Solutions Of Final Exam - Quantum Physics II | PHYS 487, Exams of Quantum Physics

Download Solutions Of Final Exam - Quantum Physics II | PHYS 487 and more Exams Quantum Physics in PDF only on Docsity! Physics 487 Final Exam December 16, 2005 Solution This is a 3-hour exam (8:00 - 11:00). It is closed book (open notes). Please do all 7 problems. Note that they have different point values. Math that may be useful: xe ax dx! = e ax a 2 ax "1( ) x 2 e ax dx! = x 2 e ax a " 2e ax a 3 ax "1( ) P 0 cos!( ) = 1 P 1 cos!( ) = cos! P 2 cos!( ) = 1 2 3cos 2 ! "1( ) 1. [5 points] Suppose that, in a variational calculation of the ground state energy, the trial wave function ! "( ) happens to be the true ground state wave function for some value, α0, of the parameter α. Show that in this case your estimate of the ground state energy will be the true ground state energy. All wave functions can be written as the superposition of energy eigenstates: ! = c i i " ! i . Therefore, the expectation value of the energy, E = c i 2 E i i ! " E0 . E equals E0 only when every ci is zero except c0. This happens when α = α0. Because this is the smallest possible E , this is the answer that the variational calculation will obtain. The answer is unique, because energy eigenstates are linearly independent, That is, it cannot happen for any other α , unless ψ happens to be the same function for two values of α. 2. [15 points] Use the trial wave function, ! x( ) = Ae"# x , to estimate the ground state energy of the 1-D harmonic oscillator, Ĥ = p̂ 2 2m + 1 2 m! 2 x̂2 . Hint: Don’t forget about the kink at x = 0. Normalization requires that 2 A 2 e !2" x dx 0 # $ = 1 2A 2 2" = 1 A = " We must do two integrals. (The green terms result from the kink at x = 0.) 1 2m p̂ 2 = ! ! 2 2m " d 2" dx 2 dx !# # $ = ! ! 2% 2 2m 2A 2 e !2% x dx 0 # $ + ! 2 2m 2%A2 = ! ! 2% 2 2m + ! 2% 2 m = ! 2% 2 2m 1 2 m! 2 x2 = 1 2 m! 2 2A2 x2e"2# xdx 0 $ % = m! 2# 2 8# 3 = m! 2 4# 2 Minimize ! 2 ! 2 2m + m" 2 4! 2 with respect to α: ! 2 = m" 2! , so E 0 = 1 2 !! , a bit higher than the actual ground state energy. 5. [10 points] Use the WKB approximation to calculate the transmission probability of a particle (mass m and energy E) through a square barrier of height V > E and width L. Why is this not the exact solution? That is, what approximation was made? T ! e"2# where ! " 1 ! 2m V # E( )dx = 0 L $ L 2m V # E( ) ! . We are ignoring reflection from the far end (the growing exponential solution). This only holds if ! ! 1 . 6. [15 points] Consider an infinitely deep 2-D square well of side L. a. What is the energy of the first excited state (not the ground state)? What is the degeneracy (i.e., how many states have this energy)? Write the wave functions. b. A time independent perturbation is applied, V '(x, y) = A! x " 3L 4( )! y " 3L 4( ) . This is a bump on the diagonal (the dot in the figure). Calculate the first order perturbation theory correction to the first excited state wave functions and energies. a. The first excited state energy is E = 5E g = 5! 2 2mL 2 . There are two states: 1 = 2 L sin 2! x L " #$ % &' sin ! y L " #$ % &' and 2 = 2 L sin ! x L " #$ % &' sin 2! y L " #$ % &' . b. There is degeneracy, so we need to diagonalize the 2×2 V’ matrix: V 11 ' = 4A L 2 ! x " 3L 4 ( )! y " 3L 4 ( )sin2 2# x L $ %& ' () sin 2 # y L $ %& ' () dx 0 L * dy 0 L * = 4A L 2 sin 2 3# 2 $ %& ' () sin 2 3# 4 $ %& ' () = 2A L 2 V 12 ' = 4A L 2 ! x " 3L 4 ( )! y " 3L 4 ( )sin 2# x L $ %& ' () sin # y L $ %& ' () sin # x L $ %& ' () sin 2# y L $ %& ' () dx 0 L * dy 0 L * = 2A L 2 V 21 ' = 4A L 2 ! x " 3L 4 ( )! y " 3L 4 ( )sin # x L $ %& ' () sin 2# y L $ %& ' () sin 2# x L $ %& ' () sin # y L $ %& ' () dx* dy* = 2A L 2 V 22 ' = 4A L 2 ! x " 3L 4 ( )! y " 3L 4 ( )sin2 # x L $ %& ' () sin 2 2# y L $ %& ' () dx 0 L * dy 0 L * = 2A L 2 All four matrix elements are the same, so we must diagonalize 1 1 1 1 ! "# $ %& . Its eigenvalues and eigenvectors are: λ = 2: even = 1 2 1 1 ! "# $ %& , and λ = 0: odd = 1 2 1 !1 " #$ % &' . The odd superposition doesn’t have an energy shift, because it has a node at the bump. The even superposition has an energy shift, !E even = 4A L 2 . x y L 0 L V = 0 V = ∞ 7. [15 points] A particle of mass m and kinetic energy E, scatters from a spherically symmetric potential. Assume that we know the l = 0 and l = 1 phase shifts: δ0 = δ1 ≡ δ = 5°. All other phase shifts are zero. a. Calculate the total scattering cross section. The total cross section is just the sum of the partial wave cross sections: ! tot = 4" k 2 2l +1( )sin2 # l l $ = 4" k 2 1sin 2 5° + 3sin 2 5°( ) = 4.38 k 2 b. Calculate the ratio of the differential scattering cross section at θ = 90° to the scattering at θ = 0°. That is, d! d" 90 0 d! d" 0 0 . For differential cross sections, the partial waves might interfere (add amplitudes). Each partial wave amplitude is: fl !( ) = 2l +1( )alPl cos!( ) = 1 k 2l +1( )ei"l sin "l( )Pl cos!( ) At θ = 0, we have (every Legendre polynomial = 1 at θ = 0): f 0( ) = 4 k ei! sin !( ) . Similarly, at θ = 90°, we have (P1 = 0 at θ = 90°): f 90°( ) = 1 k ei! sin !( ) . The ratio of differential cross sections is the ratio of the f 2 , namely 1/16.