Crystal Structure of Graphite, Graphene and Silicon
Dodd Gray, Adam McCaughan, Bhaskar Mookerji∗ 6.730—Physics for Solid State Applications (Dated: March 13, 2009) We analyze graphene and some of the carbon allotropes for which graphene sheets form the basis. The real-space and reciprocal crystalline structures are analyzed. Theoretical X-ray diffrac- tion (XRD) patterns are obtained from this analysis and compared with experimental results. We show that staggered two-dimensional hexagonal lattices of graphite have XRD patterns that differ significantly from silicon standards.
The wide-variety of carbon allotropes and their associ- spacing between sheets c = 6.71A.˚ The sheets align such ated physical properties are largely due to the flexibility that their two-dimensional hexagonal lattices are stag- of carbon’s valence electrons and resulting dimensionality gered, either in an ABAB pattern or an ABCABC pat- of its bonding structures. Amongst carbon-only systems, tern. The ABAB alignment is shown in Figure 1, which two-dimensional hexagonal sheets—graphene—forms of indicates four atoms per unit cell labeled A, B, A’, and the basis of other important carbon structures such as B’, respectively. The primed atoms A–B on one graphene graphite and carbon nanotubes. (:: Say something about layer are separated by half the orthogonal lattice spacing interesting band structure here) from the A’–B’ layer; BB’ atomic pairs differ from their In the following, we will examine the planar lat- corresponding AA’ pairs in their absence of neighboring tice structure of graphene and its extension to higher- atoms in adjacent layering planes. The coordinates of dimensional lattice structures, such as hexagonal these atoms forming the basis are given by: graphite. We first analyze the lattice and reciprocal- c space structures of two-dimensional hexagonal lattices of ρA = (0, 0, 0) ρA′ = 0, 0, carbon, and use the resulting structure factors to esti- 2 mate the x-ray diffraction (XRD) intensities of graphite. a 1 a a c ρB = , 1, 0 ρB′ = − , − , (1.1) We conclude by comparing its calculated XRD spectra to 2 √3 2√3 2 2 experimental spectra of graphene and crystalline silicon. With respect to an orthonormal basis, the primitive lattice vectors are given by, 1. PRELIMINARY QUESTIONS a = a √3/2, 1/2, 0 a = a = 2.46A˚ 1 − | 1| 1.1. Lattice Structure a2 = a √3/2, 1/2, 0 a2 = a = 2.46A˚ | | Our discussion of the crystal structure of graphite fol- a3 = c (0, 0, 1) a3 = c = 6.71A˚. (1.2) lows partially from D.D.L. Chung’s review of graphite [1]. | | When multiple graphene sheets are layered on top of each The magnitudes of the primitive lattice vectors corre- other, van der Walls bonding occurs and the three di- spond to the lattice constants parallel and perpendicu- mensional structure of graphite is formed with a lattice lar to the graphene sheet. The corresponding ABCABC layer forms a rhombohedral structure with identical lat- tice spacing parallel and orthogonal to the layer.
1.2. Reciprocal Lattice Structure
Recall that the reciprocal lattice vectors bi are defined as a function of the primitive lattice vectors ai such that a a b 2 3 1 = 2π a a× a FIG. 1: In-plane structure of graphite and reciprocal lattice 1 3 3 a· ×a vectors [1]. 3 1 b2 = 2π × a2 a3 a1 a· ×a b 1 2 3 = 2π a a× a (1.3) 3 · 1 × 2 ∗Electronic address: [email protected], [email protected], mook- [email protected], The reciprocal lattice vectors for graphite are then 2
of points in space using the specified lattice vectors and atom bases. Normals generated from lattice vector sums were used to extract planes and display them. Varying colors were used to depict vertical spacing between adja- cent planes.
2.2. Structure Factors and X-Ray Diffraction Intensities
Calculate the structure factor for all the 2 reciprocal lattice vectors Kl < 16 (2π/a) .
The structure factor is calculated as
n −iKi·ρi FIG. 2: Reciprocal lattice planes [001], [110], and [111]. Mp (Ki) = fc (Ki) e Xj=1 given by, where fc is the structure factor of Carbon and ρi are the b 2π 1 b 2π 2 basis vectors of our lattice. We find that only four unique, 1 = , 1, 0 1 = K a √3 − | | a √3 non-zero values of Mp ( i) occur in the reciprocal lattice. 2π 1 2π 2 Each of these corresponds to the height of a diffraction b2 = , 1, 0 b2 = intensity peak and their relative values are referenced in a √3 | | a √3 Table I. 2π 2π b = (0, 0, 1) b = . (1.4) 3 c | 3| c Calculate the ratio of the intensities ex- pected for the following lines of the diffraction The reciprocal lattice plane generated by the b and b 1 2 pattern with respect to the [111] line: [100], vectors forms the outline of the first Brillouin zone, as [200], [220], [311], and [400]. depicted in Figure 1. The intersection of the the planes kz = 2π/c with the plane forms a hexagonal prism of ± Including the structure factor, there are other several height 4π/c. factors contributing to the intensities of the diffraction peaks [3]: 1.3. Atomic form factors 1. The Lorenz correction is a geometric relation al- tering the intensity of an x-ray beam for different As carbon is the only element present in graphene and scattering angles θ. graphite, the atomic form factor is uniform across the entire crystal, and thus can be factored out when calcu- 2. The multiplicity factor p is defined as the num- lating the structure factor. Thus the atomic form factor ber of different planes having the same spacing has no effect on the relative intensities of x-ray diffrac- through the unit cell. Systems with high symme- tion occuring in different planes of graphite. According tries will have different planes contributing to the to the NIST Physics Laboratory, the atomic form factor same diffraction, thereby increasing the measured of carbon varies from 6.00 to 6.15 e/atom with incident intensity. radiation ranging from 2 to 433 KeV [2]. 3. Temperature, absorption, polarization each con- tribute higher-order corrections ultimately ignored 2. X-RAY DIFFRACTION in our calculation. These include Doppler broad- ening from thermal vibrations in the material, ab- sorption of x-rays, and the polarization of initially 2.1. Planes in the Reciprocal Lattice unpolarized x-rays upon elastic scattering.
Provide pictures of the crystal and of the These factors all contribute to the relative intensity of a reciprocal lattice in the [100], [110], and [111] [hkl] diffraction peak given by planes. Indicate the vertical positions of 2 atoms with respect to the plane. 2 1 + cos 2θ I[hkl] (θ)= p Ma (Km) . (2.1) | | sin2 θ cos θ Pictures of the crystal and of the reciprocal lattice in the [100], [110], and [111] planes are included in Fig- Using the formula from the previous question to calcu- ure 2. In MATLAB, the crystal was represented as a set late the ratios of the structure factors in the given planes, 3
diffraction?
The crystal structure of another common semiconduc- tor material, silicon (Si) is featured in Figure 3. Silicon forms a diamond cubic crystal structure with a lattice spacing of 5.42A.˚ This crystal structure corresponds to a face-centered cubic Bravais lattice whose unit-cell basis contains 8 atoms located at vector positions, a d = ~0 d = (1, 3, 3) 0 4 4 a a d = (1, 1, 1) d = (2, 2, 0) 1 4 5 4 a a d = (3, 3, 1) d = (2, 0, 2) 2 4 6 4 a a d = (3, 1, 3) d = (0, 2, 2) . (2.2) 3 4 7 4 FIG. 3: Crystal structure of silicon. The structure factor contributing to its X-ray diffrac- tion pattern is given by TABLE I: X-Ray diffraction intensities for graphite and sili- con [4, 5]. Structure factors are included in parentheses. n (j) −iKm·dj Ma (Km) = f (Km) e Si C (Graphite) a ◦ ◦ Xj=1 2θ ( ) Exp. 2θ ( ) Calc. Exp. = f(1+( 1)h+k +( 1)k+l +( 1)h+l [111] 28.47 100.00 — (3) — − − − h+k+l 3h+k+l 3h+3k+1 [100] — — 42.77 4(3) 3.45 +( 1) +( i) +( i) − − − [200] — — — (1) — +( i)h+3k+1) [002] — — 26.74 106(16) 100 − = f 1+( 1)h+k +( 1)k+l +( 1)h+l [220] 47.35 64.31 — (1) — − − − [311] 37.31 37.31 — (3) — 1+( i)h+k+l . (2.3) [400] 69.21 9.58 — (1) — · − This term undergoes a number of simplifications based we obtain the following results presented in Table I. Cal- on the parity of its Miller indices. If [hkl] are all even culations show that planes [100], [200], [220] and [400] ex- and are divisible by 4, then Ma (Km) = 8f. If they are hibit relative diffraction intensities 1/3 that of the [111] not divisible by 4 or have mixed even and odd values, and [311] planes. The [002] plane exhibits the highest in- then Ma (Km) = 8f. Lastly if [hkl] are all odd, then tensity diffraction, 16/3 that of the [111] and [311] planes. Ma (Km) = 4f (1 i). ± We also found several non-zero structure factors that are The experimental X-ray diffraction intensities from not present in the experimental data. these contributions are listed in Table I. The intensity values for silicon were measured with respect to a ref- erence value I/I0 = 4.7, which is a direct ratio of the 2.3. Crystal Structure of Silicon strongest line of the sample to the strongest line of a ref- erence sample α–Al2O3. The number of visible peaks and What are the ratios if the material were the relative intensities between them suggest that silicon Si? How could you use this information to and graphite can be easily distinguished from each other distinguish Si from your material by x-ray using an x-ray diffraction experiment.
[1] D. Chung, Journal of Material Science 37, 1475 (2002). [5] (????), URL http://rruff.info/graphite/display= [2] (????), URL http://physics.nist.gov/PhysRefData/. default/. [3] B. Cullity, Elements of X-Ray Diffraction (Addison- Wesley, 1978). [4] E. E. B. P. J. d. G. C. H. S. C. M. Morris, H. Mc- Murdie, Standard X-ray Diffraction Powder Patterns, vol. Monograph 25, Section 13 (National Bureau of Standards, 1976). Phonon Spectra of Graphene
Dodd Gray, Adam McCaughan, Bhaskar Mookerji∗ 6.730—Physics for Solid State Applications (Dated: April 17, 2009) We use a Born model to calculate the phonon dispersion of graphene by accounting for stretching (αs) and bending (αφ) interactions between nearest neighbors. Our model describes four in-plane vibrational modes to whose dispersion relations we fit experimental lattice mode frequencies, yielding force constants αs = 445N/m and αφ = 102N/m. Our model also reasonably accounts for graphene’s macroscopic properties, particularly sound speeds and elastic constants.
The lattice dynamical properties of graphene form the atom at R and the atom at R + p is basis of understanding the vibrational spectra of carbon- E based allotropes of various geometries, such as graphite [R, R + p]= Es + Eφ or carbon nanotubes. We can use an analytical descrip- 1 = α p (u[R + p] u[R]) 2 tions of micromechanical behavior to better understand 2 s| · − | the acoustic and optical properties of these materials. 1 + α u[R + p] u[R] 2 p (u[R + a] u[R]) 2 . In the following, we calculate the in-plane vibrational 2 φ | − | − | · − | spectrum of graphene and its contributions to macro- (1.1) scopic elastic and thermodynamic quantities. We first discuss the force parameters of our Born model and de- The model assumes that the bond is only rive a general potential to calculate the dynamical ma- slightly displaced from equilibrium. How trix of a primitive cell. Using our dispersion relations at would you modify the model to make the high-symmetry points, we can determine the vibrational bond energy more realistically dependent on density of states, in-plane sound velocity, and elastic con- displacement from equilibrium - what or- stants. We will briefly touch upon weak out-of-plane der would the corrections be, and of what vibrational modes and its contributions to graphene’s sign? Justify your answer physically; include macroscopic properties. sketches if appropriate. The energy in the bond between the atom at lattice vector Ri and the atom at lattice vector Ri+a is generally 1. BACKGROUND FOR THE BOHR MODEL estimated using a Taylor expansion to the second order as 1.1. Parameters of Bohr Model V (u[Ri, t])) = ∂2V How many force constants are required for V0 + u [R , t], ∂u [R , t]∂u [R + a, t] n i each bond? Why? n m n i m i eq X X (1.2)
Two force constants, αs and αφ, are required to model where the position of the atom at Ri + a is fixed, n and each bond. Let αs represent the restoring force seen when m index all dimensions being considered, V0 is the bond a bond is stretched and let αφ represent the restoring energy seen with zero displacement and V is the potential force seen when a bond is bent away from the axis along defined as the sum of all bond energies in the entire lattice which it is normally aligned. V = + (E[R, R + a ]+ E[R, R a ])+ , (1.3) ··· n − n ··· n What is the energy of a single bond in the X Born model? where this is the slice of the lattice potential related to an atom at R and the number of different vectors an The energy in a single bond is the sum of the stretching between the atom at R and atoms coupled to it depends on which nth nearest neighbor model is used to model and bending energies, Es and Eφ. If p is a vector along which the bond is aligned in equilibrium and R is a lattice the lattice. vector, then the energy contained in a bond between the Note that there is no first order term in this expan- sion. A first derivative of potential energy would imply a net force, so this term must equal zero as the Taylor coefficients are evaluated at equilibrium. ∗Electronic address: [email protected], [email protected], mook- To make the Born model more accurate, we would need [email protected], to take into account higher order terms from the Taylor 2 expansion of the potential. We know that the third order term is non-zero because the curve we are attempting to fit is not even around the equilibrium point. Based on the fact that the function has higher curvature to the left of the equilibrium point than it does to the right, it seems that the sign of the third order term would be negative.
1.2. Nearest Neighbor Couplings FIG. 1: Hexagonal crystal structure of a graphene primitive If you use the only nearest neighbor cou- cell and its neighbors; First Brillouin zone of graphene and plings, how many force constants will your its symmetry points. model require for your material? How large Born force model and, and then briefly discuss the out- will the dynamical matrix be? What if you of-plane vibrational modes. used nearest neighbor and next-nearest neigh- bor couplings? Two force constants are needed to model graphene 2.1. Lattice and Reciprocal Space Structures whether a nearest neighbor coupling or nearest and next nearest neighbor model is used, as long as the model is [Label] all the atoms in the basis and all their two-dimensional. As mentioned before, these are αs and nearest neighbors. For each atom labeled A- αφ, representing restoring forces due to bond stretching H, [verify] the lattice vectors Rp to each unit and bond bending respectively. As graphene is a two cell. dimensional material with a two atom basis, the dynam- ical matrix will be four-by-four. A third force constant A graphene sheet forms a two-dimensional hexagonal αz is required to account for out-of-plane phonon modes. crystal lattice with a primitive cell containing two atoms The size of the dynamical matrix is not affected by the (A and B), depicted in Figure 1. Based on our previ- number of force constants used in our model, however if ous discussion of the Born model, let us assume that a third dimension is added to account for out of plane the primitive cell interacts with nearest neighbors and vibrations then the dynamical matrix will be six-by-six. next-nearest neighbors with spring constants αs and αφ, respectively, and that atoms of the primitive cell have a lattice spacing given by a = 1.42A˚ and a primitive lat- 1.3. Elastic Properties tice constant 2.46A.˚ Figure 1 also depicts graphene’s first Brillouin zone and its symmetry points located at kΓ = k k How many independent elastic constants (0, 0), M = 2π/a√3, 0 , and K = 2π/a√3, 2π/3a . The lattice contains two sublattices 0 and 1, which does your material possess? What are they (give numbers)? Why will a nearest neighbor differ by their bond orientations. The first atom A in approach not provide the most general solu- the primitive cell has three first neighbors in the other tion for a cubic material? sublattice 1 with relative unit vectors Graphene has two elastic constants, λ and µ. The 1 √3 1 √3 e = (1, 0) e = , e = , . measured values of transverse and longitudinal phonon B C −2 2 D −2 − 2 3 1 ! ! velocities in graphene are vt = 14 10 ms− and vl = 3 1 · (2.1) 21.7 10 ms− respectively [1]. Given the density ρ = 2 · 3√3a 1 7 2 and six next-nearest neighbors in the same sublattice 0 2MC [ 2 ]− = 7.6 10− kgm− , where the mass of · 26 with relative unit vectors carbon MC = 1.99 10− kg and the lattice constant 10 · a = 1.42 10− m, the Lam´e coefficients are calculated · √3 1 √3 1 using that vt = µ/ρ and vl = (λ + 2µ)/ρ to be µ eE = , eF = , eG = (0, 1) 29.3N/m and λ 72.2N/m. It is worth noting that these≃ 2 2! 2 −2! ≃p p macroscopic elastic constants are not usually calculated √3 1 √3 1 or measured for graphene because it is a two-dimensional eH = , eI = , eJ = (0, 1) . material with single atom thickness. 2 2! − 2 −2! − (2.2)
2. CONSTRUCTION OF THE DYNAMICAL MATRIX 2.2. Born Force Model
We will first determine the phonon dispersion relations [Using a Born force model, find] a general of the in-plane vibrational modes in the context of the expression for the potential energy of all the 3
TABLE I: Calculated nearest neighbor and next-nearest neighbor forces for graphene. The first (second) row of a given planar index specifies the force interaction between the A (B) atom and one of its neighbors.
A B CD EFGHIJ
xx 3αs/2+3αφ αs αs/4 αs/4 3αs/4 3αs/4 0 3αφ/4 3αφ/4 0 − − − − − − − αs 3αs/2+3αφ — — αs/4 αs/4—— — — − − − xy, yx 0 0 √3αs/4 √3αs/4 √3αφ/4 √3αφ/4 0 √3αφ/4 √3αφ/4 0 − − − 0 0 —— √3αs/4 √3αs/4— — — — − yy 3αs/2+3αφ 0 3αs/4 3αs/4 αφ/4 αφ/4 αφ αφ/4 αφ/4 αφ − − − − − − − − 0 3αs/2+3αφ — — 3αs/4 3αs/4— — — — − −
′′ atoms in the crystal in terms of their displace- In this explanation, Vij are second-partial derivatives of ment from equilibrium. our Born model potential at equilibrium, Rp are the rel- ative lattice vectors described in Equations 2.1 and 2.2, The total elastic potential energy for the all the atoms u 1 2 1 2 T in the crystal in terms of their displacements from and = ux, uy, ux, uy . The force constants from tak- ing the appropriate second derivatives is given in Table I. equilibrium is given by summing the nearest-neighbor stretching interactions, The dynamical matrix is then given by
1 2 V (R)= α e (u1 [R] u2 [R + eB]) s 2 s| B · − | A0 B0 C D 1 e R R e 2 + αs C (u1 [ ] u2 [ + C]) B0 A1 D B1 2 | · − | D (k)= . (2.7) 1 2 C∗ D∗ A0 B0 + αs eD (u1 [R] u2 [R + eD]) , (2.3) D B B0 A1 2 | · − | ∗ 1∗ with the nearest neighbor bending interactions
Vφ If we let (AA) and (BB) represent the 0 and 1 sublattices, respectively, these matrix elements represent 1 2 2 = α (u1 [R] u2 [R + eB]) e (u1 [R] u2 [R + eB]) 2 φ − | −| B · − | 1 2 2A0 = Dxx (AA)= Dxx (BB) + α (u1 [R] u2 [R + eC]) e (u1 [R] u2 [R + eC]) 2 φ | − | −| C · − | B0 = Dxy (AA)= Dyx (AA)= Dxy (BB)= Dyx (BB) 1 2 + αφ (u1 [R] u2 [R + eD]) eD (u1 [R] u2 [R + eD]) C. 0 = D (AB)= D (BA) 2 | − |−| · − | xx xx∗ (2.4) D0 = Dxy (AB)= Dyx (AB)= Dxy∗ (BA)= Dyx∗ (BA) A1 = Dyy (AA)= Dyy (BB)
2.3. Dynamical Matrix B1 = Dyy (AB)= Dyy∗ (BA) . (2.8)
Use the expression for the potential energy As a sample calculation, consider the element Dxx given to determine the force on a given atom in by A0, the crystal in terms of its displacement and its neighbors displacements. Check your an- √3 + ky a swer by directly calculating the force from the ′′ ′′ i 2 kxa 2 A0 = V (AA)+ V (AE) e− “ ” spring constants and displacements. xx xx √3 ky a ′′ i“ 2 kxa 2 ” ′′ iky a Verify from the potential, by explicitly taking + Vxx (AF ) e− − + Vxx (AG) e− √3 ky a √3 + ky a the derivatives, the factors in the matrix. ′′ i“ 2 kxa 2 ” ′′ i“ 2 kxa 2 ” + Vxx (AH) e − + Vxx (AI) e From the potential energy, the force on a given atom + V ′′ (AJ) eiky a in the crystal is given in terms of a harmonic oscillator xx √3 + ky a √3 ky a force expression 3αs 3αs i“ 2 kxa 2 ” 3αs i“ 2 kxa 2 ” = + 3αφ e− e− − k 2 − 4 − 4 fi = Dij ( ) uj, (2.5) √3 ky a √3 ky a 3αφ i kxa 3αφ i kxa+ + 0 e “ 2 − 2 ” e “ 2 2 ” + 0 where Dij is the dynamical matrix is given by − 4 − 4 √ k ˜ R ik Rp ′′ R ik Rp 3 3 1 Dij ( )= Dij ( p) e− · = Vij ( p) e− · . = α + 3α 1 cos k a cos k a 2 s φ − 2 x 2 y Rp Rp " ! # X X (2.6) (2.9) 4
TABLE II: Calculated and measured [2] macroscopic prop- erties of graphene.
Quantity Measured Calculated Lattice Mode Frequencies (meV) Γ M K Γ M K ωLO 194.8 175.1 155.36 170.3 182.9 197.4 ωLA 0 169.84 155.36 0 151.5 156.4 ωTO 194.8 169.84 155.36 170.3 180.6 156.4 ωTA 0 50.03 125.07 0 66.5 99.8 ωZO 0 75 60 — — — ωZA 110 52 60 — — — Sound Velocities (km/s) vLA 21.7 13.12 vTA 14 6.21 vZA — — Elastic Constants (10 GPa) C11 106 2 131.0 ± C12 28 2 72.3 ± C44 0.43 0.05 — ±
A similar calculation for the remainder of the matrix elements gives us
3 √3 1 A0 = α + 3α 1 cos k a cos k a 2 s φ − 2 x 2 y " ! #
√3 kya B0 = α √3sin k a sin − φ 2 x 2 ! i kxa 1 i kxa kya C = α e− “ √3 ” + e “ 2√3 ” cos − s 2 2 √3 i kxa kya D = i α e “ 2√3 ” sin − 2 s 2 3 √3 1 A1 = α + α 3 2 cos (k a) cos k a cos k a 2 s φ − y − 2 x 2 y " ! # 3 i kxa kya B1 = α e “ 2√3 ” cos . (2.10) −2 s 2 FIG. 2: Calculated and experimental phonon dispersion curves for graphene. (a) Fitted experimental dispersion 3. MODEL OPTIMIZATION AND curves using inelastic x-ray scattering in graphite [2]; (b) (c) COMPARISON TO MACROSCOPIC calculated dispersion relations using our fitted force constants PROPERTIES (αs = 445N/m and αφ = 102N/m) and suggested force con- stants (αs = 1 and αφ =0.25). 3.1. Comparison with Published Theoretical and Experimental Data
Plot the phonon dispersion in appropriate optic and lowest acoustic modes at X and units along the Γ X , X L, and Γ L L. Provide drawings of the atomic motion − − − directions. of these modes. How many modes are there at Γ? For these values of force constants and masses, determine the atomic displacements for all the modes at Γ, and for the highest 5
3.2. Phonon Dispersion Relations 4. Many qualitative features of our calculated disper- sion relations are consistent with the experimental and The phonon dispersion relation ω = ω (k) is deter- ab initio data in Figure 2 [4]. Note degeneracies at the mined from the eigenvalue equation Γ point and linear dispersion relations for small displace- ments from Γ. Hexagonal symmetry of the graphene lat- 1 2 M− D (k) ǫ = ω ǫ. (3.1) tice also accounts for the measured and calculated de- generacies at the M and K points. The largest discrep- Note additionally, that while graphene is two- ancies between experimental data and our Born model dimensional, it possesses miniscule elastic movement in are for the TO and LO modes, which require accounting the vertical direction, leading to a very small, but non- of electron-phonon interactions in ab initio models for zero elastic constant C44. We can define force constants accurate prediction [5]. βs and βφ for the nearest and next-nearest interactions of Theoretical sound velocities for both longitudinal and the out-of-plane vibrational modes. It has been shown [3] optical polarizations are estimated by calculating the that the phonon dispersion relations for the out-of-plane slopes of the acoustical phonon dispersion curves of both optical and acoustic vibrational modes are δω polarizations near Γ and taking c = δk . The elastic constants C11 and C12 are determined from the phonon ω = √u v, (3.2) ZO,ZA ± sound velocities as where vLA = C11/ρ vTA = (C11 C12) /ρ, (3.4) − 3kxa √3kya p p u = 2βφ cos √3kya + 2 cos cos 3βswhere ρ is the mass density of graphene. " 2 2 !# − 1/2 √3k a 3k a √3k a v = β 1 + 4 cos2 y + 4 cos x cos y , 3.3. Density of States s 2 2 2 " ! # (3.3) Plot the total density of states (histogram method, include all modes) versus frequency. where the out-of-plane spring constants are given by βs = 1.176 and βs = 0.190. These dispersion relations Figure 3 shows the overall density of states and those are decoupled− from the in-plane vibrational modes. By for individual modes. Experimental data for graphene is diagonalizing the Hermitian dynamical matrix in Equa- unavailable, although the our calculation seemed qualita- tion 2.7, we can add these dispersion relations a sub- tively consistent with experimental data for graphite [4] space. From empirical data (see II), we know that elastic out-of-plane motion is several orders of magnitude less than the in-plane motion. Our own plots of Equation 3.3 3.4. Specific Heat yielded high lattice mode frequencies, contradicting this intuition. As such, we have omitted its inclusion in our Calculate the specific heat of your material determination of macroscopic elastic constants. versus temperature using (a) your calculated Figure 2 compares the measured dispersion relation density of states, (b) a Debye model, and (c) of graphene against our calculated dispersion relations a combined Debye-Einstein model (Debye for using the suggested force constants (αs = 1 and αφ = acoustic modes, Einstein for optical modes). 0.25) and our fitted force constants (αs = 445N/m and Plot your results for temperatures between 0K αφ = 102N/m). The optimum force constants are found and 3ΘD. Comment on the strengths and for our model by parameterizing the force constants and weaknesses of your model. comparing the results of ω at the relevant zone edges. Two force constants are chosen, and ω is calculated for The specific heat for graphene is calculated using the each of the three zone edges, Γ, M, and K. An RMS combined density of states (3), Debye model, and com- error is generated by subtracting the experimental values bined Density of States (optical branches) and Debye from the calculated values, and iteratively the optimum model (acoustic branches). The two models result in rad- values for the force constants are found by searching for ically different specific heats, perhaps because the De- the lowest RMS error value. bye model does not account for the optical modes, upon The relative atomic displacements for the acoustic and which the specific heat significantly depends. The plots optical modes at the Γ, K, and M symmetry points for the specific heats in these models is given in Figure 4. is shown in Figure 5.There are a total of four in-plane In the Debye model calculation, the longitudinal acous- modes (acoustical and optical, transverse and longitudi- tic velocity is used as the speed of sound instead of the nal) present at Γ, however both the optical and acousti- inverse addition of longitudinal and tranverse velocities, cal pairs are degenerate. The atomic displacement pat- because in the calculated phonon dispersion relation the terns associated with the various eigenmodes at different transverse acoustic line is by far the most deviant from points of high symmetry in K space are shown in Figure experimental value. In addition, it should be noted that 6
use of the longitudinal acoustic velocity provides a rea- sonably accurate ΘD of 2200 K (with experimental data suggesting a ΘD of 2400 K [6].
FIG. 3: Calculated total and individual-mode density of states.
[1] J. Hone, Book Series Topics in Applied Physics 80, 273 [5] L. M. Woods and G. D. Mahan, Phys. Rev. B 61, 10651 (2001). (2000). [2] V. K. Tewary and B. Yang, Physical Review B (Condensed [6] W. DeSorbo and W. W. Tyler, The Journal of Chemi- Matter and Materials Physics) 79, 075442 (pages 9) cal Physics 21, 1660 (1953), URL http://link.aip.org/ (2009), URL http://link.aps.org/abstract/PRB/v79/ link/?JCP/21/1660/1. e075442. [3] L. A. Falkovsky, Soviet Journal of Experimental and Theoretical Physics 105, 397 (2007), arXiv:cond- mat/0702409. [4] M. Mohr, J. Maultzsch, E. Dobardˇzi´c, S. Reich, I. Miloˇsevi´c, M. Damnjanovi´c, A. Bosak, M. Krisch, and C. Thomsen, Physical Review B (Condensed Matter and Materials Physics) 76, 035439 (pages 7) (2007), URL http://link.aps.org/abstract/PRB/v76/e035439. 7
FIG. 4: Calculated specific heats using calculated density of states, Debye model, and combined Debye-Einstein model. 8
(a)Γ point. (b)K point.
(c)M point.
FIG. 5: Atomic displacements for the eigenmodes of graphene at the 5(a) Γ, 5(b) K,and 5(c) M points [4]. Electronic Band Structure of Graphene
Dodd Gray, Adam McCaughan, Bhaskar Mookerji∗ 6.730—Physics for Solid State Applications (Dated: April 17, 2009)
The band structure properties of graphene form the basis of understanding the electronic spectra of carbon- based allotropes of various geometries, such as graphite or carbon nanotubes. We can use an analytical and nu- merical descriptions of carbon’s valence electrons to bet- ter understand the conduction and semiconductor prop- erties of these materials.
1. BAND STRUCTURE BACKGROUND FIG. 1: Hexagonal crystal structure of a graphene primitive QUESTIONS cell and its neighbors; First Brillouin zone of graphene and its symmetry points. How many extended orbital basis func- tions will you have for your material? Why? interactions between the 1s and 2s/2pi orbitals. Being independent of the valence orbitals, the two extra rows of A graphene sheet forms a two-dimensional hexagonal the Hamiltonian are block diagonalized from other eight crystal lattice with a primitive cell containing two atoms rows. (A and B); its lattice structure of graphene and its first Brillouin zone is shown in Figure 1. The electron con- 2 2 2 figuration of free carbon atoms is 1s 2s 2p . For the 2. CONSTRUCTION OF HAMILTONIAN two atoms in the basis, there are four valence orbitals, MATRIX yielding eight extended orbital basis functions for our material. Because of its planar structure, atoms undergo bonding with four hybridized sp2-wave functions of the What are the atomic configurations of the form two atoms in your material? Which orbitals do you expect to play a significant role in 1 bonding? 2s + √2 2pi (i = x, y, z) . (1.1) √3 | i | i The electron configuration of free carbon atoms is Conceivably one could also construct the 2 2 2 1s 2s 2p . The valence electrons in the 2s, 2px, 2py, LCAO wave function out of core orbitals as and 2pz orbits play a significant role in bonding. well as valence orbitals. How many extended orbital basis functions would you have to use for your material? How large would your Draw all the atoms in the basis and all Hamiltonian matrix be in this case? How their nearest neighbors with appropriate or- do you expect your results would differ from bitals on each atom. Label the orbitals accord- those you would get with just valence orbitals? ing to their lattice vector, basis vector, orbital What if we used higher (totally unoccupied) type, and the type atom they are associated orbitals, too? How many orbitals per atom with (cation or anion). would we have to use to get an exact band structure? Why? The lattice structure of graphene and its first Brillouin zone is shown in Figure 1. The lattice contains two sub- If we were to add the 1s orbital to our LCAO basis set, lattices 0 and 1, which differ by their bond orientations. we would have five orbitals per atom in a two atom basis, The first atom A in the primitive cell has three first neigh- yielding 10 LCAO wavefunctions and a 10x10 Hamilto- bors in the other sublattice 1 with relative unit vectors nian matrix. The additional of this orbital to our basis has a negligible effect due to the statistically insignificant 1 √3 1 √3 eB = (1, 0) eC = , eD = , . −2 2 ! −2 − 2 ! (2.1) ∗Electronic address: [email protected], [email protected], mook- [email protected], and six next-nearest neighbors in the same sublattice 0 2 with relative unit vectors
√3 1 √3 1 eE = , eF = , eG = (0, 1) 2 2! 2 −2! √3 1 √3 1 eH = , eI = , eJ = (0, 1) . 2 2! − 2 −2! − (2.2)
The carbon basis atom A at cell vector R has orbitals (a)Four sp2 hybridized orbitals in carbon bonding. 2s (R) and 2pi (R) (i = x, y, z), and orbitals from |three nearesti neighbors| i shown in Figure 1,
A R e A R e B : 2s ( + B) and 2pi ( + B) (2.3) A R e A R e C : 2s ( + C) and 2pi ( + C) (2.4)
A R e A R e D : 2s ( + D) and 2pi ( + D) , (2.5)
and the carbon basis atom B at cell vector R has orbitals 2s (R) and 2pi (R) (i = x, y, z), and orbitals from |three nearesti neighbors,| i
B B A : 2s (R eB) and 2p (R eB) (2.6) − i − B R e B R e E : 2s ( C) and 2pi ( C) (2.7) − − B R e B R e (b)The rotation of 2p and (c)Sample matrix elements for F : 2s ( D) and 2pi ( D) . (2.8) x − − σ-bond band parameter σ-bonding. [. . . ] approximate all the nearest neigh- overlaps for the 2s and 2p (i = x, y, z) orbitals. bor interactions. Is this approximation rea- i sonable? Calculate values for Ess, Esp, Exx, FIG. 2: Orbital overlaps in graphene. Figure 2(b) shows how and Exy in terms of Vssσ, Vspσ, etc to project 2px along the σ and π components, and the non- vanishing (1-4) and vanishing (5-8) elements of the Hamilto- In examining the orbital interactions contributing to nian matrix. Figure 2(c) shows examples of Hamiltonian ma- the matrix elements of the Hamiltonian, we will refer A ˆ B A ˆ B trix elements for σ orbitals ˙2s ˛ H ˛2px ¸ and ˙2px ˛ H ˛2py ¸, ˛ ˛ ˛ ˛ to several figures taken from Saito and Dresselhaus that respectively. Figures taken from [1]. depict carbon’s valence orbital overlaps [1]. The hy- bridized orbitals contributing to the LCAO approxima- 2 The matrix elements for the Bloch orbitals between tion are given by the sp orbitals, the A and B atoms of the basis are obtained by tak- ing the components of the 2px and 2py orbitals in the σ A,B 1 A,B 2 A,B and π basis. From rotating the orbitals, we have one of φ1 = 2s 2px √3 ± 3 eight overlap configurations given by Figure 2. For exam- E r A,B 1 A,B 1 A,B 1 A,B ple, the wavefunction 2px is determined by projecting φ2 = 2s 2p 2p | i √3 ∓ √6 x ± √2 y it onto the σ and π basis such that E A,B 1 A,B 1 A,B 1 A,B π π φ3 = 2s 2p 2p 2p = cos 2p + sin 2p . (2.11) √3 ∓ √6 x ± √2 y | xi 3 | σi 3 | πi E A,B A,B φ4 = 2p . (2.9) Using these rotation transformations, we can determine E the energies given in Equation 2.10 in terms of Vijk over-
Figure 2 shows that all nearest neighbor bonding inter- lap terms. For example, the third energy overlap term actions are described by each of shown in Figure 2(b) is given by
A B ˆ e A B 3 ikxa/2√3 iky a/2 Ess = 2s (0) H 2s ( B) 2p Hˆ 2p = (V + V ) e− e x y 4 ppσ ppπ E = 2sA (0) Hˆ 2pB (e ) sp x B 3 2 3 2 ikxa/ √ iky a/ A ˆ B e (Vppσ + Vppπ) e− e− Exx = 2px (0) H 2px ( B) − 4
A B √ ˆ e 3 ikxa/2√3 kya Exy = 2px (0) H 2py ( B) , (2.10) = i (V + V ) e− sin . 2 ppσ ppπ 2 (2.12) where all remaining orbital overlaps going to zero. 3
The remaining overlap terms are summarized in Equa- tion 2 in the next section of this paper. Finally, in our LCAO approximation, note that A ˆ B e B ˆ A e 2s (0) H 2px ( B) = 2s (0) H 2px ( B) A ˆ A B e ˆ B e 2s (0) H 2s (0) = 2s ( B ) H 2s ( B ) . (2.13)
The first is valid because of the uniformity of our medium: valence orbitals will be the same regardless of where the electrons are located. Permutations of orbitals leave energy overlaps invariant because of the symmetry FIG. 3: The energy dispersion relations for graphene are of the interactions. The latter expression is also valid, shown through the whole region of the Brillouin zone. The as orbital self-interactions will remain the same amongst lower and the upper surfaces denote the valence π and the ∗ identical atoms. In the case, both basis atoms are carbon. conduction π energy bands, respectively. The coordinates of high symmetry points are Γ = (0, 0), K = (0, 2π/3a), and Construct a set of extended atomic or- M = (2π/√3a, 0). The energy values at the K, M, and Γ bitals. Be explicit about your choice of phase points are 0, t, and 3t, respectively (http://www.iue.tuwien. factors. Write your trial wave function as a ac.at/phd/pourfath/node18.html). linear combination of these extended orbitals. Recall that for the tight-binding solution that we use with the trial wave function A2 2 = diag (Es, Ep, Ep, Ep) , (2.15) ψ (r) = c [R ] φ (r R ) , (2.14) × | i α l | α − l i α R and X Xl where α denotes both each type of wave function for each Vssσg0 Vspσg1 Vspσg2 0 atom and also the type of atom in the basis, and Rl Vspσg1 Vppσg3 + Vppπg4 (Vppσ + Vppπ) g5 0 B2 2 = . denotes the direct lattice vectors. Therefore, our LCAO × Vspσg2 (Vppσ + Vppπ) g5 Vppσg3 + Vppπg4 0 trial wave function is given by hybridized wave functions 0 0 0 Vppπg0 for the atoms in the primitive basis (2.15) The phase factors in B2 2 are A R i A R × ψ = aA 2s ( ) + bA 2pi ( ) | i = i x,y,z ik RB ik C 1 ik RC 1 ik RC X g0 =1+ e− · + e− · g1 = 1 e− · e− · B i B − 2 − 2 + aB 2s (R + eB) + b 2p (R + eB) B i √ i=x,y,z 3 ik RC ik RC 1 ik RC 1 ik RC X g2 = e− · e− · g3 =1+ e− · + e− · (2.15) 2 − 4 4 √ 3 ik RC ik RC 3 ik RC ik RC g4 = e− · e− · g5 = e− · e− · . 4 − 4 − Find the Hamiltonian matrix for the near- est neighbor approximation. Conduction in the plane is limited entirely to π bond- For a systems Hamiltonian operator Hˆ , the Hamilto- ing, so the graphene band structure of graphene is limited nian matrix in the LCAO method is to a subset of the matrix given in Equation 2
ik Rp H (k)= H˜ (Rp) e− · , (2.15) Ep Vppπg0 H2D = . (2.13) R Vppπg0∗ Ep Xp where the matrix elements in right side of the equality are 3. BAND CALCULATIONS
H˜ β,α (Rn, Rm)= φβ (r Rn) Hˆ φα (r Rm) . | − i | − (2.15)i Look up a real energy band diagram for Assuming only nearest neighbor interactions and that your material. Include a copy in your report. S (k), the Hamiltonian matrix of a two-dimensional hexagonal crystal of carbon atoms is given by a block An accepted electronic band energy diagram is shown diagonal matrix in Figure 3.
A2 2 B2 2 Write a matlab program to plot the free- H3D = × × (2.15) B2† 2 A2 2 electron band structure for your material × × 4
FIG. 4: Free-electron band structure for graphene.
along the same directions as used in the en- ergy band you found in the literature. Indi- cate the degeneracies of the different bands. Where is the Fermi level located? How does the free electron band structure compare to the real band structure? Our calculated free-electron band structure for graphene is included in Figure 4. In this model, we as- sume that the electron experiences no potential energy, and therefore has energies described by ~2k2 E (k)= . (3.0) 2m This is applied in the reduced-zone scheme, where the FIG. 5: Energy dispersion relations for graphene. evaluation to reciprocal lattice vectors K in the first Bril- louin zone. Subsequently, the energies are evaluated at they correspond to physically? Using Har- K for risons Solid State Table (attached) find nu- merical values for E , E , E , E , E , ~2 sa sc pa pc ss 2 E , E , and E . Compare your calculated E (k)= (k′ + K) , (3.0) sp xx xy 2m energies at the Γ point with values from the literature. where k′ is restricted to the symmetries of the first Bril- louin zone. For the free electron model of graphene using the six Our calculated total density of states for the conduc- nearest neighbors in reciprocal space, there are six de- tion and valence bands is included in Figure 5. Denoting generacies at the K point, two double degeneracies and t = Eppπ as the tight-binding energy from the valence one triple degeneracy at both the Γ and M points. We orbitals, the band structure depicted in Figures 5 and 6 see two degenerate pairs along K to Gamma, one along Γ is given analytically by to M and three along M to K. For this model, the Fermi energy is found exactly where the conduction and valence √3kxa kya 2 kya E± = t 1 + 4 cos cos + 4 cos bands touch at the K point. It is clear when comparing 2D v ± u 2 ! 2 2 this band structure to those found in the literature that u the free electron model is not a good approximation for t (3.0) electron transport in graphene. We see several band de- after diagonalizing Equation 2. Dresselhaus cites t = 3.033eV, differing from optimized value of t = 8.1eV generacies that do not match the actual band structure − − and the general shapes are very different. (see Matlab code)[1]. Algebraically diagonalize the Hamiltonian Write a Matlab program to plot the LCAO matrix at the Γ point. What are the differ- energy bands along the same directions as ent energies and eigenvectors, and what do above, along with the approximate location of 5
FIG. 6: Graphene energy dispersion from π-bonding. FIG. 7: Valence and conduction band edges for graphene. the Fermi level. How do your results com- pare (qualitatively) with the band structure you found in the literature? If you wish, op- timize the matrix elements for your material.
The LCAO energy dispersion is shown in Figure 6. Because this figure is based on a known analytic solution, the results compare almost exactly in qualitative features to the known literature [3].
Where are the valence band maximum and the conduction band minimum located? What is the energy gap? Is your material direct or indirect?
The valence band maximum and the conduction band FIG. 8: Energy contours for graphene. minimum are degenerate at the K-point, yielding a zero energy band gap. Because the points are coincident in the edges. Fit these points using a quadratic poly- plane, the material is direct. Notably, because the energy nomial (be sure to think about your results dispersion around the K point is linear, corresponding to from D1 when you do this). Use your results dispersion relation of a relativistic Dirac fermion [3]. to find the effective masses for both the va- lence bands and the conduction band. How do your results compare with results from the 4. EFFECTIVE MASSES, CONSTANT ENERGY literature? How would you improve your re- SURFACES, AND DENSITY OF STATES sults?
Plot constant energy contours near the va- The energy surface near the conduction band edges lence and conduction band edges for appropri- form cones that touch each other precisely at the K points ate planes. For example, if your minimum is as shown in Figure 6, making graphene effectively a zero- at kmin along Γ X, you should probably plot bandgap material. This differs greatly from the form of − an energy contour for the kx ky plane, and cubic semiconductors, which have band edges that can be for the plane parallel to k k− that contains approximated by parabaloids. Since the band structure y − z kmin. is resultant from a 2x2 Hamiltonian, its scale is com- pletely determined by the energy offset (which we set to A close up of the conduction and valence band edges zero) and a scaling factor ηppπ. The results are as close and a contour plot is shown in Figures 7 and 8, respec- as they possibly can be for our near-neighbor approxima- tively. tion, as we initially scaled our band structure along its high-symmetry points to that of the literature. Given the Solve for the energy at a number of k linear shape of the dispersion relation around the band points near the valence and conduction band edges, an electron in graphene for our approximation has 6
Our calculated total density of states for the conduc- tion and valence bands is included in Figure 8. Our elec- tronic specific heat of graphene is largely linear, matching well with the literature. However, the electronic specific is significantly smaller than its phonon counterpart. Ac- cordingly, the specific heat is completely dominated by the phonon specific heat by two orders of magnitude [2]. Our calculated total density of states for the conduc- tion and valence bands is included in Figure 10.
FIG. 9: Electronic density of states for graphene. an effective mass of zero everywhere around the band FIG. 10: Electronic heat capacity for graphene. edge except exactly where the bands meet, where it has an infinite effective mass.
Plot the total density of states (histogram Use the band structure you found in the method, include all bands) versus energy. Use literature to discuss the characteristics of your calculated effective masses to determine your material. What electronic/optical appli- an approximate expression for the density of cations would your material be good/bad for? states near the valence and conduction band Why? edges. How does this calculation compare with the total D.O.S.?
Our calculated total density of states for the conduc- Because the expression of the π band in graphene is ex- tion and valence bands is included in Figure 9. As shown, act, our band structure calculation is exactly similar to our generated density of states defies simple expression, the those found in the literature. Figure 6 shows the up- as unlike a cubic system we cannot represent the den- per π∗-energy anti-bonding and lower π-energy bonding sity of states as a straightforward √E relation. As shown bands that are degenerate at the K-point; the two elec- in Figure 9, our calculated DOS matches the shape of trons in the π band are fully occupy the lower π band. experimental data for the conduction band well. From the degeneracy at the K-point, the density of states at the Fermi level is zero, making graphene a zero band Using your total D.O.S., calculate the gap semiconductor [3]. The planar π-bonding bestows electronic specific heat of your material as a graphene with extremely high room temperature electron function of temperature. Compare this with mobility, allowing a conductivity greater than silver. The your calculations for the phonons, and com- linear shape of the energy dispersion formed the basis of ment. frequency multiplier chip produced by MIT [3].
[1] R. Saito, G. Dresselhaus, and M. S. Dresselhaus, Phys- (2001). ical Properties of Carbon Nanotubes (World Scientific [3] A. H. C. Neto, F. Guinea, N. M. R. Peres, K. S. Novoselov, Publishing Company, 1998), ISBN 1860940935, URL and A. K. Geim, Reviews of Modern Physics 81, 109 http://www.amazon.ca/exec/obidos/redirect?tag= (pages 54) (2009), URL http://link.aps.org/abstract/ citeulike09-20%&path=ASIN/1860940935. RMP/v81/p109. [2] J. Hone, Book Series Topics in Applied Physics 80, 273 Part IV: Electronic Band Structure of Graphene Continued
Dodd Gray, Adam McCaughan, Bhaskar Mookerji∗ 6.730—Physics for Solid State Applications (Dated: May 8, 2009)
1. A. OPTICAL PROPERTIES this formulation is then broken down into its Bloch constituents and further simplified. Orienting the sheet Utilizing the (optimized) LCAO band- of graphene in the x-y plane, it is pointed out that for structure from Part III, use the lowest energy two lattice atoms Rj1 and Rj2 direct bandgap (from valence band to conduc- tion band) to estimate the momentum matrix r R ∂ r R element. <φ( j1) φ( j2)> = 0 (1.4) − |∂z | − The band structure around graphene’s bandedge (the since the integrand is an odd function of z. Thus we K point) is a pair of intersecting cones. As such, the can restrict the atomic dipole vector to the x-y plane, double gradient of the band structure in that region is ie D =(dx,dy, 0). The matrix element P D is further · discontinuous, yielding an effective mass m∗ of zero at the simplified using a Taylor expansion around the K point K point and infinity in the immediate vicinity. Looking (kx0,ky0) until the arrival of the result at the momentum matrix element calculation outlined in Lecture 27: 3M pcv = P D = (py(kx kx0) px(ky ky0)) (1.5) | | · 2k − − − 2 1 1 2 pcv = + 2 | | (1.1) where M is the optical matrix element for two nearest- m∗ m m Eg neighbor atoms separated by b1 1 1 we see that m∗ = E = at the K point, which gives g ∞ ∂ us intuition that something special is happening at that M = <φ(R + b1) φ(R)> (1.6) location. If we think about the band structure, we real- |∂x| ize that the matrix element must be zero at the K point, The above can be numerically evaluated for a tight- since the valence and conduction bands are degenerate binding model including 2s and 2p orbitals, giving a value there–there is no gap across which to absorb a photon. of M =0.206 [1] However, if we move away from the K point and no longer have a linear dispersion relation, Eg becomes nonzero as Utilize the optimized LCAO bandstructure does pcv , but the effective mass becomes zero. Clearly and momentum matrix element from A1 to | | this equation, which was formulated for zinc-blende crys- estimate the optical absorption. Plot the ab- tals cannot help us with determining the momentum ma- sorption coefficient for your material from the trix element aside from suggesting. Since 1.1 is not di- direct bandedge to 5 kBT above the bandedge. rectly useful in analytically finding pcv for our conic bandstructure approximation, we instead| | turned to the Using the absorption coefficient formula derived in Lec- literature. ture 27, we arrive at Within the literature [1][2], the momentum matrix elements are derived explicitly, by the dot product of a 2 P D πq cµ0 3M 2 2 photon polarization vector and the dipole vector α(ω)= 2 (py∆kx px∆ky) pr(hω Eg) where m0ωn | 2k | − | − (1.7) where pr(hω Eg) is the reduced density of states we calculated in Part− II. D <Ψf Ψi> (1.2) ≡ |∇| As shown in Figure 1, the absorption is zero directly and at the K points (as our intuition suggested) and grows as k moves away in most directions. Notice the line of zero absorption, however, for k directions associated with pcv = P D = P <Ψf Ψi> (1.3) (py∆kx px∆ky) = 0. | | · · |∇| − Compare your estimate for the interband absorption coefficient with literature values. ∗Electronic address: [email protected], [email protected], mook- What are the major sources of discrepancy be- [email protected], tween your theory and experiment? 2
Utilize the phonon dispersion from Part II to estimate the phonon occupancy for all modes within the Brillouin zone. Assume a lattice temperature of 300K.
Considering the lattice temperature is so high, we can approximate the phonon occupancy function in the Boltzmann limit, ie
k T N(w)= B (2.5) !w Using the LCAO density-of-states and the FIG. 1: Absorption spectrum for graphene for polarization P results from B1 and B2, estimate the longi- = (0,1) [1]. Darker areas correspond to stronger absorption tudinal acoustic phonon scattering for your material. Compare your result with literature Considering that our method of derivation was almost values. identical to that of the literature, it matches *very* well. In many of the articles we reviewed, we found that the op- Beginning with Eq (2.76) from Lundstrom, modified tical absorption was solely shaped by a density of states, for 2D (after doing the integral present in Eq (2.75) the scaled by constants that appeared to be of approximately only difference is the removal of a factor of β) the same form as those in Equation 1.7 1 Ω βmax 1 1 = (Nβ + )Cββdβ (2.6) τ 4π2 2 ∓ 2 2. B. ELECTRON-PHONON SCATTERING (βmin and since we’re calculating the acoustic deformation Utilize the LCAO bandstructure from Part potential III to estimate the deformation potential for electrons near the conduction bandedge and 2 holes at the valence bandedge. Compare the πm∗DA Cβ = pΩ (2.7) estimated deformation potentials with the lit- !ρvs erature values for the longitudinal acoustic deformation potential. we’re operating in the Boltzmann limit as well so
We can calculate the deformation potential directly 1 Ω βmax from the bandstructure since we have it in analytic form = N C βdβ (2.8) from Part III, as τ 4π2 ωs β (βmin
√ kBTL kBTL 3kxa kya 2 kya Nω = = (2.9) E± = t 1 + 4 cos cos + 4 cos s ! ! 2D ! ωs βvs ± " $ 2 % 2 2 " & ' & ' # (2.1) we find that where t =8.1eV , a =1.42A˚, and A = a2. At the 2π 2π bandedge point K, kx = and ky = 3 To perform √3a a 1 m D2 k T βmax the calculation we took a small increment and divide it = ∗ A B L dβ (2.10) τ 4π!2c p by the change in atomic areal lattice spacing. l (βmin m D2 k T 2p = ∗ A B L (2.11) 4π!2c p ! ∂En(K) ∂En(K) a l DA = eqA = eq (2.2) 2 ∂A | ∂a | · 2 DAkBTL m∗ = 2 (2.12) E(a + da) E(a) a 1 a 2!cl π! − = ∆E (2.3) 2 ≈ da · 2 2 da DAkBTL = g2D(E) (2.13) = 14.15eV (2.4) 2!cl
7 the resulting calculation is in very good accordance We can calculate cl using ρ =6.5 10− and using the · with experimental values from graphene devices grown longitudinal sound velocity vs = 13.12km/s from Part on SiO2, DA(emp) = 18 1eV [3]. III of the project, and TL is 300K. The first part of the ± 3
2 3 cl = v ρ =8.51 10− kg/s (2.14) s ∗ · Within the relaxation time approximation in the Boltzmann limit, estimate the heavy- hole mobility. Compare your result with lit- erature values. What are the major sources of discrepancy between your theory and experi- ment?
The verdict is still out on the methodology to de- termine the mobility of graphene. Measured values of graphene mobility show that it has extremely high mo- bilities, theoretically infinite at the Dirac points, but that these values are limited by acoustic phonon scat- tering [3]. Graphene has been experimentally shown to have hole mobilities commonly in the range of 3400-4400 V/cm2s [4] and as high as 15,000 V/cm2s [5]. Current re- FIG. 2: Density of states histogram calculated in Part III search suggests that graphene mobility is limited by the density of charged impurities within the lattice, and so final equation form yields a constant factor to propor- the key to increasing mobility for use in higher-frequency tion g2D(E), which is the density of states calculated via electronics is to reduce the number of impurities when histogram method in Part III and shown in Figure 2. building graphene components. [6]
[1] R. Saito, A. Grneis, G. Samsonidze, G. Dresselhaus, M. Dresselhaus, A. Jorio, L. Canado, M. Pimenta, and A. S. Filho, Applied Physics A: Materials Science & Pro- cessing 78, 1099 (2004), URL http://dx.doi.org/10. 1007/s00339-003-2459-z. [2] A. K. Gupta, O. E. Alon, and N. Moiseyev, Physical Review B 68, 205101 (2003), copyright (C) 2009 The American Physical Society; Please report any problems to [email protected], URL http://link.aps.org/abstract/ PRB/v68/e205101. [3] J. Chen, C. Jang, S. Xiao, M. Ishigami, and M. S. Fuhrer, Nat Nano 3, 206 (2008), ISSN 1748-3387, URL http:// dx.doi.org/10.1038/nnano.2008.58. [4] Y. Q. Wu, P. D. Ye, M. A. Capano, Y. Xuan, Y. Sui, M. Qi, J. A. Cooper, T. Shen, D. Pandey, G. Prakash, et al., Applied Physics Letters 92, 092102 (2008), URL http://link.aip.org/link/?APL/92/092102/1. [5] A. K. Geim and K. S. Novoselov, Nat Mater 6, 183 (2007), ISSN 1476-1122, URL http://dx.doi.org/10. 1038/nmat1849. [6] E. H. Hwang, S. Adam, and S. D. Sarma, Physical Review Letters 98, 186806 (2007), URL http://link.aps.org/ abstract/PRL/v98/e186806.