Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p...

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4 ( ) ( ) ( ) 3 3 3 3 2 1 1 3 2 1 1 3 2 1 1 3 6 1 3 1 2 3 6 1 1 K K i K i K K K K K K K i K i K e e K E K E e e K E φ φ φ ψ φ φ φ ψ φ φ φ ψ π π π π + + = + = + + = Problem 5.2 (Free electron energy bands for a FCC lattice in 3D) a) and b) a π 4 a π 4 a π 4 1 4

Transcript of Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p...

Page 1: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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

Homework 5 Solutions
Problem 5.1
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Page 2: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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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

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Page 3: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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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

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Page 4: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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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

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Page 5: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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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.

Problem 5.3
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The next page has all the matrix elements andpictorial representation of each of them.
Page 6: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy
Page 7: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

1

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

Problem 5.4
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Page 8: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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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.

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Page 9: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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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:

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Page 10: Problem 5.2 (Free electron energy bands for a FCC lattice ...and let the energies of the 2s and 2p orbitals of the carbon atom be . E. 2S. and . E. 2P, respectively. The average energy

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

9