p268r-ps5-solutionsv2 - Phys 268r Fall 2013 Problem Set 5 Solutions Corrected v2 B Halperin Solutions to Problem Set 5 Problem 1 a We have the

# p268r-ps5-solutionsv2 - Phys 268r Fall 2013 Problem Set 5...

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Phys 268r Fall 2013 Problem Set 5 Solutions - Corrected v2 B. Halperin Solutions to Problem Set 5 Problem 1. a) We have the Hamiltonian H = v~ τ · ~ p + V ( ~ r ), and we can write out the momentum term explicitly: ~ τ · ~ p = 0 p x - ip y p x + ip y 0 ! . We write the wave function as a spinor: Ψ = ψ A ψ B ! , and letting k E/v , we find a system of equations: A = ( p x - ip y ) ψ B B = ( p x + ip y ) ψ A . (1.1) We can combine these equations this to find k 2 Ψ = ( p 2 x + p 2 y )Ψ, which has clearly has solutions which are plane waves, with arbitrary direction of ~ k and arbitrary form of the spinor. However, the original pair of equations (1.1) is more restrictive. For E > 0, the direction of ~ k determines the spinor up to an overall phase factor, so we have, for a given ~ k = k (cos θ, sin θ ) Ψ ~ k = A ~ k e i ~ k · ~ r φ A φ B ! , where the spinor is an eigenstate of ~ k · ~ τ with eigenvalue 1. To be specific,we choose φ A = 1 / 2 , φ B = ( e / 2) For x > 0, we have vk 0 = V - E > 0, which again allows plane wave solutions at wave vectors ~ k 0 . Now, however, the solution with energy E has a velocity in the direction opposite to ~ k 0 , so it is convenient to write ~ k 0 = - k 0 (cos θ 0 , sin θ 0 ), where θ 0 is the direction of propagation. The plane wave solutions with energy E may then be written A ~ k 0 e i ~ k 0 · ~ r Φ 0 , where Φ = (1 , e 0 ) T / 2. This spinor is an eigenstate of ~ k 0 · ~ τ with eigenvalue -1. Now, suppose we have an incident wave with incident angle θ . The momentum k y is conserved by the Hamiltonian, so the transmitted and reflected waves must have the same k y as the incident wave. This means that - k 0 sin θ 0 = k sin θ . For the transmitted wave, we require - π/ 2 < θ 0 < π/ 2, which means that the sign of θ 0 is opposite to the sign of θ . Thus if the 1
Phys 268r Fall 2013 Problem Set 5 Solutions - Corrected v2 B. Halperin incident wave is moving in a direction between x and y , the transmitted wave will be moving in a direction between x and - y . This means that the planar boundary acts as a focusing lens when V is greater than E . Now we can use conservation of energy to see that vk 0 = V 0 - vk . There will be an acceptable solution | sin θ 0 | < 1 0 if and only if ( k/k 0 ) sin θ < 1 . If V 0 > 2 E , we have k 0 > k , so this condition is fulfilled for any angle of incidence. However if E < V 0 < 2 E , there will be no transmitted wave (total reflection), unless the incident angle is smaller than a critical value, given by θ c = sin - 1 [( V 0 - E ) /E ] . (1.2) b) Since the Hamiltonian is a first-order differential operator, the boundary condition at x = 0 is that the wave functions should match at the boundary. If the amplitudes of the incident, reflected and transmitted waves are denoted respectively by A, B, C , then the matching component for the upper component of the spinor is A + B = C , while the lower component gives Ae + Be r = Ce 0 (1.3) where θ r = π - θ is the direction of propagation of the reflected wave.

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