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04hw - Solutions to problem set 4(141A Sp07 1 1 From ISSP...

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Solutions to problem set 4 (141A Sp07) 1 1. From ISSP page 91: - m n ω 2 u n = C ( u n + 1 - 2 u n + u n - 1 ) For n > 0 we have - m ω 2 e - kn = C ( e - k ( n + 1) - 2 e - kn + e - k ( n - 1) ) , ie. - m ω 2 = C ( e - k - 2 + e k ) . For n = 0, we have - M ω 2 = C ( e - k (1) - 2 + e - k |- 1 | ) = C (2 e - k (1) - 2) = 2 C ( e - k - 1) Therefore 2 C M ( e - k - 1) = C m ( e - k - 2 + e k ) 2 C M (1 - e k ) = C m (1 - 2 e k + e 2 k ) Let x = e k ; then 1 - 2 x + x 2 - (1 - x ) 2 m M = 0 x 2 - 2 1 - m M x + 1 - 2 m M = 0 So x = 1 - m M ± 1 - m M 2 - 1 - 2 m M = 1 - m M ± m M . There is a trivial solution with x = 1, ie k = 0 and ω = 0 (this corresponds to a bulk motion of the lattice.) The non-trivial solution has x = 1 - 2 m / M , ie k = ln(1 - 2 m / M ) . Remember that the displacement function looks like u n = e - kn for n > 0. For M > 2 m , k will be negative and so the displacement will blow up. Therefore there is no localized mode for M > 2 m . For M < 2 m , the argument of the log is negative. Recall that ln( -| y | ) = ln( e i π | y | ) = i π + ln | y | . Therefore the real part of k is ln(2 m / M - 1). For 2 m / M - 1 < 1 ie m < M this is still negative; however for M < m it is positive.
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