December 4, U(x) = U 0 cos 4 πx 8

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1 PHZ66: Fll 013 Problem set # 5: Nerly-free-electron nd tight-binding models: Solutions due Wednesdy, 11/13 t the time of the clss Instructor: D L Mslov mslov@physufledu Rm 11 Office hours: TR 3 pm- pm Plese help your instructor by doing your work netly Every (lgebric) finl result must be sup December, 013 P1 [0 points ] Consider 1D electron system subject to wek periodic potentil [ U(x) = U 0 cos πx 3 ], 8 Find nd plot the dispersions of energy bnds in the first Brillouin zone Plot the energies in units of π /m, the wve numbers in units of 1/, nd ssume tht U 0 = 01 in these units Solution Using the trigonometric identity cos x = (1 + cos(x))/, the potentil cn be re-written s ( 1 πx U(x) = U 0 cos + 1 ) ( πx 1 cos = U 0 8 eiπx/ + 1 ) 16 eπix/ + cc There re two non-zero hrmonics with mplitudes u 1 = U 0 / nd u = U 0 /16 with wvenumbers q = π/ nd q = π/ Scttering t hrmonic with q = π/ lifts denegercy of the freeelectron sttes with energies ε 1 = k /m nd ε = (k π/) /m The degenercy points re k = π/ The energy levels ner these points re given by ε = ε 1 + ε (ε 1 ε ) + ( ) U0 Likewise, scttering t hrmonic with q = π/ lifts denegercy of the free-electron sttes with energies ε 1 = k /m nd ε 3 = (k π/) /m The degenercy points re k = π/ The energy levels ner these points re given by ε = ε 1 + ε 3 (ε 1 ε 3 ) + ( ) U0 16 1

2 P [0 points ] Consider D electrons subject to wek periodic potentil ( U(x, y) = U 0 cos πx + cos πy ) [8 points ] Find the energy bnds ner the edges of the first Brillouin zone (Hint: the edges of the Brillouin zone re the Brgg plnes where the eigensttes re doubly degenerte) Plot the bnds nd isoenergetic surfces Solution The non-zero hrmonics of the potentil hve wvenumbers (b, q y = 0) nd (q x = 0, q y = b), where b = π/, with mplitudes U 0 / Consider one of the edges, eg, k x = π/; by symmetry, the energy spectrum will be the sme t other edges The two free-electron sttes with energies ε 0 k x,k y = ( kx + ky) /m nd ε 0 kx π/,k y = ( (k x π/) + ky) /m re degenerte t this edge The energy levels ner the edge re given by ε = ε0 k x,k y + ε 0 k x π/,k y ) ) (ε 0kx,ky ε 0kx π/,ky + ( ) U0 b) [1 points ] Repet the nlysis of prt ) for regions ner the corners of the first Brillouin zone, where there re four degenerte eigensttes Solution The bsis must now include four degenerte sttes with energies ε 0 k x,k y, ε 0 k x b,k y, ε 0 k x b,k y q y, nd ε 0 k x b,k y b Using the fct the only non-zero hrmonics of the periodic potentil re U qx=b,0 = U qx=0,q y=b = U 0 /, we obtin equtions for the weighing fctors of the four degenerte sttes [cf Eq (919) in AM]: (ε ε 0 k x,k y )c kx,k y = U (c k x b,k y + c kx,k y b) (ε ε 0 k x b,k y )c kx b,k y = U (c k x,k y + c kx b,k y b) (ε ε 0 k x,k y b)c kx,k y b = U (c k x,k y + c kx b,k y b) (ε ε 0 k x b,k y b)c kx b,k y b = U (c k x b,k y + c kx,k y b) (1) Inverting the mtrix nd mesuring the wvenumbers in units of 1/ nd energies in units of /m, we obtin for the four energy levels where ε 1 = E 0 + R + ε = E 0 R + ε 3 = E 0 + R ε = E 0 R E 0 = kx + ky π(k x + k y ) + π R = 1 [ U0 + 3π (kx + ky π(k x + k y ) + π ) (U0 (π k x ) )(U0 + 16π (π k y ) ) P3 [0 points ] The Kronig-Penney model llows one to consider both the nerly-free-electron nd tight-binding limits Previously, we focused on the ltter Now, consider the (repulsive) Dirc-Kronig-Penney model, which, s we showed erlier, gives the following implicit eqution for the energy bnds: cos k = cos q + u sin q q, () ] 1/

3 where q = me/ nd u = mu 0 / > 0 Assuming tht u 1, find the solution of Eq () ner the Brgg points k π/, π/ (Reminder: the sign mens tht k is close yet not equl to the wvenumber of the Brgg point) Compre your result with tht of the nerly-free-electron model Solution The mximum vlue of the rhs of Eq () (= 1 + u) occurs t q = 0 The minimum vlue occurs ner q = π To see this, substitute q = π + δ nd expnd the rhs to second order in δ: sin q R cos q + u = 1 + δ q uδ π + R/ δ = 0 gives δ = u/π, nd the minimum vlue of R is R min = 1 u /π < 1 Since cos k 1, Eq () hs no soluton in the region round the minimum of R, where R min R 1 This region corresponds to gp in the energy spectrum To find the energy levels ner the gp, substitute k = π + p into Eq () nd expnd cos k in p s cos(π + p) = 1 + p / The resulting eqution δ π uδ p = 0 hs two roots δ = u ( u ) p π + π Returning to the originl vribles, we obtin for the energy levels where the lst line is vlid to first order in u (k E = π 1 m + u ) ( u ) π π 1 + π (k π 1 m + u ) ( u ) π π 1 + π, (3) Recll the result of the NFE model (with digonl mtrix elements U 11 nd U equl to zero) E = ε k + ε k π/ + U 1 (εk ε k π/ ) + U 1, () where ε k = k /m The difference of the free-electron energies cn be re-written s ε k ε k π/ = π ( ) k m π 1, while in their sum we cn replce k by π/: ε k + ε k π/ = π m Now it is obvious tht the NFE result is equivlent to Eq () P [15 points ] Find the dispersions of energy bnds for fcc nd bcc lttices tking into ccount hopping between the nerest nd next-to-nerest neighbors Solution:

4 The tight binding dispersion reltion is: E k = E 0 t Rnn e i k R nn t () cse of bcc 8 nn: (1, +1, +1), (1, +1, 1), (1, 1, +1), (1, 1, 1) 6 nnn: (1, 0, 0), (0, 1, 0), (0, 0, 1) R nnn e i k Rnnn (5) R bcc nn e i bcc k R nn = e (1/)i(+k x+k y+k z) + e (1/)i( kx+ky+kz) +e (1/)i(+kx+ky kz) + e (1/)i( kx+ky kz) +e (1/)i(+kx ky+kz) + e (1/)i( kx ky+kz) +e (1/)i(+kx ky kz) + e (1/)i( kx ky kz) = cos[(/)(k x + k y + k z )] + cos[(/)(k x k y + k z )] + cos[(/)(k x + k y k z )] + cos[(/)(k x k y k z )] = 8 cos(k x /) cos(k y /) cos(k y /) (6) R bcc nnn e i bcc k R nnn = e ik x + e ikx + e iky + e iky + e ikz + e ikz = cos(k x ) + cos(k y ) + cos(k z ) (7) (b) cse of fcc 1 nn: (+1, 0, 1), (0, +1, 1), ( 1, 0, 1), (0, 1, 1), (+1, +1, 0), (+1, 1, ), ( 1, +1, 0), ( 1, 1, 0) ; 6 nnn: (1, 0, 0), (0, 1, 0), (0, 0, 1); (sme s bcc) R fcc nn e i fcc k R nn = cos[(/)(kx + k y )] + cos[(/)(k x k y )] + cos[(/)(k x + k z )] + cos[(/)(k x k z )] + cos[(/)(k y + k z )] + cos[(/)(k y k z )] = [cos(k x /) cos(k y /) + cos(k y /) cos(k z /) + cos(k z /) cos(k x /)] (8) R fcc nnn e i fcc k R nnn = R bcc nnn e i bcc k R nnn (9) P5 [5 points ] Bilyer grphene consists of two grphene lyers stcked on top of ech other in Bernl mnner, ie, such tht only one of the two inequivlent lttice sites, eg, A, hs the nerest neighbor long the

5 verticl (see Fig 1) The tight-binding Hmiltonin in the nerest-neighbor-hopping pproximtion reds Ĥ = γ 0 â m,rˆb m,r γ 1 â 1,Râ,R + Hc (10) m,r,r R,R where â m nd ˆb m with m = 1, re the electron nnihiltion opertors for sites A nd B in the bottom (1) nd top () plnes, correspondingly, sum over R nd R includes only the nerest neighbors, nd prmeters γ 0 nd γ 1 re rel (FYI, γ 0 3 ev nd γ 1 0 ev) The plnes re seprted by distnce c ( 3Å) Find the energy spectrum for rbitrry k in the Brillouin zone Plot the isonergetic contours Anlyze the spectrum ner the K nd K points Solution The unit cell of grphene bilyer consists of four toms The electron stte is -spinor 1 ψ = b 1, b where 1 b re the mplitudes of finding n electron on site A 1 B In bsis of ψ sttes, the Hmiltonin cn be written s mtrix 0 S γ 1 0 Ĥ = S γ S, (11) 0 0 S 0 where S is the mtrix element connecting A nd B sites in the sme plne The top left nd bottom right blocks re the sme s for monolyer Digonlizing Hmiltonin in Eq (11), we obtin for the energy levels ε = γ 1 + S γ 1 + γ 1 S 3 S (1) Recll tht S = (3/)γ 0 k v F k ner the K nd K points, where k is mesured from either of these two points In the low-energy limit S γ 1, two out of the four brnches in Eq (1) pproch finite vlue (equl to γ 1 ), wheres the other two vnish Expnding these lst two dispersions in Tylor series, we find ε = S /γ 1 = ( v F /γ 1 )k k /m, where m γ 1 /v F Therefore, the low energy excittions in bilyer grphene re mssive electrons nd holes Degenercy of the spectrum t K nd K points is not lifted by inter-lyer coupling but the functionl form of the dispersion round the degenercy point is modified from liner to qudrtic

6 Figure 1: Crystl structure of bilyer grphene, from A H Cstro Neto, F Guine, N M R Peres, K S Novoselov, A K Geim, Rev Mod Phys 81, 109 (009)