SOLUTIONS for Homework #3

Transcription

1 SOLUTIONS for Hoework #3 1. In the potential of given for there is no unboun states. Boun states have positive energies E n labele by an integer n. For each energy level E, two syetrically locate classical turning points x ± = ±x 0 (E) are the points where the classical oentu for otion with given E, vanishes, p(x; E) = [E U(x)], (1) p(x ± ; E) = 0 U(x ± ) = E. () The approxiate quantization rule reas x p(x; E) = πn h, (3) where the integral runs over the classical perio of otion, or, in our case of an even potential, U(x) = U( x), four ties fro x = 0 to the turning point x = x 0, 4 x0 x E U(x) = πn h. (4) 0 This equation eterines energy levels E n for large n 1, in the valiity region of the seiclassical quantization. For our potential U(x) it is convenient to change the coorinate variable introucing x = [(E/α)η] 1/s. Then the upper liit x 0 1, an the quantization conition (4) takes the for ( ) 4 1/s E 1 E I s = πn h, I s = η η (1 s)/s 1 η. (5) s α 0 The integral here is a nuber of the orer of 1 which epens on the potential power s. Therefore the energy spectru is given by E n = (C s n) s/(s+), (6) where the energy scaling is eterine by the constant paraeter C s = π hsα1/s I s, (7) The integral I s is the Euler integral of the first orer, or the Beta-function, an can be expresse via the Gaa-functions, I s = Γ(1/s)Γ(3/) Γ[(3/) + (1/s)]. (8) 1

2 For the haronic oscillator potential U(x) = (1/)ω x, we have s =, α = (1/)ω, Γ(1/) = π, Γ(3/) = 1 Γ(1/), Γ() = 1, (9) so that I = π E n = C n = hωn. (10) The ore precise quantization rule woul contain (n + 1/) instea of n in the right han sie of eq. (3); this woul lea to the exact result for the haronic oscillator E n = hω(n + 1/) an to better approxiations for other values of s.. Let the typical raii for the two electrons be r 1 an r. In the groun state their typical oenta are, accoring to the uncertainty relation, p 1 h/r 1 an p h/r. The iniu repulsion energy for the two electrons can be roughly estiate as e / r 1 r ax = e /(r 1 + r ). Then the energy of the groun state can be written as E(r 1, r ) = h ( 1 r r ) Ze ( 1r1 + 1r ) + e r 1 + r. (11) Obviously, the electrons are equivalent (they shoul have opposite spin projections but the sae orbital wave functions). Therefore, in the groun state it shoul be r 1 = r r. The energy becoes a function of r, E(r, r) = h Z (1/4) e. (1) r r The iniu of this function is reache at r = a B 1 Z (1/4), a B = h e, (13) as if each electron woul feel the Coulob fiel of the effective charge Z eff = Z 1 4. (14) The total two-electron energy (1) for this raius is equal to ouble energy of a single-electron orbit in a hyrogen-like fiel of the effective charge (13), E = e4 h Z eff = Z eff Ry; (15) recall that 1 Ry (Ryberg)=e 4 / h =13.6 ev. Now we preict bining energies (in Ry) 1.1 (H ), 6.1 (He), 15.1 (Li + ), 8.1 (Be ++ ), 45.1 (B +++ ), an 66.1 (C ++++ ), in agreeent with ata uch better than one woul expect for such a siple estiate.

3 3. a. Using the Schröinger equations for two wave functions with the sae Hailtonian Ĥ, i h Ψ 1 = Ĥ Ψ 1, i h Ψ an taking the ifference of these equations, we obtain = ĤΨ, (16) i h (Ψ 1Ψ ) = Ψ 1(ĤΨ ) (Ĥ Ψ 1)Ψ. (17) In the coorinate representation the potential ters in the Hailtonian Ĥ = ˆK + U cancel if the potential U(r) is real, U = U. The reaining kinetic ter ˆK = ˆp = h = ˆK (18) is real as well. Introucing the transition ensity an the transition current j 1 = we coe to the continuity equation ρ 1 Ψ 1Ψ (19) h i [Ψ 1 Ψ ( Ψ 1)Ψ ], (0) ρ 1 + iv j 1 = 0. (1) The stanar equation correspons to the iagonal case, Ψ 1 = Ψ. b. Two stationary wave functions Ψ 1 an Ψ escribe the states with certain energies E 1 an E, respectively. Their tie epenence is given by Ψ 1, (r, t) = ψ 1, (r)e (i/ h)e1.t. () The coorinate aplitues ψ 1, are the eigenfunctions of the sae Hailtonian, Ĥψ 1, = E 1, ψ 1,. (3) The continuity equation of point a can be written as i h ρ 1 = (E E 1)ψ 1ψ e (i/ h)(e E 1 )t. (4) Our first assuption shoul be that the energy values E 1 an E are real. Then eq. (4) eans that the transition ensity ρ 1 oscillates in tie with the transition frequency ω 1 = (E E 1 )/ h; the expectation value of the ensity ρ 11 is siply constant in tie for a stationary state Ψ 1. Now let 3

4 us integrate both parts of eq. (4) over the entire available volue V. The left han sie, accoring to the continuity equation, reuces to i h 3 r ρ 1 = i h 3 r ivj 1. (5) The volue integral in eq. (5) can be converte into the surface integral A j1, the flux of the transition current through the surface area A. Now we ake the secon assuption that this flux vanishes. This happens in particular if the wave functions ψ 1 an ψ, along with their graients, fall off at the reote bounaries of the volue sufficiently fast. If this is the case, eqs. (5) an (4) lea to the conclusion that (E 1 E ) 3 r ψ1ψ = 0. (6) If the energies E 1 an E o not coincie, the corresponing coorinate eigenfunctions are orthogonal, 3 r ψ 1ψ = 0. (7) For coinciing energies we only extract that the integral of ρ 1 oes not change in tie, 3 r ψ 1ψ = const. (8) If there is no egeneracy so that there exists only one function ψ corresponing to given energy, its noralization is tie-inepenent, 3 r ψ = const. (9) 4. The Ehrenfest equations of otion for the expectation value of a tie inepenent operator Ô in the syste with hailtonian Ĥ are For a free particle in one iension We nee the coutators ı h Ô = [Ô, Ĥ]. (30) t Ĥ = ˆp (ˆp = ˆp x, [ˆx, ˆp] = i h). (31) [ˆx, ˆp ] = i hˆp, (3) [ˆx, ˆp ] = ˆx[ˆx, ˆp ] + [ˆx, ˆp ]ˆx = i h(ˆxˆp + ˆpˆx). (33) 4

5 Using these rules, we obtain the equations of otion for the ean values: i h ˆx = [ˆx, Ĥ] = i h ˆp t (an analog of the velocity efinition v = p/); ˆx t = ˆp (34) t ˆx = 1 (ˆxˆp + ˆpˆx); (35) t ˆp = t ˆp = 0. (36) The last result, eq. (36), eans that the oentu istribution oes not change in free otion, in concorance with physical arguents. The conservation of ˆp is the sae as the conservation of ean energy. Finally, 1 ˆxˆp + ˆpˆx = t i h [ˆxˆp + ˆpˆx, Ĥ] = ˆp. (37) Now we can solve the equations of otion for the expectation values. Fro eq. (37) we obtain ˆxˆp + ˆpˆx = ˆp t + ˆxˆp + ˆpˆx 0, (38) where the last ite is eterine by the initial conitions. Eq. (35) now gives ˆx = ˆp t + ˆxˆp + ˆpˆx 0 t + ˆx 0, (39) whereas eq. (34) efines the analog of the unifor otion, ˆx = ˆp t + ˆx 0. (40) Cobining those results, we can calculate the uncertainty of the position as a function of tie: ( x) = ˆx ˆx (41) ( x) = ( x) [ ˆxˆp + ˆpˆx 0 ˆx 0 ˆp 0 ] t + ( p) t. (4) After a very long tie interval, one will see only ballistic spreaing, ( x) ( p) t, (43) the packet is broaening because of the sprea of velocities v p/ in the initial state. 5

6 5. a. The equations of otion for the expectation values of the position an oentu are linear an siilar to classical Newton equations: ˆp ˆx = t, (44) t ˆp = ω ˆx. (45) The general solution escribes oscillations with frequency ω, ˆx = A cos(ωt) + B sin(ωt), ˆp = C cos(ωt) + D sin(ωt). (46) Fro equations of otion we obtain an fro the initial conitions Thus, the solution is C = ωb, D = ωa, (47) ˆx 0 = A, ˆp 0 = C. (48) ˆx = ˆx 0 cos(ωt) + ˆp 0 sin(ωt), (49) ω ˆp = ˆp 0 cos(ωt) ω ˆx 0 sin(ωt). (50) b. The equations of otion for quaratic coponents of the Hailtonian, ˆK = ˆp, Û = 1 ωˆx, (51) can be easily erive with the help of the coutators, t ˆK = ω ˆxˆp + ˆpˆx, (5) ω Û = ˆxˆp + ˆpˆx. (53) t Of course, energy is conserve, t ˆK + Û = Ĥ = 0. (54) t For the operator in the right han sie parts of eqs. (53) an (54) we fin t ˆxˆp + ˆpˆx = 4 ˆK Û. (55) 6

7 Taking the secon tie erivative we coe to ( ) t + 4ω ˆK Û = 0. (56) The general solution correspons to the oscillation with a ouble frequency, ˆK Û = A cos(ωt) + B sin(ωt). (57) Reebering that Ĥ = ˆK + Û = ˆK + Û 0, (58) we fin separately the expectation values of kinetic an potential energy, ˆK = 1 [ ˆK ] + Û 0 + A cos(ωt) + B sin(ωt), (59) Û = 1 [ ˆK ] + Û 0 A cos(ωt) B sin(ωt), (60) To fin the constant coefficients A an B, we apply the initial conitions: A = ˆK Û 0, B = ω ˆxˆp + ˆpˆx 0, (61) where the last equation follows fro eqs. (5) an (59). With all these results, ˆx = 1 { ω Û 0[1 + cos(ωt)] + ˆK 0 [1 cos(ωt)] + ω } ˆxˆp + ˆpˆx 0 sin(ωt). (6) Siilarly, { ˆp = Û 0[1 cos(ωt)] + ˆK 0 [1 + cos(ωt)] ω } ˆxˆp + ˆpˆx 0 sin(ωt). (63) c. Collecting our previous calculations we fin the ean square eviation of the coorinate ( x) = ( x) 0 cos (ωt)+ ( p) 0 ω sin (ωt)+ ˆxˆp + ˆpˆx 0 ˆx 0 ˆp 0 sin(ωt), ω (64) as in the textbook. For ω 0 we arrive at the liit of free otion; using sin x/x 1 for x 0, eq. (64) becoes ( x) = ( x) 0 + ( p) 0 t + ˆxˆp + ˆpˆx 0 ˆx 0 ˆp 0 t. (65). With the use of eq. (50) we fin the ean square eviation of the oentu ( p) = ( p) 0 cos (ωt)+ ω ( x) 0 sin (ωt) ω [ ˆxˆp+ˆpˆx 0 ˆx 0 ˆp 0 ] sin(ωt). (66) 7