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HOMEWORK CH6 Hasnaa Aldawsari 442203265 4. Densities of states in low dimensions: Find the density of k and energy states for an ideal noninteracting Fermi gas in one and two dimensions. density of k and energy states in one dimension π·πβ = 2 ( 1 π 1 ) = 2π π π·(β) = β«[ππβ ] πΏ(β β β°βπ0 ) β«[ππβ ] β 2 1 β = β« ππβ π·πβ = β« ππβ π£ π β π ππβ = 2 ππ π·(β) = β«[ππβ ] πΏ(β β β°πβ0 ) = Where β°πΉ = ππΉ = β2 ππΉ 2 2π β ππΉ 2 = 1 β 1 β π 1 2π β β« 2 ππ πΏ(β β β°πβ0 ) = β« 2 β 2 πβ°πΉ πΏ(β β β°πβ0 ) = π 0 π 0 2β β°πΉ πβ β°πΉ 2πβ°πΉ β2 1 2π π π β2πβ°πΉ πππΉ β = = =β 2 β πβ°πΉ 2 ββ2πβ°πΉ 2β β°πΉ ββ2πβ°πΉ density of k and energy states in two dimension π·πβ = 2 ( 1 π 1 2 ) = 2( ) 2π 2π π·(β) = β«[ππβ ] πΏ(β β β°βπ0 ) β«[ππβ ] β 2 1 2 β = β« ππβ π·πβ = 2 ( ) β« ππβ π£ 2π β π ππβ = 2ππππ π·(β) = β«[ππβ ] πΏ(β β β°πβ0 ) = 2 ( 1 2 β ) β« 2ππππ πΏ(β β β°πβ0 ) 2π 0 = 2( 1 2 β π 1 2 π β2πβ°πΉ β2πβ°πΉ 0 ) β« 2π β 2 πβ°πΉ πΏ(β β β°πβ ) = 2 ( ) 2 π β 2 2π β 2β β°πΉ 2π β 2β β°πΉ 0 = 2( 1 2 π ππ ) 2π 2 = 2 2π β β β2 ππΉ 2 2πβ°πΉ π€βπππ β°πΉ = β ππΉ 2 = 2π β2 ππΉ = 1 2π π π β2πβ°πΉ πππΉ β = = =β 2 β πβ°πΉ 2 ββ2πβ°πΉ 2β β°πΉ ββ2πβ°πΉ 5. Fermi pancakes: Consider a thin layer of silver, 106 Â wide and 106 Â long along x and y. (a) Take the layer to be 4.1 Â thick along z- Treat the layer as a free Fermi gas, demanding that the wave function vanish at the boundaries along the z, direction. Find the difference between the energies of the lowest- and highestoccupied single-particle states, and compare this difference to the bulk Fermi energy. The density of electrons π = 0.0586 π΄Μβ3 The bulk Fermi energy β°πΉ = 2 2 β2 ππΉ 2 = 36.46(π β π΄Μ3 )3 ππ£ = 36.46(0.0586 π΄Μβ3 β π΄Μ3 )3 ππ£ = 5.50ππ£ 2π 1 π€βπππ ππΉ = (3π 2 π)3 The k vector of single particle states ππΉ = 2π (π , π , π ) πΏ π₯ π¦ π§ The Fermi energy of single particle β°πΉ = β2 ππΉ 2 β2 2π 2 2 = ( ) (ππ₯ + ππ¦2 + ππ§2 ) 2π 2π πΏ The energy of a single electron in layer 2 β2 2π π 2 β°πΉ = [( ) (ππ₯2 + ππ¦2 ) + ( ) ππ§2 ] 2π πΏπ₯,π¦ πΏπ§ The lowest energy state is (0,0,1) and for large ππ₯ , ππ¦ , we will reach a state (ππ₯ , ππ¦ ,1) whose energy is the same as the state (0, 0, 2). β° (ππ₯ , ππ¦ , 1) = β° (0,0,2) 2 β2 2π β2 π 2 β2 π 2 ( ) (1)2 = 0 + ( ) (2)2 ( ) (ππ₯2 + ππ¦2 ) + 2π πΏπ₯,π¦ 2π πΏπ§ 2π πΏπ§ 2 β2 2π β2 π 2 β2 π 2 2 2 2 ( ) (2) β ( ) (1)2 ( ) (ππ₯ + ππ¦ ) = 2π πΏπ₯,π¦ 2π πΏπ§ 2π πΏπ§ 2 β2 2π β2 π 2 ( ) β3 ( ) (ππ₯2 + ππ¦2 ) = 2π πΏπ₯,π¦ 2π πΏπ§ β2 π 2 1 2 ( ) β 3 3 ( 2π πΏπ§ πΏπ§ ) 2 2 = (ππ₯ + ππ¦ ) = 2 β 2π 2 1 2 ( ) 4 ( 2π πΏπ₯,π¦ πΏπ₯,π¦ ) (ππ₯2 + ππ¦2 ) = 3πΏ2π₯,π¦ 4πΏ2π§ The number of electrons 2 π =2× π(ππ₯ 2 + ππ¦ 2 ) πcircle ππ 2 π 2π =2× = 2 × =2× ( ) (ππ₯2 + ππ¦2 ) 2 2 2 ππ πΏ 2π 2π 2π π₯,π¦ (πΏ ) (πΏ ) (πΏ ) π₯,π¦ π₯,π¦ π₯,π¦ = 2π(ππ₯2 + ππ¦2 ) 2π 2 πΏ Where ππ₯ 2 + ππ¦ 2 = ( ) (ππ₯2 + ππ¦2 ) The actual number of available electrons in the silver film is ππΏ2π₯,π¦ πΏπ§ 2π(ππ₯2 + ππ¦2 ) = ππΏ2π₯,π¦ πΏπ§ πππ (ππ₯2 + ππ¦2 ) = π π , 3πΏ2π₯,π¦ 4πΏ2π§ ππΏ2π₯,π¦ πΏπ§ 3πΏ2π₯,π¦ β€ 2π 4πΏ2π§ 2ππΏ3π§ β€1 3π 0.86 β€ 1 So ,all electrons stay in the energy levels with ππ§ = 1 2 2 β2 2π β2 π 2 β2 2π β2 π 2 β°βππβ (ππ₯ , ππ¦ , 1) = ( ) (1)2 = ( ) ( ) (ππ₯2 + ππ¦2 ) + ( ) (ππ₯2 + ππ¦2 ) + 2π πΏπ₯,π¦ 2π πΏπ§ 2π πΏπ₯,π¦ 2π πΏπ§ 2 β2 2π β2 π 2 β2 π 2 β°πΉ = β°βππβ (ππ₯ , ππ¦ , 1) β β°πππ€ (0,0,1) = ( ) β ( ) ( ) (ππ₯2 + ππ¦2 ) + 2π πΏπ₯,π¦ 2π πΏπ§ 2π πΏπ§ 2 = 2 β2 2π β2 2π ππΏ2π₯,π¦ πΏπ§ β2 = πππΏπ§ ( ) (ππ₯2 + ππ¦2 ) = ( ) 2π πΏπ₯,π¦ 2π πΏπ₯,π¦ 2π π 16 2 (6.5821 × 10 ππ£ β π ) = × 3.14 × 0.0586π΄Μβ3 × 4.1π΄Μ = 5.75 0.511 × 106 ππ£ Μ 2 (3 × 108 × 1010 π΄βπ ) (b) Repeat the previous problem with a layer 8.2 Â thick along z The lowest energy state is (0,0,1) and for large ππ₯ , ππ¦ , we will reach a state (ππ₯ , ππ¦ ,3) whose energy is the same as the state (0, 0, 4). β° (ππ₯ , ππ¦ , 3) = β° (0,0,4) 2 β2 2π β2 π 2 β2 π 2 ( ) (3)2 = 0 + ( ) (4)2 ( ) (ππ₯2 + ππ¦2 ) + 2π πΏπ₯,π¦ 2π πΏπ§ 2π πΏπ§ 2 β2 2π β2 π 2 β2 π 2 ( ) (4)2 β ( ) (3)2 ( ) (ππ₯2 + ππ¦2 ) = 2π πΏπ₯,π¦ 2π πΏπ§ 2π πΏπ§ 2 β2 2π β2 π 2 2 2 + π = ( ) β7 ( ) (ππ₯ π¦) 2π πΏπ₯,π¦ 2π πΏπ§ β2 π 2 1 2 ( ) β 7 7 ( 2π πΏ πΏπ§ ) = (ππ₯2 + ππ¦2 ) = 2 π§ β 2π 2 1 2 ( ) 4( ) 2π πΏπ₯,π¦ πΏπ₯,π¦ (ππ₯2 + ππ¦2 ) = 7πΏ2π₯,π¦ 4πΏ2π§ 2π(ππ₯2 + ππ¦2 ) = ππΏ2π₯,π¦ πΏπ§ πππ (ππ₯2 + ππ¦2 ) = π π, 7πΏ2π₯,π¦ 4πΏ2π§ ππΏ2π₯,π¦ πΏπ§ 7πΏ2π₯,π¦ β€ 2π 4πΏ2π§ 2ππΏ3π§ β€3 7π 2.9 β€ 3 So ,all electrons stay in the energy levels with ππ§ = 3 The total number of electrons is π = 2π(ππ₯2 + ππ¦2 ) ππΏ2π₯,π¦ πΏπ§ = π1 + π2 + π3 π1 πππ ππ§ = 1, π2 πππ ππ§ = 2, π3 πππ ππ§ = 3 2 2 β2 2π π1 β2 π 2 β2 2π π2 β2 π 2 + ( ) (1)2 = + ( ) (2)2 ( ) ( ) 2π πΏπ₯,π¦ 2π 2π πΏπ§ 2π πΏπ₯,π¦ 2π 2π πΏπ§ 2 β2 2π π3 β2 π 2 = + ( ) (3)2 ( ) 2π πΏπ₯,π¦ 2π 2π πΏπ§ π1 + (1)2 = π2 + (2)2 = π3 + (3)2 2 β2 2π π1 β2 π 2 + ( ) (1)2 = 5.57 ( ) 2π πΏπ₯,π¦ 2π 2π πΏπ§ (6.5821 × 10β16 ππ£ β π )2 π 2 (8.2) 6 0.511 × 10 ππ£ 2× Μ 2 (3 × 108 × 1010 π΄βπ ) (6.5821 × 1016 ππ£ β π )2 π (106 )2 0.511 × 106 ππ£ 2 Μ (3 × 108 × 1010 π΄βπ ) 5.57 β π1 = β2 π 2 5.57 β 2π (πΏ ) (1)2 π§ β2 2π ( ) 4ππ πΏπ₯,π¦ = 2 = β2 π 2 5.57 β 2π (πΏ ) π§ β2 π π πΏ2π₯,π¦ = 5.57 β 0.56 = 2.09 × 1011 2.40 × 10β11 π2 = π1 + (1)2 β (2)2 = π1 β 3 = 2.09 × 1011 β 3 = 2.09 × 1011 π3 = π1 β 8 = 2.09 × 1011 2 β°πΉ = β°βππβ (ππ₯ , ππ¦ , 3) β β°πππ€ (0,0,1) = 2 β2 2π π3 β2 π 2 β2 π 2 2 (3) + ( ) β ( ) (1)2 ( ) 2π πΏπ₯,π¦ 2π 2π πΏπ§ 2π πΏπ§ 2 β2 2π π3 β2 π 2 β2 2π π3 π 2 = + ( ) 8= + ( ) 8] ( ) [( ) 2π πΏπ₯,π¦ 2π 2π πΏπ§ 2π πΏπ₯,π¦ 2π πΏπ§ = (6.5821 × 10β16 ππ£ β π )2 1 2 π 2 [( ) 2π × 2.09 × 1011 + ( ) 8] = 5.64 ππ£ 6 6 Μ 0.511 × 10 ππ£ 10 8.2π΄ 2× Μ 2 (3 × 108 × 1010 π΄βπ )