Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p...
Transcript of Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p...
4
( )
( )
( )3
3
3
3211
3211
3211
36
13
12
36
11
KKi
Ki
K
KKKK
KKi
Ki
K
eeKE
KE
eeKE
φφφψ
φφφψ
φφφψ
ππ
ππ
++=⇒
−+=⇒
++−=⇒
−−
Problem 5.2 (Free electron energy bands for a FCC lattice in 3D)
a) and b)
aπ4
aπ4
aπ4
1
4
5
Problem 5.2 (Ethene (or Ethylene) molecule: LCAO)
a) As discussed in the lectures, the pz-orbitals (the ones that stick out of the plane of the molecule) are anti-symmetric with respect to reflections about the x-y plane. On the other hand, px-orbitals, py-orbitals, and s-orbitals are all symmetric with respect to reflections about the x-y plane. Consequently, the energy matrix elements of pz-orbitals with all other orbitals will be zero. b) Each carbon atoms is sp2 hybridized. c) The trial solution could be: or in terms of the carbon sp2 orbitals: d) Let the energy of the s-orbital of the hydrogen atom be E1S and let the energies of the 2s and 2p orbitals of the carbon atom be E2S and E2P, respectively. The average energy of the carbon sp2 orbital is given by the expression:
⎟⎟⎠
⎞⎜⎜⎝
⎛ +=
32 22
2ps
spEE
E
Each hydrogen atoms will make a sigma bond with the sp2-orbital of the carbon atom pointing towards it. This will result in a sigma bonding orbital and a sigma anti-bonding orbital with energies equal to:
22
2121
22)A( η+⎟
⎟⎠
⎞⎜⎜⎝
⎛ −+⎟
⎟⎠
⎞⎜⎜⎝
⎛ += spssps
CHEEEE
E degeneracy = 4
22
2121
22)B( η+⎟
⎟⎠
⎞⎜⎜⎝
⎛ −−⎟
⎟⎠
⎞⎜⎜⎝
⎛ += spssps
CHEEEE
E degeneracy = 4
where:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +−=
3
2 σση ppsp VV
The sp2-orbitals of the two carbon atoms will make a sigma bond. This will result in a sigma bonding orbital and a sigma anti-bonding orbital with energies equal to:
λ+= 2)A( spCC EE degeneracy = 1
C C
H H
H H
θ =120o
θ =120o
σ
σ
σ
σ σ π
x
y
( ) ( ) ( ) ( ) ( )∑ −+−+−+∑ −==
++=
6
524222
4
11
mm
Cpymm
Cpxmm
Csm
jj
Hsj rrcrrcrrcrrcr
rrrrrrrrrφφφφψ
( ) ( ) ( ) ( ) ( )∑ −+−+−+∑ −==
++=
6
534221
4
11
mm
Cmm
Cmm
Cm
jj
Hsj rrcrrcrrcrrcr
rrrrrrrrrϕϕϕφψ
6
λ−= 2)B( spCC EE degeneracy = 1 where:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ++−=
3222 σσσλ spppss VVV
e) The solution is: f) The pz-orbitals on carbon atoms make a π-bond with each other. The corresponding energies of the bonding and anti-bonding orbitals are:
πpppCC VEE += 2)A( degeneracy = 1
πpppCC VEE −= 2)B( degeneracy = 1
g) The energy level diagram is shown below. The specific alignments are rough – my best guesses.
2:E2pz
1:ECC(A)
1:ECC(B)
4:E2p
2:E2s
4:E1s
6:Esp2
4:ECH(A)
1:ECC(A)
4:ECH(B)
1:ECC(B)
( ) ( ) ( )622521 rrbrrbr Cpz
Cpz
rrrrr−+−= φφψ
7
6:E2p
2:E2s
4:E1s
6:Esp2 4:ECH(A-σ)
1:ECC(A-σ)
4:ECH(B-σ)
1:ECC(B-σ)
1:ECC(B-π)
1:ECC(A-π) 2:E2pz
1
6.2 a) ppspps EHEHEHEHEHEH 665544332211 ,,,,,
b)
¸¸¹
·¨¨©
§�
¸¹·
¨©§ �� ¸
¹·¨
©§ ���
32
321321
..16
...15
...14
23
23
21
21
nkinkisp
nkinkinkisp
nkinkinkiss
eeVH
eeeVHeeeVH
&&&&
&&&&&&&&&&&&
V
VV
c) ¸¸¹
·¨¨©
§�� ¸
¹·
¨©§ ��� 32321 ..
34...
24 23
23
21
21 nkinki
spnkinkinki
sp eeVHeeeVH&&&&&&&&&&
VV
d) ¸¹·
¨©§ ��¸
¹·
¨©§ ���
¸¹·
¨©§ ���¸
¹·
¨©§ ��
32321
32321
.....36
.....25
43
43
41
41
43
43
41
41
nkinkipp
nkinkinkipp
nkinkipp
nkinkinkipp
eeVeeeVH
eeVeeeVH
&&&&&&&&&&
&&&&&&&&&&
VS
SV
e) There will be 6 bands; 3 will be completely filled and 3 completely empty at zero temperature.
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ECE407 Homework 5 Solutions (By Farhan Rana) Problem 5.1 (Energy bands for a hexagonal lattice in 2D)
a) The K-point (0 , a34π ) is strongly coupled to two other equivalent K-points in the FBZ, as shown in the picture above (all three K-points have been labeled with subscripts 1,2, and 3 for convenience). The equivalent three K-points are related to each other by a reciprocal lattice vector, and the free-electron states with these wavevectors have the same energy. The wavevectors of the other two K-points (2 and 3) are ( a32π , a32π− ) and ( a32π− , a32π− ). b) 3211 321 KKKK ccc φφφψ ++=
After plugging into the solution into the Schrodinger equation and multiplying by the bras corresponding to the three plane wave states one by one we obtain three equations which in the matrix form are:
( )( )
( )( )
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
3
2
1
1
3
2
1
11
11
11
222222
ccc
KEccc
KeVVVKeVVVKe
r
Where ( )Ke equals the free electron energy at (all the equivalent) K-points. Also note that the vector ( )21 bb
rr+ points from K3 to K2 and so in the expression for the potential the term with the wavevector
equal to this vector couples points K3 and K2. The energies of the 3 bands at the K1-point are found to be:
( ) ( ) ( ) 211211 VKeKEKE −== ( ) ( ) 113 VKeKE +=
aπ4
1br
2br
a34π
xk
yk
Γ
K1
M1
K2 K3
M2
2
Note that two bands are still degenerate and one has a higher energy. The corresponding wavefunctions are:
( )
( )
( )3
2
2
3211
311
211
13
12
11
KKKK
KKK
KKK
KE
KE
KE
φφφψ
φφψ
φφψ
++=⇒
−=⇒
−=⇒
c) The M1-point ( a32π ,0) is coupled to the equivalent M2-point ( a32π− ,0). The subscripts 1 and 2 have been added for clarity. d) 211 21 MMM cc φφψ +=
After plugging into the solution into the Schrodinger equation and multiplying by the bras corresponding to the two plane wave states one by one we obtain two equations which in the matrix form are:
( )( ) ( ) ⎥
⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
2
11
2
1
1
12
2cc
MEcc
MeVVMe
The energies of the 2 bands at the M1-point are found to be:
( ) ( ) 2111 VMeME −= ( ) ( ) 2112 VMeME +=
. The corresponding wavefunctions are:
( )
( )2
2
212
11
11
211
MMM
MMM
ME
ME
φφψ
φφψ
+=⇒
−=⇒
e) Notice the points where the bandgaps open up (M-point) and where they don’t (K-point) between the first and the second energy bands. Note also that the exact shape of the bands comes from my imagination – so don’t pay too much to it except to note where the bandgaps open up and where they don’t.
3
f) and g) and h) Solution is obvious if you take the expression for the potential and write the vectors
21, bbrr
in terms of their x- and y-components. i) 3211 321 KKKK ccc φφφψ ++=
After plugging into the solution into the Schrodinger equation and multiplying by the bras corresponding to the three plane wave states one by one we obtain three equations which in the matrix form are:
( )( )
( )( )
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 3
2
1
1
3
2
1
11
11
11
222222
ccc
KEccc
KeViVViKeVVVKe
r
Notice how the entries (2,3) and (3,2) of the matrix have changed. The energies of the 3 bands at the K1-point are found to be:
( ) ( ) 23 111 VKeKE −= ( ) ( )KeKE =12
( ) ( ) 23 113 VKeKE += Note that all bands are now non-degenerate at the K-point. The corresponding wavefunctions are:
4
( )
( )
( )3
3
3
3211
3211
3211
36
13
12
36
11
KKi
Ki
K
KKKK
KKi
Ki
K
eeKE
KE
eeKE
φφφψ
φφφψ
φφφψ
ππ
ππ
++=⇒
−+=⇒
++−=⇒
−−
Problem 5.2 (Free electron energy bands for a FCC lattice in 3D)
a) and b)
aπ4
aπ4
aπ4
1
4