Subbands

1. Subbands • So far, we saw how to calculate bands for solids • Boundary conditions we used were artificial periodic bcs • Now we’ll see what happens if you have real boundaries • Quantization of bands along the confinement direction  Subbands • If we confine all 3 directions, the levels are fully quantized, as in an atom • Can then calculate “density of states” and “number of modes”

2. dE dk x x x x x x x x x Where are the states? E dE k k For 1D parabolic bands, DOS peaks at edges ~1/(E-Ec)

3. dE dk . . . . . . . . . . . . . . . . . . . . . . . . Increasing Dimensions E dE dE # k points increases with diameter squared D ~ q(E-Ec) 2pkdk k k In higher dimensions, DOS has complex shapes

4. From E-k to Density of States Σ1 dNs = D(E)dE = 2  2 (dk/[2p/L]) for each dimension k For E = Ec + ħ2k2/2mc • D = (Wmc/2p2ħ3)[2mc(E-Ec)]1/2 in 3-D = (Smc/2pħ2)q(E-Ec) in 2-D = (mcL/pħ)/√2mc[E-Ec] in 1-D

5. Hard to draw a 3-D paraboloid ! ħ2(kx2 + ky2 + kz2) E = +Ec 2mc 3-D DOS E E ~ (E-Ec)1/2 ky kz Ec kx DOS

6. 9ez 4ez ez ħ2(kx2 + ky2) E = +Ec + p2ez 2mc Subband bottoms quantized due to confinement along z ez ≈ ħ2p2/2mcd2 (like modes in a waveguide) Quasi 2D  subbands E ~ Spq(E-Ec-p2ez) ky kx DOS (p=0,1,2,3,…) Quasi-2D d

7. ħ2(kx-k0);2 ħ2(ky2+kz2) E1 = + +Ec 2ml 2mt Thin Films ml=0.91m0, mt=0.19m0

8. d Quantizing k along confinement direction Source: Dragica Vasileska, ASU

9. ħ2(kx2 + ky2) E = +Ec 2mc (Bottommost subband only, included in Ec) 2-D  step function DOS E E ~ q(E-Ec) ky kx DOS 2D

10. ħ2kx2 E = +Ec + p2ez + q2ey 2mc Confinement in 2 directions, 1D subbands Quasi 1D E E kx DOS Quasi-1D (p,q=0,1,2,3,…)

11. Silicon Nanowire Wang, Rahman, Ghosh, Klimeck, Lundstrom, APL ‘05

12. Carbon Nanotube (12,0) (9,0) Kienle, Cerda, Ghosh, JAP ‘06

13. ħ2kx2 E = 2mc 1-D ~ 1/(E-Ec)1/2 E E kx DOS

14. General Results NT(E) = Saq(E-Ea) As each mode starts, you get a new step function D(E) = dNT(E)/dE = Sad(E-Ea)

15. Separable Problems Convolve individual DOS to sum over its modes e(m,n) = ex(n)+ey(m) D(E) = Sm,nd(E-ex(n)–ey(m)) = ∫dE’Dx(E’)Dy(E-E’) = Sm,n∫dE’d(E-ey(n)–E’)d(E’-ex(m))

16. 3D Bulk Solid Quasi-2D wire 2D well 0-D artifical molecule (quantum dot) Quasi 0-D Quantum Dot Quasi 1D wire 1D wire Evolution of DOS E Ec DOS

17. From graphite to nanotube Animation: Dr. Shigeo Maruyama Real boundary conditions (periodic) along circumference

18. Carbon Nanotube Electronics lap ~ 1 mm lop ~ 10 nm Near ballistic operation Ion ~ 3000mA/mm High-k (HfO2) S ~ 65mV/dec Javey et al, Nano Lett ‘04 Yao et al, PRL2000 • Max Current • 4G0ħw ~ 25 mA • Slope 4G0 • (~6.5 kW)

19. 6 BZ vertices, each shared by 3 unit cells Can translate three BZs (each 1/3rd valley) using RLVs Only 2 distinct BZ valleys K1 + K2 K1 + K2 K2 K1 K1 K2 K2 = (p/a)x - (p/b)y K1 = (p/a)x + (p/b)y Graphene BZ (0,2p/3b) (p/a,p/3b) (-p/a, p/3b) (-p/a,-p/3b) (p/a,-p/3b) (0,-2p/3b)

20. Semi-metallic bands

21. ±t √[1 + 4cos(3kxa0/2)cos(kya0√3/2) + 4cos2(kya0√3/2)] Recap: Graphene bandstructure ky (0,2p/3b) kx (0,-2p/3b) E(k) = Periodic boundary conditions quantize ky If ky goes through above BZ points  metallic, else semiconducting

22. ±|h0|, h0 = -t(1 + 2eikxacoskyb) Recap: Graphene Bandstructure E(k) = (0,2p/3b) At kx=0, ky = ±2p/3b, E = 0 (Same for other BZs) The question is whether these BZ points are included in the set of quantized k’s (0,-2p/3b)

23. Quantizing k along circumference

24. Rolling up graphene rc = ma1 + na2 gives the circumference vector

25. Choosing the Chiral vector Different Circumference vectors give different tubes

26. Graphene subbands  CNT bands rc k.rc = m(kxa + kyb) + n(kxa-kyb) = 2pn BZ (0,±2p/3b) is included in these k points if (m-n) = 3n Then we have a metallic nanotube Otherwise semiconducting

27. Chirality controls metallicity SWNT as molecular interconnects: • Cylindrical boundary conditions define a tube: • Chiral indices (n,m) determine the band structure‡: |n-m| = 0,3,6,… , metallic; otherwise semiconducting. Reference ‡ J.W. Mintmire et al., J. Phys. Chem. Sol.54(12) 1835-1840, 1993.

28. -iat(-1) (-at). ±|h0|, h0 = -t(1 + 2eikxacoskyb) Linearize around minima E(k) = Linearize around BZ points (kx = 0,ky = ±2p/3b) by Taylor expansion h0 ≈ ∂h0/∂kx|0kx + ∂h0/∂ky|2p/3b(ky-2p/3b) Zigzag tube (m,0), m(kxa+kyb) = 2pn E = √[En2 + (takx)2], En = takn, kn = (2p/3b).(3n/2m – 1) D(E) = Sn (2L/pat).[E/√(E2 – En2)] = L/pħv

29. Nanotubes are quasi 1-D D(E) = Sn (2L/pat).[E/√(E2 – En2)] E DOS Zigzag MetallicTube (n = 3m), allowed ks include BZ Looks like our quasi-1D results

30. Gap Nanotubes are Quasi 1-D E DOS Zigzag SemiconductingTube (n ≠ 3m)

31. +ve states filled by left contact’s chemical potential Only these states contribute to net current, interval (m1-m2) -ve states filled by right contact’s chemical potentail Current per mode I = (-q/L) Σvx(kx)>0 = -q ∫(dkx/2p) [1/ħ.∂En(kx)/∂kx] = -q/ħ∫dEn vx(kx) = -q/ħ(m1 – m2)= -q2V/h m1 m2 Note: here we assume band velocity v determines I instead of injection velocity g/ħ

32. Mode Counting E W L Modes confined along W DE = ħ2p2/2mcW2 M = Int √(EF – EC)/DE = Int(kFW/p) = Int(2W/lF) Gmax = (2q2/h)M ns = (mc/pħ2)(EF-EC) RminW = 16.28 kW/√ns kx

33. What if conduction were not perfect? If transmission isn’t perfect, T < 1 needs to be included Zero-bias conductance G = 2q2/h x T x M (modes are like “levels”) (eg. Nanotube T=1, M=2) Corresponding I-V I = 2q/ħ∫dE.T(E).[f1(E)-f2(E)] Landauer Theory In ch. 1 we saw how to calculate T(E) = (g1g2/g)D(E)

34. Summary Confinement in a solid breaks its bands into subbands Counting subbands gives us the density of states DOS Each subband mode carries a current –q/ħ

35. µ1 µ2 H + U What next? We have been building up our device model Numbers (e,g,U)  Matrices (H, S, U) Rate equations  NEGF formalism So far we derived [H] for atoms, molecules, solids and confined solids (creating levels, bonds, bands and subbands) The next chapter will concentrate on [U]