Question

5) Nearly free electron model: We have seen that (6.43) gives an approximate solution of the Schrödinger for a weak potential with Fourier components U u U. The biggest deviation from the energies of free electrons was found at the Brillouin zone boundary (k = n/a). Using (6.43), show that ck-g is much larger than cg if k = a/a. Neglect then c and use (6.43) to show that the size of the gap opening at the Brillouin zone boundary is 2U.

Answer 1

when introduced Energy of a particle in a periodic system we start from free particles ( w o ) and a small periodic potential into the system - P(x) - ikx Now, we A a are small going vex to is use added perturbation Theory- to Hamiltonian - Then {"), (7248031 0()| 1663) ple) = First order correction I 147 (til volu; o Ele) = 3 kHelvez) dy> S iti j-li Vix "V (2 ta) Lovce: É um eine ) nu-00 iGix - 2 Voe where, Gi= 2in a na-oo, too

so ver va IGE since. V is real ie. von = V(o) we can also write - VerEVGene = V+ 2 Va Cos(678) [va oval 0 GO Gto Now for free electrons : - ikx 412)Le - iux So, el I Cike I Vo + I va G70 5, ) = cales lelety + Katie zviete okay? or Ell) Celkx I volelky + -ike igre Icelke ve uto selarde = o[ato or EC1) = Vo * order or change in energy is So, first Constant. does not D E show any change in energy.

Now, K4 114&> ľ ERK 5 Vo katika kesi estos > G7O ktk where, hk zm 20 on. Keles, Ticiasclay: Va fi cick-wit mo da Vook', ktG G - EEG when to we apply a potential V(x) weak periodic free electron gas El = constant = v. = o. [(2) I I Veid? G (Ex - Exta) y = { 14> {atulelat va 14k)

Elk) : EC + E 1 € = tik? 2W 9th ---- G:29 In both case. E blows up. а The reason can not is apply b = Int are non degenerate degenerate. So, we perturbation Theory. To will apply degenerate consider only perturbation those levels theory which we are degenerate At ka I, we will consider k = I & k = - At wel we will considor k= I & k-1 Let Iks and lk) are degenerate. H2 Hot V(x) = - Ź + Even dhe 2 G im making of the linean energy of combination the states of lk) and formed 16'7 is by E. H1-4) E14)

147 = - a1k> + STK'> alus + blk+G) Hotva)] 147 €147 La Elk> + av(x)16) + b Extalk+G) + b (x) 1k+G) = at lk) + BEIK+G) Taking inner product with (kl - or, or, Taking a Ei + b (ki Valkta)= aĒ acEr-E) + BLklvolkta) a( -E) + b va 20 - inner product with Ikta) - alktal vext) k> + b Excta = b Exx a (ktalvik> + b(Epta- E) = 0 ce® Ox, a Na + b( ticke-E) - 0 - ® and y is non-zero, then we have- of a E-E Von 1 LE+l Val & & Er-Nol 01, (:- € ) ( Gite- ) - Ivaje = 0. since & Eura E = E + lval { = 2 Verya
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