# sln5 - 7224 Homework Solution 5 Problem 1 The superposition...

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7224 Homework Solution 5 Problem 1 The superposition of the mode u ( x,y ) = e i q 1 · x + φ 1 + e i q 2 · x + φ 2 + e i q 3 · x + φ 3 + c.c. with the wave vectors q 1 , q 2 and q 3 q 1 = (1 , 0) , q 2 = ( - 1 2 , 3 2 ) , q 3 = ( - 1 2 , - 3 2 ) . (a) Change the origin of the coordinates x ( x,y ) x 0 ( x 0 ,y 0 ) q 1 · x + φ 1 = q 1 · x 0 q 2 · x + φ 1 = q 2 · x 0 We can get x 0 = x - φ 2 q 1 y - φ 1 q 2 y q 1 y q 2 x - q 2 y q 1 x = x + φ 1 , y 0 = y - φ 2 q 1 x - φ 1 q 2 x q 1 y q 2 x - q 2 y q 1 x = y + φ 1 3 + 2 φ 2 3 . The superposition of the modes in the new coordinate u ( x 0 ,y 0 ) = e i q 1 · x 0 + e i q 2 · x 0 + e i q 3 · x 0 + φ 1 + φ 2 + φ 3 + c.c. The φ 0 1 and φ 0 2 are zero, so that the form of the pattern is determined by a single phase variable φ 0 3 = φ 1 + φ 2 + φ 3 = φ . (b) φ = 0 -10 -5 0 5 10 -10 -5 0 5 10 1

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φ = π/ 4 -10 -5 0 5 10 -10 -5 0 5 10 φ = π/ 3 -10 -5 0 5 10 -10 -5 0 5 10 φ = π/ 2 -10 -5 0 5 10 -10 -5 0 5 10 2
φ = π -10 -5 0 5 10 -10 -5 0 5 10 Only when φ = 0 or π one can get hexagonal patterns. (c) When

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## This note was uploaded on 08/25/2011 for the course PHYS 7268 taught by Professor Staff during the Spring '11 term at Georgia Institute of Technology.

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sln5 - 7224 Homework Solution 5 Problem 1 The superposition...

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