228
(e) We have
Problems and Solutions in Quantum Mechanics
2 2 2 v⊥ (t) − v t = v⊥ (t) − v⊥ t + v||
Now substitute into the transverse term on the right-hand side the solved expression from (b) for the transverse velocity operator: 2 2 '2 & v⊥ (t) − v⊥ t = v⊥ (0) − v⊥ 0 sin2 ωt + v⊥ (0) − v⊥ 0 × Bˆ cos2 ωt 2 = v⊥ (0) − v⊥ 0 Thus, we obtain
or, equivalently,
(v⊥ − v⊥ t )2 t = (v⊥ − v⊥ 0 )2 0 (v⊥ − v t )2 t = (v⊥ − v 0 )2 0
From the expressions for the transverse and parallel components of the position operator, 1 1 v⊥ (0) sin ωt − v(0) × Bˆ cos ωt ω ω r|| (t) = r|| (0) + v|| (0) t
r⊥ (t) = R +
we can write
2 2 2 r⊥ (t) − r t = r⊥ (t) − r⊥ t + r|| t
The transverse term is 2 r⊥ (t) − r⊥ t
2 ' 1 1 & ˆ v⊥ (0) − v⊥ 0 sin ωt − v⊥ (0) − v⊥ 0 × B cos ωt = R − R0 + ω ω or, remembering that we are considering an operator equation, 2 r⊥ (t) − r⊥ t = (R − R0 )2 1 & + 2 v⊥ (0) − v⊥ 0 sin ωt ω '2 − v⊥ (0) − v⊥ 0 × Bˆ cos ωt & 1 + (R − R0 ) · v⊥ (0) − v⊥ 0 sin ωt ω ' − v⊥ (0) − v⊥ 0 × Bˆ cos ωt 1 & v⊥ (0) − v⊥ 0 sin ωt + ω ' − v⊥ (0) − v⊥ 0 × Bˆ cos ωt · (R − R0 )