4604_Exam3_solutions_fa05

4604_Exam3_solutions_fa05 - PHY4604 Final Exam Solutions...

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PHY4604 Final Exam Solutions Department of Physics Page 1 of 8 PHY 4604 Final Exam Solutions Monday December 5, 2005 (Total Points = 100) Problem 1 (10 points): Circle true or false for following (1 point each). (a) (True or False) If the expectation value of the operator O op is a real number then O op is hermitian. (b) (True or False) If | ψ > is a (normalized) “ket” vector, then the operator O op = | ψ >< ψ | has the property that (O op ) 2 = 1. (c) (True or False) If the operator U op is unitary then, its inverse is equal it its hermitian conjugate. (d) (True or False) If P op is the parity operator and if ) 1 ( 2 1 op op P O ± = ± , then ± ± = op op O O 2 ) ( . (e) (True or False) One can simultaneously know the precise values of commuting observables. (f) (True or False) The overall wave function (space × spin) for two identical spin 3/2 particles is always symmetric under the interchange of the two particles. (g) (True or False) A µ - particle has the same charge as an electron but its mass is 207 times greater, hence the ground state energy of a muonic atom ( µ - and a proton) is lower than the ground state of hydrogen. (h) (True or False) In SU(2), 3 × 3 = 4 + 3 + 2. (i) (True or False) If one neglects the electron-electron interactions then the ground state of helium would have a degeneracy of 2. (j) (True or False) The maximum number of electrons that can be in the n th energy level of an atom is n 2 .
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PHY4604 Final Exam Solutions Department of Physics Page 2 of 8 Problem 2 (30 points): Consider a bead of mass m that slides frictionlessly around a circular wire ring of radius R . The bead is constrained to lie on the circle. This is similar to the free particle with Hamiltonian 2 2 2 2 2 2 2 2 2 2 2 θ d d mR ds d m m p H h h = = = . where s = R θ is the distance around the circle. We look for stationary state solutions of the form h / ) ( ) , ( iEt e t = Ψ ψ with the time-independent Schrödinger equation given by E d d mR = 2 2 2 2 2 h . Consider separately the solutions corresponding to clockwise, + , and counter-clockwise, , movement around the wire. The probability density 2 | | ) ( ± ± = ρ is the probability of finding the particle between θ and θ +d θ and is normalized as follows: 1 ) ( 2 0 = ± π d . (a) (20 points) Find the stationary states ± (with the correct normalization) and the corresponding allowed energies. (Hint: require ) ( ) 2 ( = + since s and s+2 π R correspond to the same point on the wire) Answer: in n e m 2 1 ) ( = ± , 2 2 2 2 mR n E n h = ± (n = 0, 1, 2, …) Solution: We see that α 2 2 2 = d d with 2 2 2 2 h E mR = and 2 2 2 2 mR E h = with solution t i t i Be Ae + + = ) ( For α > 0 the first term corresponds to counter-clockwise movement and the second term corresponds to clockwise movement. We also know that αθ i i i i Be Ae Be Ae + + + + + = = + = + ) ( ) 2 ( ) 2 ( ) 2 ( which implies that 0 ) 1 ( ) 1 ( 2 2 = + + + i i i i e Be e Ae . This equation implies that 1 2 = ± i e which implies that α = n = 0, ±1, ±2, …and hence in n e + = 2 1 ) ( and 2 2 2 2 mR n E n h = + (n = 0, 1, 2, …) in n e + = 2 1 ) ( and 2 2 2 2 mR n E n h = (n = 0, 1, 2, …).
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4604_Exam3_solutions_fa05 - PHY4604 Final Exam Solutions...

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