Homework 3 Solutions - Quantum Mechanics | PHYS 5250, Assignments of Quantum Mechanics

Download Homework 3 Solutions - Quantum Mechanics | PHYS 5250 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics - 1: HW 3 Solutions Leo Radzihovsky Paul Martens October 2, 2007 1 Problem 1 Consider a particle in the ground state of a box of length a. The wave-function of this state is ψ0 (x, 0) = {√ 2 a cos ( π ax ) x ∈ [ −a2 , a2 ] 0 else (1) 1.1 Momentum Distribution To get the momentum distribution we need to decompose the positions states into momentum eigen state φk (x) = e ıkx. We find the coefficient for each term by taking the Fourier transform. ||ψ0 (x)〉 = ∫ |k〉 〈k|ψ0 (x)〉 dk (2) 〈k|ψ1〉 = √ 2 a ∫ a 2 − a2 cos (π a x ) e−ıkx = 1 2 √ 2 a ∫ a 2 − a2 [ eı( π a −k)x + e−ı( π a +k)x ] = 1 2 √ 2 a [ 1 ı ( π a − k ) ( eı( π a −k) a2 − e−ı(πa −k) a2 ) + 1 −ı ( π a + k ) ( e−ı( π a +k) a2 − eı(πa +k) a2 ) ] = √ 2 a [ 1 π a − k cos ( ka 2 ) + 1 π a + k cos ( ka 2 )] 〈k|ψ0〉 = √ 2 a 2π/a ( π a )2 − k2 cos ( ka 2 ) = Ψ̃ (k, 0) (3) The φk ’s are normalized and orthogonal, so the Probability is just |〈k|ψ0〉|2 8π2 a3 cos2 ( ka 2 ) (( π a )2 − k2 )2 (4) 1 1.2 adiabatic expansion Suppose the potential is shut off suddenly then each momentum eigen state will evolve with a factor of e−ı p2 2m~ t. So our time dependent wave-function function will just be the sum of these. Ψ (x, t) = ∫ dk 2π Ψ̃ (k, 0) eıkxe−ı ~ 2k2 2m t/~ Ψ (x, t) = ∫ dk 2π √ 2 a 2π a cos ( ka 2 ) cos (kx) ( π a )2 − k2 e−ı ~k2 2m t (5) In this last part we used parity to argue that inside the integral we can replace eıkx by cos kx and not change the result. 1.3 ground state after adiabatic expansion Consider the case where the length of the box is doubled very quickly. The probability that the we end up in the new ground state is just. |〈Ψ′0|Ψ0〉| 2 (6) Where 〈x|Ψ′〉 = √ 2 a cos (πx 2a ) (7) so, c1 = 〈Ψ′0|Ψ0〉 = 2 × 2 a √ 2 ∫ a 2 0 cos ( π 2a x ) cos (π a x ) = 8 3π (8) therefore P0−>0′ = ( 8 3π )2 (9) 2 Problem 2 2.1 Show that 〈ψ |H|ψ〉 ≥ E0 This important result can be proved by decomposing our general state into energy eigen states of the Hamiltonian. |Ψ〉 = ∑ Cn |En〉 (10) Because this state is normalized ∑ |Cn|2 = 1. Also note that 〈En|En′〉 = δnn′ . The second statement means that we can write the energy of our state 〈Ψ |H |Ψ〉 = ∑ |Cn|2En = E0 + ∑ |Cn|2 n ≥ E0 (11) 2 we get the restriction on k ~k mU0 = cot δk (28) δk is arbitrary, so we still have a continuous energy spectrum E = ~ 2k2 2m . The first derivative of the even wave functions are discontinuous at x = 0 and we have a phase shift δk. 3.3 Positive δ 3.3.1 No bound state If U0 → −U0, then κ < 0, which gives us e|κx| solutions. These don’t make sense, so there are no bound solutions. 3.3.2 new phase shift As before odd states don’t change ψ (odd) k (x) = sin kx (29) Even solutions are shifted away from the origin instead of toward the origin ψ (even) k (x) = { cos (kx+ δk) x > 0 cos (kx− δk) x < 0 (30) where cot δk = − ~ 2k mU0 (31) 4 Problem 4 Consider the potential V (x) = { 0 |x| ≤ a V0 |x| > a (32) 4.1 The Parity Operator Consider the parity operator P . Pψ (x) = ψ (−x) (33) and the Hamiltonian H = − ~ 2 2m d2 dx2 + V (x) (34) 5 The commutator [P,H ]ψ = (PH −HP )ψ = H (−x)ψ (−x) −Hψ (−x) = H (x)ψ(−x) −H (x)ψ(−x), because V (−x) = V (x) [P,H ] = 0 (35) suppose ψE is an energy eigenfunction. Then HψE = EψE (36) PHψE = EPψE HPψE = EPψE using the commutator H (PψE) = E (PψE) (37) There is no degeneracy in the 1-d case so this state must be cψE . P 2 = I (38) so λp = ±1 ⇒ c = ±1 (39) Pψ±E = ±ψ±E (40) where ψ+E is even and ψ − E is odd. 4.2 Energy Spectrum We’ll match the BC at x = a. Either side should give the same result. 4.2.1 Even solutions the general form of the even bound solutions is ψeven (x) =    Aeκx x < −a B cos kx −a < x < a Ae−κx x > a (41) We match both the wavefucntion and it’s first derivative at the boundary. B cos ka = Ae−κa (42) −Bk sin ka = −Aκe−κa (43) these give us the relation k tan ka = κ (44) and A B = eκa cos ka (45) 6 4.2.2 Odd Solution ψodd (x) =    −Ceκx x < −a D sin ka −a < x < a Ce−κx x > a (46) matching wave-function and it’s first derivative. D sin ka = Ce−κa (47) Dk cos ka = −Cκe−κa (48) This leads to the relation k cot ka = −κ (49) note that ~ 2k2 2m = E (50) ~ 2κ2 2m = V0 −E (51) so k2 + κ2 = 2mV0 ~2 (52) 4.3 Graphical solution for energy spectrum To find the energies of our particle in a box we look at the intersection of β = α tanα (53) β2 + α2 = 2mV0a 2 ~2 (54) for even solutions and β = −α cotα (55) β2 + α2 = 2mV0a 2 ~2 (56) for odd solutions Note there is always an even bound state, but if 2mV0a 2 ~2 < π 2 (57) Then there will be no odd bound solutions. On the figure there is only one even solutions for V0 = V01, and three solutions (two even and one odd) for V0 = V02. 4.4 A sketch of low energy solutions The three lowest energy solutions are plotted below. 7 where k = √ 2mE ~ k′ = √ 2m (E − V0) ~ (72) we can relate the coefs. via B.C.’s { cos ka = C sin k′a+D cos k′a −k sin ka = Ck′ cos k′a−Dk′ sin k′a (73) Solve as matrix equation ( sin k′a cos k′a k′ cos k′a −k′ sin k′a ) ( C D ) = ( cos ka −k sin ka ) (74) we find the inverse M−1 = ( sin k′a 1k′ cos k ′a cos k′a − 1k′ sin k′a ) (75) We find C(even) = cos ka sin k′a− k k′ sin ka cos k′a (76) D(even) = cos ka cos k′a+ k k′ sin ka sin k′a (77) cot δ ≡ −D (even) C(even) (78) 4.8.2 odd ψ (x) = { sin kx |x| ≤ a C sin k′x+D cos k′x = A cos (k′x+ δ) |x| > a (79) where once again k = √ 2mE ~ k′ = √ 2m (E − V0) ~ (80) we can relate the coefs. via B.C.’s { sin ka = C sin k′a+D cos k′a k cos ka = Ck′ cos k′a−Dk′ sin k′a (81) Solve as matrix equation ( sin k′a cos k′a k′ cos k′a −k′ sin k′a )( C D ) = ( sin ka −k cos ka ) (82) 10 We find C(odd) = sin ka sin k′a+ k k′ cos ka cos k′a (83) D(odd) = sin ka cos k′a+ k k′a − k k′ cos ka sin k′a (84) tan δ = D(odd) C(odd) (85) 4.9 Reducing to δ First we shift the energy so that Voutside = 0, so that κ = √ 2mE ~2 . In this limit ka→ 0. We treat bound and unbound cases separately. 4.9.1 bound k tan ka→ k2a = κ (86) ( 2mV0 ~2 ) a = κ (87) κ = mU0 ~2 (88) Ebound = − mU20 2~2 (89) For our normalization factors A B = eκa cos ka→ 1 (90) 1 = ∫ |ψ|2 dx (91) = 2A2 ∫ ∞ 0 e−2κxdx = A2 κ (92) A = √ κ (93) 4.9.2 unbound from part h. in the limit ka ∼ 1/√a→ 0. For even D(even) → 1 + k 2a k′ k′a ≈ 1 (94) cot δeven = − D(even) C(even) → ~ 2k′ mU0 (95) (96) 11 For odd C(odd) → ∞ (97) D(odd) → 0 (98) C(odd) D(odd) = cot δ0 → ∞ ⇒ δ = 0 (99) This limit reduces to the δ case as expected. 5 Problem 5 Compute j = − ı~ 2m (ψ∗∇ψ − ψ∇ψ∗) (100) for ψ = Aeıpx/~ +Be−ıpx/~ (101) ∇ψ = ıp ~ ( Ae−ıpx/~ −Beıpx/~ ) (102) so j = − ı~ 2m ıp ~ [( A∗e−ıpx/~ +B∗eıpx/~ ) ( Aeıpx/~ −Be−ıpx/~ ) + C.C. ] = p 2m [ |A|2 − |B|2 +AB∗e2ıpx/~ −A∗Be−2ıpx/~ + C.C ] (103) j = p m [ |A|2 − |B|2 ] (104) 6 Problem 6 6.1 Wave packet Oscillations consider the wave-function ψ (x, 0) = ( 1 πx20 ) 1 4 e − (x−a) 2 2x20 (105) 6.1.1 time evolution |ψ (t)〉 = ∞∑ n=0 Cn |En〉 e− ı ~ Ent (106) where Cn = 〈En|ψ (0)〉 (107) 12 The solution is ψ~n = 1 π1/2x0 Hn1 ( x x0 ) Hn2 ( y x0 ) √ 2n1n1! √ 2n2n2! e −x 2+y2 2x20 (135) E~n = ~ω0 (n1 + n2 + 1) (136) note that there are degenerate energy values. For the n-th, energy levels there are n+ 1 ways to add up n1 and n2 to get n. n = 0 + n n = 1 + (n− 1) ... n = n+ 0 g2d (n) = n+ 1. (137) 6.2.2 3-d In this case H = Hx +Hy +Hz ⇒ ψE (x, y, z) = ψE1 (x)ψE2 (y)ψE3 (z) (138) similar to the 2-d case. Our solution to the 3-d case is ψ~n (~r) = Nn1Nn2Nn3Hn1 ( x x0 ) Hn2 ( y x0 ) Hn3 ( z x0 ) e −x 2+y2+z2 2x2 0 (139) E~n = ~ω0 ( n1 + n2 + n3 + 3 2 ) (140) g3d (n) = n∑ n2=0 n−n2∑ n3=0 (141) = n∑ n2=0 (n− n2 + 1) = n (n+ 1) + n+ 1 − n∑ n2=0 n2 ︸ ︷︷ ︸ 1 2 n(n+1) g3d (n) = 1 2 (n+ 1) (n+ 2) (142) 15 6.3 particle in ~B-field 6.3.1 from H to Schroed. For a free particle in a magnetic field. H = (p− qA)2 2m (143) note that if we define ~p′ ≡ ~p− e ~Ac this looks like a free particle. Given that B = B0ẑ we have some choice in our quantity for A. I will choose A so that it points in the ŷ direction. ~A = B0xŷ (144) we solve the eigenvalue problem. ~ 2 2m [ − ∂ 2 ∂x2 + ( −ı ∂ ∂y − qB ~ x )2 − ∂ 2 ∂z2 ] ψE = EψE (145) this is separable ψE (x, y, z) = ψEx (x)ψEy (y)ψEz (z) (146) Note there is no explicit z or y dependence. So the y and z parts are just plane wave solutions. − ~ 2 2m ψ′′Ex (x) + ( x `2 − ky )2 ψEx (x) = ExψEx (x) (147) The total energy of this solution E = Ex + ~ 2 2m ( k2z ) (148) define xk ≡ `2ky `2 ≡ h 2πgB (149) then − ~ 2 2m ψ′′Ex (x) + ~ 2 2n`4 (x− xk)2 ψEx (x) = ExψEx (x) (150) The solution to this equation ψn,ky ,kz (x, y, z) = Nne ıkyy+ıkzzHn ( x− xk ` ) e− 1 2 (x−xk) 2 `2 (151) En,ky ,kz = ~ω ( n+ 1 2 ) + ~ 2 2m k2z (152) ~ωc ≡ ~ 2 m`2 = ~ 2qB m~ (153) ωc = qB m (154) 16