Exam 1 Solutions for Introduction to Quantum Mechanics | PHY 4604, Exams of Physics

Download Exam 1 Solutions for Introduction to Quantum Mechanics | PHY 4604 and more Exams Physics in PDF only on Docsity! PHY4604 Fall 2007 R. D. Field Department of Physics Page 1 of 7 Exam 1 Solutions PHY 4604 Exam 1 Solutions (Total Points = 100) Problem 1 (20 points): Circle true or false for following (2 point each). (a) (True or False) One of the “breakthroughs” that lead to quantum mechanics was the idea of associating differential operators with the dynamical variables. (b) (True or False) Solutions of Schrödinger’s equation of the form )()(),( txtx φψ=Ψ correspond to states with definite energy E. (c) (True or False) Solutions of Schrödinger’s equation of the form )()(),( txtx φψ=Ψ correspond to states in which the probability density 2|),(|),( txtx Ψ=ρ is independent of time. (d) (True or False) The wave function Ψ(x,t) must vanish in a region of infinite potential. (e) (True or False) If Pop is the parity operator, Popψ(x) = ψ(-x), then Pop2 = Pop. (f) (True or False) In quantum mechanics particles can enter the “classically forbidden” region where V0 > E (i.e. KE < 0). (g) (True or False) The operator AopA↑op is hermitian. (h) (True or False) If Aop and Bop are hermitian then AopBop is also hermitian. (i) (True or False) The commutator operator [(px)op,(x2)op] is equal to xih2− . (j) (True or False) In position-space the commutator operator [(px)op,sin(kx)] is equal to )cos(kxih− . Problem 2 (40 points): Consider an electron with mass me confined within an infinite square well defined by V(x) = 0 for 0 < x < L, V(x) = +∞ otherwise. (a) (2 points) Using Schrödinger’s equation calculate the allowed stationary state eigenfunctions ψn(x), where the complete wavefunctions are given by h/)(),( tiEnn nextx −=Ψ ψ , and normalize the eigenfunctions V = +infinity V = +infinity Infinite Square Well 0 L x PHY4604 Fall 2007 R. D. Field Department of Physics Page 2 of 7 Exam 1 Solutions so that the probability of finding the electron somewhere in the box is one. Answer: )/sin(2)( Lxn L xn πψ = Solution: For the region outside of 0 < x < L 0)( =xψ and inside the region )()( 2 2 22 xE xd xd me ψψ =− h or )()( 22 2 xk xd xd ψψ −= with em kE 2 22h = The most general solution is of the form )cos()sin()( kxBkxAx +=ψ . The boundary condition at x = 0 gives 0)0( == Bψ and the boundary condition at x = L gives 0)sin()( == kLALψ which implies that πnkL = with n = 1, 2, 3,… Thus, )/sin()( LxnAxn πψ = with 2 222 2mL nEn hπ = . The normalization is arrived at by requiring that 24 )2sin( 2 )(sin)/(sin1)()( 2 0 2 0 2 2 0 22 LAyy n LAdyy n LAdxLxnAdxxx nnL nn =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −==== ∫∫∫ +∞ ∞− ∗ ππ ππ πψψ Thus, LA /2= . These states are called stationary because the probability density and all the expectation values are independent of time. (b) (3 points): Show that the allowed energy levels of the system are, En = E0 n2, where )2/( 2220 mLE πh= is the ground state energy and n = 1, 2, 3, …. Why is n = 0 excluded as a possible energy level? Solution: We see from above that 0 2 2 222 2 En mL nEn == hπ with 2 22 01 2mL EE hπ== . The state with n = 0 correstions to ψ0(x) = 0 which is not normalizable and corresponds to ρ0(x) = 0 which means there is no particle in the square well. (c) (10 points): Consider the operator, O = (x)op(px)op (i.e. the product of the position operator times the momentum operator). Is O an hermitian operator? Calculate the expectation value of the operator O for the nth stationary state (i.e. calculate <ψn|O|ψn>). Answer: O is not hermitian and 2/hixp nx =>< . Solution: We see that xxxx xpxpxpxpO ≠=== ↑↑↑↑ )()( since (x)op and (px)op do not commute. Also, ( ) 24 2cossin 4 1)sin( 22 2 )/2sin( 2 2)/cos()/sin(2 )()()())(( 2 0 2 0 2 00 h hhh hh h i n niyyy n idyyy n L L n L i dxLxnx L n L idxLxnLxnx L n L i dx dx xdxxidxxxpxxp n n LL n nnopxnnx =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−=−⎟ ⎠ ⎞ ⎜ ⎝ ⎛−=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛−= =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛−=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛− −==>< ∫ ∫∫ ∫∫ +∞ ∞− ∗ +∞ ∞− ∗ π π ππ π πππππ ψψψψ π π where I used θθθ 2sincossin 21= and I let y = 2nπ/L. PHY4604 Fall 2007 R. D. Field Department of Physics Page 5 of 7 Exam 1 Solutions ( ) π π πππ π π π 2 )2( 4 )cos()sin( 4 )sin( 4 )/2sin( 2 2 2 2 02 22 0 2 2 0 LLyyyLdyyyLdxLxx L −=−=−== ∫∫ and ( ) π π π ππ π π π 2 )2)24(( 8 )cos()2()sin(2 8 )sin( 8 )/2sin( 3 2 3 3 2 0 2 3 32 0 2 3 3 0 2 LL yyyyLdyyyLdxLxx L −=−−−= −−== ∫∫ The probability, 2E P , of measuring E2 is 0|| 222 == cPE . Problem 3 (40 points): Suppose that particles with energy E > V0 enter from the left and travel to the right and encounter both a delta-function potential and a step-down potential at x = 0 as follows: ⎩ ⎨ ⎧ +− = )( 0 )( 0 xV xV αδ 0 0 ≥ < x x where V0 and α are positive (real) constants. Let E Vr 01+= and E m mE mR 2 2 2 2 2 h h h αα == . (a) (10 points) Calculate the quantum mechanical reflection probability, PR, and express your answer in terms of r and R. Answer: 22 22 )1( )1( Rr RrPR ++ +− = Solution: We look for solutions of the time-independent Schrödinger equation )()()()( 2 2 22 xExxV dx xd m ψψψ =+− h or )())(( 2)( 22 2 xxVEm dx xd ψψ −−= h with h/)(),( iEtextx −=Ψ ψ . In the region x < 0 (left region) for E > 0 and V(x) = 0 we have )()(2)( 222 2 xkxmE dx xd ψψψ −=−= h with 2 2 h mEk = and m kE 2 22h = The most general solution is ikx L ikx LL eBeAx −+ +=)(ψ , In the region x > 0 (right region) we have iqx R iqx RR eBeAx −+ +=)(ψ , where rkVEmq =+= 2 0 )(2 h Since there are no particles entering from the right in the “right” region we set BR = 0. The boundary conditions at x = 0 are )0()0( RL ψψ = which implies that (1) AL + BL = AR. -V0 αδ(x) Delta-Function + Step-Down E x = 0 x PHY4604 Fall 2007 R. D. Field Department of Physics Page 6 of 7 Exam 1 Solutions Also, )0(2)()( 2 R x L x R m dx xd dx xd ψαψψ εε h =− −=+= which implies RLLR A mikBikAiqA 2 2 h α =+− or (2) ( ) RRLL AiRrAk mi k qBA +=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=− 2 2 h α Adding (1) + (2) yields ( ) RL AiRrA ++= 12 which implies ( )iRr AA LR ++ = 1 2 And subtracting (1)-(2) gives ( ) RL AiRrB −−= 12 which implies ( ) ( ) ( )iRr AiRrAiRrB RRL ++ −− =−−= 1 112 1 . The reflection probability is 22 22 2 2 )1( )1( || || Rr Rr A B P Lm k Lm k R ++ +− == h h . (b) (10 points) Calculate the quantum mechanical transmission probability, PT, and express your answer in terms of r and R. Answer: 22)1( 4 Rr rPT ++ = Solution: The transmission probability is 222 2 )1( 4 || || Rr r A A P Lm k Lm q T ++ == h h . (c) (4 points) Show that PT + PR = 1. Solution: Adding PR and PT gives 1 )1( )1( )1( 4)1( )1( 4 )1( )1( 22 22 22 22 2222 22 = ++ ++ = ++ ++− = ++ + ++ +− =+ Rr Rr Rr Rrr Rr r Rr RrPP TR . (d) (4 points) What is the transmission probability, PT1, for the case R = 0 (i.e. no delta function)? What is the numerical value of PT1 for the case V0 = 3E (with R = 0)? Answer: 889.0 9 8 )1( 4 221 ≈⎯→⎯+ = =rT r rP Solution: For R = 0 the transmission probability becomes 889.0 9 8 )21( 8 )1( 4 2221 ≈=+ ⎯→⎯ + = =rT r rP . Note that V0/E = 3 implies that r = 2. (e) (4 points) What is the transmission probability, PT2, for the case V0 = 0 (i.e. no step)? What is the numerical value of PT2 for the case R = 2 (with V0 = 0)? PHY4604 Fall 2007 R. D. Field Department of Physics Page 7 of 7 Exam 1 Solutions Answer: 2 1 4 4 222 ⎯→⎯+ = =RT R P Solution: For r = 0 the transmission probability becomes 2 1 44 4 4 4 222 =+ ⎯→⎯ + = =RT R P . Note that V0 = 0 implies that r = 1. (f) (8 points) What is the the numerical value of overall transmission probability, PT, for the case V0 = 3E and R = 2? Does PT = PT1PT2? Answers: 615.0 13 8 ≈=TP , TTT PPP ≠≈= 44.09 4 21 . Solution: For r = 2 and R = 2 the transmission probability becomes 615.0 13 8 49 8 )1( 4 2,222 ≈=+ ⎯⎯⎯ →⎯ ++ = == RrT Rr rP . We that TTT PPP ≠≈=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= 44.0 9 4 2 1 9 8 21 .