PHY4604 Introduction to Quantum Mechanics Fall 2004 Practice Exam Solutions Dec. 13, 2004


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1 PHY4604 Introduction to Quantum Mechanics Fall 2004 Practice Exam Solutions Dec. 1, 2004 No materials allowed. If you can t remember a formula, ask and I might help. If you can t do one part of a problem, solve subsequent parts in terms of unknown answer define clearly. All parts 10 pts., max=120. Problem 1 required, attempt 2 of remaining problems; circle which ones you want graded. a 0 Possibly helpful formulae and constants ˆL + ψ lm = h ll + 1) mm + 1)ψ lm+1 ˆL ψ lm = h ll + 1) mm 1)ψ lm 1 E n = me4 1 2 h 2 n 2 E n = hω n + 1 ) 2 ˆL 2 = ˆL + ˆL + ˆL 2 z hˆl z = ˆL ˆL+ + ˆL 2 z + hˆl z Hψ = Eψ Hψ = i h ψ t H = h2 2m 2 + V r) dx x n e x = n! sin 2 ax dx = a2 cosa 2 ) sina 2 ) 2a ) 2 1 ψ 0 x) = e 1 x 2 x 0 π 1/2 x 0 ψ 1 x) = ψ 2 x) = ψ x) = ) ) 2 1 2x e 1 x 2 x 0 2π 1/2 x 0 x 0 ) ) 1 x π 1/2 x x0 0 ) 1 x x π 1/2 x x0 0 x 0 e 1 2 ) ) 2 x x 0 e 1 2 ) 2 x x 0 1
2 1. Short answer. Must attempt only) 4 of 6. a) Explain what is meant by a 2p state of an atomic electron. The 2 refers to the principal quantum number n = 2, so state has energy E 2 = 1 Ryd/4, and p means total angular momentum quantum number l = 1. b) What is the degeneracy of the 1st excited state E = 5/2) hω) of the isotropic D simple harmonic oscillator? The D simple harmonic oscillator has energies E nx + E ny + E nz, where each E n is hωn + 1/2), so E tot = hωn x + n y + n z + /2). The first excited state has one quantum of excitation in either x, y, or z, so degeneracy =. c) Sketch the first eigenfunctions of the 1D infinite square well with V = 0 for a x a and V = otherwise. Label them according to their parity. Ground state ψ 1 and 2nd excited state ψ are both even functions of x even parity, in other words parity eigenvalue π = +1. 1st excited state ψ 2 is odd function odd parity, in other words parity eigenvalue π = 1. 2
3 d) State whether the following operators are selfadjoint, antiselfadjoint, unitary, or none of the above and why. i. ˆLx L x = r p) x is a product of selfadjoint operators representing observables, and therefore also selfadjoint. It is also an observable itself, of course. ii. xˆp x xˆp x) = x ˆp x = xˆp x. Selfadjoint. iii. d dx Operator is antiselfadjoint as discussed in class. Proof: definition of adjoint is O χ, ψ) = χ, Oψ) χ, ψ. For d/dx we have χ, d d x ψ) = = χ x) dψ dx dx d ) dx χ x) ψx)dx, where I ve performed an integration by parts and assumed that χ, ψ 0 at. From defintition, ) d = d dx dx iv. a + a a, a are raising & lowering operators for the 1D simple harmonic oscillator.) Since a) = a, this is selfadjoint. e) If a particle is in the state ψ, and n is the nth eigenvector of ˆQ corresponding to eigenvalue q n, what is the probability of measuring q? ψ 2.
4 f) Identify: i. Photoelectric effect Photons of wavelength ν are shone on a metal surface, and found to kick out an electron only if ν is greater than some threshold value, independent of intensity of light. Evidence Einstein) for quantum nature of light, E = hν. ii. DavissonGermer effect DavissonGermer experiment 1927) proved wave nature of electron hypothesized by de Broglie by diffracting electrons from a crystal. iii. Ehrenfest theorem says that expectation values of quantum mechanical observables obey classical equations of motion. For example, d p dt = ī h [H, p] = ī h i hdv dx = d V dx, where the right hand side is now just the classical force on a particle, i.e. this is Newton s law on the average. iv. StefanBoltzmann law Total radiation energy density emitted from a blackbody at temperature T is T Hydrogenic orbitals. An electron moving in the Coulomb field of a proton is in a state described by the wave function ignoring spin) Ψr, θ, φ) = 1 10 [ψ 100 r) + ψ 11 r, θ, φ)] 1) a) What is the expectation value of the energy? H = 1 10 E 1 + 9E ) = 1 Ryd /2 ) = 1 5 Ryd b) What is the expectation value of ˆL 2? L 2 = h2 ) = 1 5 h2 4
5 c) Is the wavefunction an eigenstate of parity? Yes or no? Explain either answer. The parity of the hydrogenic wavefunctions is given by 1) l. Therefore the wavefunction given is an admixture of an even parity and an odd parity wavefunction meaning it is not an eigenstate of the parity operator. d) What is the expectation value of the operator φ in this state? Easy way: recall L z = i h φ. The expectation value of L z is therefore φ = 9/10)i. L z = 1 10 ψ ψ 11, L z ψ ψ 11 )) = 9 h/10,. Electron in hydrogenic state. The electron in a hydrogen atom occupies a state ψ = R 21 r) 1 2 Y Y 1 1 2) where Y 0 1 = 1 1 R 21 r) = 4π cos θ, Y 1 1 = )/2 ) r e r/2a 0, ) 2a 0 a 0 8π sin θeiφ 4) a) What values) could a measurement of the zcomponent of the orbital angular momentum, ˆL z yield, and what is the probability of each? What is the expectation value of ˆL z in this state? m is either 0 or 1, so L z can be either 0 or h, with probability 1/ or 2/, respectively. L z = 1 = 2 h/, ψ ψ 211, L z ψ ψ 211 ) ) 5
6 b) Calculate the average distance of the electron from the nucleus in this state. r = 1 ψ ψ 211, rψ ψ 211 ) ) = 1 ψ 210, r ψ 210 ) + 2 ψ 211, r ψ 211 ) = )/2 ) ) r 2 dr r 1 1 r 0 2a 0 a 0 = a 0 dy y 5 exp y 24 0 = 5a 0 } {{ } e r/2a 0 c) What is the expectation value of ˆL x in this state? L x = 1 ψ ) L+ + L 2ψ 211, ψ ) 2ψ 211 ) 2 = h ψ ψ 211, 2ψ ψ ) 2ψ = h 2 + 2) = 2 h/ 6 d) If you measured the zcomponent of the angular momentum and the distance of the electron from the origin r simultaneously, what is the probability density for finding ˆL z with eigenvalue zero at a distance r? P m=0 r) = r 2 1 R2 21r) 6
7 4. Scattering potential. For a 1D potential as shown in the figure and E < V 0, a) write down the Schrödinger equation and its general solution in the regions I,II, and III assuming the particle is incident from the left. Hψ = Eψ, H = h2 2m 2. with ψ I = Ae ipx/ h + Be ipx/ h, ψ II = Ce qx/ h + De qx/ h ψ III = F e ipx/ h p = 2mE q = 2mV 0 E) b) Write down the matching conditions at the boundaries x = a, b. Require continuity of ψ and its derivatives at the interfaces: ψ I a) = ψ II a) ψ II b) = ψ III b) ψ Ia) = ψ IIa) ψ IIb) = ψ IIIb), in other words Ae ipa/ h + Be ipa/ h = Ce qa/ h + De qa/ h Ce qb/ h + De qb/ h = F e ipb/ h Ae ipa/ h Be ipa/ h )ip = Ce qa/ h De qa/ h )q qce qb/ h De qb/ h ) = F ipe ipb/ h 7
8 c) Do not solve for all coefficients, but reduce the problem to a single equation determining the eigenvalues. This is tedious, not a good exam question. 2iap/ h p + iq B/A = e p iq or F/A = 4e ipa/ h e ipb/ h e qa+b)/ h pq iq p)ipe 2qa/ h ipe 2qb/ h + qe 2qa/ h + qe 2qb/ h ) d) Sketch the probability of finding the electron in all three regions. 8
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