Energy Bands Energy Bands and Band Gap. Phys463.nb Phenomenon


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1 Phys463.nb 49 7 Energy Bnds Ref: textbook, Chpter 7 Q: Why re there insultors nd conductors? Q: Wht will hppen when n electron moves in crystl? In the previous chpter, we discussed free electron gses, where we mde three pproximtions Ignore the lttice bckground Ignore the interctions between electrons Ignore the interctions between phonons nd electrons In this chpter, we will tke the lttice bckground into considertion nd see wht will hppen Energy Bnds nd Bnd Gp Phenomenon Consider 1D solid. In the bsence of lttice bckground, the kinetic energy of one electron cn tke ny positive vlues (ϵ = p 2 /2 m > ). In the presence of lttice bckground, the kinetic energy (s function of the momentum) breks into pieces. Ech piece is known s n energy bnd. Between two energy bnds, there my be forbidden region, which the energy of n electron cn never enter. This forbidden region is clled the bnd gp. Fig. 1. The dispersion reltion for () free electron gs nd (b) electrons moving in crystl.
2 5 Phys463.nb Fig. 2. The probbility density for the ψ + nd ψ  wves Why re there two possible energy t k = π /? Here we consdier the gp tht ppers t the momentum k = ±π/. At these two momentum points, there re two linerindependent plne wves 1 ei π/ x nd 1 ei π/ x. We cn form two stnding wves using these two plne wves ψ + (x) = ei π/ x i π/ x + e 2 ψ  (x) = ei π/ x i π/ x  e 2 i The probbility density is = 2/ cos π x = 2/ sin π x (7.1) (7.2) ρ ± = 2 ψ ± (x) 2 cos2 π x for + = 2 sin2 π x for  (7.3) In the bsence of lttices, these two stnding wves hve exctly the sme energy ϵ = ħ2 k 2 ħ 2 ± π 2 m = 2 2 m = ħ2 π2 2 m 2 So, we hve ϵ + = ϵ  = ħ 2 π 2 /2 m 2 However, in the presence of the lttice, the ψ + wve hs lower energy thn ψ , becuse the electron hs higher probbly to pper ner n ion for ψ +. This is the reson why t momentum k = π/ we hve two possible energies (point A nd B in figure 1). Similr bnd gps will rise t k = n π/ for ny integer n. (7.4)
3 Phys463.nb 51 The momentum region n π < k < (n  1) π nd (n  1) π < k < n π is clled the nth Brillouin zones (This is the sme Brillouin zones s we lerned in the reciprocl lttice). In side the of these Brillouin zones, the energy is smooth function nd this smooth function is clled the nth bnd. At ech boundry of the Brillouin zones, the energy curve shows jump nd thus n energy gp opens up Mgnitude of the bnd gp The size of the gp cn be estimted s the following E g = U(x) ρ + (x) d x  U(x) ρ  (x) d x (7.5) where U(x) is the potentil energy from the ions. Becuse the lttice is periodic, U(x) is periodic function U(x + ) = U(x) where is the lttice constnt. In ddition, with out loss of generlity, we cn ssume tht U(x) is n even function U (x) = U(x). For such n even periodic function, we cn write it s Fourier series. nd U(x) = U 2 + U n cos 2 π n n U n = 2 U(x) cos 2 π n x d x The energy gp t k = ±π/ is E g = x U(x)[ρ + (x)  ρ  (x)] d x = 2 π U(x) cos 2 x  sin2 The energy gp t k = ±n π/ is E g = π x d x = 2 U(x) cos 2 π (7.6) (7.7) x d x = U 1 (7.8) U(x)[ρ + (x)  ρ  (x)] d x = 2 U(x) cos 2 n π x  sin2 n π x d x = 2 U(x) cos 2 π n x d x = U n (7.9) 7.2. Bloch wves The effect of the lttice potentil: the Hmiltonin nd the Schrodinger eqution The Hmiltonin H = p2 ħ2 + U(x) =  2 m 2 m x 2 +U(x) (7.1) The Schrodinger eqution  ħ2 2 m x 2 ψ(x) + U(x) ψ(x) = ϵ ψ(x) (7.11) This eqution is in generl hrd (or impossible) to solve nlyticlly. However, F. Bloch mnged to prove very importnt theorem, which sttes tht the solution to this eqution must tke the following form: ψ k (x) = u k (x) e i k x Here, u(x) is periodic function u k (x + ) = u k (x), which hs the sme period s the lttice. This conclusion is true in ny dimensions. For exmple, in 3D, we hve ψ r = u k r exp i k r where u r is 3D periodic function u k r = u k r + T where T is ny lttice vector. This type of wves re known s the Bloch wve. They hve some nice properties The plne wves re specil type of Bloch wves with the function u(x) = constnt. (7.12)
4 52 Phys463.nb The Bloch wve is NOT n eigenstte of the momentum opertor. In other words, it doesn t hve welldefined momentum (unless u(x) = constnt). However, the prmeter k in the Bloch wve behves very similrly to the momentum. We cll ħ k the lttice momentum. Lttice momentum is not the momentum, but it is conserved in lttice The proof of the Bloch theorem Define the trnsltion opertor T ψ(r) = ψ(r + ) (7.13) where is the lttice constnt. This opertor shift our quntum system in rel spce by the lttice constnt. Under trnsltion by, quntum opertor O chnges s O T O T 1. Therefore, for the Hmiltonin T H(x) T 1 = H(x + ) (7.14) Becuse nd H(x) =  ħ2 H(x + ) =  ħ2 2 m x 2 +U(x) (7.15) 2 m x 2 +U(x + ) =  ħ2 we know immeditely tht H (x) = H(x + ). And thus 2 m x 2 +U(x) (7.16) T H(x) T 1 = H(x + ) = H(x) (7.17) In other words, T H(x) = H(x) T (7.18) so [T, H] = (7.19) If two opertors commute, we know tht we cn find common eigensttes of these two opertors. For T nd H, this mens tht we cn find set of (complete orthonorml) bsis ψ k (x), such tht H ψ k (x) = ϵ ψ k (x) nd T ψ k (x) = e i k ψ k (x) (7.2) Here, we used the fct tht H is n Hermitin opertor, so tht its eigenvlues ϵ must be rel. For T, becuse it is unitry opertor, its eigenvlue must be complex with bsolute vlue 1. Define u k (x) = e i k x ψ k (x) (7.21) It is esy to prove tht u k (x) is periodic function u k (x) = u k (x + ). u k (x + ) = e i k (x+) ψ k (x + ) = e i k (x+) T ψ k (x) = e i k (x+) e i k ψ k (x) = e i k x ψ k (x) = u k (x) (7.22) Therefore, ψ k (x) = u k (x) e i k x (7.23) where u k (x) is periodic function Properties of Bloch wves nd crystl momentum Eigen wve functions for single prticle moving in periodic potentil: Bloch wves ψ n k (x) = u n k(x) e i k x (7.24) Here k is known s the crystl momentum. The corresponding eigenenergy is ϵ n,k, which stisfies
5 Phys463.nb 53 H ψ n,k (x) = ϵ n,k ψ n,k (x) (7.25) vlue of k? The crystl moment re only welldefined modulo G, where G is reciprocl vector. Therefore, we cn limit the crystl momentum to the first Brillouin zone without loss of informtion (e.g. for 1D system, π/ k < π/). Proof : Suppose we hve Bloch wve with crystl momentum k ψ n k (x) = u n k (x) e i k x (7.26) Define k ' = k + G ψ n k (x) = u n k (x) e i k x = u n k'g (x) e i (k'g ) x = u n k'g (x) e i G x e i k' x (7.27) Define u n k' (x) = u n, k'g (x) e i G x, ψ n k (x) = u n k'g (x) e i G x e i k' x = u n k' (x) e i k' x (7.28) It is esy to check tht u n,k' (x) = u n,k' (x + T) where T is ny lttice vector. u n,k' (x + T) = u n, k'g (x + T) e i G (x+t) = u n, k'g (x) e i G x = u n,k' (x) (7.29) Therefore, ψ n k (x) = u n k' (x) e i k' x is lso Bloch wve function with crystl momentum k ' ψ n k (x) = u n k' (x) e i k' x = ψ n,k' (x) (7.3) Conclusion: s long s k nd k ' differs by reciprocl lttice vector G, we cn write Bloch wvefunction with crystl momentum k s Bloch wvefunction with crystl momentum k '. In other words, crystl momentum is only well defined modulo G Fig. 3. Dispersion reltion ϵ n (k) shown in extended zone, reduced zone nd periodic zone. (Figure 4 in text book pge 225).
6 54 Phys463.nb Reduced Brillouin zone Since the crystl momentum is only well defined modulo G, we cn limit the momentum to the first Brillouin zone (e.g. for 1D system, π/ k < π/ ) without loss of informtion. We cn plot the eigenenergy ϵ n,k s function of k in the first Brillouin zone. This wy of plotting ϵ n,k is known s the reduced Brillouin zone (See figure b bove) The bnd index n is (from bottom to top) n = 1, 2, 3, Periodic Brillouin zone There is nother wy of plotting ϵ n,k. We cn strt from the reduced Brillouin zone nd then we plot ϵ n,k s periodic function of k, repeting the sme curve in the second, third,... Brillouin zone (See figure c bove) Folding, Reduced Brillouin zone nd extended Brillouin zone for free prticles without lttices We know tht plne wves is specil cse of Bloch wves (where the periodic potentil is V = ). Therefore, we cn present the dispersion of free prticle ϵ = ħ 2 k 2 /2 m in the sme wy (in the reduced BZ nd periodic BZ). The cn be chieve using the following procedures: strt from the dispersion ϵ(k) = ħ 2 k 2 /2 m, then move the curve in the regions (2 n  1) π/ < k < 2 n π/ to π/ < k < nd move the curve in the regions 2 n π/ < k < (2 n + 1) π/ to < k < π/. So we get the figure below. Fig. 4. Reduced zone for free prticle. (textbook p 177 Fig. 8) This construction is known s Brillouin zone folding. Similrly, the reverse procedure is known s unfolding. We cn unfold the reduced Brillouin zone to get the extended Brillouin zone, going bck to the cse  < k < + nd get one single curve with k 2 /2 m for free prticle.
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