Final Exam with Solutions | Quantum Mechanics I | PHY 4604, Exams of Quantum Mechanics

Download Final Exam with Solutions | Quantum Mechanics I | PHY 4604 and more Exams Quantum Mechanics in PDF only on Docsity! PHY-4604 Final Exam: Fall 2006 (Solve at least 4 from these 5 problems) Name_______________________ Panther ID ______________________ 1. The operators associated with the radial component of the momentum rp and the radial coordinator r is denoted by ˆ rp and r̂ , respectively. Their action on a radial wave function ( )R r are given by ˆ ( ) (1/ )( / )[ ( )]rp R r i r r rR r= − ∂ ∂ and ˆ ( ) ( )rR r rR r= . a) Find the commutator [ ˆ rp , r̂ ] and rp r∆ ∆ where 22ˆ ˆ ˆr r rp p p∆ ≡ − and 22ˆ ˆr r r∆ ≡ − . b) Show 2 2 2 2 ˆ ( / )( )rp r rr ∂ = − ∂ Solution: (a) Since ˆ ( ) ( )rR r rR r= and ˆ ( ) (1/ )( / )[ ( )] 1 ( )( ) rp R r i r r rR r dR ri R r i r dr = − ∂ ∂ = − − and since 2ˆ ˆ[ ( )] (1/ )( / )[ ( )] ( )2 ( ) rp rR r i r r r R r dR ri R r i r dr = − ∂ ∂ = − − the action of the commutator [ ˆ rp , r̂ ] on a function ( )R r is given by ˆ ˆ ˆ ˆ ˆˆ[ , ] ( ) ( ) ( ) ( ) ( )2 ( ) ( ) ( ) r r rp r R r p r rp R r dR r dR ri R r i r ihR r i r dr dr i R r = − = − − + + = − so that we have ˆ ˆ[ , ]rp r i= − Using the uncertainty relation for a pair of operators  and B̂ (Eq (3.62) in the text book) or 1 ˆ ˆ[ , ] 2 A B A B∆ ∆ ≥ , we can write 1 2r p r∆ ∆ ≥ (b) The action of 2ˆ rp on ( )R r gives 2 2 2 2 1 1 1ˆ ( ) { [ ( ( ))]} [ ( )]rp R r i r i rR r rR rr r r r r r ∂ ∂ ∂ = − − = − ∂ ∂ ∂ hence 2 2 2 2 ˆ ( / )( )( )rp r rr ∂ = − ∂ 2. A particle of mass m, which moves freely inside an infinite potential well of length a, has the initial wave function at t = 0 as 3 3 1 5( ,0) sin( ) sin( ) sin( ) 5 5 A x x xx a a a aa a π π πψ = + + , where A is a real constant a) Find A so that ( ,0)xψ is normalized. b) If measurements of the energy are carried out, what are the values that will be found and what are the corresponding probabilities? Calculate the average energy (the expectation value) c) Find the wave function ( , )x tψ at any later time t. d) Determine the probability of finding the system at a time t in the state 5 2 5( , ) sin( )exp( )iE txx t a a πϕ −= ; now determine the probability of finding it in the state 2 2 2( , ) sin( ) exp( )iE txx t a a πχ −= , where both E5 and E2 is the energy eigenvalue, respectively. Solution: As the functions 2( ) sin( )n n xx a a πϕ = are orthonormal, so 0 2 sin( )sin( ) a n m nm n x m x dx a a a π πϕ ϕ δ| = =∫ it is more convenient to write ( ,0)xψ in terms of ( )n xϕ as 1 3 5 3 3 1 5( ,0) sin( ) sin( ) sin( ) 5 5 3 1( ) ( ) ( ) 102 10 A x x xx a a a aa a A x x x π π πψ ϕ ϕ ϕ = + + = + + (a) The normalization of ( ,0)xψ yields 2 3 11 2 10 10 Aψ ψ= | = + + or 6 5 A = hence 1 3 5 3 3 1( ,0) ( ) ( ) ( ) 5 10 10 x x x xψ ϕ ϕ ϕ= + + ------------ (1) (b) Since 2 2 2 2 ˆ 2n n n nE ma πϕ ϕ= | Η | = , the values of energy measurements along with the corresponding probabilities are: 2 2 1 1 1 2 ˆ 2 E ma πϕ ϕ= | Η | = 2 1 1 1 3( ) ( ,0) 5 P E xϕ ψ= | = 2 2 3 3 3 2 9ˆ 2 E ma πϕ ϕ= | Η | = 2 3 3 3 3( ) ( ,0) 10 P E xϕ ψ= | = 2 2 5 5 5 2 25ˆ 2 E ma πϕ ϕ= | Η | = 2 5 5 5 1( ) ( ,0) 10 P E xϕ ψ= | = And the average energy is 2 2 1 3 5 2 3 3 1 29 5 10 10 10n nn E P E E E E ma π = = + + =∑ (c) As the initial state ( ,0)xψ is given by Eq (1), the wave function ( , )x tψ at any later time t is