Final Report:Unfolding Supercell Band Structure:

Application to a One-Band Tight BindingModel

Jared BlandMentor: Prof. Krakauer

August 26, 2014

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1 Introduction

Typical disruptions of the periodic structure of crystals are due to atomicsubstitutions, vacancies, or other defects. It is sometimes convenient to modelsuch disorder using “supercells” that impose artificial periodicity on a largerscale. For example, isolated vacancies(a missing atom) in a three-dimensionalaluminum crystal, which contains one Al atom per primitive unit cell mightbe modeled by a 3x3x3 supercell containing 26 Al atoms and a single va-cancy. The electron eigenstates of the supercell often retain some ”hidden”order reflecting the original pristine crystal. I have implemented a theoreticalmethod to extract this hidden order from the calculated supercell electroniceigenstates. Here I present representative calculations for a one-dimensionallattice, treated with a one-band tight-binding Hamiltonian.

1.0.1 Positions in the crystal

Main concepts used in my work, and the related literature, are the notionsof primitive and unit cells. The primitive unit cell is the smallest repeat-able structure within a crystal. The primitive unit cell is one atom in amonatomic crystal, and two adjacent atoms in a diatomic crystal. A su-percell is composed of several primitive cells. Similar notation is used forsupercell and primitive cells; the notation is summarized more completely inthe next section.

We present a way to index atoms in the crystal. We let m and m′ indicatethe atom within a supercell, thus, since each supercell is a replica of theothers, m and m′ provide a way of discussing particular atoms within asupercell. Further, R and R′ are vectors that point to specific supercellswithin the crystal, and τm and τm′ are vectors that point to the m and m′

atoms, respectively. This effect is actually easiest in two dimensions. Weshow an image of a two dimensional crystal with a nine atom supercell inFig. 1. As you can see, R and R′ do not necessarily point to the samesupercell, but they may indeed be the same. For our two-atom model, wehave a repeating structure of two atom supercells. The structure for fouratoms is similar. Fig. 1 and 2 show the vector indexing of the supercellsin the lattice. Fig. 1 illustrates this in two-dimensions for ease of viewing,while Fig. 2 is restricted to the one-dimensional case.

To use the supercell for calculations, we take the Fourier transform ofthe lattice. In the reciprocal (imaginary) space, the supercell is smaller than

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Figure 1: Position in a two dimensional crystal. The R and R′ vectors pointto supercells, m and m′ index the atoms in the supercell, and τm and τm′

point to atoms within the supercell.

Figure 2: Position in a one dimensional crystal using the same notation asFig. 1. Here m and m′ take on combinations of values of A and B.

the primitive cell, which follows from properties of the Fourier transform.Rather than dealing with positions ~x in the real space, we move to ~k inthe reciprocal space. Further, if we have a vector in the supercell, we maytranslate ~K into ~k using one of the ~G, which translate from the supercellzone to the primitive zone. When we restrict ourselves to ~k ∈ [−π, π) (since

we are in one dimension) we find that the ~G satisfying ~k = ~K + ~G is unique.Due to the periodicity, we may always shift by 2π to return the vector to theinterval.

In higher dimensions, we utilize matrix transformations of linearly inde-pendent basis vectors to form the relationship between the primitive– andsuper–cells. If using an orthogonal basis to move in the crystal, the processvery closely follows the one dimensional case, but we are not guaranteed or-thogonality. Details of this process are given in the paper from Allen et al.[3].

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1.1 Crystal Structures – Primitive & Supercell Nota-tion

We borrow notation from Allen et al. [3]. Let N be the transformation

matrix that produces the ~Ais; i.e. ~Ai =∑

j Nij ~aj. Below is the majorityof the notation used to describe the positions in the lattice, including thoseused on the reciprocal lattice.

~ai Primitive cell lattice vector~r =

∑i ni~ai Primitive cell lattice vector

~R =∑

imi~Ai Supercell lattice vector

~g =∑

i ni~bi Primitive cell reciprocal lattice vector

~G =∑

imi~Bi Supercell reciprocal lattice vector~k Primitive cell reciprocal wave vector~K Supercell reciprocal wave vector

N = det(N) Number of primitive cells in the supercell

2 Mathematical Formulation

2.1 Periodic Potentials & Bloch’s Theorem

We examine crystals that exhibit periodic potentials; that is

V (x+R) = V (x) (1)

for some vector R. We expect that the electron wavefunctions will also beperiodic, but with a possible added phase shift. This result is called Bloch’stheorem (after Felix Bloch). For an atom chain with a spacing of a thatrepeats after n atoms we have R = m(na), with m ∈ Z.

The Schrodinger equation is then

Hψn,k(x) =

(− ~2

2m∇2 + V (x)

)ψn,k(x). (2)

The solutions under the periodic potential may be chosen such that for eacheigenstate there is an associated wave vector k such that

ψn,k(r +R) = eik·Rψn,k(r), (3)

which is the same result given in [1].

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2.2 Tight-binding Hamiltonians

Constructing the Hamiltonians follows the tight-binding method. We assumethat there is only one orbital per atomic site, and that the orbitals are or-thogonal to one another. Additional orbitals increase the number elementsin the matrix representation of the Hamiltonian, but they do not change theoverall problem. The first step is to expand the wavefunction ψ in our orbitalbasis, in which each element satisfies Bloch’s theorem [1]:

ψk(r) =∑R

eik·Rχ(r −R). (4)

Since the basis is orthogonal we have∫χ∗(r − R′)χ(r − R) d r = δR,R′ . We

have assumed that the basis is normalized; if it is not orthonormal, thenthe Graham-Schmidt process may be used to create an orthonormal basis,and we would expand in terms of that new orthonormal basis. Using thenearest neighbor approximation, the elements of the Hamiltonian matrix aresimplified. The only atoms that contribute to the Hamiltonian at a site arethe atom at that site and the adjacent atoms. In other words, and introducingt and to as parameters:

∫χ(r −R′)Hχ(r −R) d r =

−to R = R′

t |R−R′| = a

0 otherwise

. (5)

If we add some disorder to the crystal, then certain elements will have avalue of t′ rather than t. It is important to recognize that t0, t and t′, andthe like are simply notationally convenient. For actual calculations we willfind the allowed energies in terms of the actual energy divided by t and writeeverything else in terms of the ratio with t.

2.3 Unfolding and Projection Weights

The algorithm for unfolding is given in Allen et al; we will only give a briefoverview here [3]. We calculate the wavefunction in the ~K basis, then project

it onto the ~k basis. This is done by translating the ~K vectors according to~k = ~K + ~G. The projection operator needs to remove the bands that shouldnot appear, else the structure would just be replicated in the primitive cell’srepresentation in the reciprocal space. The projection operator allows us to

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assign weights to each band as it is translated into the primitive cell, whichwe then use while plotting to remove unwanted translated points. We use nas the band index for the primitive cell, and J for the supercell band index.

The overall procedure is as follows:

1. Solve the eigenvalue problem for the basis

The coefficients of the eigenvectors serve as our orbitals

2. Find the weights using the projection and translation operators

Translation: T (~ri)f(~x) = f(~x+ ~ri)

Projection: P ( ~K → ~K + ~G) = 1N∑N

i T (~ri)e−i( ~K+ ~G)·~ri , and

Weight:

W ~K,J(~G) =〈 ~KJ |P ( ~K → ~K + ~G)| ~KJ〉 (6)

=1

N

N∑j=1

e−i(~Kj+ ~Gj)·~rj〈 ~KJ |T (~ri)| ~KJ〉 (7)

3. Plot the values translated and reflecting the weights of the translation

Darker values are closer to a weight of 1, and lighter values have aweight closer to 0

3 Application to a One-Dimensional Crystal

Since we will calculate in one dimension, we will remove the vector notationfor simplicity.

It is easier to understand the general formulas after applying them to atoy model. If we assume a perfect monatomic crystal, one of the simplestcases is a two atom supercell, but with slightly different atoms. Since thisrepeats indefinitely, this yields a structure with two alternating atoms.

3.1 Two atom unit cell

There are several good reasons to study the two atom supercell: the energiesof the hamiltonian can be solved exactly, the atomic orbital basis is small,and it is notationally easier. These all come from the fact that we only have

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two atomic orbitals. The hamiltonian is a 2 × 2 matrix, meaning a seconddegree characteristic polynomial for the eigenvalues, and the summations areonly over two values.

We will detail the exact solution of the two atom supercell. First, let’sexamine what this looks like:

Figure 3: Two-atom supercell with asymmetry between the atoms. One atomcouples more strongly to the left than to the right by a factor of T .

We’ll call the atoms A and B for convenience. We begin by expanding ψ:

ψm,k(r) =∑m

cm(k)χm,k(r) (8)

=cAχA,k(r) + cBχB,k(r), (9)

where cm is the coefficient of the normalized wavefunction in the basis andχm,k(r) is the mth atomic orbital at the site, r.

Next, we expand χA,k and χB,k using Bloch’s theorem:

χA,k(r) =1√N

∑R

eikRχA(r −R), (10)

where we sum over all possible R, and N is 2 since we move a distance of 2ato be in the next supercell. The expansion for χB,k is similar.

Using |ψ〉 = cA|χA,k〉 + cB|χB,k〉, the proceeding step is to apply theexpansion to the eigenvalue problem:

Hψ =Eψ (11)

〈χA,k|H(cA|χA,k〉+ cB|χB,k〉) =E〈χA,k|ψ〉 (12)

〈χB,k|H(cA|χA,k〉+ cB|χB,k〉) =E〈χB,k|ψ〉, (13)

〈χA,k|H(cA|χA,k〉+ cB|χB,k〉) =EcA (14)

〈χB,k|H(cA|χA,k〉+ cB|χB,k〉) =EcB. (15)

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The last two lines lend themselves to be put into matrix notation; labeling〈χA,k|H|χA,k〉 = HA,A, 〈χA,k|H|χB,k〉 = HA,B, 〈χB,k|H|χA,k〉 = HB,A, and

〈χB,k|H|χB,k〉 = HB,B. In the matrix representation:[HA,A HA,B

HB,A HB,B

] [cAcB

]= E

[cAcB

]. (16)

The reason for labeling the energy E will become apparent soon.Let’s calculate the matrix elements. The matrix element for HA,A is given

by the following:

HA,A =〈χA,k|H|χA,k〉 (17)

=

(1

N∑R

eikR)(

1

N∑R′

e−ikR′)〈χA(r −R′)|H|χA(r −R)〉, (18)

but the only nearest neighbors are B atoms. Using the nearest neighborapproximation, any orbitals other than R = R′ will be orthogonal; i.e.

〈χA(r −R′)|H|χB(r −R)〉 = δR,R′〈χA(r −R)|H|χA(r −R)〉. (19)

This yields

HA,A =1

N

N∑1

eik(R−R′)〈χA(r −R)|H|χA(r −R)〉

∣∣∣∣R=R′

(20)

=〈χA(r −R)|H|χA(r −R)〉 (21)

=E0, (22)

where we called this energy E0. It is the on-site energy; the energy fromthe orbital at the site of interest. A careful reader will notice that when wetook the difference of R and R′ that it simplified our sum, ridding us of afactor of 1

N . Thus, we examine R = R − R′, or as other authors write, callR′ = 0. To justify this: the crystal is infinite, and we may arbitrarily chooseour origin. A similar derivation for HB,B yields E0 as well. This is a commonsubstitution [1]. We may choose the on-site energies to be different; howeverthis does not drastically change our results.

Now we may calculate the off-diagonal elements, HA,B and HB,A. Sincethe hamiltonian is hermitian, we have HB,A = H∗A,B. We begin by expanding

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in the same way: Using R = R−R′ as before:

HA,B =〈χA,k|H|χB,k〉 (23)

=

(1

N∑R

eikR)(

1

N∑R′

e − ikR′)〈χA(r −R′)|H|χB(r −R)〉 (24)

=∑R

eikR〈χA(r)|H|χB(r −R)〉, (25)

which when we use the nearest neighbor approximation gives two non-zeroterms in the sum; the terms corresponding to the adjacent B atoms. We getR = 0 (for the B atom in the same supercell), and R = 2a (for the B atom inthe supercell to the left). This gives, if we allow different energies for the two,HA,B = −t− t′ei2ka where −t is the energy coupled within the supercell, and−t′ is the energy coupling with the atom in the other supercell. Hermiticityrequires HB,A = −t− t′e−i2ka.

Thus, the hamiltonian is:[E0 −t− t′ei2ka

−t− t′e−i2ka E0

] [cAcB

]= E

[cAcB

]. (26)

We may manipulate this slightly, subtracting E0 from both sides and makingthe substitution of E = E − E0 and t′ = Tt we have:[

0 −1− Tei2ka−1− Te−i2ka 0

] [cAcB

]=E

t

[cAcB

]. (27)

If we choose to subtract the E/t from both sides, we have

(E/t)2 = 1 + T 2 + 2T cos(2ka) (28)

as a characteristic equation after reducing the algebra. We can plot theenergy as a unitless proportion to the energy t. With this, we have twosolutions, a positive and negative. These are our two bands that we need tounfold. These will be investigated for varying values of T . First, we need tofind the weights for unfolding.

The weights are given by

WK(G) =1

N

(∑m,m′

C∗m′(k)Cm(k)eiK·R)×

( N∑j=1

e−i(K+G)·rj∫χ∗m′(r − τm′)χm(r − τm −R + rj) d r

),

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where we have used the substitution of R′ = 0 and R = R − R′ in the samefashion as earlier. We have two values of rj: r1 = 0 and r1 = a. From theorthogonality condition, when using r1 we have that τm′ = τm and m′ = m.When examining r1, we have that if m′ = A, then m = B and vice versa.Further, when m′ = B then R = 2a and we gain a phase shift. Performingthe sum,

WK(G) =1

2

[C∗A,JCA,J + C∗B,JCB,J

+ C∗A,JCB,Je−i(K+G)a + C∗B,JCA,Je

i(K+G)2ae−i(K+G)a

]=

1

2[1 + 2|CA,J ||CB,J | cos((K +G)a− φA + φB)],

where φA and φB are the phases from the coefficients.For the two atom model, G takes on two possible values 0 and π/a. In

this case, we can explicitly write out the weights:

WK(0) =1

2(1 + 2|CA,J ||CB,J | cos(x− φA + φB)) (29)

WK(π

a) =

1

2(1 + 2|CA,J ||CB,J | cos(x− φA + φB + π)) (30)

=1

2(1− 2|CA,J ||CB,J | cos(x− φA + φB)), (31)

where x = Ka and ranges from −0.5 to 0.5. The weights, indeed, sum to 1as expected.

Let us examine the plots for different cases. Fig. 4 shows the primitivecell band structure for the 1-band tight binding model, T = 1. Two-atomsupercell calculations, with T=1.5, are unfolded into the primitive Brillouinzone in Fig. 5. Note the overall similarity to the perfect T = 1 results, exceptfor the opening of gaps at ±π/2a and increase in overall band width.

When we unfold the structure for T 6= 1 we have more interesting results.For T = 1.5 (shown in Fig. 5) we have two different atoms, and the functionE(k) looks very similar to results found for phonons by Allen et al. [3] ForT values larger than 1 the bands gap is larger. Fig. 6 shows the results forT = 0, the atoms to do not affect the site energies of the nearest neighbor,and just have the energies from the individual orbitals, without the couplingeffects from adjacent atoms.

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Figure 4: (Left) We find the same energies as that for the primitive cell– aninfinite crystal with only one atom.

Figure 5: (Right) The bandstructure for T = 1.5. The result is analogous tothe results found by Allen et al [3] for phonons.

3.2 Four atom supercell

Following the same procedure as for the two atom case but instead of m,m′ ∈{A,B} we have values of 1,2,3,4. We factor out t, subtract E0, and only alterthe fourth atom. The Hamiltonian for this arrangement is shown below:

H4 =

0 −1 0 −Te−i4ka−1 0 −1 00 −1 0 −T

−Tei4ka 0 −T 0

. (32)

Rather than solving for the eigenvalues and eigenvectors explicitly as be-fore we allow a computer program to find them. Solving the eigenvectorsby hand involves two square-roots and a plus-or-minus sign for each. Theeigenvectors are also more complicated. Our goal is to generalize the model,and after degree four polynomials we have no general solution for the charac-teristic equation. Similarly, we retain complex vectors and let the computerhandle the algebra; the weights are purely real and in [0, 1].

Let us look at situations. The four atom supercell with T = 1 yields thesame cosine shape. Fig. 7 and 8 give the folded and unfolded band structures

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Figure 6: The T = 0 case. The energies from each orbital are separated anddo not yield the familiar cosine shape.

for T = 1.3. We get a similar band gap, but with additional breaks in theallowed energy curve.

3.3 Results for a Large Supercell

The goal of unfolding is to have a continuous function of E in terms of thequantum number k, but even our unfolding for K ∈ [−π/N , π/N ] is discrete.The larger the supercell, the more “folds” we get, and as the supercell be-comes very large the majority of the points are within a small neighborhoodof the K = 0 point. Therefore, for very large supercells, it is sufficient to sim-ply unfold K = 0 supercell results, since the supercell zone is so small withnegligible band dispersion. Results for a 100 atom supercell with T = 1.5(all nearest neighbors are t, except one pair in the supercell with t’=1.5t) areshown in Fig. 9. Note the overall similarity to the perfect T = 1 case, exceptfor the appearance of “interface states” above the band near k = ±π/a andbelow the band near k = 0.

4 Summary

Crystals with a structure repeating after m atoms were investigated, with oneatom possibly different from the others. I implemented the unfolding strategy

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Figure 7: (Left) The folded band structure for an asymmetric four-atomsupercell with T = 1.3.

Figure 8: (Right) The unfolded band structure for an asymmetric four-atomsupercell with T = 1.3. Note the additional bands compared to the two-atomcase.

given by Allen et al, 2013 for one dimensional crystals using unfolding weightsto remove unwanted points.[2] The Hamiltonians for states satisfying Blochstheorem were found using the tight-binding method with orthogonal atomic-like orbitals with one atomic orbital per site.

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Figure 9: Band structure of 100 atom supercell with one atom altered. Notethe similarity with the primitive cell structure, but with added “interface”states near k = 0 and k = ±π/a.

References

[1] N. Ashcroft, N. Mermin. Solid State Physics, Cengage Learning, 1976.

[2] V. Popescu, A. Zunger. Extracting E versus ~k effective band structurefrom supercell calculations on alloys and impurities, Physical Review B,2012.

[3] P.B. Allen, T. Berlijn, D.A. Casavant, J.M Soler. Recovering hiddenBloch character: Unfolding electrons, phonons, and slabs, Physical Re-view B, 2013.

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