Space group symmetry - Chris Marianetti...Space group symmetry elements have a point and translation...

Space group symmetry

So far we have covered point group symmetry and translationsymmetry seperately.

We need to combine them → space group.

While there are an infinite number of point groups, there areonly a finite number that are consistent with translation→ 32 crystallographic point groups

We focused on a few point groups (all crystallographic) in ourexamples: Ci ,C3v ,C4v .

Highest symmetry point group is Oh (cubic group, h = 48).

There are 230 space groups (http://www.cryst.ehu.es/).

Space group symmetry elements have a point and translationcomponent: {Ri , t + τ} (where τ is additional shift).

11/12/2020 at 14:24:48 slide 1

Symmorphic space group

We will only concern ourselves with symmorhphic, as opposedto non-symmorhphic, space groups (73/230 are symm).

Non-symmorhphic space groups (e.g. diamond structure) areslightly more complicated and can be left for another day.

Symmorphic are those with all operations of the form: {Ri , t}→ no glide or screw operations.

A sufficient condition to be symmorhphic is:

Only one species, all on lattice sites.If more than one species, each must form a lattice (e.g. NaCl).At least one point in space has full point symmetry of lattice.

Basic strategy is to start with irreps of translation (ie.k-points) and then use whatever point symmetry remains.

11/12/2020 at 14:24:48 slide 2

Some notes on vector notation for real vectors

Consider usual column vectors vi , and orthogonal operationsO, where Ovi is some new column vector.

When taking a dot product, we can explicitly use matrixnotation, or the dot symbol will account for transpose.

v1 · v2 = v2 · v1 = vᵀ1 v2 = vᵀ

2 v1 = |v1||v2| cos(θ)

We can insert identity 1 = OᵀO in different notations.

vᵀ1 v2 = vᵀ

1 OᵀOv2 = (Ov1)ᵀOv2 ↔ v1 · v2 = (Ov1) · (Ov2)

We have chosen most quantities to be row vectors, such as t,T , k , K , etc, and so k · t = ktᵀ, etc.

Recall equality

K · T = (kb) · (ta) = KT ᵀ = kbaᵀtᵀ = 2πktᵀ = 2πk · t

11/12/2020 at 14:24:48 slide 3

Recall 2d transformations of Cartesian space for C4v

Recall all operations that leave square invariant.a

b

x

y

Below are matrices which transform Cartesian space.

1 =

[1 00 1

]Rπ/2 =

[0 1−1 0

]Rπ =

[−1 00 −1

]R3π/2 =

[0 −11 0

]

Rma =

[0 11 0

]Rmb =

[0 −1−1 0

]Rmx =

[1 00 −1

]Rmy =

[−1 00 1

]

These matrices can be used to build any representation.

Same symbol (e.g. Rmx) used for different reps (e.g.triangluar, square, hexagon molecule).

11/12/2020 at 14:24:48 slide 4

Final points on notation

Given point operations in usual column format, then we haveRT ᵀ; changing column vector to row format gives TRᵀ.

A column operation R operators on a row vector as TRᵀ.

We normally prefer to use lattice coordinates, not Cartesian.

Recall that Ti = ti a, all row ordered. Then we have:

RjT ᵀ = Rj aᵀtᵀ = rjtᵀ rj = Rj aᵀ

We have introduced notation r for operation transforminglattice coordinate.

A column operation r operates on row vector as trᵀj .

11/12/2020 at 14:24:48 slide 5

Point operations on irreps on translation

Recall vectors that transform like irreps of translation

|uαk 〉 =∑

te i2πk·t |uαt 〉 =

∑

t1t2

e2πi(k1t1+k2t2)|uα(t1,t2)〉

〈uαk | =∑

te−i2πk·t〈uαt | =

∑

t1t2

e−2πi(k1t1+k2t2)〈uα(t1,t2)|

Recall that k and t are row vectors.Effect of point group operation centered at some point:

R|uαk 〉 =∑

te i2πkt

ᵀR|uαt 〉 =

∑

te i2πkrᵀ r tᵀRTtR−1RT−1

t |uαt 〉

=∑

te i2π(r kᵀ)ᵀ r tᵀRTtR−1R|uα0 〉 =

∑

te i2π(r kᵀ)ᵀ r tᵀTr tᵀR|uα0 〉

=∑

t′e i2π(krᵀ)·t′Tt′R|uα0 〉 =

∑

t′e i2π(krᵀ)·t′Tt′R|uα0 〉

There are two possible scenarios:

krᵀ = k + G krᵀ = k ′11/12/2020 at 14:24:48 slide 6

Point operations on irreps on translation

If krᵀ = k + G , then we know that the irrep of translationgroup is unchanged:

Γkrᵀ(t) = e−i2πkrᵀ·t = e−i2π(k+G)·t = e−i2πk·t = Γk(t)

In this case, we only need to worry about R|uα0 〉

R|uαk 〉 =∑

β

Γ(R)α,β|uβk 〉

In the latter case, we actually transform the vector to anentirely different k-point:

R|uαk 〉 =∑

β

Γ(R)α,β|uβk ′〉

In any case, given that [R, V ] = 0, we know that:

V |φnk〉 = mω2n,k |φnk〉 → V R|φnk〉 = mω2

n,k R|φnk〉 = mω2n,k |φnk ′〉 → ωn,k = ωn,krᵀ

11/12/2020 at 14:24:48 slide 7

Implications of point symmetry

We see that if R maps the k-point to a distinct one, it must

have the same Eigenvalue: ω(n)k = ω

(n)krᵀ = ω

(n)k ′

Is this a degeneracy? Yes, as the Eigenfunctions are related bya symmetry operation (see Cornwell, 1997).

Let us return to our 1d chain example (point group Ci )ω

k12−1

2

2√

γm

Given that σkᵀ = −kᵀ, we find that the mirror impliesωk = ωk

11/12/2020 at 14:24:48 slide 8

Implications of point symmetry

The maximum degeneracy can be the order of the pointgroup, in the case that the given k-vector is mapped todistinct ones.

In practice, this is not normally called a degeneracy, eventhough it is. Dresselhaus goes so far as to say that it is not.

The important takeaway message is that we only need tocompute a subset of k-point in FBZ that are not related by apoint group operation.

This gives rise to the Irreducible Brilluoin Zone (IBZ)

Once Eigenfunctions in IBZ are known, the rest of FBZ canbe constructed using point operations.

11/12/2020 at 14:24:48 slide 9

Star of the wave vector k and the IBZ

Given a wave vector k , we can form a set by acting with allpoint operations → “identity star of wave vector”.

This is also sometimes called an “identity orbit”.

Consider two stars for the 1d chain (point group Ci ).

In the second case, there is ony one wavevector in the stargiven that the other is equivalent by a reciprocal lattice vector.

The IBZ (green) is simply a collection of one member of allpossible stars in the FBZ (light blue).

The smaller the star, the higher the point symmetry of thatk-point. i.e. the more Ri where krᵀi = k + GAll Ri that leave k invariant is known as “little group” ofk-point. Always subgroup of crystal PG in symmorphic case.

11/12/2020 at 14:24:48 slide 10

IBZ for square lattice and the “little group”

Γ

Σ

R

X∆

Shaded green region is IBZ.

All points in FBZ can be generated frompoints in IBZ using an element of C4v

Must find point symmetry for every pointin IBZ (aka “little group” of k-point) .

Most generally, this info given by “Wycoffpositions” (real or reciprocal).

Why do we care about the little group? These operationsleave k-vector invariant so we may further block diagonalizeour Hamiltonian.

Little group of k is group of Dk (ie. elements that commute)

11/12/2020 at 14:24:48 slide 11

Symmetry of k-points in IBZ

Γ

Σ

R

X∆

Which elements R of C4v

satisfy krᵀ = k + GΓ→ C4v

R → C4v

X → {1,Rπ, σx , σy} → C2v

∆→ {1, σx} → Ci

Σ→ {1, σx} → Ci

All other points → {1}

http://www.cryst.ehu.es/

All of this information haspreviously been tabulated.

11/12/2020 at 14:24:48 slide 12

Square Lattice - Dispersion

We can already understand origin of degeneracy.

Only C4v has multidimensional irreps.

0

0.5

1

1.5

2

2.5

ω

γd=0.1

γσ=1

γπ=1/3

Γ ΓX R

11/12/2020 at 14:24:48 slide 13


How do |uxk〉 and |uyk〉 transform at each k-point?

Γ(C4v ),∆(Ci ),X (C2v ),R(C4v ),Σ(Ci )

0

0.5

1

1.5

2

2.5

ω

γd=0.1

γσ=1

γπ=1/3

Γ ΓX R

D(k1,k2) = 2

γσ(1− cos 2πk1) + γπ(1− cos 2πk2)+γd [2− cos 2π(k1 + k2)− cos 2π(k1 − k2)]

γd [− cos 2π(k1 + k2) + cos 2π(k1 − k2)]

γd [− cos 2π(k1 + k2) + cos 2π(k1 − k2)]γπ(1− cos 2πk1) + γσ(1− cos 2πk2)+γd [2− cos 2π(k1 + k2)− cos 2π(k1 − k2)]

11/12/2020 at 14:24:48 slide 14


0

0.5

1

1.5

2

2.5

ω

γd=0.1

γσ=1

γπ=1/3

Γ ΓX R

D(k1,k2) = 2

γσ(1− cos 2πk1) + γπ(1− cos 2πk2)+γd [2− cos 2π(k1 + k2)− cos 2π(k1 − k2)]

γd [− cos 2π(k1 + k2) + cos 2π(k1 − k2)]

γd [− cos 2π(k1 + k2) + cos 2π(k1 − k2)]γπ(1− cos 2πk1) + γσ(1− cos 2πk2)+γd [2− cos 2π(k1 + k2)− cos 2π(k1 − k2)]

DR = 2

[2γσ + 2γπ 0

0 2γσ + 2γπ

]

D∆ = 2

[γσ(1− cos 2πk∆) + γd [2− 2 cos 2π(k∆)] 0

0 γπ(1− cos 2πk∆) + γd [2− 2 cos 2π(k∆)]

]

11/12/2020 at 14:24:48 slide 15

Recall character tables...

C4v1 Rπ Rπ/2R3π/2 RmxRmy RmaRmb

C1 C2 C3 C4 C5

A1 1 1 1 1 1B1 1 1 −1 1 −1A2 1 1 1 −1 −1B2 1 1 −1 −1 1E 2 −2 0 0 0

C2v1 Rπ Rmx Rmy

C1 C2 C3 C4

A1 1 1 1 1B1 1 −1 1 −1A2 1 1 −1 −1B2 1 −1 −1 1

1 σ

A 1 1B 1 −1

11/12/2020 at 14:24:48 slide 16

Visualizing the Γ-point of square lattice

Left panel: |uxΓ〉; Right panel: |uyΓ〉; modulation e0

11/12/2020 at 14:24:48 slide 17

Visualizing the R-point of square lattice

Left panel: |uxR〉; Right panel: |uyR〉; modulation e iπ(t1+t2)

11/12/2020 at 14:24:48 slide 18

Visualizing the X-point of square lattice

Left panel: |uxX 〉; Right panel: |uyX 〉; modulation e iπt1

11/12/2020 at 14:24:48 slide 19

Visualizing the Y-point of square lattice

Left panel: |uxY 〉; Right panel: |uyY 〉; modulation e iπt2

11/12/2020 at 14:24:48 slide 20

Visualizing the ∆ = (1/3, 0) of square lattice

Left panel: |ux( 1

3,0)〉; Right panel: |uy

( 13,0)〉; modulation e i

2π3t1

11/12/2020 at 14:24:48 slide 21

Visualizing the Σ = (1/3, 1/3) of square lattice

Left panel: |ux( 1

3, 1

3)〉; Right panel: |uy

( 13, 1

3)〉; mod. e i

2π3

(t1+t2)

11/12/2020 at 14:24:48 slide 22

Visualizing the Σ = (1/3, 1/3) (irrep) of square lattice

Left panel: |uAk 〉; Right panel: |uBk 〉

11/12/2020 at 14:24:48 slide 23

Graphene

Let us return to graphene to do a major example.

Triangular lattice point symmetry is C6v .

How do we know graphene is symmorphic?

a1

a2

b1

b2

M

∆

K

Σ

Γ

Consider K = ( 23 ,

13 ); what is LG?

At K-point, Krᵀ = K + G for all R in C3v

Symmetrize in-plane displacements at K-point.

11/12/2020 at 14:24:48 slide 24

Recall in-plane dispersion relation

0

500

1000

1500

cm

-1

Γ ΓM K

b1

b2

M

∆

K

Σ

Γ

Symmetry will explain degeneracy at K -point.

Recall definition of translationally symmetrized basis:

|u(a,α)(k1,k2)〉 =

∑

t1t2

e i2π(k1t1+k2t2)|u(a,α)(t1,t2)〉

|u(a,α)

( 23, 1

3)〉 =

∑

t1t2

e i2π3

(2t1+t2)|u(a,α)(t1,t2)〉

11/12/2020 at 14:24:48 slide 25

Recall D( 23 ,

13 ) for NN and NNN springs in graphene

d1 and d2 are linear combinations of γσ, γπ, γ′σ, γ′π

D( 23, 1

3) =

d1 0 −d2 00 d1 0 d2

−d2 0 d1 00 d2 0 d1

+ i

0 0 0 −d2

0 0 −d2 00 d2 0 0d2 0 0 0

d1 =3

2

(γσ + γπ + 3γ′σ + 3γ′π

), d2 =

3

4(γσ − γπ)

d1 = 65.5950 eV/A2 and d2 = 12.8213 eV/A2

Taking the Eigenvals of this we get

λ =(39.9525, 65.595, 65.595, 91.2375

)

divide by Carbon mass, take square root, and convert to cm-1:

ω =(951.1, 1218.6, 1218.6, 1437.2

)

Prove why we get a 2-fold and two 1-folds.

Are zeros in above D( 23, 1

3) required by symmetry?

11/12/2020 at 14:24:48 slide 26

Irreducible representation vectors of little group

Need to keep big picture in mind.

Already symmetrized according to translation (i.e. |u(a,α)k 〉)

Now we know that Dk is invariant to little group.

Symmetrize |u(a,α)k 〉 according to little group.

How? The same recipe as always.

Represent little group elements in basis of choice.

Find composition by tracing representation.

Project irrep vectors using projection operator.

11/12/2020 at 14:24:48 slide 27

Decompose basis at K-point: little group C3v

Recall the character table of C3v

C3v 1 R2π/3R4π/3 σy σaσb

A1 1 1 1A2 1 1 −1E 2 −1 0

We have a 4-d representation for each R:

|uA,xK 〉, |uA,yK 〉, |uB,xK 〉, |uB,yK 〉

〈u(a,α)K |R|u(b,β)

K 〉 =1

N2

∑

t1t2t′1t′2

e i2π[ 23

(t1−t′1)+ 13

(t2−t′2)]〈u(a,α)(t′1,t

′2)|T(t1,t2)R|u(b,β)

(0,0) 〉

Let us quickly find the composition. We know tr(1) = 4, solet us make R2π/3 and σy .I will use the C3v subgroup of C6v , located at a1.Only need to know following for R ∈ {R2π/3, σy}:

R|uA,x0 〉, R|uA,y0 〉, R|uB,x0 〉, R|uB,y0 〉11/12/2020 at 14:24:48 slide 28

Schematic of phases for K-point

Each k has corresponding supercell: recall e i2πk·t .

Red lines give K-cell: S1 = (1, 1),S2 = (1, 2)

Supercell must have 2 linearly ind. vecs where k · Si ∈ Z

e0 e0

ei4π3ei

4π3

ei2π3 ei

2π3

σy

σa

σb

11/12/2020 at 14:24:48 slide 29

Primitive cells overlayed with K-cell

(0,0)

(1,1)

(1,0)(0,1)

σy

σa

σb

11/12/2020 at 14:24:48 slide 30

Decompose atom (site) representation at K-point

Decompose atoms and polarizations separately: R = Rs ⊗ Rp

Rs2π/3|uA(0,0)〉 = |uA(1,0)〉 = T(1,0)|uA(0,0)〉, Rs

2π/3|uB(0,0)〉 = |uB(1,1)〉

Let us get diagonal elements for Rs2π/3

〈uAK|Rs2π/3|uAK〉 = 〈uAK|T(1,0)|uAK〉 = e−i2π( 2

3, 1

3)·(1,0) = e i2π

−23 = e i

2π3

〈uBK|Rs2π/3|uBK〉 = 〈uBK|T(1,1)|uBK〉 = e−i2π( 2

3, 1

3)·(1,1) = e i2π

−13 = e i

4π3

We then get Tr(Rs2π/3) = −1

2 +√

32 i− 1

2 −√

32 i = −1

We can immediately deduce that tr(σsy ) = 0

σsy |uA(0,0)〉 = |uB(0,0)〉 σs

y |uB(0,0)〉 = |uA(0,0)〉

We conclude χs = (2,−1, 0) = E

11/12/2020 at 14:24:48 slide 31

Decompose polarizations at K-point

Representations for polarizations can be written seperately:

σpy =

[−1 00 1

]Rp

2π/3 =

[cos θ − sin θsin θ cos θ

]=

[−1

2 −√

32√

32 −1

2

]

We have Tr(σpy

)= 0 and Tr

(Rp

2π/3

)= −1

Decomposition is χp = (2,−1, 0) = E

Total decomposition is

χ = χs ∗ χp = (4, 1, 0)→ Γ = E ⊗ E = A1 ⊕ A2 ⊕ E

This explains 2-fold and two 1-folds at the K-point.

Now construct the actual matrices so we can project out.

11/12/2020 at 14:24:48 slide 32

Representation matrices for atoms

We already deduced Rs2π/3 and σs

y , which generate group.

Rs2π/3 =

[e i

2π3 0

0 e−i2π3

]σsy =

[0 11 0

]

Generate remaining group elements.

Rs4π/3 =

(Rs

2π/3

)2=

[e−i

2π3 0

0 e i2π3

]

σsa = Rs

4π/3σsy =

[0 e−i

2π3

e i2π3 0

]

σsb = Rs

2π/3σsy =

[0 e i

2π3

e−i2π3 0

]

11/12/2020 at 14:24:48 slide 33

Representation matrices for polarizations

We already deduced Rp2π/3 and σp

y , which generate group.

Rp2π/3 =

[−1

2 −√

32√

32 −1

2

]σpy =

[−1 00 1

]

Generate remaining group elements.

Rp4π/3 =

(Rp

2π/3

)†=

[−1

2

√3

2

−√

32 −1

2

]

σpa = Rp

4π/3σpy =

[12

√3

2√3

2 −12

]

σpb = Rp

2π/3σpy =

[12 −

√3

2

−√

32 −1

2

]

11/12/2020 at 14:24:48 slide 34

Full representation matrices

We now have the representations for the atoms and thepolarizations, and we can build the full representation usingthe Kronecker Product.

R = Rs ⊗ Rp v = v s ⊗ vp

Given that we are doing this by hand, it is easier to work withthe matrices seperately.

Let us now project out the irreps A1 ⊕ A2 ⊕ E

A possible seed is |uAK〉 ⊗ |x〉 =

[10

]⊗[

10

]= |s〉

Let us tabulate how the atom and polarization reps operateon these seeds, by simply extracting the first column fromeach matrix.

11/12/2020 at 14:24:48 slide 35

Tabulation of projected seeds for atoms/polarizations

First tabulate site operations on site seed.

C s2π/3|uAK〉 =

[10

]e i

2π3 C s

4π/3|uAK〉 =

[10

]e−i

2π3

σsy |uAK〉 =

[01

]σsa|uAK〉 =

[01

]e i

2π3 σs

b|uAK〉 =

[01

]e−i

2π3

Now tabulate polarization operations on polarization seed.

Cp2π/3|x〉 =

[−1

2√3

2

]Cp

4π/3|x〉 =

[−1

2

−√

32

]

σpy |x〉 =

[−10

]σpa |x〉 =

[12√3

2

]σpb |x〉 =

[12

−√

32

]

We can now just add the character weighted Kroneckerproduct of each pair of vectors.

Do not forget the identity operator.11/12/2020 at 14:24:48 slide 36

Projecting out A1 irrep

PA1 |s〉 =

[10

]⊗[

10

]+

[10

]⊗[−1

2√3

2

]e i

2π3 −

[10

]⊗[

12√3

2

]e−i

2π3 +

[01

]⊗[−10

]+

[01

]⊗[

12√3

2

]e i

2π3 +

[01

]⊗[

12

−√

32

]e−i

2π3

=

1000

+

−12√3

200

e i

2π3 −

12√3

200

e−i

2π3 −

0010

+

0012√3

2

e i

2π3 +

0012

−√

32

e−i

2π3 → |uA1

K 〉 =1

2

1i−1i

K

11/12/2020 at 14:24:48 slide 37

Projecting out A2 irrep

PA2 |s〉 =

[10

]⊗[

10

]+

[10

]⊗[−1

2√3

2

]e i

2π3 −

[10

]⊗[

12√3

2

]e−i

2π3 −

[01

]⊗[−10

]−[

01

]⊗[

12√3

2

]e i

2π3 −

[01

]⊗[

12

−√

32

]e−i

2π3

=

1000

+

−12√3

200

e i

2π3 −

12√3

200

e−i

2π3 +

0010

−

0012√3

2

e i

2π3 −

0012

−√

32

e−i

2π3 → |uA2

K 〉 =1

2

1i1−i

K

11/12/2020 at 14:24:48 slide 38

Projecting out E irrep

PE |s〉 =

2000

+

12

−√

32

00

e i

2π3 +

12√3

200

e−i

2π3 → |uE (1)

K 〉 =1√2

1−i00

K

We can rotate this to find second one, or just guess based onorthogonality.

|uE (2)

K 〉 =1√2

001i

K

11/12/2020 at 14:24:48 slide 39

Summary of Symmetrized Modes at K

The four symmetrized modes are thus:

|uA1K 〉 =

1

2

1i−1i

K

|uA2K 〉 =

1

2

1i1−i

K

|uE (1)

K 〉 =1√2

1−i00

K

|uE (2)

K 〉 =1√2

001i

K

11/12/2020 at 14:24:48 slide 40

Block diagonalizing

So we deduced that:

|uA1K 〉 =

1

2

(|u(A,x)

K 〉+ i|u(A,y)K 〉 − |u(B,x)

K 〉+ i|u(B,y)K 〉

)

Now we can write second derivative of A1 mode at K

〈uA1K |D( 2

3, 1

3)|uA1

K 〉 = d1 + 2d2 =3

2

(2γσ + 3γ′σ + 3γ′π

)

= 91.2376 eV/A2

ω(A1)K =

√d1 + 2d2

m=

√3

2m

√2γσ + 3γ′σ + 3γ′π = 1437.2 cm−1

11/12/2020 at 14:24:48 slide 41

Block diagonalizing Dk

We can now stack the irreps to form the unitarytransformation and block diagonalize.

U = 12

1 1√2 0

i i −i√2 0

−1 1 0√2

i −i 0 i√2

DdK = U†DKU

DK = UDdKU†

Recall original Dynamical matrix.

D( 23, 1

3) =

d1 0 −d2 00 d1 0 d2

−d2 0 d1 00 d2 0 d1

+ i

0 0 0 −d2

0 0 −d2 00 d2 0 0d2 0 0 0

Block diagonalizing, we then have:

DdK =

dA1 0 0 00 dA2 0 00 0 dE 00 0 0 dE

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Irreducible derivatives

Had we not been given Hamiltonian, we could deduceirreducible derivatives.

DdK =

dA1 0 0 00 dA2 0 00 0 dE 00 0 0 dE

There are three irreducible derivatives at the K-point.

Rotating back from irrep basis to naive basis, we have:

DK =1

4

2dE + dA1 + dA2 2idE − idA1 − idA2 −dA1 + dA2 −idA1 + idA2

−2idE + idA1 + idA2 2dE + dA1 + dA2 −idA1 + idA2 dA1 − dA2

−dA1 + dA2 idA1 − idA2 2dE + dA1 + dA2 −2idE + idA1 + idA2

idA1 − idA2 dA1 − dA2 2idE − idA1 − idA2 2dE + dA1 + dA2

We see that the dynamical matrix is full in the naive basis.

If we added sufficiently long range springs, we would see allelements fill out as above.

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Picturing A1 modes

|uA1Kc〉

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Picturing A1 modes

|uA1Ks〉

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Picturing A2 modes

|uA2Kc〉

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Picturing A2 modes

|uA2Ks〉

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Summary of symmorphic space group symmetrization

We have completed the process of fully symmetrizing thedynamical matrix using point symmetry.

The dynamical matrix is thus fully block diagonalized.

Additionally, we can write down the space group irreduciblesecond derivatives of the potential energy at a given k apriori.

This might still seem disatisfying given that there are aninfinite number of k-points.

Just as we can truncate the range of the springs in real space,we can truncate the resolution of k-space.

This is naturally done simply by controlling the finite size ofour translation group via N, and hence the number ofk-points in the FBZ.

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Irreducible derivatives of finite space group

N will set the discretization of the FBZ.

For lower symmetry systems, we could individually vary Ni .

For a tetragonal crystal, we could individually set N1 = N2

and N3.

More advanced schemes are possible, but the translationgroup must always remain consistent with point symmetry.

Let us explore uniform grids of k-points for the triangularlattice.

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Uniform grid for triangular lattice

b1

b2

M

K

Γ

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Uniform grid for triangular lattice

M

K

Γ

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Supercell for given k-point

An N × N supercell will be commensurate with all k points.

However, a smaller supercell may suffice.

The supercell of the K-cell was already given.

Supercell must have 2 linearly ind. vecs Si where k · Si ∈ ZWe are looking for a a set of vectors for which the phasepattern will be left invariant.

e−i2πk·Si = 1

Therefore, the phase at a given lattice point will be invariantto Si . Consider site t for k :

e−i2πk·(t+Si ) = e−i2πk·t

In dimension d , one needs to find d linearly independentsupercell vectors.

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Minimal unit cell for M= (12 ,

12) : s1 = (1, 1) s2 = (1, 1)

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Minimal unit cell for K= (23 ,

13) : s1 = (1, 2) s2 = (1, 1)

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Minimal unit cell for ∆ = (13 ,

13) : s1 = (2, 1) s2 = (1, 2)

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Summary of Little groups and irreps for Grapheneb1

b2

M

∆

K

Σ

Γ

Γ : C6v M : C2v K : C3v ∆3 : Cs

In-plane displacements transform like:

Γ : E1 M : B1 ⊕ B2 K : E ∆3 : A⊕ B

Atoms transform like:

Γ : A1 ⊕ B1 M : A1 ⊕ B1 K : E ∆3 : A⊕ A

Forming the full representation, we have:

Γ : E2 M : A1 ⊕ A2 ⊕ B1 ⊕ B2 K : A1 ⊕ A2 ⊕ E ∆3 : 2A⊕ 2B

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Space group symmetry - Chris Marianetti...Space group symmetry elements have a point and translation...

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Transcript of Space group symmetry - Chris Marianetti...Space group symmetry elements have a point and translation...