummin during the Fall '08 term at University of Cincinnati.

Page1 / 13

#12.3.7 - Quantum Mechanics - 1: HW 6 Solutions Leo...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online