hw2_sol - Physics 240B, Spring 2008 Homework 2 Solutions...

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Physics 240B, Spring 2008 Homework 2 Solutions Jesse Nofsinger April 6, 2008 Solutions are in general not the original work of the author. Problem 1) The Hamiltonian is given by H = 1 2 m b p p x + eH 2 c y P 2 + p p y + eH 2 c x P 2 + p 2 z B (1) In the Heisenberg picture, the time evolution of the operators is given by: ˙ x = i ¯ h [ H, x ] = 1 m p p x + eH 2 c y P (2) ˙ y = i ¯ h [ H, y ] = 1 m p p y eH 2 c x P (3) ˙ z = i ¯ h [ H, z ] = p z m (4) ˙ p x = i ¯ h [ H, p x ] = eH 2 mc p p y eH 2 c x P (5) ˙ p y = i ¯ h [ H, p y ] = eH 2 mc p p x + eH 2 c y P (6) ˙ p z = i ¯ h [ H, p z ] = 0 (7) Introduce the cyclotron frequency ω c = | e | H mc ,then, ¨ x = ω c p y 2 m + ω 2 c 4 x ω c 2 ˙ y = ω c ˙ y (8) ¨ y = ω c p x 2 m + ω 2 c 4 y + ω c 2 ˙ x = ω c ˙ x (9) ¨ z = 0 (10) The solutions of these equations are: 1
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x = r 0 cos ( t + φ 0 ) (11) y = r 0 sin ( ω c t + φ 0 ) (12) z = z 0 + p z t (13) Problem 2) The Hamiltonian in a magnetic feld is given by: H = 1 2 m p p e c A P 2 (14) ThereFore the operator k is given by: k = 1 ¯ h p p e c A P (15) Then k × k = 1 ¯ h 2 ± p × p e c p × A e c A × p + e 2 c 2 A × A ² (16) The frst and last terms vanish, leaving k × k = e ¯ h 2 c ( p × A + A × p ) (17) Now use p = i ¯ h r and the relation r × ( A ψ ) = ( r ψ ) × A + ψ ( r × A ) (18) and we arrive at k × k ψ = e i ¯ hc ( r × A ) ψ (19) Or k × k = ie ¯ hc H (20) Problem 3) We give a semiclassical derivatino oF the de Haas-van Alphen e±ect starting with the equation oF motion oF the conduction electrons in a magnetic feld H .
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hw2_sol - Physics 240B, Spring 2008 Homework 2 Solutions...

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