HW-10-solution PDF

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Solution of HW #10...

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Homework #10 1. Square lattice, free electron energies. (a) Show for a simple square lattice (two dimensions) that the kinetic energy of a free electron at a corner of the first Brillouin zone is higher than that of an electron at the midpoint of a side face of the zone by a factor of 2. (b) What is the corresponding factor for a simple cubic lattice (three dimensions)? (c) What bearing might the result of (b) have on the conductivity of divalent metals? Solution: (a)

At

the

corner

of

the

first

Brillouin

zone,

r ⎛π π ⎞ k = ⎜ , ⎟, ⎝a a ⎠

thus

h2 k 2 h2π 2 2 2 h2π 2 ( 1 1 ) = + = . 2m 2ma 2 ma 2 r ⎛π ⎞ h2π 2 2 h 2π 2 2 At the midpoint of a side face, k = ⎜ ,0⎟ , thus E(1,0 ) = ( 1 0 ) + = . 2ma 2 2 ma 2 ⎝a ⎠ Therefore E (1,1) / E (1,0 ) = 2 . r ⎛π π π ⎞ (b) For a simple cubic lattice, at the corner of the first Brillouin zone, k = ⎜ , , ⎟ , ⎝a a a⎠ 2 2 2 2 3h π hπ thus E(1,1,1) = (12 + 12 + 12 ) = . 2 2ma 2 ma 2 r ⎛π h2π 2 2 h 2π 2 ⎞ 2 At the midpoint of a side face, k = ⎜ ,0,0⎟ , thus E(1,0,0 ) = ( 1 0 ) + = . 2ma 2 2 ma 2 ⎝a ⎠ Therefore E(1,1,1) / E (1,0,0 ) = 3 . E(1,1) =

(c) For a divalent metal, the electron density is n = 2 / a3 . Then the Fermi energy is h2 h 2π 2 h2 2/3 π . EF = (3π 2n ) 2 / 3 = ( 6 / ) 1 . 53 = 2m 2ma 2 2ma 2 Thus E (1,0, 0 ) < EF < E (1,1,1) . The Fermi sphere is greater than the first Brillouin zone in the [100] direction, but smaller than the first Brillouin zone in the [111] direction. The electrons should start to fill in the second Brillouin zone before filling up the first Brillouin zone.

(c) The electrons start to fill in the 2nd Brillouin zone before completely filling out the 1st Brillouin zone. Then both the first and second bands are not completely filled so that the material will be a metal. 2. Free electron energies in reduced zone. Consider the free electron energy bands of an fcc crystal lattice in the approximation of an empty lattice, but in the reduced zone scheme in which all k’s are transformed to lie in the first Brillouin zone. Plot roughly in the [111] direction the energies of all bands up to six times the lowest band energy at the zone boundary at k=(2π/a)(1/2,1/2,1/2). Let this be the unit of energy. This problem shows why band edges need not necessarily be at the zone center. Several of the degeneracies (band crossings) will be removed when account is taken of the crystal potential. Solution: h 2k 2 h 2 2 ( k + k 2y + k 2z ) . E= = 2m 2m x r 2π By adding a reciprocal lattice vector, G = ( n1, n2, n3) , where n1 , n2 , n3 are all even a r r or all odd, we fold the k vector into k ' in the first Brillouin zone along [111] r r r π π π direction, i.e., k = k ' +G and kx ' = k y ' = kz ' ≡ β . Because − ≤ k x ' = k y ' = k z' ≤ , a a a there is − 1 ≤ β ≤ 1 . Then h 2 ⎡⎛ 2 π n1 ⎞ ⎛ 2πn2 ⎞ ⎛ 2πn3 ⎞ ⎟ ⎟ + ⎜ ky ' + ⎟ + ⎜k z ' + ⎢⎜ kx ' + a ⎠ ⎝ a ⎠ ⎝ a ⎠ 2m ⎢⎣ ⎝ 2

E=

2

2

⎤ ⎥ ⎥⎦

2 2 2 h2 ⎡⎛ π 2π n1 ⎞ ⎛ π 2 πn2 ⎞ ⎛ π 2πn3 ⎞ ⎤ + + + + + β β β ⎟ ⎥ ⎟ ⎜ ⎟ ⎜ ⎢⎜ a ⎠ ⎝a a ⎠ ⎝a a ⎠ ⎥⎦ 2 m ⎢⎣⎝ a h2 π 2 3 β2 + 4 β (n1 + n 2 + n 3 ) + 4(n12 + n 22 + n 23 ) = 2 2 ma

=

[

]

h 2π 2 × 2ma 2

( n1 , n2 , n3 )

E=

( 000)

3β 2

(1 1 1 ), ( 1 1 1 ), ( 1 1 1) (11 1 ), (1 1 1), ( 1 11) (00 2), ( 020), ( 2 00) ( 002), (020), ( 200) (111) ( 1 1 1) ( 004) (00 4 )

3 β 2 − 4 β + 12 3 β 2 + 4 β + 12 3β 2 − 8β + 16 3β 2 + 8β + 16 3β 2 + 12 β + 12 3 β 2 − 12 β + 12 3β 2 + 16 β + 64 > 6 × 3 3β 2 − 16 β + 64 > 6 × 3

3. Kronig-Penney model. (a) For the delta-function potential and with P...