4 Problems on Quantum Mechanics 1 - Exam 2 Solutions | PHY 4604, Exams of Physics

Download 4 Problems on Quantum Mechanics 1 - Exam 2 Solutions | PHY 4604 and more Exams Physics in PDF only on Docsity! PHY4604 Exam 2 Solutions Department of Physics Page 1 of 9 PHY 4604 Exam 2 Solutions (Total Points = 100) Problem 1 (20 points) Problem 1a (2 points): The wave function for an electron in a state with zero angular momentum: (circle the correct answer) (a) is zero everywhere (b) is spherically symmetric (c) depends on the angle from the z axis (d) depends on the angle from the x axis (e) is spherically symmetric for some shells and depends on the angle from the z axis for others Problem 1b (2 points): The magnitude of the orbital angular momentum of an electron in an atom is what multiple of h ? (l = 0, 1, 2, …) (circle the correct answer) (a) 1 (b) ½ (c) )1( +ll (d) 12 +l (e) 2l Problem 1c (2 points): An electron is in a quantum state for which the magnitude of the orbital momentum is h26 . How many allowed values of the z-component of the angular momentum are there? (circle the correct answer) (a) 4 (b) 5 (c) 7 (d) 8 (e) 9 I goofed up this problem. The correct answer is 17. Problem 1d (2 points): If the wave function ψ is spherically symmetric then the radial probability density is given by: (circle the correct answer) (a) ψπ 24 r (b) 2||ψ (c) 22 ||4 ψπr (d) 2||4 ψπ (e) 2||4 ψπr Problem 1e (3 points): An electron in an atom is in a state with l = 3 and ml = 2. The angle between L r and the z-axis is given by: (circle the correct answer) (a) 48.2º (b) 60º (c) 30º (d) 35.3º (e) 54.7º Problem 1f (3 points): SU(2) is the group of 2×2 matrices, U, where: (a) 1== ↑↑ UUUU (b) ↑= UU (c) det(U)=1 (d) ↑− = UU 1 (e) TUU =* (Note: circle all the correct answers) Problem 1g (3 points): If H is an Hermitian operator then: (a) 1== ↑↑ HHHH (b) ↑= HH (c) <H> ≥ 0 (c) <H2> ≥ 0 (e) its eigenvalues are real (Note: circle all the correct answers) Problem 1h (3 points): If L v is the orbital angular momentum operator then: (a) 0=∆∆ yx LL (b) ||2 ><≥∆∆ zyx LLL h (c) 2h≥∆∆ yx LL (d) 0 2 ≥∆∆ LLz (Note: circle all the correct answers) PHY4604 Exam 2 Solutions Department of Physics Page 2 of 9 Problem 2 (30 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 sustem. 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 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       >= 0 1 | 0ψ . 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 = ½ and you get energy E1 with probability P1 = ½ and the average energy is <E> = ε0. Solution: We see that PHY4604 Exam 2 Solutions Department of Physics Page 5 of 9 2222 0 )/( 0 )/( 0 / 0 /||/ )( 2 )/( 2 2)/)(/( 2 2 )/( 1 )/( 1 222 222 )( ap aa ap aa aipaip aa aipaip adxeAdxeA dxeeAdxeeAdxeeAp xxxx xx xaipxaip axixpaxixpxaixp x xx xxx h hh hhhhh hhhhh hhh hh hhh + = + = +−+ =       +− + + =+= +== ∫∫ ∫∫∫ ∞+ +−− ∞+ +− ∞− +− +∞ −− +∞ ∞− −− πππ πππ πππ φ Thus, 222 3 2 ))(( )(2|)(|)( ap app x xx h h + == π φρ and we see that 1 )(2 )(2 ))(( )(2|)(| 3 3 222 3 2 == + = ∫∫ +∞ ∞− +∞ ∞− a adp ap adpp x x xx h h h h π ππ φ , where I used 3 0 222 4)( 1 b dp bp xx π = +∫ +∞ . (d) (6 points). Compute <px>, <px2>, and ∆px using the momentum space wave function φ(px). Answer: 0>=< xp , 22 )( apx h>=< , and apx h=∆ . Solution: We see that 0 ))(( )(2|)(| 222 3 2 = + =>=< ∫∫ +∞ ∞− +∞ ∞− x x x xxxx dpap padpppp h h π φ 2 3 222 23 222 )( )(2 )(2 ))(( )(2|)(| a a a dp ap padpppp x x x xxxx h h h h h == + =>=< ∫∫ +∞ ∞− +∞ ∞− π π π φ where I used b dp bp p x x x 4)(0 222 2 π = +∫ ∞ . We see that appp xxx h=><−><=∆ 22 (e) (2 points). What is ∆x ∆px? Is it consistent with the uncertainty principle? Answer: 22 2 hh >=∆∆ xpx Solution: We see that hhh 2 2 2 1 2 1 ===∆∆ a a px x . PHY4604 Exam 2 Solutions Department of Physics Page 6 of 9 Problem 4 (25 points): Suppose we have two vector operators opJ )( 1 r and opJ )( 2 r with 0])(,)[( 21 =opop JJ rr and each of the vectors obey the same SU(2) “lie algebra”: opkijkopjopi JiJJ )(])(,)[( 111 ε= and opkijkopjopi JiJJ )(])(,)[( 222 ε= . The states |j1m1> are the eigenkets of opJ )( 2 1 and opzJ )( 1 and the states |j2m2> are the eigenkets of opJ )( 2 2 and opzJ )( 2 as follows: >>= >+>= 111111 111111 2 1 ||)( |)1(|)( mjmmjJ mjjjmjJ opz op >>= >+>= 222222 222222 2 2 ||)( |)1(|)( mjmmjJ mjjjmjJ opz op Also we know that >±±−+>= >±±−+>= ± ± 1|)1()1(|)( 1|)1()1(|)( 222222222 111111111 mjmmjjmjJ mjmmjjmjJ op op where opyopxop JiJJ )()()( 111 ±= ± and opyopxop JiJJ )()()( 222 ±= ± . Now consider the vector sum of the two operators, opopop JJJ )()()( 21 rrr += or opiopiopi JJJ )()()( 21 += for i = 1,2, 3. (a) (5 points): Show that opzopzopopopopopop opopopopopopopopop JJJJJJJJ JJJJJJJJJJJ )()(2)()()()()()( )()(2)()()()()()()( 212121 2 2 2 1 21 2 2 2 12121 2 ++++= ⋅++=+⋅+=⋅= +−−+ rrrrrrrr Solution: First we note that )( )( 112 1 1 112 1 1 −+ −+ −= += JJJ JJJ iy x and )( )( 222 1 2 222 1 2 −+ −+ −= += JJJ JJJ iy x Hence, zz zz zzyyxx JJJJJJJJ JJJJJJJJJJJJ JJJJJJJJJJJJJ 212121 2 2 2 1 2122112 1 22112 12 2 2 1 212121 2 2 2 121 2 2 2 1 2 2 2))(())(( 2222 ++++= +−−−++++= ++++=⋅++= +−−+ −+−+−+−+ rr (b) (5 points): Evaluate the following in SU(2). 3 × 2 = 4 × 3 = 5 × 3 = 5 × 4 = 2 × 3 × 4 = Answers: 3 × 2 = 4 + 2 (j1 = 1, j2 = ½, j = 3/2, 1/2) 4 × 3 = 6 + 4 + 2 (j1 = 3/2, j2 = 1, j = 5/2, 3/2, 1/2) 5 × 3 = 7 + 5 + 3 (j1 = 2, j2 = 1, j = 3, 2, 1) 5 × 4 = 8 + 6 + 4 + 2 (j1 = 2, j2 = 3/2, j = 7/2, 5/2, 3/2, 1/2) 2 × 3 × 4 = 2 × (6 + 4 + 2) = 2×6 + 2×4 + 2×2 = 7 + 5 +5 + 3 + 3 + 1 where I used 2 × 6 = 7 + 5 (j1 = 1/2, j2 = 5/2, j = 3, 2) PHY4604 Exam 2 Solutions Department of Physics Page 7 of 9 2 × 4 = 5 + 3 (j1 = 1/2, j2 = 3/2, j = 2, 1) 2 × 2 = 3 + 1 (j1 = 1/2, j2 = 1/2, j = 1, 0) (c) (15 points): Now consider the case where j1 = 1 and j2 = ½ (i.e. 3 × 2) and define the states as follows: >−=> >=> >=> − 11|| 10|| 11|| 111 110 111 Y Y Y and >−=↓> >=↑> 2 1 2 1 2 2 1 2 1 2 || || Now consider the two superposition states 21102111 ||3 2|| 3 1| ↑>>+↓>>>≡+ YY and 21102111 ||3 1|| 3 2| ↑>>−↓>>>≡− YY . Calculate the following and express your answer in terms of |+> and |->: (1) >++ |)( 22 2 1 JJ (2) >−+ |)( 2 2 2 1 JJ (3) >+++ +−−+ |)2( 212121 zz JJJJJJ (4) >−++ +−−+ |)2( 212121 zz JJJJJJ (5) >+|zJ (6) >−|zJ (7) >+| 2J (8) >−|2J Are the states >±| eigenstates of the J2 and Jz and if so what are their eigenvalues? Answer: (1) >+>=++ ||)( 411 2 2 2 1 JJ (2) >−>=−+ ||)( 411 2 2 2 1 JJ (3) >+>=+++ +−−+ ||)2( 212121 zz JJJJJJ (4) >−−>=−++ +−−+ |2|)2( 212121 zz JJJJJJ (5) >+>=+ || 21zJ (6) >−>=− || 21zJ (7) >+>=+ || 4 152J (8) >−>=− || 4 32J The state |+> correspomds to j = 3/2 m = ½ and |-> corresponds to j = ½ and m = ½. Solution: We know that 110 2 1110 2 1 111 2 1111 2 1 |210|)11(110|| |211|)11(111|| >>=+>==> >>=+>==> YJYJ YJYJ 24 3 2 1 2 1 2 1 2 1 2 1 2 12 22 2 2 24 3 2 1 2 1 2 1 2 1 2 1 2 12 22 2 2 ||)1(|| ||)1(|| ↓>>=−+>=−=↓> ↑>>=+>==↑> JJ JJ 010|010|| |111|111|| 11111 11111111 >=>==> >>=>==> zz zz JYJ YJYJ 22 1 2 1 2 1 2 1 2 1 2 1 222 22 1 2 1 2 1 2 1 2 1 2 1 222 |||| |||| ↓>−>=−−>=−=↓> ↑>>=>==↑> zz zz JJ JJ 11111101 11111 |211|)10(0)11(110|| 011|| >>=+++>==> >==> ++ ++ YJYJ JYJ 11011111 |210|)11(1)11(111|| >>=−−+>==> −− YJYJ 22 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 222 2 1 2 1 222 ||)1()1(|| 0|| ↑>>=+−++>=−=↓> >==↑> ++ ++ JJ JJ 0|| ||)1()1(|| 2 1 2 1 222 22 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 222 >=−=↓> ↓>>=−−−+>==↑> −− −− JJ JJ and hence