Quantum Mechanics I - Homework 3 Solutions | PHY 851, Assignments of Quantum Mechanics

Download Quantum Mechanics I - Homework 3 Solutions | PHY 851 and more Assignments Quantum Mechanics in PDF only on Docsity! SOLUTIONS for Homework #3 1. In the potential of given form there is no unbound states. Bound states have positive energies En labeled by an integer n. For each energy level E, two symmetrically located classical turning points x± = ±x0(E) are the points where the classical momentum for motion with given E, p(x;E) = √ 2m[E − U(x)], (1) vanishes, p(x±;E) = 0 ; U(x±) = E. (2) The approximate quantization rule reads∮ dx p(x;E) = 2πnh̄, (3) where the integral runs over the classical period of motion, or, in our case of an even potential, U(x) = U(−x), four times from x = 0 to the turning point x = x0, 4 √ 2m ∫ x0 0 dx √ E − U(x) = 2πnh̄. (4) This equation determines energy levels En for large n 1, in the validity region of the semiclassical quantization. For our potential U(x) it is convenient to change the coordinate variable introducing x = [(E/α)η]1/s. Then the upper limit x0 → 1, and the quantization condition (4) takes the form 4 s √ 2mE ( E α )1/s Is = 2πnh̄, Is = ∫ 1 0 dη η(1−s)/s √ 1− η. (5) The integral here is a number of the order of 1 which depends on the potential power s. Therefore the energy spectrum is given by En = (Csn)2s/(s+2), (6) where the energy scaling is determined by the constant parameter Cs = πh̄sα1/s 2 √ 2mIs , (7) The integral Is is the Euler integral of the first order, or the Beta-function, and can be expressed via the Gamma-functions, Is = Γ(1/s)Γ(3/2) Γ[(3/2) + (1/s)] . (8) 1 For the harmonic oscillator potential U(x) = (1/2)mω2x2, we have s = 2, α = (1/2)mω2, Γ(1/2) = √ π, Γ(3/2) = 1 2 Γ(1/2), Γ(2) = 1, (9) so that I2 = π 2 ; En = C2n = h̄ωn. (10) The more precise quantization rule would contain (n + 1/2) instead of n in the right hand side of eq. (3); this would lead to the exact result for the harmonic oscillator En = h̄ω(n + 1/2) and to better approximations for other values of s. 2. Let the typical radii for the two electrons be r1 and r2. In the ground state their typical momenta are, according to the uncertainty relation, p1 ∼ h̄/r1 and p2 ∼ h̄/r2. The minimum repulsion energy for the two electrons can be roughly estimated as e2/|r1 − r2|max = e2/(r1 + r2). Then the energy of the ground state can be written as E(r1, r2) = h̄2 2m ( 1 r21 + 1 r22 ) − Ze2 ( 1 r1 + 1 r2 ) + e2 r1 + r2 . (11) Obviously, the electrons are equivalent (they should have opposite spin projections but the same orbital wave functions). Therefore, in the ground state it should be r1 = r2 ≡ r. The energy becomes a function of r, E(r, r) = h̄2 mr2 − 2e2Z − (1/4) r . (12) The minimum of this function is reached at r = aB 1 Z − (1/4) , aB = h̄2 me2 , (13) as if each electron would feel the Coulomb field of the effective charge Zeff = Z − 1 4 . (14) The total two-electron energy (12) for this radius is equal to doubled energy of a single-electron orbit in a hydrogen-like field of the effective charge (13), E = −me 4 h̄2 Z2eff = −2Z2eff Ry; (15) recall that 1 Ry (Rydberg)=me4/2h̄2=13.6 eV. Now we predict binding energies (in Ry) 1.12 (H−), 6.12 (He), 15.12 (Li+), 28.12 (Be++), 45.12 (B+++), and 66.12 (C++++), in agreement with data much better than one would expect for such a simple estimate. 2 Using these rules, we obtain the equations of motion for the mean values: ih̄ d dt 〈x̂〉 = [x̂, Ĥ] = ih̄ 〈p̂〉 m ; d〈x̂〉 dt = 〈p̂〉 m (34) (an analog of the velocity definition v = p/m); d dt 〈x̂2〉 = 1 m 〈(x̂p̂+ p̂x̂); (35) d dt 〈p̂〉 = d dt 〈p̂2〉 = 0. (36) The last result, eq. (36), means that the momentum distribution does not change in free motion, in concordance with physical arguments. The con- servation of 〈p̂2〉 is the same as the conservation of mean energy. Finally, d dt 〈x̂p̂+ p̂x̂〉 = 1 ih̄ 〈[x̂p̂+ p̂x̂, Ĥ] = 2 m 〈p̂2〉. (37) Now we can solve the equations of motion for the expectation values. From eq. (37) we obtain 〈x̂p̂+ p̂x̂〉 = 2 〈p̂ 2〉 m t+ 〈x̂p̂+ p̂x̂〉0, (38) where the last item is determined by the initial conditions. Eq. (35) now gives 〈x̂2〉 = 〈p̂ 2〉 m2 t2 + 〈x̂p̂+ p̂x̂〉0 m t+ 〈x̂2〉0, (39) whereas eq. (34) defines the analog of the uniform motion, 〈x̂〉 = 〈p̂〉 m t+ 〈x̂〉0. (40) Combining those results, we can calculate the uncertainty of the position (∆x)2 = 〈x̂2〉 − 〈x̂〉2 (41) as a function of time: (∆x)2 = (∆x)20 + 1 m [〈x̂p̂+ p̂x̂〉0 − 2〈x̂〉0〈p̂〉0] t+ (∆p)2 m2 t2. (42) After a very long time interval, one will see only “ballistic” spreading, (∆x)2 ≈ (∆p) 2 m2 t2, (43) the packet is broadening because of the spread of velocities ∆v ∼ ∆p/m in the initial state. 5 5. a. The equations of motion for the expectation values of the position and momentum are linear and similar to classical Newton equations: d dt 〈x̂〉 = 〈p̂〉 m , (44) d dt 〈p̂〉 = −mω2〈x̂〉. (45) The general solution describes oscillations with frequency ω, 〈x̂〉 = A cos(ωt) +B sin(ωt), 〈p̂〉 = C cos(ωt) +D sin(ωt). (46) From equations of motion we obtain C = mωB, D = −mωA, (47) and from the initial conditions 〈x̂〉0 = A, 〈p̂〉0 = C. (48) Thus, the solution is 〈x̂〉 = 〈x̂〉0 cos(ωt) + 〈p̂〉0 mω sin(ωt), (49) 〈p̂〉 = 〈p̂〉0 cos(ωt)−mω〈x̂〉0 sin(ωt). (50) b. The equations of motion for quadratic components of the Hamiltonian, K̂ = p̂2 2m , Û = 1 2 mω2x̂2, (51) can be easily derived with the help of the commutators, d dt 〈K̂〉 = −ω 2 2 〈x̂p̂+ p̂x̂〉, (52) d dt 〈Û〉 = ω 2 2 〈x̂p̂+ p̂x̂〉. (53) Of course, energy is conserved, d dt 〈K̂ + Û〉 = d dt 〈Ĥ〉 = 0. (54) For the operator in the right hand side parts of eqs. (53) and (54) we find d dt 〈x̂p̂+ p̂x̂〉 = 4〈K̂ − Û〉. (55) 6 Taking the second time derivative we come to( d2 dt2 + 4ω2 ) 〈K̂ − Û〉 = 0. (56) The general solution corresponds to the oscillation with a double fre- quency, 〈K̂ − Û〉 = A cos(2ωt) +B sin(2ωt). (57) Remembering that 〈Ĥ〉 = 〈K̂ + Û〉 = 〈K̂ + Û〉0, (58) we find separately the expectation values of kinetic and potential energy, 〈K̂〉 = 1 2 [ 〈K̂ + Û〉0 +A cos(2ωt) +B sin(2ωt) ] , (59) 〈Û〉 = 1 2 [ 〈K̂ + Û〉0 −A cos(2ωt)−B sin(2ωt) ] , (60) To find the constant coefficients A and B, we apply the initial conditions: A = 〈K̂ − Û〉0, B = − ω 2 〈x̂p̂+ p̂x̂〉0, (61) where the last equation follows from eqs. (52) and (59). With all these results, 〈x̂2〉 = 1 mω2 { 〈Û〉0[1 + cos(2ωt)] + 〈K̂〉0[1− cos(2ωt)] + ω 2 〈x̂p̂+ p̂x̂〉0 sin(2ωt) } . (62) Similarly, 〈p̂2〉 = m { 〈Û〉0[1− cos(2ωt)] + 〈K̂〉0[1 + cos(2ωt)]− ω 2 〈x̂p̂+ p̂x̂〉0 sin(2ωt) } . (63) c. Collecting our previous calculations we find the mean square deviation of the coordinate (∆x)2 = (∆x)20 cos 2(ωt)+ (∆p)20 m2ω2 sin2(ωt)+ 〈x̂p̂+ p̂x̂〉0 − 2〈x̂〉0〈p̂〉0 2mω sin(2ωt), (64) as in the textbook. For ω → 0 we arrive at the limit of free motion; using sinx/x→ 1 for x→ 0, eq. (64) becomes (∆x)2 = (∆x)20 + (∆p)20 m2 t2 + 〈x̂p̂+ p̂x̂〉0 − 2〈x̂〉0〈p̂〉0 m t. (65) d. With the use of eq. (50) we find the mean square deviation of the momentum (∆p)2 = (∆p)20 cos 2(ωt)+m2ω2(∆x)20 sin 2(ωt)−mω 2 [〈x̂p̂+p̂x̂〉0−2〈x̂〉0〈p̂〉0] sin(2ωt). (66) 7