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10/05/2015 PHY 752 Fall 2015 -- Lecture 18 1

PHY 752 Solid State Physics11-11:50 AM MWF Olin 103

Plan for Lecture 18:

Reading: Chapter 6 in GGGPP Electronic properties of selected materials

1. Ionic crystals – Ewald summation and binding energy

2. Band structure

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10/05/2015 PHY 752 Fall 2015 -- Lecture 18 3

Ionic solids

Example -- NaCl

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10/05/2015 PHY 752 Fall 2015 -- Lecture 18 4

Ewald summation methods -- motivation

( , ; )0

Consider a collection of point charges { } located at points { }.

The energy to separate these charges to infinity( } is

.4 | |

1

i i

i

i j

i j i j i j

q

q qW

r

r

r r

Here the summation is over all pairs of ( , ),

excluding . It is convenient to sum over all particles

and divide by 2 to compensate for the double counting:

i j

i j

, ;0

1.

8 | |

i j

i j i j i j

q qW

r r

The energy scales as the number of particles

. As N , the ratio

Here the summation is over all pairs

/ remains well-defined

in principle, but difficult to calculate in prac

of , , excluding

. W

N W N

i j

i j

tice.

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Ewald summation methods – exact results for periodic systems

2 / 2

20 0

1erfc( | |)4 42 .

8 | | 8

i Gq qW e e Q

N G

αβG τ

αβ

G 0 T αβ

τ T

τ T

See lecture notes for details.

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Summary of Ewald summation for electrostatic energy

Using identity:

Electrostatic energy becomes:

Summation in reciprocal space

Summation in real space

N

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10/05/2015 PHY 752 Fall 2015 -- Lecture 18 7

Some details for singular term

2

1

2sing

s

30 0

in

0

g diverges, representing the Coulomb

interaction of an infinite amount of charge. In this case, we can

4 1 4 1

8 2 8 2

Suppose that 0.

howev

f

er

q q q qduvdv

W

uN

q

W

Q

23 3

20

2comp

0 0

ind the corresponding energy of a neutral system, where we subtract a

compensating uniform charge density with

1 1 1' 2 ' '

8 8

Q

W Q Qd r

Nd r r dr

N

r r'

2

2

2sing comp

0

2 2

0

0

0 0

2' ' ' '

8

2 4 = ' '

8 8

W W Qr dr r dr

N N

Q Qr dr

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2 / 2

20 0

1erfc( | |)4 42 .

8 | | 8

i Gq qW e e Q

N G

αβG τ

αβ

G 0 T αβ

τ T

τ T

Ewald expansion formula:

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Example for CsCl structure

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10/05/2015 PHY 752 Fall 2015 -- Lecture 18 10

2 / 2

20 0

1erfc( | |)4 42 .

8 | | 8

i Gq qW e e Q

N G

αβG τ

αβ

G 0 T αβ

τ T

τ T

Evaluation of summation

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Maple expressionsSee Maple sheet

2

0

2

0

For this case, we find:

4.070723105

8

In terms of Madelung constant:

where for CsCl, 1.76

4

27

M

M

W e

N a

W e

N R

lattice constant

nearest neighbor3

For CsCl, 2

4.070723105 1.7627

2

R a

a R

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10/05/2015 PHY 752 Fall 2015 -- Lecture 18 13

Summary of Coulomb interaction energiesNote: In the rest of the lecture notes, we will resume the cgs Gaussian units used in your textbook

2

Coul

Repul

Coulombic interaction:

Convenient form for quantum repulsion:

M

n

eU N

NR

R

U

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Simple interaction model for ionic crystals

Bulk modulus

For NaClstructure

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Band structurefor NaCl

Filled Cl bands

Empty Na bands

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10/05/2015 PHY 752 Fall 2015 -- Lecture 18 16

How do you know if a material is ionic?

Atomic configurations (neutral)Na: 1s2 2s2 2p6 3sCl: 1s2 2s2 2p6 3s2 3p5

6 valence electrons

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Band structurefor NaCl

Filled Cl bands;accommodate 6 electrons

Empty Na bands

Cl 3p5 3p6

Na 3s1 3s0

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Electronic structure of Li2SnO3

Note: Valence bands must accommodate 72 valence electrons from 8 Li 2s1 , 4 Sn 5s25p2, and 12 O 2p4 atoms per unit cell

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Partial density of states for Li2SnO3

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Diamond lattice (2 C atoms per unit cell)

Example of a non-ionic material

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Ref. PRB 2, 2054 (1970)

Note: Valence bands must accommodate 8 valence electrons from two C 1s22s22p2

atoms per unit cell

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Crystal structure of graphite (4 atoms per unit cell)

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Ref. PRB 26, 5382 (1982)

Band diagram for graphite

Note: Valence bands must accommodate 16 valence electrons from four C 1s22s22p2 atoms per unit cell