Solutions to Exam 2 in PHY4604: Quantum Mechanics, Exams of Physics

Download Solutions to Exam 2 in PHY4604: Quantum Mechanics and more Exams Physics in PDF only on Docsity! PHY4604 Exam 2 Solutions Department of Physics Page 1 of 7 PHY 4604 Exam 2 Solutions (Total Points = 100) Problem 1 (10 points): Circle true or false for following (1 point each). (a) (True or False) Bohr’s model of the atom assumes the orbital angular momentum is quantized according to hnL = , with n = 1, 2, 3, …, which is also true quantum mechanically. (b) (True or False) The eigenvalues of hermitian operators are real numbers. (c) (True or False) Eigenfunctions of hermitian operators corresponding to different eigenvalues are orthogonal. (d) (True or False) If an observable, O, commutes with the Hamiltonian and if O does not depend explicitly on time, then <O> is zero. (e) (True or False) If [A, B] = 0, then ∆A∆B ≥ 0. (f) (True or False) All quantum operators can be expressed in terms of functions or differentials. (g) (True or False) If the three operators J1, J2, and J3, all commute with each other then they are said to form a “Lie Algebra”. (h) (True or False) SU(2) is the group of all unitary 2 × 2 matrices with determinant equal to one. (i) (True or False) The spin 2 (i.e. s = 2) gravaton has five possible spin states. (j) (True or False) In SU(2), 2 × 2 = 4. PHY4604 Exam 2 Solutions Department of Physics Page 2 of 7 Problem 2 (30 points): Consider the (one dimensional) wave function at t = 0 given by 22)( ax Ax + =ψ , where A and a are real constants. (a) (5 points) Find the normalization constant A such that 1|)(| 2 =∫ +∞ ∞− dxxψ and sketch the probability density .|)(|)( 2xx ψρ = Answer: π 32aA = Solution: We see that 1 4 2 )( 12 )( 1|)(| 3 2 0 222 2 222 22 == + = + = ∫∫∫ +∞+∞ ∞− +∞ ∞− a Adx ax Adx ax Adxx πψ and hence π 32aA = , where I used )/(tan 2 1 )(2)( 1 1 3222222 axaxaa xdx xa −+ + = +∫ and 30 222 4)( 1 a dx xa π = +∫ ∞ . Thus, 222 3 2 )( 12|)(|)( ax axx +       == π ψρ (b) (5 points) Compute <x> and <x2> and ∆x using the position space wave function ψ(x). Answer: 0>=< x , 22 ax >=< , and ∆x = a Solution: We see that 0 )( |)(| 222 22 = + =>=< ∫∫ +∞ ∞− +∞ ∞− dx ax xAdxxxx ψ 2 2 222 2 2 222 2 2222 4 2 )( 2 )( |)(| a a Adx ax xAdx ax xAdxxxx == + = + =>=< ∫∫∫ +∞ ∞− +∞ ∞− +∞ ∞− πψ , where I used )/(tan 2 1 )(2)( 1 22222 2 ax axa xdx xa x −+ + −= +∫ and adxxa x 4)(0 222 2 π = +∫ ∞ . We see that axxx =><−><=∆ 22 . (c) (10 points). Find the momentum space wave function at t = 0, where ∫ +∞ ∞− −= dxexp xixpx h h /)( 2 1)( ψ π φ and verify that it is properly normalized. Sketch the probability density .|)(|)( 2xx pp φρ = ρ(x) -5.0 0.0 5.0 PHY4604 Exam 2 Solutions Department of Physics Page 5 of 7 (d) (6 points) What is the probability density, ρ(r,θ,φ) , for finding the electron at (r,θ,φ) no matter what its spin? Is this density spherically symmetric (i.e. independent of θ and φ)? Answer: 2|)(| 4 1)( rRr π ρ = spherically symmetric Solution: We see that 22222222 |)(| 4 1)cos(sin|)(| 4 1sin|)(| 4 1cos|)(| 4 1 ),,(),,()( rRrRrRrR rrr π θθ π θ π θ π φθρφθρρ =+=+= += −+ Problem 4 (30 points): Consider a spin ½ system described by the Hamiltonian: 222 0 )(4 zyx aSSSH ++= where a is a real constant and σr r h 2=S with       = 01 10 xσ       − = 0 0 i i yσ       − = 10 01 zσ . (a) (10 points) Find the energy levels and the corresponding eigenkets of the system. How many energy levels are there? How many eigenkets are there? What is the ground state energy, E0? Answer: There is one energy level 20 )4/2( haE += , but there are two eigenkets both with the same energy (i.e. degeneracy of 2)       + => β α βα χ 2210 |||| 1| and       −+ => ∗ ∗ α β βα χ 2220 |||| 1| . Solution: We see that       + + =            −+=−+=++= 2 2 222222 )4/2(0 0)4/2( 10 01 )4( 4 13)4(4)(4 h h h a a aSaSaSSSH zzyx where I used       =      += 10 01 10 01 )1( 24 32 2 1 2 12 hhS       =      −       − = 10 01 10 01 10 01 2 4 12 4 12 hhzS The energy levels are the solution of 0 )4/2(0 0)4/2( 2 2 = −+ −+ λ λ h h a a which gives 2)4/2( ha+=λ and 2)4/2( ha+=λ . There is only one energy 20 )4/2( haE += . We see that       =      =            β α β α β α 0 0 0 0 0 0 0 E E E E E and hence E0α = E0α and E0β = E0β which is satisfied for any choice of α and β, however there are only two independent eigenkets. So let me take one eigenket to be PHY4604 Exam 2 Solutions Department of Physics Page 6 of 7       + => β α βα χ 2210 |||| 1| and then       −+ => ∗ ∗ α β βα χ 2220 |||| 1| . We see that ( ) 0 |||| 1| 221002 =      − + =>< β α αβ α χχ b (b) (10 points) If the Hamiltonian in (a) is changed to xbSHH += 0 , where b is a positive real constant, what are the energy levels and the corresponding eigenkets of this new system. How many energy levels are there? How many eigenkets are there? What is the ground state energy? Express your answer in terms of the ground state energy, E0, from (a). How does adding this additional term shift the energy levels of H0? Answer: Now there are two energies. The ground state is )2/(01 bEE h−= and the 1 st excited state is )2/(02 bEE h+= with eigenkets       − >= 1 1 2 1| 1χ and       >= 1 1 2 1| 2χ . Adding this new term splits the degeneracy from ∆E = 0 in (a) to bE h=∆ . Solution: We see that       =      + + =+= 0 0 2 2 0 2/ 2/ )4/2(2/ 2/)4/2( Eb bE ab ba bSHH x h h hh hh The energy eigenvalues are determined from the determinant 0 2/ 2/ 0 0 = − − λ λ Eb bE h h which yields 0)2/()( 220 =−− bE hλ or )2/()( 0 bE h±=− λ and )2/(01 bEE h−= and )2/(02 bEE h+= . The eigenkets are determined from       =      + + =            β α βα βα β α )2/( 2/ 2/ 2/ 2/ 0 0 0 0 0 bE Eb bE Eb bE hm h h h h . For the ground state we have 2/2/ 00 ααβα bEbE hh −=+ which means that β = - α and hence       − >= 1 1 2 1| 1χ . For the first excited state we have 2/2/ 00 ααβα bEbE hh +=+ that β = α and hence       >= 1 1 2 1| 2χ . Double check       =      =            1 1 2 1)2/( 2/ 2/ 2 1 1 1 2/ 2/ 2 1 0 0 0 0 0 m hm mh hm mh h bE Eb bE Eb bE . PHY4604 Exam 2 Solutions Department of Physics Page 7 of 7 (c) (10 points) Suppose that at t = 0 the system described by the Hamiltonian in (b) is in the state       >=>= + 0 1 |)0(| χχ . What is >)(| tχ ? If I measure the energy of the state >)(| tχ , what values might I get and what is the probability of getting these values. What is the expectation value of Sz in the state >)(| tχ ? Answer:       − >= − )2/sin( )2/cos( )(| /0 bti bt et tiE hχ , )cos(21 btSz h>=< . Measure energy )2/(01 bEE h−= with probability ½ and energy )2/(01 bEE h+= with probability ½. Solution: We see that ( )       +      − =>+>=      >=>= + 1 1 2 1 1 1 2 1|| 2 1 0 1 |)0(| 21 χχχχ and hence ( ) ( )       − =      +− + = >+>= >+>>= − −+ −+− +−−− −− )2/sin( )2/cos( 2 || 2 1 || 2 1)(| / 2/2/ 2/2// 2 /)2/( 1 /)2/( 2 / 1 / 0 0 00 21 bti bt e ee eee ee eet tiE ibtibt ibtibttiE tbEitbEi tiEtiE h h hhhh hh χχ χχχ . The expectation value of Sz is given by ( ) ( ) ( ) ( ) )cos(1)2/(cos2 )2/(sin)2/(cos )2/sin( )2/cos( )2/sin()2/cos( )2/sin( )2/cos( 10 01 )2/sin()2/cos()(||)( 2 12 2 1 22 2 1 2 1 2 1 btbt btbt bti bt btibt bti bt btibttStS zz hh hh h =−= −=      =       −       − >=>=<< χχ