Exam 2 Solutions - Introduction to Quantum Mechanics | PHY 4604, Exams of Physics

Download Exam 2 Solutions - 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 8 Exam 2 Solutions PHY 4604 Exam 2 Solutions (Total Points = 100) Problem 1 (20 points): Circle the correct answer for following (4 points each). (1a) An electron is in a one-dimensional infinite square well with zero potential energy in the interior and infinite potential energy at the walls. A graph of its wave function ψ(x) versus x is shown. What is the value of the quantum number n? (circle the correct answer) (a) 0 (b) 2 (c) 4 (d) 6 (e) 8 (1b) An atom is in a state with orbital quantum number l = 2. What are the possible values of the magnetic quantum number m. (circle the correct answer) (a) 1, 2 (b) 0, 1, 2 (c) 0, 1 (d) 1, 0, -1 (e) 2, 1, 0, -1, -2 (1c) An electron is in a quantum state for which the magnitude of the orbital momentum is h103 . How many allowed values of the z-component of the angular momentum are there? (circle the correct answer) (a) 9 (b) 10 (c) 17 (d) 18 (e) 19 (1d) An electron is in a quantum state for which there are six allowed values of the z-component of the angular momentum. What is the magnitude of the angular momentum vector? (circle the correct answer) (a) h6 (b) 2/5h (c) h6 (d) h103 (e) 2/35h (1e) What is 2 × 2 × 2 × 2 in SU(2)? (circle the correct answer) (a) 5 + 3+ 3+ 3 + 1 +1 (b) 16 (c) 9 + 5 + 2 (d) 6 + 5 + 3 + 2 (e) 10 + 6 Problem 2 (40 points): Consider a spin ½ system described by the Hamiltonian: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = 01 10 εε εε i i H where ε0 and ε1 are real positive constants. (a) (6 points) Find the energy levels of the system. How many energy levels are there? What is the ground state energy, E0, and the first excited state energy, E1? Answer: There are two energy levels: E0 = ε0 – ε1, E1 = ε0 + ε1. Solution: The energy levels are the solution of 0 01 10 = −− − λεε ελε i i PHY4604 Fall 2007 R. D. Field Department of Physics Page 2 of 8 Exam 2 Solutions which yields 0)( 21 2 0 =−− ελε , which implies that 10 ελε ±=− and hence 10 εελ ±= . There are two energy levels, E0 = ε0 – ε1 and E1 = ε0 + ε1. (b) (6 points) What are the (normalized) eigenkets corresponding to the ground state, |E0>, and the first excited state , |E1>? Answer: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − >= 12 1| 0 i E ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ >= 12 1| 1 i E . Solution: For the ground state we see that ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− + =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − b a bai bia b a i i )( 10 01 10 01 10 εε εε εε εε εε and hence ε0a +iε1b = (ε0-ε1)a, which implies that a = -ib and hence if b = -1 then a = i and ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − >= 12 1| 0 i E . For the first excited state we see that ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− + =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − b a bai bia b a i i )( 10 01 10 01 10 εε εε εε εε εε and hence ε0a +iε1b = (ε0+ε1)a, which implies that a = ib and hence if b = 1 then a = i and ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ >= 12 1| 1 i E . (c) (6 points) Now suppose that at t = 0 the system is in the state ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + >= )12( )21(| 0 iNψ . What is the normalization N? Answer: 6/1=N Solution: We see that ( ) ( ) 6)12()21( )12( )21()12()21(|1 2222200 NN iiN =−++=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + −+−>==< ψψ . Hence 6/1=N . (d) (10 points) Suppose that at t = 0 the system is in the state ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + >= )12( )21(| 0 iNψ . If you measure the energy of the state |ψ0>, what are the possible values you might get, and what is the probability of getting each of them? What is the expectation value of the energy for this state (i.e. the average energy <E>)? Answer: You get energy E0 with probability P0 = 1/3 and you get energy E1 with probability P1 = 2/3 and the average energy is 1310 εε +>=< E . Solution: We see that PHY4604 Fall 2007 R. D. Field Department of Physics Page 5 of 8 Exam 2 Solutions (8) 0| 2 3 2 3 >=−−S (c) (20 points) Using the results from (a) and (b) construct the (4×4) spin matrices Sx, Sy, and Sz, for a particle of spin 3/2 and verify that ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ =++= 1000 0100 0010 0001 2 4 152222 hzyx SSSS . Answer: ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = 0300 3020 0203 0030 2 h xS , ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − − = 0300 3020 0203 0030 2 hiSy , ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − = 3000 0100 0010 0003 2 h zS Solution: Let ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = 4321 4321 4321 4321 dddd cccc bbbb aaaa Sz h From part (a) we see that >>= 2 3 2 3 2 3 2 3 2 3 || hzS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 0 0 1 2 3 0 0 0 1 1 1 1 1 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a1 = 3/2, b1 = 0, c1 = 0, and d1 = 0. Also, >>= 2 1 2 3 2 1 2 1 2 3 || hzS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 0 1 0 2 1 0 0 1 0 2 2 2 2 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a2 = 0, b2 = 1/2, c2 = 0, and d2 = 0. Also, >−−>=− 2 1 2 3 2 1 2 1 2 3 || hzS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ −= ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 1 0 0 2 1 0 1 0 0 3 3 3 3 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a3 = 0, b3 = 0, c3 = -1/2, and d3 = 0. Also, >−−>=− 2 3 2 3 2 3 2 3 2 3 || hzS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ −= ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 1 0 0 0 2 3 1 0 0 0 4 4 4 4 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa PHY4604 Fall 2007 R. D. Field Department of Physics Page 6 of 8 Exam 2 Solutions which implies that a4 = 0, b4 = 0, c4 = 0, and d4 = -3/2. Hence, ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − = ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − = 3000 0100 0010 0003 2 000 000 000 000 2 3 2 1 2 1 2 3 h hzS and ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − − = 9000 0100 0010 0009 4 000 000 000 000 000 000 000 000 2 2 3 2 1 2 1 2 3 2 3 2 1 2 1 2 3 22 hhzS . Now let ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ =+ 4321 4321 4321 4321 dddd cccc bbbb aaaa S h From part (b) we see that 0| 2323 >= +S and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 0 0 0 0 0 0 1 1 1 1 1 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a1 = 0, b1 = 0, c1 = 0, and d1 = 0. Also, >>=+ 23232123 |3| hS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 0 0 1 3 0 0 1 0 2 2 2 2 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a2 = 3 , b2 = 0, c2 = 0, and d2 = 0. Also, >>=−+ 2 1 2 3 2 1 2 3 |2| hS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 0 1 0 2 0 1 0 0 3 3 3 3 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a3 = 0, b3 = 2, c3 = 0, and d3 = 0. Also, >−>=−+ 2 1 2 3 2 3 2 3 |3| hS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 1 0 0 3 1 0 0 0 4 4 4 4 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a4 = 0, b4 = 0, c4 = 3 , and d4 = 0. Hence, PHY4604 Fall 2007 R. D. Field Department of Physics Page 7 of 8 Exam 2 Solutions ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ =+ 0000 3000 0200 0030 hS . We can determine S- from S+ as follows ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ == ↑+− 0300 0020 0003 0000 )( hSS Or we can do it the long way by letting ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ =− 4321 4321 4321 4321 dddd cccc bbbb aaaa S h From part (b) we see that >>=− 2 1 2 3 2 3 2 3 |3| hS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 0 1 0 3 0 0 0 1 1 1 1 1 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a1 = 0, b1 = 3 , c1 = 0, and d1 = 0. Also, >−>=− 2 1 2 3 2 1 2 3 |2| hS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 1 0 0 2 0 0 1 0 2 2 2 2 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a2 = 0, b2 = 0, c2 = 2, and d2 = 0. Also, >−>=−− 2 3 2 3 2 1 2 3 |3| hS and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 1 0 0 0 3 0 1 0 0 3 3 3 3 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a3 = 0, b3 = 0, c3 = 0, and d3 = 3 . Also, 0| 2 3 2 3 >=−−S and hence ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎜ ⎝ ⎛ 0 0 0 0 1 0 0 0 4 4 4 4 4321 4321 4321 4321 d c b a dddd cccc bbbb aaaa which implies that a4 = 0, b4 = 0, c4 = 0, and d4 = 0. Hence,