PH575 Spring Lecture #28 Nanoscience: the case study of graphene and carbon nanotubes.

Transcription

1 PH575 Spring 2014 Lecture #28 Nanoscience: the case study of graphene and carbon nanotubes.

2 Nanoscience scale nm "Artificial atoms" Small size => discrete states Large surface to volume ratio Bottom-up vs. top-down construction ~pcii/nanocrystals.htm

3 A close look on single quantum dots A. Zrenner Walter Schottky Institut, Technische Universität München, D Garching, Germany The Journal of Chemical Physics, Vol. 112, No. 18, pp , 8 May 2000 Fig. 1. AFM images of self assembled In 0.4 Ga 0.6 As QDs grown by MBE on a GaAs substrate at a temperature of 530 C. Nominally 7.5 monolayers of In 0.4 Ga 0.6 As, slightly above the critical thickness for Stranski Krastanow growth have been deposited under the condition of nonrotated substrate. The resulting gradient in In 0.4 Ga 0.6 As coverage leads to a variation of the QD surface density across the wafer (upper part: ~100 µm 2, lower part: ~20 µm 2 ). Fig. 2. Power dependent PL spectra from a single QD isolated by near field spectroscopy through a nano-aperture. At low excitation power PL only the single exciton decay from the s-shell is observed (1X). At elevated PL occupancies with two and more excitons are realized. The sequential biexciton decay leads to the appearance of the biexciton line (21) in the p-shell (2X). The decay of higher occupancies (for example 32, 43) leads to new emission lines in the spectral region of the p-shell and additional, further renormalized lines in the region of the s-shell

4 Carbon nanotube Gold electrode

5 The measured vibration amplitude revealed an exceptionally high elastic Young's modulus of about N/m 2 (or one TPa) - about five times the value for steel. This results in talk of a "space elevator", to tether an orbiter to earth asp?articleid=1137

6 graphene GRAPHENE Si Graphene nanofabric. SEM micrograph of a strongly crumpled graphene sheet on a Si wafer. Note that it looks just like silk thrown over a surface. Lateral size of the image is 20 microns. Si wafer is at the bottom-right corner. Image "graphene molecule.jpg" is copyright of Chris Ewels ( Fig. 1. Phonon dispersion of graphene calculated within a valence force field (VFF) model (Aizawa). ~vpopov/phdsgraphene.htm

7 A CNT is made by cutting a strip of graphene so that the wrapping vector, shown in red at left, wraps around the circumference of the tube. There are two special cases: the armchair and the zig-zag CNT. The general case is called the chiral nanotube.

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9 CNT towers 500 µm 30 µm 1 µm

10 Approximate 1-D band structure and density of states of (a) metallic CNT and (b) semiconducting CNT. The band gap of a semiconducting tube is inversely proportional to its diameter E g = 4hv F /3d

11 Differential conductance of a CNT. Note how the conductance jumps every time the peak in the DoS lines up with the electron supply reservoir

12 E( k x,k ) y = ±γ cos 1 2 k x a 0 ( 3 )cos 2 k y a 0 ( ) + 4 cos 2 ( k x a 0 ) 2

13 The program Identify unit cell and reciprocal lattice Set up LCAO wavefunction Set up Hamiltonian and Schrödinger s equation Project onto basis states Set up coefficient matrix and solve eigenvalue determinant Plot dispersion relation Find Fermi energy For nanotubes Set up boundary constraints

14 The graphene lattice has 2 inequivalent C atoms per unit cell. Each one is bonded to 3 other C atoms in the plane via sp 2 hybrid orbitals, and each has one electron in a p z orbital perpendicular to the plane. The C-C bond distance is Å.

15 Reciprocal lattice a i b j = 2πδ ij ( ) = 2π ( ) = 0 a 1 b 1 = 3a x 2 1y a 2 b 1 = 3a 0 1 b x 2 1y b 1x = 2π / 3a 0 b 1y = 2π / 3a 0 b 1 = b 2 = 2π 1, 3a 0 1 2π 1, 3a 0 3 ; b ; b 2 = 4π 3a 0 = 4π 3a 0 The reciprocal lattice is triangular. The Wigner-Seitz cell (1st Brillouin zone) is shaded. There are 6 equivalent K points (What are they?).

16 The program ü Identify unit cell and reciprocal lattice Set up LCAO wavefunction Set up Hamiltonian and Schrödinger s equation Project onto basis states Set up coefficient matrix and solve eigenvalue determinant Plot dispersion relation Find Fermi energy For nanotubes Set up boundary constraints

17 Set up LCAO wave function 2 atoms per unit cell Ĥ Ψ k = E( k ) Ψ k R ( ) Ψ k = e i k R φ x R φ( x ) = b 1 φ ( 1 x) + b2 φ ( 2 x) Hamiltonian Ĥ = ˆp 2 2m + V at R Ĥ = ˆp 2 2m + V at x x2 V at x x 1 R ( ) + V at x x2 R ( ) ( x x ) 1 + ( ) + V at x x1 R R 0 ( ) + V ( at x x2 R )

18 Ĥ Ĥ φ 2 x ( ) = ˆp 2 2m + V at + V at x x2 = ε 1 x = ε 1 x x x 1 ( ) x ( ) ( ) + V at x x1 R R 0 ( ) + Vat x x2 ( ) + ΔU1 ( x) ( x) = ε2 φ 2 ( x) + ΔU2 φ 2 ( x) ( ) + V ( at x x2 R ) ( ) + V at x x1 R R 0 x ( ) ( ) + V ( at x x2 R ) x ( ) Onsite matrix element ε 1 = ε 2 Set to zero by choice Ĥ φ j = ΔU j φ j ΔU 1(2) Can be considered the perturbing potential for 1 (2)

19 The program ü Identify unit cell and reciprocal lattice ü Set up LCAO wavefunction ü Set up Hamiltonian and Schrödinger's equation Project onto basis states Set up coefficient matrix and solve eigenvalue determinant Plot dispersion relation Find Fermi energy For nanotubes Set up boundary constraints

20 Ĥ Ψ k = E( k ) Ψ k Now project onto basis states φ j Ĥ Ψ k = φ j E( k ) Ψ k φ j ΔU j Ψ k = E( k ) φ j Ψ k This will be step 2 This will be step 1

21 Step 1 φ j Ψ k Step 2. Ψ k = e i k R φ x R Ψ k R = b 1 + b 2 γ 0 α ( k ) ( ). ( ) ΔU 1 Ψ k = ΔU 1 e i k R φ x R ΔU 1 Ψ k φ j ΔU j Ψ k R = b 2 γ 1 α ( k )

22 Step 1 φ j Ψ k... R ( ) Ψ k = e i k R φ x R Ψ k = b 1 + b 2 γ 0 α ( k )

23 Step 1 φ j Ψ k R ( ) Ψ k = e i k R φ x R R { ( ) + b 2 φ ( 2 x R ) } = e i k R b 1 x R ( ) φ1 ( x) + b2 φ ( 1 x) φ2 ( x) = b 1 x R=0 ( ) + b 2 e i k a 1 x ( ) φ1 x + a1 ( ) ( ) φ2 x + a1 +b 1 e i k a 1 x ( ) φ1 x + a2 R= a 1 ( ) + b 2 e i k a 2 x ( ) ( ) φ2 x + a2 +b 1 e i k a 2 x R= a 2

24 Ψ k = b 1 +b 2 φ ( 1 x ) φ ( 2 x ) +b 2 e i k a 1 φ ( 1 x ) φ ( 2 x + a1 ) +b 2 e i k a 2 φ ( 1 x ) φ ( 2 x + a2 ) Ψ k = b 1 + b 2 γ 0 1+ e i k a 1 + e i k a 2 Ψ k ( ) = b 1 + b 2 γ 0 α k φ 2 Ψ k = b 2 + b 1 γ 0 α * ( k ) End step 1

25 Step 2 φ j ΔU j Ψ k... R ( ) ΔU 1 Ψ k = ΔU 1 e i k R φ x R ΔU 1 Ψ k = b 2 γ 1 α ( k )

26 Step 2 φ j ΔU j Ψ k R ( ) ΔU 1 Ψ k = ΔU 1 e i k R φ x R R = ΔU 1 e i k R b 1 x R { ( ) + b 2 φ ( 2 x R ) } ( ) ΔU1 φ ( 1 x) + b2 φ ( 1 x) ΔU1 φ ( 2 x) = b 1 x R=0 ( ) ΔU1 x + a1 ( ) + b 2 e i k a 1 x ( ) ( ) ΔU1 φ 2 x + a1 +b 1 e i k a 1 x ( ) ΔU1 x + a2 R= a 1 ( ) + b 2 e i k a 2 x ( ) ( ) ΔU1 φ 2 x + a2 +b 1 e i k a 2 x R= a 2

27 Define hopping (offsite) matrix element ΔU 1 Ψ k = b 2 γ 1 1+ e i k a 1 + e i k a 2 φ 2 ΔU 2 Ψ k = b 1 γ 1 1+ e +i k a 1 + e +i k a 2 γ 1 = ΔU 1 φ 2 = φ 2 ΔU 2 ( ) = b 2γ 1 α k = b 1γ 1 α * k ( ) End step 2 Now remember these were the equations we began with: φ j ΔU j Ψ k = E( k ) φ j Ψ k

28 The program ü Identify unit cell and reciprocal lattice ü Set up LCAO wavefunction ü Set up Hamiltonian and Schrödinger s equation ü Project onto basis states Set up coefficient matrix and solve eigenvalue determinant Plot dispersion relation Find Fermi energy For nanotubes Set up boundary constraints

29 Now remember these were the equations we began with: φ j ΔU j Ψ k = E( k ) φ j Ψ k And we found in step 2 and step 1 that: ΔU 1 Ψ k = b 2 γ 1 α ( k ) Ψ k = b 1 + b 2 γ 0 α ( k ) φ 2 ΔU 2 Ψ k = b 1 γ 1 α * ( k ) φ 2 Ψ k = b 2 + b 1 γ 0 α * ( k ) It s easy to put these together to get the eigenvalue equation: ( ) α γ 0 E( k ) γ 1 E k ( ) α *( γ 0 E( k ) γ ) 1 E( k ) = 0

30 The solution, in the limit of small γ 0 is ( ) 2 = γ 2 1 α ( k ) 2 E( k ) = ±γ 1 α ( k ) E k Recall the definition of α(k): ( ) = 1+ e i k a 1 + e i k a 2 α k You find α(k) and show that E ( k ) = ±γ cos( k a ) 1 + 2cos( k a ) 2 + 2cos( k [ a 2 a ]) 1 And explicitly in terms of x and y coordinates E( k x,k ) y = ±γ cos 1 2 k x a 0 ( 3 )cos 2 k y a 0 ( ) + 4 cos 2 ( k x a 0 ) 2

31 The program ü Identify unit cell and reciprocal lattice ü Set up LCAO wavefunction ü Set up Hamiltonian and Schrödinger s equation ü Project onto basis states ü Set up coefficient matrix and solve eigenvalue determinant Plot dispersion relation Find Fermi energy For nanotubes Set up boundary constraints

32 The dispersion relation for graphene! Where is the Fermi energy? Notice the K points. What is E(K x,k y )? E( k x,k ) y = ±γ cos 1 2 k x a 0 ( 3 )cos 2 k y a 0 ( ) + 4 cos 2 ( k x a 0 ) 2

33 New feature for nanotubes is that boundary constraints are important. The wave vector parallel to the axis of the tube in unconstrained (for practical purposes) The wave vector perpendicular to the axis must result in a wavelength that fits into the circumference. Armchair: w = N ( a 1 + a ) 2 3Na = mλ m k y,m = m 2π 3Na

34 3Na = mλ m k y,m = m 2π 3Na

35 E( k x,k ) y = ±γ cos 1 2 k x a ( 3 )cos 2 k y a ( ) + 4 cos 2 ( k x a ) 2 cos 2πm N E F Armchair CNTs are metallic

36 Zig-zag CNTs are metallic or semiconducting

37 General wrapping vector (but limited to vicinity of the Fermi energy): E( k ) = ± 2v F d m n 3 + p 2 + k d 2 2 w = n a 1 + m a 2 Metallic; m-n is a multiple of 3 Semiconducting; m-n is NOT a multiple of 3