d k \u03c3 where d z k m d x k 2 k 3 cos3 \u03b8 2 k x k 2 x 3 k 2 y d y k 2 k 3 sin3 \u03b8 2

D k σ where d z k m d x k 2 k 3 cos3 θ 2 k x k 2 x

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= ~ d ( ~ k ) · ~σ, where d z ( ~ k ) = m , d x ( ~ k ) = 2 k 3 cos(3 θ ) = 2 k x ( k 2 x - 3 k 2 y ) , d y ( ~ k ) = 2 k 3 sin(3 θ ) = 2 k y ( - k 2 y + 3 k 2 x ) . As in problem 4 of the previous problem set, we will argue below that the result can be obtained by mapping to a sphere. If you are interested in seeing some of the math worked out, we have included it at the end. Let ˆ d be a unit vector in the direction of ~ d . The Berry curvature F xy = a unit area on the sphere. When m > 0 and k = 0, ˆ d = ˆ z . As we increase k , ˆ d spans the upper hemisphere. In particular, if we write ~ k = ( k, θ ) and ~ d = d (sin θ 0 cos φ, sin θ 0 sin φ, cos θ 0 ) = (2 k 3 cos(3 θ ) , 2 k 3 sin(3 θ ) , m ), we see by analogy that φ = 3 θ . It is clear in general that for a Hamiltonian with k n , we will have φ = . If we map the Brillouin zone to the sphere, we see that for a fixed sign of m , we cover the hemisphere n times. We thus see immediately (see previous problem set) that σ xy = n 2 sign( m ) e 2 h . From here it follows that Δ σ xy = ne 2 /h , and we see by comparison with the previous question that in general, (Chern number) = - (Berry Phase) / 2 π . Below we show the integral done explicitly. As in the previous problem set, we write: F xy = 1 2 | ~ d ( ~ k ) | 3 ~ d ( ~ k ) · ( k x ~ d ( ~ k ) × k y ~ d ( ~ k )) = 1 2 | ~ d | 3 abc d a k x d b k y d c , 3
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Phys 268r Fall 2013 Problem Set 4 - Part I - Solutions B. Halperin where | ~ d | = p m 2 + (2 k 3 ) 2 . The non-zero contractions are ( a, b, c ) = { ( z, x, y ) , ( z, y, x ) } , which gives: F xy = 1 2( m 2
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  • Fall '13
  • BertrandI.Halperin
  • Physics, Resistance, Orthogonal matrix, Hilbert space, Unitary matrix

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