Homework 1 with Solutions - Introduction to Quantum Mechanics 1 | PHYS 5250, Assignments of Quantum Mechanics

Download Homework 1 with Solutions - Introduction to Quantum Mechanics 1 | PHYS 5250 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics - 1: HW 1 Solutions Leo Radzihovsky August 25, 2007 1 Problem 1 Harmonic oscillator dynamics in 2-d via Lagrangian and Hamiltonian formalism in Cartesian and Polar Coordinates. 1.1 Lagrangian L = 1 2 m|~̇r|2 − 1 2 mω2r2 (1) The equation of motion can be found by using the Euler-Lagrange equations d dt ∂L ∂q̇i − ∂L ∂qi (2) 1.1.1 Cartesian I set ~q = {x, y}. So q1 = x and q2 = y. In this coordinate system the Lagrangian is L = 1 2 m ( ẋ2 + ẏ2 ) − 1 2 mω2 ( x2 + y2 ) (3) Using the Euler-Lagrange equations (2) mẍ+mω2x = 0 mÿ +mω2y = 0 (4) These two equations can be expressed in a compact form by invoking vector notation ~̈r = −ω2~r (5) 1.1.2 Polar I set ~q = {r, θ}. It is important to remember that the unit vectors in this coordinate system can change with time. d dt r̂ = θ̇θ̂ d dt θ̂ = −θ̇r̂ (6) 1 These should remind you of sine and cosine for obvious reasons. We will need to know ṙ2. first we’ll need to know ddt~r d dt ~r = ṙr̂ + r ˙̂r = ṙr̂ + rθ̇θ̂ (7) Dotting the above equation with itself∣∣∣∣ ddt~r ∣∣∣∣ = ṙ2 + r2θ̇2 (8) All together the Lagrangian is L = 1 2 m ( ṙ2 + r2θ̇2 ) − 1 2 mω2r2 (9) Using the Euler-Lagrange equations (2) like before mr̈ −mrθ̇2 +mω2r = 0 mr2θ̈ + 2mrṙθ̇ = 0 (10) 1.2 Hamiltonian The Hamiltonian of a system is defined via the Lagrangian, coordinates, and conjugate momentum. H = ∑ piq̇i − L (11) where pi = ∂L ∂q̇i (12) The Hamilton equations of motions are q̇i = ∂H ∂pi ṗi = − ∂H ∂qi (13) 1.2.1 Cartesian First, I find the conjugate momenta. px = ∂L ∂ẋ = mẋ py = ∂L ∂ẏ = mẏ (14) Now I can write H in terms of p’s and q’s H = p2x 2m + p2y 2m + 1 2 mω2 ( x2 + y2 ) (15) 2 3 Problem 3 There is more than one way to do this problem. One approach is to gen- eralize what I did in class for a particle’s coordinate q(t) using the defini- tion of functional derivative δq(t′)/δq(t) = δ(t − t′) to a field ψ(~r, t), with δψ(~r′, t′)/δψ(~r, t) = δ(3)(~r − ~r)δ(t− t′), and similarly for ψ(~r, t). Another approach is to simply compute the differential of S, remembering that the boundary terms vanish by definition that fields do not vary at the boundary. S = ∫ ( −ıh̄Ψ∗∂tΨ + h̄2 2m ∇Ψ∗ · ∇Ψ + VΨ∗Ψ ) d3r dt (34) δΨ∗S = ∫ −ıh̄∂tΨ − h̄22m∇2Ψ︸ ︷︷ ︸ int. by parts +VΨ  δΨ∗d3r dt = 0 (35) δΨS = ∫ ıh̄∂tΨ∗ − h̄22m∇2Ψ∗︸ ︷︷ ︸ int. by parts +VΨ∗  δΨd3r dt = 0 (36) The above conditions are met if each integrand is = 0. This leads us to Schrödringer’s equation, and it’s conjugate − h̄ 2 2m∇ 2Ψ + VΨ = ıh̄∂tΨ (37) − h̄ 2 2m∇ 2Ψ∗ + VΨ∗ = −ıh̄∂tΨ∗ (38) 4 Problem 4 Bohr-Sommerfeld quantization states∮ orbit ~p · d~q = hn (39) If the electron is just moving in a circular orbit, then it has momentum. ~p = meωrφ̂ (40) I use our orbit, which is circular, for my contour, so d~q = rdφφ̂. After taking the integral I’m left with 2πmeωr2 = nh (41) `φ = nh̄ (42) 5 The Hamiltonian is H = `2φ 2mer2 − Ze 2 r (43) The election will orbit so that it’s energy is minimized ∂E ∂r = − n 2h̄2 mer3n + Ze2 r2n ⇒ rn = n2h̄2 mee2Z (44) Using the above relation we can write the Energy without rn dependence. En = − Z2e4me 2h̄2n2 = −1 2 mc2Z2α2 1 n (45) rn = h̄2n2 me2Z = λen 2 2πZα (46) where α = e2 h̄c ≈ 1 137 (you should memorize this number!) (47) λe = hc mc2 (48) 5 Problem 5 5.1 harmonic oscillator En = p2n 2m + m 2 ω2x2n (49) Using our rough quantization rule. pn = nh̄ xn ⇒ En = n2h̄2 2mx2n + m 2 ω2x2n (50) I find xn by minimizing the energy. −n 2h̄2 mx3n +mω2xn = 0 ⇒ xn = √ h̄n mω = x0n 1 2 , where x0 = √ h̄ mω (51) In terms of dimensional constants and n our energy is En = n2h̄2mω 2mh̄n + m 2 ω2 h̄n mω En = h̄ωn (52) 6 5.2 quartic oscillator En = p2n 2m + a 4 x4 (53) En = n2h̄2 2mx2n + a 4 x4n (54) Finding minimum energy −n 2h̄2 mx3n + ax3n = 0 ⇒ xn = ( n2h̄2 ma ) 1 6 (55) Writing energy in terms of n and constants En = n2h̄2 2m ( n2h̄2 ma ) 1 3 + a 4 ( n2h̄2 ma ) 2 3 (56) En = 3a4 ( h̄2n2 ma ) 2 3 = E1n 4 3 (57) 5.3 H-atom with Coulomb potential En = p2n 2m − Ze 2 rn (58) En = n2h̄2 2mr2n − Ze 2 rn (59) Finding minimum energy −n 2h̄2 mr3n + Ze2 r2n = 0 ⇒ rn = h̄ 2n2 me2Z (60) Writing energy in terms of n and constants, and simplifying En = −Z 2e4m 2h̄2n2 (61) 5.4 e− in the potential 1 s V0 ( x x1 )s En = n2h̄2 2mx2n + 1 s V0 ( xn x1 )s (62) Finding the minimum energy − n 2h̄2 mex3n + V0 x ( xn x1 )s−1 ⇒ xn = x1 ( n2h̄2 mex21V0 ) 1 s+2 (63) 7