Quantum Mechanics I Problem set No.1


 Elinor Knight
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1 Quantum Mechancs I Problem set No.1 Septembe0, The Least Acton Prncple The acton reads S = d t L(q, q) (1) accordng to the least (extremal) acton prncple, the varaton of acton s zero 0 = δs = t f t d t ( δq δq + δ q δ q ) = t f t ( d t δq d ) + ( δq ) t f δq dt δ q δ q t (2) snce we have q (t) s arbtrary functon. In the last term, snce we have fxed the ntal and fnal poston, δq (n) = δq ( f n) = 0. So δs = 0 requres, f 2 Posson Brackets 2.1 d = 0. (3) δq dt δ q take dervatve wth respect to t, we have, d A d t = A t + ( A ṗ + A ) q p q (4) Posson bracket of A and H ( A {A, H} = A ) q p p q (5) substtute equaton of moton nto (5), one can prove d A d t = A + {A, H} (6) t 1
2 2.2 Posson bracket between p and q j. Be aware of that one has to use a dfferent summed up ndex to get a rght answer. ( q p j {q, p } = p j p ) = δ k δ jk = δ j (7) q k p k p k q k we can show {q, q j } = {p, p j } = 0 as well k k 2.3 we can show ( q A {q, A} = q A) A = (8) q j p j p j q j p j {p, A} = A q (9) 2.4 The component of angular momentum reads, L = ɛ jk x j p k (Ensten notaton: sum over repeated ndexes x y = x y ). Accordng to (8) accordng to (9), we can show that {x, L j } = L j p = ɛ jklx k p l p = ɛ jk x k (10) {p, L j } = ɛ jk p k (11) 3 Charged Partcle n an Electromagnetc Feld 3.1 canoncal momentum p = δẋ = mẋ + q c A (12) 3.2 e.o.m of the charged partcle n EM feld follows EulerLagrangan eqn, δ x = q φ + q (ẋ A) }{{} c E d d t δẋ = mẍ + q c Ȧ }{{} q c ( A +(ẋ ) A) t (13) 2
3 accordng to ẋ A = (ẋ ) A (ẋ A), one has 3.3 mẍ = q E + q c v B (14) accordng to (13), we can read out Hamltonan H(x, p) = δd ẋ L(x, ẋ) = 1 f x 2m (p q c A)2 + qφ (15) 3.4 we can prove the new coordnate and momentum are canoncal by show they obey Posson bracket, and {Q 1, Q 2 } = 2 =1 ( Q1 x Q 2 p Q 1 p Q 2 x ) = 0 (16) {P 1, P 2 } = 0, {P, Q j } = δ j (17) the Hamltonan could be wrtten as H = 1 ( (p x + q By 2m c 2 )2 + (p y q ) Bx c 2 )2 = P2 1 2m + q2 B 2 2mc 2 Q2 1 (18) a 1d moton n the new coordnates. 3.5 we can compare the Hamltonan n the new coordnate (18) wth the standard Hamltonan of Harmonc oscllator. H = P2 2m mω2 Q 2 ω = qb (19) mc here we can see that the ω s cyclotron frequency. We have the cyclc coordnate s Q 2, and the assocated constant of moton s P 2. Snce the Hamltonan s ndependent of P 2. The partcle s movng n a crcle n homogeneous magnetc feld, whch s a 1d moton could always be expressed by a sngle momentum. 4 Symmetres and Conservaton Laws 4.1 we have r 1 = R + m 2 M r, = R m 1 M r (20) L = 1 2 MṘ2 + 1 µṙ V( r ) (21) 2 3
4 4.2 n sphercal coordnates, d r = ê r d r + ˆr θ r d θ + ê ϕ r sn θ d ϕ ṙ 2 = ṙ 2 + θ 2 + sn 2 θ ϕ 2 (22) we have Lagrangan n sphercal coordnates, L = 1 2 µ(ṙ2 + θ 2 + sn 2 θ ϕ 2 ) MṘ2 V(r) (23) the angular momentum s conserved, means the moton s always n the plane whch s perpendcular to the angular momentum, let s choose θ = π/2, such that θ = 0, so the moton stays n x y plane, the Lagrangan can be smplfed as L = 1 2 µ(ṙ2 + ϕ 2 ) MṘ2 V(r) (24) canoncal momentum and Hamltonan p r = L ṙ = µṙ, L = µr2 ϕ, p R = MṘ (25) H = p2 R 2M + 1 2µ (p2 r + L2 ) + V(r) (26) r2 4.3 cyclc coordnate s ϕ, and the assocated conserved momentum s angular momentum, L, ( L = 0). Says, n central potental, (V(r) s sotropc), angular momentum s conserved. 4.4 symmetry generator L (p, x) = ɛ jk x j p k. The assocated nfntesmal transformaton of coordnates and momentum n terms of Posson bracket δp l = θ {L, p l } = θ L x l = θ ɛ lk p k, for an observable u(x, p) under rotaton, δx l = θ {L, x l } = θ L p l = θ ɛ jl x j (27) δu = u δx l + u δp l = θ {u, L} = θ ɛ jk (x j + p j )u = u(r j x, R j p ) u(x, p) (28) x l p l x k p k and proved that L s generator of rotaton. When the observable u = H, we have [H, L] = 0, and L s conserved. 4.5 {p r, H} = r = d V e f f (r) d r one has p r s not conserved, caused by effectve central force V e f f = L2 + V(r). (29) 4
5 4.6 rewrte the RungeLenz vector, noted that A (B C) = B(A B) C(A B) M = 1 µ p L k r r = 1 µ [x p2 p(x p)] k r r (30) we can prove all component of M are constant of moton, and wrte the Hamltonan n terms of H = 1 2µ p2 V(r) frst, evaluate some dervatves, {M, H} = { 1 µ [x p 2 p (x p)] k r x, H} = j M x j p j M p j x j (31) M = p2 [ δj p p j ] k( δj x x j ) x j µ p 2 r M = 1 ( ) 2x p j δ j x p x j p p j µ = V(r) x j x j = p j p j µ = k x j r 3 and we have x x =, (δ j p p j )p p 2 j = 0, (δ j x x j )p j = (2 x p j RungeLenzvector s a constant of moton when potental 1/r. x jp (32) x p δ j )x j we can prove that 5
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