Download HOMEWORK Ch6 Hasna Aldawsari 442203265 (1)

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
𝐷𝑘⃗ = 2 (
1 𝑑 1
) =
2𝜋
𝜋
𝐷(ℇ) = ∫[𝑑𝑘⃗ ] 𝛿(ℇ − ℰ⃗𝑘0 )
∫[𝑑𝑘⃗ ] ≅
2
1
∑ = ∫ 𝑑𝑘⃗ 𝐷𝑘⃗ = ∫ 𝑑𝑘⃗
𝑣
𝜋
⃗
𝑘
𝑑𝑘⃗ = 2 𝑑𝑘
𝐷(ℇ) = ∫[𝑑𝑘⃗ ] 𝛿(ℇ − ℰ𝑘⃗0 ) =
Where ℰ𝐹 =
𝑘𝐹 =
ℏ2 𝑘𝐹 2
2𝑚
→ 𝑘𝐹 2 =
1 ∞
1 ∞
𝑚
1 2𝑚
√
∫ 2 𝑑𝑘 𝛿(ℇ − ℰ𝑘⃗0 ) = ∫ 2 √ 2 𝑑ℰ𝐹 𝛿(ℇ − ℰ𝑘⃗0 ) =
𝜋 0
𝜋 0
2ℏ ℰ𝐹
𝜋ℏ ℰ𝐹
2𝑚ℰ𝐹
ℏ2
1 2𝑚
𝑚
𝑚
√2𝑚ℰ𝐹 𝑑𝑘𝐹
→
=
=
=√ 2
ℏ
𝑑ℰ𝐹
2 ℏ√2𝑚ℰ𝐹
2ℏ ℰ𝐹
ℏ√2𝑚ℰ𝐹
density of k and energy states in two dimension
𝐷𝑘⃗ = 2 (
1 𝑑
1 2
) = 2( )
2𝜋
2𝜋
𝐷(ℇ) = ∫[𝑑𝑘⃗ ] 𝛿(ℇ − ℰ⃗𝑘0 )
∫[𝑑𝑘⃗ ] ≅
2
1 2
∑ = ∫ 𝑑𝑘⃗ 𝐷𝑘⃗ = 2 ( ) ∫ 𝑑𝑘⃗
𝑣
2𝜋
⃗
𝑘
𝑑𝑘⃗ = 2𝜋𝑘𝑑𝑘
𝐷(ℇ) = ∫[𝑑𝑘⃗ ] 𝛿(ℇ − ℰ𝑘⃗0 ) = 2 (
1 2 ∞
) ∫ 2𝜋𝑘𝑑𝑘 𝛿(ℇ − ℰ𝑘⃗0 )
2𝜋
0
= 2(
1 2 ∞
𝑚
1 2
𝑚
√2𝑚ℰ𝐹
√2𝑚ℰ𝐹
0
) ∫ 2𝜋
√ 2 𝑑ℰ𝐹 𝛿(ℇ − ℰ𝑘⃗ ) = 2 ( ) 2 𝜋
√ 2
2𝜋
ℏ
2ℏ ℰ𝐹
2𝜋
ℏ
2ℏ ℰ𝐹
0
= 2(
1 2
𝑚
𝑚𝜋
) 2𝜋 2 = 2
2𝜋
ℏ
ℏ
ℏ2 𝑘𝐹 2
2𝑚ℰ𝐹
𝑤ℎ𝑒𝑟𝑒 ℰ𝐹 =
→ 𝑘𝐹 2 =
2𝑚
ℏ2
𝑘𝐹 =
1 2𝑚
𝑚
𝑚
√2𝑚ℰ𝐹 𝑑𝑘𝐹
→
=
=
=√ 2
ℏ
𝑑ℰ𝐹
2 ℏ√2𝑚ℰ𝐹
2ℏ ℰ𝐹
ℏ√2𝑚ℰ𝐹
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
𝑛 = 0.0586 𝐴̇−3
The bulk Fermi energy
ℰ𝐹 =
2
2
ℏ2 𝑘𝐹 2
= 36.46(𝑛 ∙ 𝐴̇3 )3 𝑒𝑣 = 36.46(0.0586 𝐴̇−3 ∙ 𝐴̇3 )3 𝑒𝑣 = 5.50𝑒𝑣
2𝑚
1
𝑤ℎ𝑒𝑟𝑒 𝑘𝐹 = (3𝜋 2 𝑛)3
The k vector of single particle states
𝑘𝐹 =
2𝜋
(𝑛 , 𝑛 , 𝑛 )
𝐿 𝑥 𝑦 𝑧
The Fermi energy of single particle
ℰ𝐹 =
ℏ2 𝑘𝐹 2
ℏ2 2𝜋 2 2
=
( ) (𝑛𝑥 + 𝑛𝑦2 + 𝑛𝑧2 )
2𝑚
2𝑚 𝐿
The energy of a single electron in layer
2
ℏ2
2𝜋
𝜋 2
ℰ𝐹 =
[(
) (𝑛𝑥2 + 𝑛𝑦2 ) + ( ) 𝑛𝑧2 ]
2𝑚 𝐿𝑥,𝑦
𝐿𝑧
The lowest energy state is (0,0,1) and for large 𝑛𝑥 , 𝑛𝑦 , we will reach a state (𝑛𝑥 , 𝑛𝑦 ,1) whose energy
is the same as the state (0, 0, 2).
ℰ (𝑛𝑥 , 𝑛𝑦 , 1) = ℰ (0,0,2)
2
ℏ2 2𝜋
ℏ2 𝜋 2
ℏ2 𝜋 2
( ) (1)2 = 0 +
( ) (2)2
(
) (𝑛𝑥2 + 𝑛𝑦2 ) +
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
2𝑚 𝐿𝑧
2
ℏ2 2𝜋
ℏ2 𝜋 2
ℏ2 𝜋 2
2
2
2
( ) (2) −
( ) (1)2
(
) (𝑛𝑥 + 𝑛𝑦 ) =
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
2𝑚 𝐿𝑧
2
ℏ2 2𝜋
ℏ2 𝜋 2
( ) ∙3
(
) (𝑛𝑥2 + 𝑛𝑦2 ) =
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
ℏ2 𝜋 2
1 2
(
)
∙
3
3
(
2𝑚 𝐿𝑧
𝐿𝑧 )
2
2
=
(𝑛𝑥 + 𝑛𝑦 ) = 2
ℏ
2𝜋 2
1 2
(
)
4
(
2𝑚 𝐿𝑥,𝑦
𝐿𝑥,𝑦 )
(𝑛𝑥2 + 𝑛𝑦2 ) =
3𝐿2𝑥,𝑦
4𝐿2𝑧
The number of electrons
2
𝑁 =2×
𝜋(𝑘𝑥 2 + 𝑘𝑦 2 )
𝑉circle
𝜋𝑘 2
𝜋
2𝜋
=2×
=
2
×
=2×
(
) (𝑛𝑥2 + 𝑛𝑦2 )
2
2
2
𝑉𝑘
𝐿
2𝜋
2𝜋
2𝜋
𝑥,𝑦
(𝐿 )
(𝐿 )
(𝐿 )
𝑥,𝑦
𝑥,𝑦
𝑥,𝑦
= 2𝜋(𝑛𝑥2 + 𝑛𝑦2 )
2𝜋 2
𝐿
Where 𝑘𝑥 2 + 𝑘𝑦 2 = ( ) (𝑛𝑥2 + 𝑛𝑦2 )
The actual number of available electrons in the silver film is 𝑛𝐿2𝑥,𝑦 𝐿𝑧
2𝜋(𝑛𝑥2 + 𝑛𝑦2 ) = 𝑛𝐿2𝑥,𝑦 𝐿𝑧 𝑎𝑛𝑑 (𝑛𝑥2 + 𝑛𝑦2 ) =
𝑠𝑜 ,
3𝐿2𝑥,𝑦
4𝐿2𝑧
𝑛𝐿2𝑥,𝑦 𝐿𝑧 3𝐿2𝑥,𝑦
≤
2𝜋
4𝐿2𝑧
2𝑛𝐿3𝑧
≤1
3𝜋
0.86 ≤ 1
So ,all electrons stay in the energy levels with 𝑛𝑧 = 1
2
2
ℏ2 2𝜋
ℏ2 𝜋 2
ℏ2 2𝜋
ℏ2 𝜋 2
ℰℎ𝑖𝑔ℎ (𝑛𝑥 , 𝑛𝑦 , 1) =
( ) (1)2 =
( )
(
) (𝑛𝑥2 + 𝑛𝑦2 ) +
(
) (𝑛𝑥2 + 𝑛𝑦2 ) +
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
2
ℏ2 2𝜋
ℏ2 𝜋 2 ℏ2 𝜋 2
ℰ𝐹 = ℰℎ𝑖𝑔ℎ (𝑛𝑥 , 𝑛𝑦 , 1) − ℰ𝑙𝑜𝑤 (0,0,1) =
( ) −
( )
(
) (𝑛𝑥2 + 𝑛𝑦2 ) +
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
2𝑚 𝐿𝑧
2
=
2
ℏ2 2𝜋
ℏ2 2𝜋 𝑛𝐿2𝑥,𝑦 𝐿𝑧 ℏ2
= 𝜋𝑛𝐿𝑧
(
) (𝑛𝑥2 + 𝑛𝑦2 ) =
(
)
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑥,𝑦
2𝜋
𝑚
16
2
(6.5821 × 10 𝑒𝑣 ∙ 𝑠)
=
× 3.14 × 0.0586𝐴̇−3 × 4.1𝐴̇ = 5.75
0.511 × 106 𝑒𝑣
̇ 2
(3 × 108 × 1010 𝐴⁄𝑠)
(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 𝑛𝑥 , 𝑛𝑦 , we will reach a state (𝑛𝑥 , 𝑛𝑦 ,3) whose energy
is the same as the state (0, 0, 4).
ℰ (𝑛𝑥 , 𝑛𝑦 , 3) = ℰ (0,0,4)
2
ℏ2 2𝜋
ℏ2 𝜋 2
ℏ2 𝜋 2
( ) (3)2 = 0 +
( ) (4)2
(
) (𝑛𝑥2 + 𝑛𝑦2 ) +
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
2𝑚 𝐿𝑧
2
ℏ2 2𝜋
ℏ2 𝜋 2
ℏ2 𝜋 2
( ) (4)2 −
( ) (3)2
(
) (𝑛𝑥2 + 𝑛𝑦2 ) =
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
2𝑚 𝐿𝑧
2
ℏ2 2𝜋
ℏ2 𝜋 2
2
2
+
𝑛
=
( ) ∙7
(
) (𝑛𝑥
𝑦)
2𝑚 𝐿𝑥,𝑦
2𝑚 𝐿𝑧
ℏ2 𝜋 2
1 2
(
)
∙
7
7
(
2𝑚 𝐿
𝐿𝑧 )
=
(𝑛𝑥2 + 𝑛𝑦2 ) = 2 𝑧
ℏ
2𝜋 2
1 2
(
)
4(
)
2𝑚 𝐿𝑥,𝑦
𝐿𝑥,𝑦
(𝑛𝑥2
+ 𝑛𝑦2 ) =
7𝐿2𝑥,𝑦
4𝐿2𝑧
2𝜋(𝑛𝑥2 + 𝑛𝑦2 ) = 𝑛𝐿2𝑥,𝑦 𝐿𝑧 𝑎𝑛𝑑 (𝑛𝑥2 + 𝑛𝑦2 ) =
𝑠𝑜,
7𝐿2𝑥,𝑦
4𝐿2𝑧
𝑛𝐿2𝑥,𝑦 𝐿𝑧 7𝐿2𝑥,𝑦
≤
2𝜋
4𝐿2𝑧
2𝑛𝐿3𝑧
≤3
7𝜋
2.9 ≤ 3
So ,all electrons stay in the energy levels with 𝑛𝑧 = 3
The total number of electrons is
𝑁 = 2𝜋(𝑛𝑥2 + 𝑛𝑦2 )
𝑛𝐿2𝑥,𝑦 𝐿𝑧 = 𝑁1 + 𝑁2 + 𝑁3
𝑁1 𝑓𝑜𝑟 𝑛𝑧 = 1, 𝑁2 𝑓𝑜𝑟 𝑛𝑧 = 2, 𝑁3 𝑓𝑜𝑟 𝑛𝑧 = 3
2
2
ℏ2 2𝜋 𝑁1 ℏ2 𝜋 2
ℏ2 2𝜋 𝑁2 ℏ2 𝜋 2
+
( ) (1)2 =
+
( ) (2)2
(
)
(
)
2𝑚 𝐿𝑥,𝑦 2𝜋 2𝑚 𝐿𝑧
2𝑚 𝐿𝑥,𝑦 2𝜋 2𝑚 𝐿𝑧
2
ℏ2 2𝜋 𝑁3 ℏ2 𝜋 2
=
+
( ) (3)2
(
)
2𝑚 𝐿𝑥,𝑦 2𝜋 2𝑚 𝐿𝑧
𝑁1 + (1)2 = 𝑁2 + (2)2 = 𝑁3 + (3)2
2
ℏ2 2𝜋 𝑁1 ℏ2 𝜋 2
+
( ) (1)2 = 5.57
(
)
2𝑚 𝐿𝑥,𝑦 2𝜋 2𝑚 𝐿𝑧
(6.5821 × 10−16 𝑒𝑣 ∙ 𝑠)2
𝜋 2
(8.2)
6
0.511 × 10 𝑒𝑣
2×
̇ 2
(3 × 108 × 1010 𝐴⁄𝑠)
(6.5821 × 1016 𝑒𝑣 ∙ 𝑠)2 𝜋
(106 )2
0.511 × 106 𝑒𝑣
2
̇
(3 × 108 × 1010 𝐴⁄𝑠)
5.57 −
𝑁1 =
ℏ2 𝜋 2
5.57 − 2𝑚 (𝐿 ) (1)2
𝑧
ℏ2
2𝜋
(
)
4𝜋𝑚 𝐿𝑥,𝑦
=
2
=
ℏ2 𝜋 2
5.57 − 2𝑚 (𝐿 )
𝑧
ℏ2 𝜋
𝑚 𝐿2𝑥,𝑦
=
5.57 − 0.56
= 2.09 × 1011
2.40 × 10−11
𝑁2 = 𝑁1 + (1)2 − (2)2 = 𝑁1 − 3 = 2.09 × 1011 − 3 = 2.09 × 1011
𝑁3 = 𝑁1 − 8 = 2.09 × 1011
2
ℰ𝐹 = ℰℎ𝑖𝑔ℎ (𝑛𝑥 , 𝑛𝑦 , 3) − ℰ𝑙𝑜𝑤 (0,0,1) =
2
ℏ2 2𝜋 𝑁3 ℏ2 𝜋 2
ℏ2 𝜋 2
2
(3)
+
( )
−
( ) (1)2
(
)
2𝑚 𝐿𝑥,𝑦 2𝜋 2𝑚 𝐿𝑧
2𝑚 𝐿𝑧
2
ℏ2 2𝜋 𝑁3 ℏ2 𝜋 2
ℏ2
2𝜋 𝑁3
𝜋 2
=
+
( ) 8=
+ ( ) 8]
(
)
[(
)
2𝑚 𝐿𝑥,𝑦 2𝜋 2𝑚 𝐿𝑧
2𝑚 𝐿𝑥,𝑦 2𝜋
𝐿𝑧
=
(6.5821 × 10−16 𝑒𝑣 ∙ 𝑠)2
1 2
𝜋 2
[(
) 2𝜋 × 2.09 × 1011 + (
) 8] = 5.64 𝑒𝑣
6
6
̇
0.511 × 10 𝑒𝑣
10
8.2𝐴
2×
̇ 2
(3 × 108 × 1010 𝐴⁄𝑠)