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Homework6solutions

# Homework6solutions - Homework 6 solutions 1 Using two by...

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Homework 6 solutions: 1. Using two by two matrices the difference equation becomes: { } '† 11 [( ) ] { } nnn n hk φ αφ βφ β φ + =+ + Using where { {}{} ' ink a no e φφ = } n and { } o are 2 x 1 column vectors and the matrices are given as follows: t t ε α ⎛⎞ = ⎝⎠ and 00 0 t β = ⎜⎟ Then the dispersion relation will be given as the eigenvalues of h . '' ' 0 ] 0 ik a ik a tt e e + ' ' ' ] ik a ik a e e + = + The eigenvalues of which are given by || ( ) ( ik a ik a ik a e e e εε −− ±+ =± + + ' ) The eigenvalues are given by 1c o s ( ) 22 ka + ±= ± c o s ( ) Let then the eigenvalues are given by ' 2 = 2 cos( ) tk a ± . But now we have to figure out what values of ka to use. The range of ka will be determined by the distinct values we can get for the eigenvalues. If we had used one atom per unit cell the values of ka would have been in a range of 2 π . But because there are two branches distinct values are only given in a range of

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Homework6solutions - Homework 6 solutions 1 Using two by...

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