The solutions of the single electron Hamiltonian were shown to be Bloch wave of the form: ( ) ( ) ikr

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1 Lecture #1 Progrm 1. Bloch solutions. Reciprocl spce 3. Alternte derivtion of Bloch s theorem 4. Trnsforming the serch for egenfunctions nd eigenvlues from solving PDE to finding the e-vectors nd e-vlues of mtrix: The centrl eqution. 5. rphicl description of eigenfunctions nd eigenvlues - bnd digrms. 6. Bnd gps wht hppens ner the edge of the BZ?

2 Eigenfunctions of infinite crystl - The effect of periodic potentil The solutions of the single electron Hmiltonin were shown to be Bloch wve of the form: u r e ir = f r n, n, Following from the Bloch Theorem which sys: The eigenfunctions of H cn be chosen such tht with ech eigenfunction u ( r) wve vector such tht: u r R e u r ir ( + ) = ( ) n n,, n, is ssocited which is n equivlent form of Bloch s Theorem, when the eigenfunction is observed t distnce R it differs by phse fctor. Reciprocl spce Suppose we found set of vectors tht stisfy the following reltion: i R e = 1 tht would led to u r u r = n, + n, Eigenfunctions which differ by re identicl! These s re clled the reciprocl lttice vectors, since they occur t regulr intervls. The lttice vectors R re defined s follows: R= n 1 1+ n + n 3 3 = mb 1 1+ mb + mb 3 3 Let s be the the set of primitive vectors for the direct lttice nd the b s re the primitive vectors for the reciprocl lttice given by: , b b, b3 π π π = = = ( ) Exmple simple cubic lttice: ˆ ˆ 1 = x, = y, 3 = zˆ π π ˆ ˆ π b1 = x, b = y, b3 = zˆ For 1D lttice we would just hve for our direct lttice the number nd for the reciprocl π lttice vector the number n n= ± 1, ±, ± Plot 1D direct nd reciprocl lttice. Any point in the reciprocl spce (not necessrily lttice point) cn be expressed s: = + where : π π This rnge is clled the first Brilllouin zone, it is the Wigner Seitz cell of the reciprocl lttice.

3 Alternte derivtion of Bloch s theorem We re considering periodic potentil, x+ = x Since the function is periodic with period it cn be expnded in Fourier series where s re the set of reciprocl lttice vectors ( x) = ( ) e ix in more fmilir form, nπ nπ ( x) = Ancos x+ Bnsin x n We wnt the potentil to be rel, ix ix x = e + e = cosx ( ) ( )( ) ( ) > 0 > 0 Where we hve ssumed tht the crystl is symmetric bout x=0. We re serching for eigenfunctions of the Hmiltonin: d ix + ( ) e u( x) = Eu( x) mdx let s ssume tht the eigenfunctions cn be written s Fourier series since there is no reson to ssume tht the eigenfunctions is periodic in lttice period we expnd it in, u( x) = C( ) e ix (Born on-krmn boundry conditions) Substituting in the Hmiltonin, d ix ix ix + ( ) e C( ) e = E C( ) e mdx ix i ( + ) x C( ) e + ( ) C( ) e = E C( ) e m Equting the coefficients of the exponentils with the sme rguments, E C( ) + ( ) C( ) = 0 m Consequences: 1. A set of coupled liner lgebric equtions (sometimes clled the centrl eqution) where C s. the unnowns re the ( ). The solution of the centrl eqution for given involves in principle n infinite number of coefficients tht ll differ by reciprocl lttice vector. 3. The number of nonzero mtrix elements in the equtions re determined by the number of Fourier components tht re needed to express the potentil function. ix

4 Thus the eigenfunctions should hve the following form: u x = C e which cn be re-written: ( ) ix ( ) = ( ) i x u x e C e which is just resttement of Bloch s theorem, since ix C e = f x where f is periodic function with the periodicity of the lttice. Exmple of n explicit solution for simple cos potentil The solution of the centrl eqution for prticulr potentil igx x = cos gx= e + e ix igx This defines n infinite set of equtions (which we truncted fter 5 terms), the equtions cn hve non-trivil solutions for the C s if the determinnt vnishes: m ( ) g E ( g) E m E m ( + g) E m ( + g) E m = 0 For given there will be 5 energy eigenvlues lbeled E ( ) n which bsiclly defines the bnd digrm nd corresponding eigenvectors C s which in turn define the Bloch wve eigenfunctions. As we mentioned in principle the mtrix is infinite nd one needs to clculte ll of the C(ng) s, in prctice just few coefficients re usully sufficient. In the following exmple we clculte the coefficients nd corresponding ground stte eigenvlue (note: the Mthemtic progrm involves the genertion of tridigonl mtrix nd will be emiled to you). C(-4g) C(-3g) C(-g) C(-g) C() C(+g) C(+g) C(+3g) C(+4g) E

5 Observtions: 1. The ground stte energy is dominted by the contributions from C() where is the smllest vector i.e it lies in the first BZ.. If we were to choose reciprocl vector tht is out of the first BZ = + the form of the mtrix bove would remin unchnged nd the eigenvectors would be identicl. i.e. the ground stte clculted for nd the one clculted for +g would be the sme. 3. This mens tht the eigenvectors nd energy eigenvlues re periodic in u x u x + = nd E + = E n nd we therefore cn choose s tht re restricted to the first BZ - reduced zone scheme. This leds to description of the energy levels of n electron in periodic potentil in terms of E + = E with the periodicity of the reciprocl lttice fmily of continuous functions n( ) n( ) vector. The informtion contined in these functions is referred to s the bnd structure. u r u r n = n, + n, ε = ε n, + n, For ech n the set of electronic levels specified by ε n ( ) is clled n energy bnd. Bnd gps: wht hppens ner the edge of the BZ? The edge of the BZ is defined by: g =± review the mster eqution: E C( ) + ( ) C( ) = 0 m And consider only the first terms in the sum: The energy eigenvlues re g E m E ± g m g = ± m E = 0

6 Thus the energy hs two roots which re seprted by, eep in mind tht this result is vlid for the prticulr form of the potentil tht we used: igx igx x = cos gx= e + e So we re ble to predict the existence of bnd gp which to first order is proportionl to the mgnitude of the periodic potentil. The form of the eigenfunctions ner the bnd edge is: g igx igx ( ) = ± u x e e One solution gives the eigenfunctions ner the bottom of the gp the other gives it ner the top of the gp. The differences in energies cn be explined by looing t the chrge distribution reltive to the ion positions. πx πx i i π x u+ ( x) = e + e = cos πx πx i i π x u ( x) = e e = sin i The two stnding wves concentrte their electrons in different regions the probbility density for prticle is: π x u+ ( x) u+ ( x) = cos This function piles electrons (with negtive chrge) on the positive ions centered t x=0,,, where the potentil energy is lowest. For the other stnding wve, π x u ( x) u ( x) = sin

7 It cn be shown tht n electron is level specified by bnd n hs non-vnishing men velocity 1 vn( ) = εn( ) This is quite remrble since it sttes tht there re time independent eigenfunctions which in spite of the interction of the electron with the fixed lttice move forever without ny degrdtion of the men velocity

8 enerting bnd digrms (displying eigenfunctions grphiclly) Wht hppen s we trnsition from free electron system to one which hs periodic potentil?.