Practice Exam Solutions - Introduction to Quantum Mechanics | PHY 4604, Exams of Physics

Download Practice Exam Solutions - Introduction to Quantum Mechanics | PHY 4604 and more Exams Physics in PDF only on Docsity! PHY4604–Introduction to Quantum Mechanics Fall 2004 Practice Exam Solutions Dec. 13, 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 3 problems; circle which ones you want graded. Possibly helpful formulae and constants L̂+ψ`m = h̄ √ `(` + 1)−m(m + 1)ψ`m+1 L̂−ψ`m = h̄ √ `(` + 1)−m(m− 1)ψ`m−1 En = me4 2h̄2 1 n2 En = h̄ω ( n + 1 2 ) L̂2 = L̂+L̂− + L̂2z − h̄L̂z = L̂−L̂+ + L̂2z + h̄L̂z Hψ = Eψ Hψ = ih̄ ∂ψ ∂t H = − h̄ 2 2m ∇2 + V (r) ∫ dx xne−x = n! ∫ a 0 sin2 ax dx = a2 − cos(a2) sin(a2) 2a ψ0(x) = 1√ π1/2x0 e − 1 2 ( x x0 )2 ψ1(x) = 1√ 2π1/2x0 ( 2x x0 ) e − 1 2 ( x x0 )2 ψ2(x) = 1√ 8π1/2x0 ( 4 ( x x0 )2 − 2 ) e − 1 2 ( x x0 )2 ψ3(x) = 1√ 48π1/2x0 ( 8 ( x x0 )3 − 12 x x0 ) e − 1 2 ( x x0 )2 1 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 en- ergy E2 = −1 Ryd/4, and “p” means total angular momentum quantum number ` = 1. (b) What is the degeneracy of the 1st excited state (E = (5/2)h̄ω) of the isotropic 3D simple harmonic oscillator? The 3D simple harmonic oscillator has energies Enx + Eny + Enz , where each En is h̄ω(n+1/2), so Etot = h̄ω(nx +ny +nz +3/2). The first excited state has one quantum of excitation in either x, y, or z, so degeneracy = 3. (c) Sketch the first 3 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 ψ1 and 2nd excited state ψ3 are both even functions of x ⇒ even parity, in other words parity eigenvalue π = +1. 1st excited state ψ2 is odd function ⇒ odd parity, in other words parity eigenvalue π = −1 . 2 (c) Is the wavefunction an eigenstate of parity? Yes or no? Explain either answer. The parity of the hydrogenic wavefunctions is given by (−1)`. 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 Lz = −ih̄ ∂∂φ . The expectation value of Lz is 〈Lz〉 = 1 10 (ψ100 + 3ψ311, Lz(ψ100 + 3ψ311)) = 9h̄/10, therefore 〈 ∂ ∂φ 〉 = (9/10)i. 3. Electron in hydrogenic state. The electron in a hydrogen atom occupies a state ψ = R21(r)   √ 1 3 Y 01 + √ 2 3 Y 11   (2) where R21(r) = √ 1 3 ( 1 2a0 )3/2 ( r a0 ) e−r/2a0 , (3) Y 01 = √ 3 4π cos θ , Y 11 = − √ 3 8π sin θeiφ (4) (a) What value(s) could a measurement of the z-component of the orbital angular momentum, L̂z yield, and what is the probability of each? What is the expectation value of L̂z in this state? m is either 0 or 1, so Lz can be either 0 or h̄, with probability 1/3 or 2/3, respectively. 〈Lz〉 = 1 3 ( ψ210 + √ 2ψ211, Lz(ψ210 + √ 2ψ211) ) = 2h̄/3, 5 (b) Calculate the average distance of the electron from the nucleus in this state. 〈r〉 = 1 3 ( ψ210 + √ 2ψ211, r(ψ210 + √ 2ψ211) ) = 1 3 (ψ210, r ψ210) + 2 3 (ψ211, r ψ211) = ( 1 3 + 2 3 ) ∫ ∞ 0 r2 dr r   √ 1 3 ( 1 2a0 )3/2 ( r a0 ) e−r/2a0   2 = a0 24 ∫ ∞ 0 dy y5 exp−y ︸ ︷︷ ︸ 120 = 5a0 (c) What is the expectation value of L̂x in this state? 〈Lx〉 = 1 3 ( ψ210 + √ 2ψ211, ( L+ + L− 2 ) (ψ210 + √ 2ψ211) ) = h̄ 6 ( ψ210 + √ 2ψ211, √ 2ψ211 + 0 + √ 2ψ21−1 + √ 2 √ 2ψ210 ) = h̄ 6 (2 + 2) = 2h̄/3 (d) If you measured the z-component of the angular momentum and the dis- tance of the electron from the origin r simultaneously, what is the proba- bility density for finding L̂z with eigenvalue zero at a distance r? Pm=0(r) = r2 1 3 R221(r) 6 4. Scattering potential. For a 1D potential as shown in the figure and E < V0, (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 = − h̄2 2m ∇2. ψI = Ae ipx/h̄ + Be−ipx/h̄, ψII = Ce qx/h̄ + De−qx/h̄ ψIII = Fe ipx/h̄ with p = √ 2mE q = √ 2m(V0 − E) (b) Write down the matching conditions at the boundaries x = a, b. Require continuity of ψ and its derivatives at the interfaces: ψI(a) = ψII(a) ψII(b) = ψIII(b) ψ′I(a) = ψ ′ II(a) ψ′II(b) = ψ ′ III(b), in other words Aeipa/h̄ + Be−ipa/h̄ = Ceqa/h̄ + De−qa/h̄ Ceqb/h̄ + De−qb/h̄ = Feipb/h̄ (Aeipa/h̄ −Be−ipa/h̄)ip = (Ceqa/h̄ −De−qa/h̄)q q(Ceqb/h̄ −De−qb/h̄) = Fipeipb/h̄ 7