852quiz2_Solution - V 1 = I 1 V I 1 = V ( | 00 aA 1 , | + |...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
NAME: QUIZ 2 PHYS852 Quantum Mechanics II, Spring 2009 A two-spin quantum system has four states. Given in terms of total-spin quantum number s and its z- component m s , these states are the singlet state | 0 , 0 a , and the three triplet states | 1 , 1 a , | 1 , 0 a , and | 1 , 1 a . Let us assume that the bare Hamiltonian is H 0 = E 0 p 2 S 2 z , where S z is the z-component of the total spin. 1. Give the energy levels of the system and their degeneracies S 2 z | s,m s a = p 2 m 2 s H 0 | 0 , 0 a = 0 H 0 | 1 , 0 a = 0 H 0 | 1 , 1 a = E 0 H 0 | 1 , 1 a = E 0 E (0) 1 = 0 , d 1 = 2 E (0) 2 = E 0 , d 2 = 2 2. Write the projector I n onto each degenerate subspace I 1 = | 0 , 0 aA 0 , 0 | + | 1 , 0 aA 1 , 0 | I 2 = | 1 , 1 aA 1 , 1 | + | 1 , 1 aA 1 , 1 | 3. Assume the interaction takes the form V = V 0 ( | 1 , 0 aA 0 , 0 | + | 0 , 0 aA 1 , 0 | + | 1 , 1 aA 1 , 1 | − | 1 , 1 aA 1 , 1 | ) . Identify the ’good’ basis for perturbation theory.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: V 1 = I 1 V I 1 = V ( | 00 aA 1 , | + | 1 , aA , | ) = V p 0 1 1 0 P V 2 = I 2 V I 2 = V ( | 1 , 1 aA 1 , 1 | + | 1 , 1 aA 1 , 1 | ) = V p 1 0 0 1 P Since V 1 and V s are of familiar forms, we know that the eigenstates of V 1 are: | 1 , 1 (0) a = 1 2 ( | , a | 1 , a ) | 1 , 2 (0) a = 1 2 ( | , a + | 1 , a ) | 2 , 1 (0) a = | 1 , 1 a | 2 , 2 (0) a = | 1 , 1 a These four states form a good basis for perturbation theory....
View Full Document

This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.

Ask a homework question - tutors are online