Space group symmetry //8 at :4:4 slide So far we have covered point group symmetry and translation symmetry seperately. We need to combine them space group. While there are an infinite number of point groups, there are only a finite number that are consistent with translation crystallographic point groups We focused on a few point groups (all crystallographic) in our examples: C i, C v, C 4v. Highest symmetry point group is O h (cubic group, h = 48). There are space groups (http://www.cryst.ehu.es/). Space group symmetry elements have a point and translation component: {ˆR i, t + τ} (where τ is additional shift).
Symmorphic space group //8 at :4:4 slide We will only concern ourselves with symmorhphic, as opposed to non-symmorhphic, space groups (7/ are symm). Non-symmorhphic space groups (e.g. diamond structure) are slightly more complicated and can be left for another day. Symmorphic are those with all operations of the form: {ˆR i, t n } 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 had full point symmetry of lattice. Basic strategy is to start with irreps of translation (ie. k-points) and then use whatever point symmetry remains.
Point operations on irreps on translation //8 at :4:4 slide u α k = t e ik t u α t = t t e πi(k t +k t ) u α (t,t ) u α k = t e ik t u α t = t t e πi(k t +k t ) u α (t,t ) Effect of point group operation centered at some point: ˆR u α k = t e ik t ˆR u α t = t e ik ˆR ˆR t ˆRˆTt ˆR ˆRˆT t ut α = t e i(ˆrk) ˆR t ˆRˆT t ˆR ˆR u α = t e i(ˆrk) ˆR t ˆTˆRt ˆR u α = t e i(ˆrk) t ˆT t ˆR u α There are two possible scenarios: ˆRk = k + G ˆRk = k
Point operations on irreps on translation //8 at :4:4 slide 4 If ˆRk = k + G, then we know that the irrep of translation group is unchanged: ΓˆRk (t) = e iˆrk t = e i(k+g) t = e ik t = Γ k (t) In this case, we only need to worry about ˆR u α ˆR u α k = β Γ(ˆR) α,β u β k In the latter case, we actually transform the vector to an entirely different k-point: ˆR u α k = β Γ(ˆR) α,β u β k In any case, given that [ˆR, ˆV =, we know that: ˆV φ n k = mω n,k φn k ˆVˆR φ n k = mω n,k ˆR φ n k = mω n,k φn k ω n,k = ω n,ˆrk
Implications of point symmetry //8 at :4:4 slide 5 We see that if ˆR maps the k-point to a distinct one, it must have the same Eigenvalue: ω (n) k = ω (n) ˆRk = ω(n) k Is this a degeneracy? Yes, as the Eigenfunctions are related by a symmetry operation (see Cornwell, 997). Let us return to our d chain example (point group C i ) ω γ m π a π a k Given that σk = k, we find that the mirror implies ω k = ω k
Implications of point symmetry //8 at :4:4 slide 6 The maximum degeneracy can be the order of the point group, in the case that the given k-vector is mapped to distinct ones. In practice, this is not normally called a degeneracy, even though it is. Dresselhaus goes so far as to say that it is not. The important takeaway message is that we only need to compute a subset of k-point in FBZ that are not related by a point group operation. This gives rise to the Irreducible Brilluoin Zone (IBZ) Once Eigenfunctions in IBZ are known, the rest of FBZ can be constructed using point operations.
Star of the wave vector k and the IBZ //8 at :4:4 slide 7 Given a wave vector k, we can form a set by acting with all point operations identity star of wave vector. This is also sometimes called an identity orbit. Consider two stars for the d chain (point group C i ). In the second case, there is ony one wavevector in the star given that the other is equivalent by a reciprocal lattice vector. The IBZ (green) is simply a collection of one member of all possible stars in the FBZ (light blue). The smaller the star, the higher the point symmetry of that k-point. i.e. the more ˆR i where ˆR i k = k + G All ˆR i that leave k invariant is known as little group of k-point. Always subgroup of crystal PG in symmorphic case.
IBZ for square lattice and the little group //8 at :4:4 slide 8 Γ Σ R X Shaded green region is IBZ. All points in FBZ can be generated from points in IBZ using an element of C 4v Must find point symmetry for every point in IBZ (aka little group of k-point). Most generally, this info given by Wycoff positions (real or reciprocal). Why do we care about the little group? These operations leave k-vector invariant so we may further block diagonalize our Hamiltonian. Little group of k is group of ˆD k (ie. elements that commute)
Symmetry of k-points in IBZ Σ R Which elements ˆR of C 4v satisfy ˆRk = k + G Γ C 4v Γ X R C 4v X {E, R π, σ x, σ y } C v {E, σ x } C i Σ {E, σ x } C i All other points {E} http://www.cryst.ehu.es/ All of this information has previously been tabulated. //8 at :4:4 slide 9
Square Lattice - Dispersion //8 at :4:4 slide.5.5 We can already understand origin of degeneracy. Only C 4v has multidimensional irreps. γ σ = γ π =/ γ d =. ω.5 Γ X R Γ
Square Lattice - Dispersion //8 at :4:4 slide How do u x k and uy k transform at each k-point? Γ(C 4v ), (C i ), X (C v ), R(C 4v ), Σ(C i ).5.5 γ σ = γ π =/ γ d =. ω.5 Γ X γ σ ( cos πk ) + γ π ( cos πk )+ ˆD (k,k) = γ d [ cos π(k + k ) cos π(k k ) γ d [ cos π(k + k ) + cos π(k k ) R Γ γ d [ cos π(k + k ) + cos π(k k ) γ π ( cos πk ) + γ σ ( cos πk )+ γ d [ cos π(k + k ) cos π(k k )
Square Lattice - Dispersion //8 at :4:4 slide.5 γ σ = γ π =/ γ d =..5 ω.5 Γ X R γ σ ( cos πk ) + γ π ( cos πk )+ ˆD (k,k) = γ d [ cos π(k + k ) cos π(k k ) γ d [ cos π(k + k ) + cos π(k k ) Γ γ d [ cos π(k + k ) + cos π(k k ) γ π ( cos πk ) + γ σ ( cos πk )+ γ d [ cos π(k + k ) cos π(k k ) [ γσ + γ ˆD R = π γ σ + γ π [ γσ ( cos πk ˆD = ) + γ d [ cos π(k ) γ π ( cos πk ) + γ d [ cos π(k )
Recall character tables... E R C π R π/ R π/ R mx R my R ma R mb 4v C C C C 4 C 5 A B A B E E R C π R mx R my v C C C C 4 A B A B E σ A B //8 at :4:4 slide
Visualizing the Γ-point of square lattice //8 at :4:4 slide 4 Left panel: u x k ; Right panel: uy k
Visualizing the R-point of square lattice //8 at :4:4 slide 5 Left panel: u x k ; Right panel: uy k
Visualizing the X-point of square lattice //8 at :4:4 slide 6 Left panel: u x k ; Right panel: uy k
Visualizing the Y-point of square lattice //8 at :4:4 slide 7 Left panel: u x k ; Right panel: uy k
Visualizing the = (/, ) of square lattice //8 at :4:4 slide 8 Left panel: u x k ; Right panel: uy k
Visualizing the Σ = (/, /) of square lattice //8 at :4:4 slide 9 Left panel: u x k ; Right panel: uy k
Visualizing the Σ = (/, /) (irrep) of square lattice //8 at :4:4 slide Left panel: u A k ; Right panel: ub k
Graphene //8 at :4:4 slide Let us return to graphene to do a major example. Triangular lattice point symmetry is C 6v. How do we know graphene is symmorphic? a b Γ Σ K M a b Consider k = (, ) or K = b + b ; what is LG? At K-point, ˆRK = K + G for all ˆR in C v
Recall dispersion relation //8 at :4:4 slide 5 cm - b K 5 Γ Σ M Γ M K Symmetry will explain degeneracy at K-point. Recall definition of translationally symmetrized basis: Γ b u α (k,k ) = t t e iπ(k t +k t ) u α (t,t ) u α (, ) = t t e i π (t +t ) u α (t,t )
Recall Dynamical matrix at K-point of graphene //8 at :4:4 slide d d d ˆD (, ) = d d d d + i d d d d d d and d are linear combinations of γ σ, γ π, γ σ, γ π d = ( γσ + γ π + γ σ + γ π) d = 4 (γ σ γ π ) d = 65.595 ev/å and d =.8 ev/å Taking the Eigenvals of this we get λ = ( 9.955, 65.595, 65.595, 9.75 ) divide by Carbon mass, take square root, and convert to cm-: ω = ( 95., 8.6, 8.6, 47. ) Prove why we get a -fold and two -folds.
Decompose basis at K-point: little group C v Recall the character table of C v C v Ê ˆR π/ ˆR 4π/ ˆσ y ˆσ aˆσ b A A E We have a 4-d representation for each ˆR: u (α,r) K ˆR u (β,s) K = N u A,x K ua,y K ub,x K ub,y t t t t K e iπ[ (t t )+ (t t ) u (α,r) (t ) ˆT (β,s),t (t,t ) ˆR u (,) Let us quickly find the composition. We know tr(ê) = 4, so let us make ˆR π/ and ˆσ y. I will use the C v subgroup of C 6v, located at a Only need to know following for ˆR = {ˆR π/, ˆσ y }: ˆR u A,x ˆR u A,y ˆR u B,x ˆR u B,y //8 at :4:4 slide 4
Schematic of phases for K-point //8 at :4:4 slide 5 σ y σ a e i π e i π e i 4π e i 4π σ b e e Red lines give K-cell: S = (, ), S = (, ) Supercell must have linearly ind. vecs where K S i Z
Primitive cells overlayed with K-cell //8 at :4:4 slide 6 (, ) (, ) (,) (,) σ y σ a σ b
Decompose atom representation at K-point //8 at :4:4 slide 7 Decompose atoms and polarizations separately: ˆR = ˆR s ˆR p ˆR s π/ ua (,) = ua (,) = ˆT (,) u A (,) ˆR s π/ ub (,) = ub (, ) Let us get diagonal elements for ˆR s π/ u A K ˆR s π/ ua K = ua K ˆT (,) u A K = e iπ(, ) (,) iπ = e = e i π u B K ˆR s π/ ub K = ub K ˆT (, ) ub K = e iπ(, ) (, ) iπ = e = e i 4π We then get Tr(ˆR s π/ ) = = and We can immediately deduce that tr(ˆσ y s ) = ˆσ s y u A (,) = ub (,) ˆσs y u B (,) = ua (,) We conclude χ s = (,, ) = E
Decompose polarizations at K-point //8 at :4:4 slide 8 Representations for polarizations can be written seperately: [ [ [ ˆσ y p = ˆR p cos θ sin θ π/ = = sin θ cos θ We have Tr (ˆσ y p ) ) = and Tr (ˆR p π/ = Decomposition is χ p = (,, ) = E Total decomposition is χ = χ p χ s = (4,, ) = E E = A A E This explains -fold and two -folds at the K-point. Now let us move on to contruct the actual matrices so we can project out.
Representation matrices for atoms //8 at :4:4 slide 9 We already deduced ˆR s π/ and ˆσs y, which generate group. [e ˆR s i π π/ = e i π Generate remaining group elements. ˆσ s y = [ ) ˆR s 4π/ (ˆR = s e i π π/ = e i π [ ˆσ a s = ˆR s e i π 4π/ˆσs y = e i π ˆσ s b = ˆR s π/ˆσs y = [ e i π e i π [
Representation matrices for polarizations //8 at :4:4 slide We already deduced ˆR p π/ and ˆσp y, which generate group. ˆR p π/ = [ ˆσ p y = [ Generate remaining group elements. [ ) ˆR p 4π/ (ˆR = p π/ = ˆσ p a = ˆR p 4π/ˆσp y = [ ˆσ p b = ˆR p π/ˆσp y = [
Full representation matrices //8 at :4:4 slide We now have the representations for the atoms and the polarizations, and we can build the full representation using the Kronecker Product. ˆR = ˆR s ˆR p v = v s v p Given that we are doing this by hand, it is easier to work with the matrices seperately. Let us now project out the irreps A A E [ [ A fine seed is uk A x = = s Let us tabulate how the atom and polarization reps operate on these seeds, by simply extracting the first column from each matrix.
Tabulation of projected seeds for atoms/polarizations //8 at :4:4 slide First tabulate site operations on site seed. Ĉ s π/ ua K = [ ˆσ s y u A K = [ [ e i π Ĉ s 4π/ ua K = [ ˆσ a u s K A = e i π e i π ˆσ s b ua K = [ e i π Now tabulate polarization operations on polarization seed. [ [ Ĉ p π/ x = Ĉ p 4π/ x = [ [ [ ˆσ y p x = ˆσ a p x = ˆσ p b x = We can now just add the character weighted Kronecker product of each pair of vectors. Do not forget the identity operator.
Projecting out A irrep //8 at :4:4 slide P A s = [ = [ [ [ + π ei + [ + [ + [ [ π ei e i π [ [ e i π + π e i [ [ e i π u A K = + i i e i π + e i π
Projecting out A irrep //8 at :4:4 slide 4 P A s = [ = [ [ [ + π ei [ + [ [ [ π ei e i π e i π [ [ π e i + [ [ e i π u A K = i i e i π e i π
Projecting out E irrep //8 at :4:4 slide 5 P E s = + π ei + e i π u E () K = i We can rotate this to find second one, or just guess based on orthogonality. u E () K = i
Summary of Symmetrized Modes at K //8 at :4:4 slide 6 The four symmetrized modes are thus: u A K = u E () K = i i i u A K = u E () K = i i i
Block diagonalizing //8 at :4:4 slide 7 So we deduced that: u A K = ( u (A,x) K ) + i u (A,y) u (B,x) + i u (B,y) Now we can write the energy of the A phonon at K K K K u A K ˆD (, ) ua K = d + d = ( γσ + γ σ + γ π) = 9.76 ev/å ω (A ) d + d K = m = γσ + γ σ m + γ π = 47. cm
Block diagonalizing ˆD k We can now stack the irreps to form the unitary transformation and block diagonalize. Û = i i i ˆD d K i i i = Û ˆD K Û ˆD K = Û ˆD d KÛ Recall original Dynamical matrix. d d d ˆD (, ) = d d + i d d d d d Block diagonalizing, we then have: d A ˆD d K = d A d E d E d d //8 at :4:4 slide 8
Irreducible derivatives //8 at :4:4 slide 9 Had we not been given Hamiltonian, we could deduce irreducible derivatives. d A ˆD d K = d A d E d E There are three irreducible derivatives at the K-point. Rotating back from irrep basis to naive basis, we have: d E + d A + d A id E id A id A d A + d A id A + id A ˆD K = id E + id A + id A d E + d A + d A id A + id A d A d A 4 d A + d A id A id A d E + d A + d A id E + id A + id A id A id A d A d A id E id A id A d E + d A + d A
Real k displacements //8 at :4:4 slide 4 We can always form a basis of real displacements. Get u ī by simply taking the complex conjugate. K Real vectors formed using representation of K K. u α K c = ( u α K + uᾱ K ) = cos(k t) u α t N u α K s = i ( u α K uᾱ ) = sin(k t) u α K t N t t
Displacements along A symmetrized modes. //8 at :4:4 slide 4 Let us illustrate this with A mode. u A K c = ) ( u A K + ua K = = u A K s = = ( u a,x K c + u a,y K s u b,x K c + u b,y i ) ( u A K ua K = i i i K s ) i i ( u a,x K s u a,y K c u b,x K s u b,y K c ) K K + i i i i K K
Picturing A modes //8 at :4:4 slide 4 u A K c
Picturing A modes //8 at :4:4 slide 4 u A K s
Picturing A modes //8 at :4:4 slide 4 u A K c
Picturing A modes //8 at :4:4 slide 4 u A K s
Uniform grid of k-points //8 at :4:4 slide 44 For the translation group, we work with BvK boundary conditions and use a finite tranlsation group parameterized by N; taking N at the end. In practice, finite N will often be used due to computational limits. In the following slides, varying density of uniform grids are shown for triangular lattice.
Uniform grid for triangular lattice //8 at :4:4 slide 45 b K Γ M b
Uniform grid for triangular lattice //8 at :4:4 slide 46 K Γ M
Supercell for given k-point //8 at :4:4 slide 47 The supercell of the K-cell was already given. Supercell must have linearly ind. vecs S i where k S i Z We are looking for a a set of vectors for which the phase pattern will be left invariant. e iπk S i = Therefore, the phase at a given lattice point will be invariant to S i. Consider site t for k: e iπk (t+s i ) = e iπk t In dimension d, one needs to find d linearly independent supercell vectors.
Minimal unit cell for M= (, ) : s = (, ) s = (, ) //8 at :4:4 slide 48
Minimal unit cell for K= (, ) : s = (, ) s = (, ) //8 at :4:4 slide 49
Minimal unit cell for = (, ) : s = (, ) s = (, ) //8 at :4:4 slide 5
Summary of Little groups and irreps for Graphene //8 at :4:4 slide 5 b Γ Σ K M b Γ : C 6v M : C v K : C v : C s In-plane displacements transform like: Γ : E M : B B K : E : A B Atoms transform like: Γ : A B M : A B K : E : A A Forming the full representation, we have: Γ : E M : A A B B K : A A E : A B