Solutions to Quantum Mechanics Final Exam Problems - Prof. Richard D. Field, Exams of Physics

Download Solutions to Quantum Mechanics Final Exam Problems - Prof. Richard D. Field and more Exams Physics in PDF only on Docsity! PHY3063 Spring 2006 R. D. Field Final Exam Solutions Page 1 of 9 April 25, 2006 PHY 3063 Final Exam Solutions Problem 1 (35 points): Consider an particle with mass m confined within an infinite square well defined by V(x) = 0 for 0 < x < L, V(x) = +∞ otherwise. (a) (5 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 . Normalize the eigenfunctions so that the probability of finding the particle 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) (5 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. (c) (10 points): Consider the operator, O = (x)op(px)op (i.e. the product of the position operator times the momentum operator). Is the operator O hermitian? Calculate the expectation value of the operator Q 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, V = +infinity V = +infinity Infinite Square Well 0 L x PHY3063 Spring 2006 R. D. Field Final Exam Solutions Page 2 of 9 April 25, 2006 ( ) 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. (d) (15 points): Suppose the particle in this infinite square well has an initial wave function at t = 0 given by )/(sin)0,( 2 LxAx π=Ψ . What is the normalization A? If you measure the energy of this particle, what is the probability that you will measure the ground state energy E0? Answer: L A 3 8 = and 96067.0 81 768 21 ≈= π P Solution: The normalization is arrived at by requiring that 8 3 8 3 32 )4sin( 4 )2sin( 8 3 )(sin)/(sin1)0,()0,( 22 0 2 0 4 2 0 42 LALAyyyLA dyyLAdxLxAdxxx L ==      +−= ===ΨΨ ∫∫∫ +∞ ∞− ∗ π ππ π π π π and hence L A 3 8 = . The overlap of Ψ(x,0) with ψ1(x) is given by ππππ π πψψ π π 9 316 3 16 3 41 3 11 3 1 3 82)cos( 3 )(cos 3 82 )(sin 3 82)/(sin 3 82)0,()(| 0 3 0 3 0 3 111 ==      +−+−=      −= ==Ψ>=Ψ=< ∫∫∫ +∞ ∞− ∗ L LL yyL LL dyyL LL dxLx LL dxxxc L and the probability, P1, of measuring E0 = E1 is 96067.0 81 768 9 316|| 2 2 2 11 ≈=      == ππ cP . PHY3063 Spring 2006 R. D. Field Final Exam Solutions Page 5 of 9 April 25, 2006 22/ 0 222 2/ 0 111 2/ 0 2/ 0 2122112 2/ 0 2/ 0 2121 int 12 3 4 3 4 3 4)/2sin()/sin(2)/2sin()/sin(2 )/2sin()/sin()/2sin()/sin(4),(      =     −     −== = ∫∫ ∫ ∫∫ ∫ πππ ππππ ππππρ LL L LL L dxLxLx L dxLxLx L dxdxLxLxLxLx L dxdxxx and ( ) ( ) ( ) πππ ππ ππ ππππ ππ 3 41 3 1 )sin()sin( 3 )/cos(1)/3cos(1 )/cos()/3cos(1)/2sin()/sin(2 2/ 0 2/3 0 2/ 0 2/ 0 2/ 0 2/ 0 −=−−= −=− =−= ∫∫ ∫∫ y L Ly L LdxLx L dxLx L dxLxLx L dxLxLx L LL LL (c) (10 points): For two identical fermions (i.e. particles with half-integral spins) we must use the symmetric wavefunction ( )),(),( 2 1),( 122121 xxxxxx A αβαβαβ ψψψ −= (antisymmetric under 1↔2) Calculate LLPαβ for the state where α = 1 and β = 2 (i.e. ground state) for two indistinguishable fermions. Answer: 07.018.0 4 1 3 4 4 1)( 2 12 ≈−≈     −= πFD LLP Solution: In this case we have [ ])/sin()/sin()/sin()/sin(2),( 212121 LxLxLxLxLxx A απβπβπαπψ αβ −= and ),(),(),( 21 int 2121 xxxxxx classicalFD αβαβαβ ρρρ −= where ( )),(),(Re),( 212121int xxxxxx ∗≡ βααβαβ ψψρ (α ≠ β) and we get 07.018.0 4 1 3 4 4 1 ),(),(),()( 2 2/ 0 2/ 0 2121 int 12 2/ 0 2/ 0 212112 2/ 0 2/ 0 21211212 ≈−≈     −= −== ∫ ∫∫ ∫∫ ∫ π ρρρ L LL L classical L L FD FD LL dxdxxxdxdxxxdxdxxxP PHY3063 Spring 2006 R. D. Field Final Exam Solutions Page 6 of 9 April 25, 2006 Problem 3 (30 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 PHY3063 Spring 2006 R. D. Field Final Exam Solutions Page 7 of 9 April 25, 2006 where I used 2 × 6 = 7 + 5 (j1 = 1/2, j2 = 5/2, j = 3, 2) 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) (20 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 |±>: (1) >±|zJ (2) >±| 2 1J (3) >±| 2 2J (4) >±|2 21 zz JJ (5) >±+ +−−+ |)( 2121 JJJJ (6) >±|2J Are the states >±| eigenstates of the J and Jz and if so what are their eigenvalues? Answer: (1) >±>=± || 21zJ (2) >±>=± |2| 2 1J (3) >±>=± || 43 2 2J (4) ( )>+>±−>=± mm |2|)((|2 21233121 zz JJ (5) >+>±±>=±+ +−−+ m|||)( 32342121 JJJJ (6) >±±>=± |)(| 4 6 4 92J 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