Solutions to Final Exam - Introduction to Quantum Mechanics | Phys 451, Exams of Quantum Mechanics

Download Solutions to Final Exam - Introduction to Quantum Mechanics | Phys 451 and more Exams Quantum Mechanics in PDF only on Docsity! PHYS 451 Final Exam Dec. 9. 2008 Write your solutions clearly and place your final answers in a box. Start each problem on a new page and number. Show your work for full credit. 100 pts. #1- The harmonic oscillator is described by a hamiltionian H = p 2 2m + 1 2 m! 2 x 2 . Verify the eigenvalue equation H! 0 = E 0 ! 0 and determine E 0 . ! 0 =! m" #! $ %& ' () 1/4 e * m" ! x 2 2 !!2 2m d 2 dx 2 ! m" #! $ %& ' () 1/ 4 e ! m" ! x 2 2 + 1 2 m" 2x2 m" #! $ %& ' () 1/ 4 e ! m" ! x 2 2 = E m" #! $ %& ' () 1/ 4 e ! m" ! x 2 2 !!2 2m d 2 dx 2 !e ! m" ! x 2 2 + 1 2 m" 2x2e ! m" ! x 2 2 = Ee ! m" ! x 2 2 !!2 2m ! m" ! $ %& ' () d dx xe ! m" ! x 2 2 $ %& ' () + 1 2 m" 2x2e ! m" ! x 2 2 = Ee ! m" ! x 2 2 !!2 2m ! m" ! $ %& ' () e ! m" ! x 2 2 ! x2 m" ! $ %& ' () e ! m" ! x 2 2 $ %& ' () + 1 2 m" 2x2e ! m" ! x 2 2 = Ee ! m" ! x 2 2 !!2 2m ! m" ! $ %& ' () 1! x2 m" ! $ %& ' () $ %& ' () + 1 2 m" 2x2 = E !!!!!!*!!!! !" 2 + x 2 !" 2 m" ! $ %& ' () $ %& ' () + 1 2 m" 2x2 = E = !" 2 #2- Consider a particle in an infinite square well potential V=0 (0<x<a). A linear perturbation is applied of the form H '(x) = !x from x=0 to x=a/2. Find the first order correction to the ground state energy. sin 2! (qx) dx = x 2 " sin(2qx) 4q x sin 2! (qx) dx = x 2 4 " x sin(2qx) 4q " cos(2qx) 8q 2 #E (1) = 2 a $x sin2( % a x) dx 0 a/2 ! = 2$ a x 2 4 " xa sin(2 % a x) 4% " a2 cos(2 % a x) 8% 2 0 a/2& ' ( ( ( ( ) * + + + + = 2$ a a2 16 + a 2 8% 2 + a 2 8% 2 & '( ) *+ #E (1) = $a 1 8 + 1 2% 2 & '( ) *+ #3-A spin ½ particle is in a state ! = A 2 1" i # $% & '( . Determine the probability that the spin SY is measured to be !! / 2 . ! X + = 1 2 1 1 " #$ % &' ! X ( = 1 2 1 (1 " #$ % &' ! Y + = 1 2 1 i " #$ % &' ! Y ( = 1 2 1 ( i " #$ % &' ! Z + = 1 0 " #$ % &' ! Z ( = 0 1 " #$ % &' P = 1 2 1 i( )i 1 6 2 1! i " #$ % &' 2 = 1 12 3 + i 2 = 5 6 x=0 a/2 a ε V=0 V= ! #4- In the quantum treatment the radiation of charged particles ! p! ! ! p + q ! A , where ! A is the electromagnetic vector potential. Show that the Schrodinger equation for a free electron H! = p 2 2m e ! leads to a Hamiltonian of the form H rad = p 2 2m e ! e me ! Ai ! p . H rad ! = ! p " e ! A( ) 2 2m ! = ! p 2 2m " e ! Ai ! p 2m !" e ! pi ! A 2m + ! A 2 2m e # $% & '( ! = 1 2m ! p 2! " e 2m ! Ai ! p! !" e 2m ! pi( ! A! ) + e 2 ! A 2 2m ! H rad ! = 1 2m ! p 2! " e 2m ! Ai ! p! !" e 2m ! Ai ! p! " e 2m ! pi ! A( ) ) i ! A=0 "#$ %$ ! + e 2 ! A 2 2m ! ~0 "#% H rad ! = ! 1 2m ! p 2! " e m ! Ai ! p! #5- A hydrogen atom is in the state ! = A ! 100 + 5i! 211 + 3! 210 " 5i! 21"1[ ] . (a) What is the expectation value of the energy <E>? (b) What is the expectation value of <L2 >? (c) What is the expectation value of < LZ>? ! = 1 60 ! 100 + 5i! 211 + 3! 210 " 5i! 21"1[ ] (a)!< E >!= 1 60 E 0 + 25 60 + 9 60 + 25 60 # $% & '( E0 4 = 63 60 E0 4 = "3.57!eV (b)!< L 2 >= 1 60 0! 2 + 25 60 + 9 60 + 25 60 # $% & '( 2! 2 = 59 60 !! 2 (b)!< L Z >= 1 35 0! + 100 35 ! + 36 35 0! " 100 35 ! = !0! #6- A hydrogen atom in the stationary n=2 state with spin-orbit energy !E = ! ! 2 " L " " S . Determine the values that !E can take and label the energy states with spectroscopic notation 2S+1 L J . !E = ! ! 2 " L " " S = ! 2! 2 J 2 # L2 #S2( ) with L = 0,1 and S = 1/ 2 Coupling 1) J = " 0 + " 1 / 2 = " 1 / 2 2 S 1/2 !E 1/2 = ! 2! 2 3 / 4 # 02 # 3 / 4( )!2 = 0 2) J = " 1+ " 1 / 2 = " 3 / 2, " 1 / 2 2 P 3/2,1/2 !E 3/2 = 3 ! ! 2 (3 / 2)(5 / 2) # 2 # 3 / 4( )!2 = 15 4 # 8 4 # 3 4 $ %& ' () ! 2 = ! 2 !E 1/2 = ! 2! 2 (1/ 2)(3 / 2) # 2 # 3 / 4( )!2 = 3 4 # 8 4 # 3 4 $ %& ' () ! 2 = #!