I. ELECTRONS IN A LATTICE. A. Degenerate perturbation theory

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1 1 I. ELECTRONS IN A LATTICE A. Degeerate perturbatio theory To carry out a degeerate perturbatio theory calculatio we eed to cocetrate oly o the part of the Hilbert space that is spaed by the degeerate states. Let s look at p ear k/. The p ad p k have similar eergy eigevalues. I fact, let s just write p = k/ + δ 1 Let s write the hamiltoia just as it pertais to these two states: H eff = k/+δ k/+δ We ca write this i terms of Pauli matrices ad the idetity: k/ +kδ k/ kδ Ad the eergy eigestates are: H eff 1 k kδ + σz 8m + σ x 3 E ± = k/ ± kδ /4m + V 0 4 Ad amazigly, whe δ = 0 there is a gap - betwee the two possible states. The way to thik about it is that we had a degeeracy which we ca see by movig the right movig dispersio by a k to the left, ad the there is this crossig. It gets resolved via this aalysis. What about other places i the bad? Actually, these degeeracies are goig to plague us i every poit i the dispersio that is related to aother poit with the same eergy by some mometum which is a iteger multiple of a reciprocal lattice vector k: p = p + k 5 will give mometa that exhibit a gap. You may woder why this is true give that hte hamiltoia?? does t have terms that coect these mometa directly for > 1. The first order aswer is that the potetial will always have compoets at haigher mometa, sice o potetial is a simple harmoic. For istace, oe ca cosider a square potetial well that repeats. If we wat to kow the periodic part of the potetial we eed a Fourier series: V x = Θ x d 6 with d < 1/ ad Θx the Heaviside fuctio. We would like to write it i terms of harmoic fuctios -sum of reciprocal lattice vectors. No problem. That is really just a Fourier series: V x = e iπx V 7 with V = dxe iπx V x 8 0.5,0.5 I our case, this reduces to: V = 0.5,0.5 dxe iπx V x = eπid e πid πi = siπd π 9

2 a. b. FIG. 1. A 1d bad structure i the folded zoe represetatio with a = 1 lattice costat. a No periodic structure. b A periodic structure with harmoics fallig of as 1/ with the harmoic. So clearly every resoatig vector ca participate. The secod order aswer is that eve whe there is o direct term coectig the mometa, higher order terms i perturbatio theory will emerge to give this coectio. For istace, I secod order perturbatio theory, we expect terms that follow the effective hamiltoia: This alows for istace for a term such as: H eff = H 0 + ˆV 1 ˆV. 10 E 0 Ĥ0 H eff p = p p + V0 p / p k p k. 11 / This leads to suppressed gaps, but otheless opes a gap to the secod Brilloui Zoe. B. Effective mass If we look a the bottom of the secod bad, it looks like a quadratic dispersio. W eca eve expa the square root E ± = k/ + δ ± kδ /4m + V0 k 8m + δ + 4m /k The expressio udereath δ is the effective mass. The effective curvature of the bottom of the bad. Or the top of the valece bad for that matter. For the coductio bad we have: 1 1 m e eff = 1 m + k 4m. 13 The top of the valece bad ca accomodate holes, ad they too will have a effective mass. The value will be give with the mius choice for the square root: 1 m e eff = 1 m k 4m. 14 C. Brilloui zoes Now the meaig of the BZ s becomes clear. The 1st Brilloui zoe is where a gap opes due to sigle reciprocal lattice vectors. But the there is a secod BZ - which is out the first BZ, but cotaied withi the lies where a gap opes due to a secod order scatterig by the lattice potetial. Ad so forth.

3 3 FIG.. The Brilloui zoe boudaries of a triagular lattice. D. 1d example rehashed i the spirit of Bloch theorem We could have doe somethig a bit differet. We could have just took the asatz that Bloch theorem gives us, ad plugged it ito the Schroediger equatio. This results i: Ee ipx e ikx ψ = p + k + V x e ipx e ikx ψ 15 With V x expaded i Fourier series V x = e ikx V, we the get the followig Ee ipx e ikx ψ = p + k + m e ikxm V m e ipx e ikx ψ 16 But idetifyig the same terms o the two sides of the equatio, we fid a simple matrix relatio: Eψ = p + k + m V m ψ m. 17 This is just a effective hamiltoia, Eψ = m H m ψ m, with: H m = p + k e δ m + V m 18 Where I added a e subscript to the mass. Not to get cofused with the matrix idex! E. d example Let s do a example i d - maybe eve the most complicated bravais lattice - the triagular oe. The reciprocal lattice vectors of the triagular lattice are 3 k 1 = 4π 3a ˆx + ŷ 1 ad k = 4π 3a ŷ. This sugests that the first gaps will ope at: p = ± 1 k i + λ a j 19 with j i. This gives this parallelogram. But wait! This does t look like it has the symmetry of the lattice. It should have some symmetry for rotatio by π/3. But that is ot the case. What are we missig? We ca also use k 1 k to ope gaps. Ad ideed, i a triagular lattice, there will be Fourier compoets correspodig to this combiatio, due to symmetry. This makes the gaps appear at a mometum hexago, as we suspected the symmetry should be. These are the borders of the first Brilloui zoe. But gaps will ope at higher mometa as well. Fig. gives the boudaries of the triagular lattice Brilloui zoes.

4 4 F. Tight bidig apprximatio. To be completely hoest, it is rare that we use the weak scatterig limit. We much rather thik i terms of atomic orbitals, or similar states that are localized at hte site of the atoms, which combie to give us a bad. Cosider, for istace, the set of wave fuctios: ψ r = ψ r α α a α 0 with α beig lattice directios, ad ψ r describig a orbital localized o a atom. I the spirit of Bloch theorem, we ca try to write a plae wave out of such a orbital. For the sake of simplicity, let s stick i 1d. p = ψ p x = e ipa ψx a 1 What s the problem with these plae waves? Mostly, that they are ot really good plae waves! O the oe had, they do have the bloch form: ψ p x + a = e ipa ψx + a a = e ip+1a ψx a = e ipa ψ p x O the other had, they are ot ormalized correctly: p p = e imp p a dxψx maψ x a = e imp p a M m 3,m,m uless M m = δ m this is ot simply proportioal to δp p without p-depedece. Ad M m is the overlap matrix: M m = dxψx maψ x a 4 It is a positive defiite matrix. Which suggests how we ca proceed. Let s fid a superpositio of the ψx a which are orthogoal: φ x = m α mψx a 5 The: dxφ mxφ x = k,l α kα ml dxψx la ψx ka = α ml M lk α k = αmα 6 But with M hermitia, we ca assume that α is also hermitia, ad get: If we do this we get: α = M 1/ 7 p = ψ p x = e ipa φ x = e ipa M 1/ m ψx ma 8 Plug it ito the hamiltoia, ad we also get the result from the problem set, which is: ɛ p m e ipm a M 1/ k H kl M 1/ lm 9 The fuctios φ m x are called the Waier fuctios. They are actually orthoormal, ad take the form of local orbitals cetered about the m th atom. This is the essece of Eq. 6.

5 5 G. Lattice Symmetries Crystals are clearly highly symmetric objects. The way that Bravais lattices distiguish themselves from each other is by symmetry groups of the lattice. A square lattice, for istace, could be rotated by π/ = π/4. If the lattice is symmetric by roatatios by π/, the we say it possesses C symmetry. C is the symmetry group which is geerated by applicatio of rotatios by π/. It has elemets therefore, ad is Abelia. The rotatios must be about a particular poit. Therefore these kid of symmetries are called poit-group symmetries. What other symmetries could we expect? How about mirror? We ca have reflectios about particular axes. For a square lattice we have reflectios by a axis that is at agle νπ/4 with ν = 0, 1,, 3 to the x axis. This is the reflectio group R 4. Usually, these groups combie to give us the Dehidral group, D which has both reflectios ad rotatios. There is aother importat poit group - iversio. r r. Ideed, this is cotaied withi the reflectio group. Sometime, however, iversio may be preset, but ot the idividual mirror reflectios that give rise to it.