p268r-ps4-solutions-part1 - Phys 268r Fall 2013 Problem Set 4 Part I Solutions B Halperin Solutions to Problem Set 4 Part I Problem 1 We assume a

# p268r-ps4-solutions-part1 - Phys 268r Fall 2013 Problem Set...

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Phys 268r Fall 2013 Problem Set 4 - Part I - Solutions B. Halperin Solutions to Problem Set 4 - Part I Problem 1. We assume a perfect system, in which the resistance between any two contacts is R Q = h/e 2 . Therefore, more contacts is equivalent to more equal resistors in series, and the potential drops across each resistor are equal. We thus have in the first case that: V 6 = V 4 = 2 3 V 2 V 5 = V 3 = 1 3 V 2 . Similarly, in the second case, we have: V 3 = 1 2 V 4 V 2 = 3 4 V 4 V 6 = 1 2 V 4 V 5 = 1 4 V 4 . Note that in the second case, the lower path has more resistance, and therefore draws a smaller current. Problem 2. Given the definition of sewing matrices: B αβ ( ~ k ) < u - ~ | T | u ~ > and B βα ( - ~ k ) < u ~ | T | u - ~ >, (B Unitary), CLAIM: | u - ~ > = X β B * αβ ( ~ k ) T | u ~ > . PROOF: The set of states T | u ~ i , for 1 β n F will form an orthonormal basis for the space of vectors | u - ~ i . Therefore we can write | u - ~ i = X β T | u ~ i M βα , 1
Phys 268r Fall 2013 Problem Set 4 - Part I - Solutions B. Halperin where the unitary matrix M βα is given by M βα = h Tu ~ | u - ~

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• Fall '13
• BertrandI.Halperin
• Physics, Resistance, Orthogonal matrix, Hilbert space, Unitary matrix

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