sol09 - Solid State Theory Solution Sheet 9 FS 11 Prof. M....

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Unformatted text preview: Solid State Theory Solution Sheet 9 FS 11 Prof. M. Sigrist Exercise 9.1 Lowest Landau level in the Corbino geometry a) Because A z = 0 and ∂ z A = 0 it follows that ( ∇× A ) x = ( ∇× A ) y = 0. Furthermore, for r 6 = 0 we have ( ∇ × A ) z = 1 2 ∂ ∂x B + ν Φ πr 2 x + 1 2 ∂ ∂y B + ν Φ πr 2 y = B. (1) On the other hand, the magnetic flux through the origin is Φ = lim → Z r< B · d σ = lim → Z r< ∇ × A · d σ = lim → Z r = A · d γ = lim → 2 π 1 2 B + ν Φ π = ν Φ (2) where we used Stokes’ theorem in the second line. b) H = 1 2 m * ( p x + eB 2 c y ) 2 + ( p y- eB 2 c x ) 2 + U ( r ) = 1 2 m * p 2 x + p 2 y- eB c ( xp y- yp x ) + eB 2 c r 2 + U ( r ) = ~ 2 2 m *- 1 r ∂ r r∂ r- ∂ 2 φ r 2- i∂ φ l 2 + r 2 l 2 2 + U ( r ) = ~ 2 2 m *- 1 r ∂ r r∂ r + i∂ φ r- r 2 l 2 2 + U ( r ) (3) Using the ansatz for the wave function of the lowest Landau level we obtain the radial Schr¨ odinger equation given on the exercise sheet: ~ 2 2 m *- 1 r ∂ ∂r r ∂ ∂r + m r- r 2 l 2 2 + U ( r )- E m r α e- r 2 4 l * 2 = 0 . (4) Furthermore, we obtain r∂ r ψ m = α- r 2 2 l * 2 ψ m (5) from which it follows that 1 r ∂ r r∂ r ψ m = 1 r 2 α- r 2 2 l * 2 2 ψ m- 1 l * 2 ψ m . (6) Defining E * m = 2 m * E m / ~ 2 we can write the radial Schr¨ odinger equation as- α r- r 2 l * 2 2 + m r- r 2 l 2 2 + 1 l * 2 + C * 1 r 2 + C * 2 r 2 4 l 4- E * m ψ m = 0 (7) 1 2 4 6 8 10 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 r/l ψ m ( r ) m = 0 m = 2 m = 5 m = 9 Figure 1: Radial part of the wave function for several m ’s....
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This note was uploaded on 10/04/2011 for the course PHYS fs11 taught by Professor Sigrist during the Spring '11 term at Swiss Federal Institute of Technology Zurich.

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sol09 - Solid State Theory Solution Sheet 9 FS 11 Prof. M....

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