Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. ·...

Crystal Structure of Graphite Graphene and Silicon

Dodd Gray Adam McCaughan Bhaskar Mookerjilowast

6730mdashPhysics for Solid State Applications

(Dated March 13 2009)

We analyze graphene and some of the carbon allotropes for which graphene sheets form thebasis 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 Weshow that staggered two-dimensional hexagonal lattices of graphite have XRD patterns that differsignificantly from silicon standards

The wide-variety of carbon allotropes and their associ-ated physical properties are largely due to the flexibilityof carbonrsquos valence electrons and resulting dimensionalityof its bonding structures Amongst carbon-only systemstwo-dimensional hexagonal sheetsmdashgraphenemdashforms ofthe basis of other important carbon structures such asgraphite and carbon nanotubes ( Say something aboutinteresting band structure here)

In the following we will examine the planar lat-tice structure of graphene and its extension to higher-dimensional lattice structures such as hexagonalgraphite We first analyze the lattice and reciprocal-space structures of two-dimensional hexagonal lattices ofcarbon and use the resulting structure factors to esti-mate the x-ray diffraction (XRD) intensities of graphiteWe conclude by comparing its calculated XRD spectra toexperimental spectra of graphene and crystalline silicon

1 PRELIMINARY QUESTIONS

11 Lattice Structure

Our discussion of the crystal structure of graphite fol-lows partially from DDL Chungrsquos review of graphite [1]When multiple graphene sheets are layered on top of eachother van der Walls bonding occurs and the three di-mensional structure of graphite is formed with a lattice

FIG 1 In-plane structure of graphite and reciprocal latticevectors [1]

lowastElectronic address doddmitedu amccmitedu mook-

erjimitedu

spacing between sheets c = 671A The sheets align suchthat their two-dimensional hexagonal lattices are stag-gered either in an ABAB pattern or an ABCABC pat-tern The ABAB alignment is shown in Figure 1 whichindicates four atoms per unit cell labeled A B Arsquo andBrsquo respectively The primed atoms AndashB on one graphenelayer are separated by half the orthogonal lattice spacingfrom the ArsquondashBrsquo layer BBrsquo atomic pairs differ from theircorresponding AArsquo pairs in their absence of neighboringatoms in adjacent layering planes The coordinates ofthese atoms forming the basis are given by

ρA = (0 0 0) ρAprime =(

0 0c

2

)

ρB =a

2

(

1radic3 1 0

)

ρBprime =

( minusa

2radic

3minusa

2c

2

)

(11)

With respect to an orthonormal basis the primitivelattice vectors are given by

a1 = a(radic

32minus12 0)

|a1| = a = 246A

a2 = a(radic

32 12 0)

|a2| = a = 246A

a3 = c (0 0 1) |a3| = c = 671A (12)

The magnitudes of the primitive lattice vectors corre-spond to the lattice constants parallel and perpendicu-lar to the graphene sheet The corresponding ABCABClayer forms a rhombohedral structure with identical lat-tice spacing parallel and orthogonal to the layer

12 Reciprocal Lattice Structure

Recall that the reciprocal lattice vectors bi are definedas a function of the primitive lattice vectors ai such that

b1 = 2πa2 times a3

a1 middot a3 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1

b3 = 2πa1 times a2

a3 middot a1 times a2(13)

The reciprocal lattice vectors for graphite are then

2

FIG 2 Reciprocal lattice planes [001] [110] and [111]

given by

b1 =2π

a

(

1radic3minus1 0

)

|b1| =2π

a

2radic3

b2 =2π

a

(

1radic3 1 0

)

|b2| =2π

a

2radic3

b3 =2π

c(0 0 1) |b3| =

2π

c (14)

The reciprocal lattice plane generated by the b1 and b2

vectors forms the outline of the first Brillouin zone asdepicted in Figure 1 The intersection of the the planeskz = plusmn2πc with the plane forms a hexagonal prism ofheight 4πc

13 Atomic form factors

As carbon is the only element present in graphene andgraphite the atomic form factor is uniform across theentire crystal and thus can be factored out when calcu-lating the structure factor Thus the atomic form factorhas no effect on the relative intensities of x-ray diffrac-tion occuring in different planes of graphite Accordingto the NIST Physics Laboratory the atomic form factorof carbon varies from 600 to 615 eatom with incidentradiation ranging from 2 to 433 KeV [2]

2 X-RAY DIFFRACTION

21 Planes in the Reciprocal Lattice

Provide pictures of the crystal and of thereciprocal lattice in the [100] [110] and [111]planes Indicate the vertical positions ofatoms with respect to the plane

Pictures of the crystal and of the reciprocal lattice inthe [100] [110] and [111] planes are included in Fig-ure 2 In MATLAB the crystal was represented as a set

of points in space using the specified lattice vectors andatom bases Normals generated from lattice vector sumswere used to extract planes and display them Varyingcolors were used to depict vertical spacing between adja-cent planes

22 Structure Factors and X-Ray Diffraction

Intensities

Calculate the structure factor for all thereciprocal lattice vectors Kl lt 16 (2πa)

2

The structure factor is calculated as

Mp (Ki) = fc

nsum

j=1

(Ki) eminusiKimiddotρi

where fc is the structure factor of Carbon and ρi are thebasis vectors of our lattice We find that only four uniquenon-zero values of Mp (Ki) occur in the reciprocal latticeEach of these corresponds to the height of a diffractionintensity peak and their relative values are referenced inTable I

Calculate the ratio of the intensities ex-pected for the following lines of the diffractionpattern with respect to the [111] line [100][200] [220] [311] and [400]

Including the structure factor there are other severalfactors contributing to the intensities of the diffractionpeaks [3]

1 The Lorenz correction is a geometric relation al-tering the intensity of an x-ray beam for differentscattering angles θ

2 The multiplicity factor p is defined as the num-ber of different planes having the same spacingthrough the unit cell Systems with high symme-tries will have different planes contributing to thesame diffraction thereby increasing the measuredintensity

3 Temperature absorption polarization each con-tribute higher-order corrections ultimately ignoredin our calculation These include Doppler broad-ening from thermal vibrations in the material ab-sorption of x-rays and the polarization of initiallyunpolarized x-rays upon elastic scattering

These factors all contribute to the relative intensity of a[hkl] diffraction peak given by

I[hkl] (θ) = p|Ma (Km) |2(

1 + cos2 2θ

sin2 θ cos θ

)

(21)

Using the formula from the previous question to calcu-late the ratios of the structure factors in the given planes

3

FIG 3 Crystal structure of silicon

TABLE I X-Ray diffraction intensities for graphite and sili-con [4 5] Structure factors are included in parentheses

Si C (Graphite)

2θ () Exp 2θ () Calc Exp

[111] 2847 10000 mdash (3) mdash

[100] mdash mdash 4277 4 (3) 345

[200] mdash mdash mdash (1) mdash

[002] mdash mdash 2674 106 (16) 100

[220] 4735 6431 mdash (1) mdash

[311] 3731 3731 mdash (3) mdash

[400] 6921 958 mdash (1) mdash

we obtain the following results presented in Table I Cal-culations show that planes [100] [200] [220] and [400] ex-hibit relative diffraction intensities 13 that of the [111]and [311] planes The [002] plane exhibits the highest in-tensity diffraction 163 that of the [111] and [311] planesWe also found several non-zero structure factors that arenot present in the experimental data

23 Crystal Structure of Silicon

What are the ratios if the material wereSi How could you use this information todistinguish Si from your material by x-ray

diffraction

The crystal structure of another common semiconduc-tor material silicon (Si) is featured in Figure 3 Siliconforms a diamond cubic crystal structure with a latticespacing of 542A This crystal structure corresponds to aface-centered cubic Bravais lattice whose unit-cell basiscontains 8 atoms located at vector positions

d0 = ~0 d4 =a

4(1 3 3)

d1 =a

4(1 1 1) d5 =

a

4(2 2 0)

d2 =a

4(3 3 1) d6 =

a

4(2 0 2)

d3 =a

4(3 1 3) d7 =

a

4(0 2 2) (22)

The structure factor contributing to its X-ray diffrac-tion pattern is given by

Ma (Km) =

nsum

j=1

f (j)a (Km) eminusiKmmiddotdj

= f(1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

+(minus1)h+k+l

+ (minusi)3h+k+l

+ (minusi)3h+3k+1

+(minusi)h+3k+1

)

= f(

1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

)

middot(

1 + (minusi)h+k+l

)

(23)

This term undergoes a number of simplifications basedon the parity of its Miller indices If [hkl] are all evenand are divisible by 4 then Ma (Km) = 8f If they arenot divisible by 4 or have mixed even and odd valuesthen Ma (Km) = 8f Lastly if [hkl] are all odd thenMa (Km) = 4f (1 plusmn i)

The experimental X-ray diffraction intensities fromthese contributions are listed in Table I The intensityvalues for silicon were measured with respect to a ref-erence value II0 = 47 which is a direct ratio of thestrongest line of the sample to the strongest line of a ref-erence sample αndashAl2O3 The number of visible peaks andthe relative intensities between them suggest that siliconand graphite can be easily distinguished from each otherusing an x-ray diffraction experiment

[1] D Chung Journal of Material Science 37 1475 (2002)[2] () URL httpphysicsnistgovPhysRefData[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 volMonograph 25 Section 13 (National Bureau of Standards1976)

[5] () URL httprruffinfographitedisplay=

default

Phonon Spectra of Graphene

(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-planevibrational modes to whose dispersion relations we fit experimental lattice mode frequencies yieldingforce constants αs = 445Nm and αφ = 102Nm Our model also reasonably accounts for graphenersquosmacroscopic properties particularly sound speeds and elastic constants

The lattice dynamical properties of graphene form thebasis of understanding the vibrational spectra of carbon-based allotropes of various geometries such as graphiteor carbon nanotubes We can use an analytical descrip-tions of micromechanical behavior to better understandthe acoustic and optical properties of these materials

In the following we calculate the in-plane vibrationalspectrum of graphene and its contributions to macro-scopic elastic and thermodynamic quantities We firstdiscuss the force parameters of our Born model and de-rive a general potential to calculate the dynamical ma-trix of a primitive cell Using our dispersion relations athigh-symmetry points we can determine the vibrationaldensity of states in-plane sound velocity and elastic con-stants We will briefly touch upon weak out-of-planevibrational modes and its contributions to graphenersquosmacroscopic properties

1 BACKGROUND FOR THE BOHR MODEL

11 Parameters of Bohr Model

How many force constants are required foreach bond Why

Two force constants αs and αφ are required to modeleach bond Let αs represent the restoring force seen whena bond is stretched and let αφ represent the restoringforce seen when a bond is bent away from the axis alongwhich it is normally aligned

What is the energy of a single bond in theBorn model

The energy in a single bond is the sum of the stretchingand bending energies Es and Eφ If p is a vector alongwhich the bond is aligned in equilibrium and R is a latticevector then the energy contained in a bond between the

lowastElectronic address doddmitedu amccmitedu mook-erjimitedu

atom at R and the atom at R + p is

E[RR + p] = Es + Eφ

=1

2αs|p middot (u[R + p] minus u[R]) |2

+1

2αφ

(

|u[R + p] minus u[R]|2 minus |p middot (u[R + a] minus u[R])|2)

(11)

The model assumes that the bond is onlyslightly displaced from equilibrium Howwould you modify the model to make thebond energy more realistically dependent ondisplacement from equilibrium - what or-der would the corrections be and of whatsign Justify your answer physically includesketches if appropriate

The energy in the bond between the atom at latticevector Ri and the atom at lattice vector Ri+a is generallyestimated using a Taylor expansion to the second orderas

V (u[Ri t])) =

V0 +sum

n

sum

m

(

part2V

partun[Ri t]partum[Ri + a t]

)

eq

un[Ri t]

(12)

where the position of the atom at Ri + a is fixed n andm index all dimensions being considered V0 is the bondenergy seen with zero displacement and V is the potentialdefined as the sum of all bond energies in the entire lattice

V = middot middot middot+sum

n

(E[RR + an] + E[RR minus an])+middot middot middot (13)

where this is the slice of the lattice potential related toan atom at R and the number of different vectors an

between the atom at R and atoms coupled to it dependson which nth nearest neighbor model is used to modelthe lattice

Note that there is no first order term in this expan-sion A first derivative of potential energy would implya net force so this term must equal zero as the Taylorcoefficients are evaluated at equilibrium

To make the Born model more accurate we would needto take into account higher order terms from the Taylor

2

expansion of the potential We know that the third orderterm is non-zero because the curve we are attempting tofit is not even around the equilibrium point Based onthe fact that the function has higher curvature to the leftof the equilibrium point than it does to the right it seemsthat the sign of the third order term would be negative

12 Nearest Neighbor Couplings

If you use the only nearest neighbor cou-plings how many force constants will yourmodel require for your material How largewill the dynamical matrix be What if youused nearest neighbor and next-nearest neigh-bor couplings

Two force constants are needed to model graphenewhether a nearest neighbor coupling or nearest and nextnearest neighbor model is used as long as the model istwo-dimensional As mentioned before these are αs andαφ representing restoring forces due to bond stretchingand bond bending respectively As graphene is a twodimensional material with a two atom basis the dynam-ical matrix will be four-by-four A third force constantαz is required to account for out-of-plane phonon modesThe size of the dynamical matrix is not affected by thenumber of force constants used in our model however ifa third dimension is added to account for out of planevibrations then the dynamical matrix will be six-by-six

13 Elastic Properties

How many independent elastic constantsdoes your material possess What are they(give numbers) Why will a nearest neighborapproach not provide the most general solu-tion for a cubic material

Graphene has two elastic constants λ and micro Themeasured values of transverse and longitudinal phononvelocities in graphene are vt = 14 middot 103 msminus1 and vl =217 middot 103 msminus1 respectively [1] Given the density ρ =

2MC [ 3radic

3a2

2]minus1 = 76 middot 10minus7kgmminus2 where the mass of

carbon MC = 199 middot 10minus26 kg and the lattice constanta = 142 middot 10minus10 m the Lame coefficients are calculatedusing that vt =

radic

microρ and vl =radic

(λ + 2micro)ρ to be micro ≃293Nm and λ ≃ 722Nm It is worth noting that thesemacroscopic elastic constants are not usually calculatedor measured for graphene because it is a two-dimensionalmaterial with single atom thickness

2 CONSTRUCTION OF THE DYNAMICAL

MATRIX

We will first determine the phonon dispersion relationsof the in-plane vibrational modes in the context of the

FIG 1 Hexagonal crystal structure of a graphene primitivecell and its neighbors First Brillouin zone of graphene andits symmetry points

Born force model and and then briefly discuss the out-of-plane vibrational modes

21 Lattice and Reciprocal Space Structures

[Label] all the atoms in the basis and all theirnearest neighbors For each atom labeled A-H [verify] the lattice vectors Rp to each unitcell

A graphene sheet forms a two-dimensional hexagonalcrystal lattice with a primitive cell containing two atoms(A and B) depicted in Figure 1 Based on our previ-ous discussion of the Born model let us assume thatthe primitive cell interacts with nearest neighbors andnext-nearest neighbors with spring constants αs and αφrespectively and that atoms of the primitive cell have alattice spacing given by a = 142A and a primitive lat-tice constant 246A Figure 1 also depicts graphenersquos firstBrillouin zone and its symmetry points located at kΓ =(0 0) kM =

(

2πaradic

3 0)

and kK =(

2πaradic

3 2π3a)

The lattice contains two sublattices 0 and 1 which

differ by their bond orientations The first atom A inthe primitive cell has three first neighbors in the othersublattice 1 with relative unit vectors

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

and six next-nearest neighbors in the same sublattice 0with relative unit vectors

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

22 Born Force Model

[Using a Born force model find] a generalexpression 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 planarindex specifies the force interaction between the A (B) atom and one of its neighbors

A B C D E F G H I J

xx 3αs2 + 3αφ minusαs minusαs4 minusαs4 minus3αs4 minus3αs4 0 minus3αφ4 minus3αφ4 0

minusαs 3αs2 + 3αφ mdash mdash minusαs4 minusαs4 mdash mdash mdash mdash

xy yx 0 0radic

3αs4 minusradic

3αφ4radic

3αφ4 0radic

3αφ4 minusradic

3αφ4 0

0 0 mdash mdash minusradic

3αs4radic

3αs4 mdash mdash mdash mdash

yy 3αs2 + 3αφ 0 minus3αs4 minus3αs4 minusαφ4 minusαφ4 minusαφ minusαφ4 minusαφ4 minusαφ

0 3αs2 + 3αφ mdash mdash minus3αs4 minus3αs4 mdash mdash mdash mdash

atoms in the crystal in terms of their displace-ment from equilibrium

The total elastic potential energy for the all the atomsin the crystal in terms of their displacements fromequilibrium is given by summing the nearest-neighborstretching interactions

Vs (R) =1

2αs|eB middot (u1 [R] minus u2 [R + eB]) |2

+1

2αs|eC middot (u1 [R] minus u2 [R + eC]) |2

+1

2αs|eD middot (u1 [R] minus u2 [R + eD]) |2 (23)

with the nearest neighbor bending interactions

Vφ

=1

2αφ

(

(u1 [R] minus u2 [R + eB]) |2 minus |eB middot (u1 [R] minus u2 [R + eB]) |2)

+1

2αφ

(

| (u1 [R] minus u2 [R + eC]) |2 minus |eC middot (u1 [R] minus u2 [R + eC]) |2)

+1

2αφ

(

| (u1 [R] minus u2 [R + eD]) | minus |eD middot (u1 [R] minus u2 [R + eD]) |2)

(24)

23 Dynamical Matrix

Use the expression for the potential energyto determine the force on a given atom inthe crystal in terms of its displacement andits neighbors displacements Check your an-swer by directly calculating the force from thespring constants and displacements

Verify from the potential by explicitly takingthe derivatives the factors in the matrix

From the potential energy the force on a given atomin the crystal is given in terms of a harmonic oscillatorforce expression

fi = Dij (k) uj (25)

where Dij is the dynamical matrix is given by

Dij (k) =sum

Rp

Dij (Rp) eminusikmiddotRp =sum

Rp

Vprimeprime

ij (Rp) eminusikmiddotRp

(26)

In this explanation Vprimeprime

ij are second-partial derivatives ofour Born model potential at equilibrium Rp are the rel-ative lattice vectors described in Equations 21 and 22

and u =(

u1x u2

y u1x u2

y

)T The force constants from tak-

ing the appropriate second derivatives is given in Table I

The dynamical matrix is then given by

D (k) =

A0 B0 C D

B0 A1 D B1

Clowast Dlowast A0 B0

Dlowast Blowast1 B0 A1

(27)

If we let (AA) and (BB) represent the 0 and 1 sublatticesrespectively these matrix elements represent

A0 = Dxx (AA) = Dxx (BB)

B0 = Dxy (AA) = Dyx (AA) = Dxy (BB) = Dyx (BB)

C0 = Dxx (AB) = Dlowastxx (BA)

D0 = Dxy (AB) = Dyx (AB) = Dlowastxy (BA) = Dlowast

yx (BA)

A1 = Dyy (AA) = Dyy (BB)

B1 = Dyy (AB) = Dlowastyy (BA) (28)

As a sample calculation consider the element Dxx givenby A0

A0 = Vprimeprime

xx (AA) + Vprimeprime

xx (AE) eminusi

ldquo radic3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AF ) eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AG) eminusikya

+ Vprimeprime

xx (AH) eildquo radic

3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AI) eildquo radic

3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AJ) eikya

=3αs

2+ 3αφ minus 3αs

4eminusi

ldquo radic3

2kxa+

kya

2

rdquo

minus 3αs

4eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ 0 minus 3αφ

4eildquo radic

3

2kxaminus

kya

2

rdquo

minus 3αφ

4eildquo radic

3

2kxa+

kya

2

rdquo

+ 0

=3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

(29)

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 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

ωZO 0 75 60 mdash mdash mdash

ωZA 110 52 60 mdash mdash mdash

Sound Velocities (kms)

vLA 217 1312

vTA 14 621

vZA mdash mdash

Elastic Constants (10 GPa)

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

C44 043 plusmn 005 mdash

A similar calculation for the remainder of the matrixelements gives us

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

3 minus 2 cos (kya) minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

3 MODEL OPTIMIZATION AND

COMPARISON TO MACROSCOPIC

PROPERTIES

31 Comparison with Published Theoretical and

Experimental Data

Plot the phonon dispersion in appropriateunits along the Γ minus X X minus L and Γ minus Ldirections

For these values of force constants andmasses determine the atomic displacementsfor all the modes at Γ and for the highest

FIG 2 Calculated and experimental phonon dispersioncurves for graphene (a) Fitted experimental dispersioncurves using inelastic x-ray scattering in graphite [2] (b) (c)calculated dispersion relations using our fitted force constants(αs = 445Nm and αφ = 102Nm) and suggested force con-stants (αs = 1 and αφ = 025)

optic and lowest acoustic modes at X andL Provide drawings of the atomic motionof these modes How many modes are thereat Γ

5

32 Phonon Dispersion Relations

The phonon dispersion relation ω = ω (k) is deter-mined from the eigenvalue equation

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

Note additionally that while graphene is two-dimensional it possesses miniscule elastic movement inthe vertical direction leading to a very small but non-zero elastic constant C44 We can define force constantsβs and βφ for the nearest and next-nearest interactions ofthe out-of-plane vibrational modes It has been shown [3]that the phonon dispersion relations for the out-of-planeoptical and acoustic vibrational modes are

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

where the out-of-plane spring constants are given byβs = minus1176 and βs = 0190 These dispersion relationsare decoupled from the in-plane vibrational modes Bydiagonalizing the Hermitian dynamical matrix in Equa-tion 27 we can add these dispersion relations a sub-space From empirical data (see II) we know that elasticout-of-plane motion is several orders of magnitude lessthan the in-plane motion Our own plots of Equation 33yielded high lattice mode frequencies contradicting thisintuition As such we have omitted its inclusion in ourdetermination of macroscopic elastic constants

Figure 2 compares the measured dispersion relationof graphene against our calculated dispersion relationsusing the suggested force constants (αs = 1 and αφ =025) and our fitted force constants (αs = 445Nm andαφ = 102Nm) The optimum force constants are foundfor our model by parameterizing the force constants andcomparing the results of ω at the relevant zone edgesTwo force constants are chosen and ω is calculated foreach of the three zone edges Γ M and K An RMSerror is generated by subtracting the experimental valuesfrom the calculated values and iteratively the optimumvalues for the force constants are found by searching forthe lowest RMS error value

The relative atomic displacements for the acoustic andoptical modes at the Γ K and M symmetry pointsis shown in Figure 5There are a total of four in-planemodes (acoustical and optical transverse and longitudi-nal) present at Γ however both the optical and acousti-cal pairs are degenerate The atomic displacement pat-terns associated with the various eigenmodes at differentpoints of high symmetry in K space are shown in Figure

4 Many qualitative features of our calculated disper-sion relations are consistent with the experimental andab initio data in Figure 2 [4] Note degeneracies at theΓ point and linear dispersion relations for small displace-ments from Γ Hexagonal symmetry of the graphene lat-tice also accounts for the measured and calculated de-generacies at the M and K points The largest discrep-ancies between experimental data and our Born modelare for the TO and LO modes which require accountingof electron-phonon interactions in ab initio models foraccurate prediction [5]

Theoretical sound velocities for both longitudinal andoptical polarizations are estimated by calculating theslopes of the acoustical phonon dispersion curves of bothpolarizations near Γ and taking c = δω

δk The elasticconstants C11 and C12 are determined from the phononsound velocities as

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

where ρ is the mass density of graphene

33 Density of States

Plot the total density of states (histogrammethod include all modes) versus frequency

Figure 3 shows the overall density of states and thosefor individual modes Experimental data for graphene isunavailable although the our calculation seemed qualita-tively consistent with experimental data for graphite [4]

34 Specific Heat

Calculate the specific heat of your materialversus temperature using (a) your calculateddensity of states (b) a Debye model and (c)a combined Debye-Einstein model (Debye foracoustic modes Einstein for optical modes)Plot your results for temperatures between 0Kand 3ΘD Comment on the strengths andweaknesses of your model

The specific heat for graphene is calculated using thecombined density of states (3) Debye model and com-bined Density of States (optical branches) and Debyemodel (acoustic branches) The two models result in rad-ically different specific heats perhaps because the De-bye model does not account for the optical modes uponwhich the specific heat significantly depends The plotsfor the specific heats in these models is given in Figure 4

In the Debye model calculation the longitudinal acous-tic velocity is used as the speed of sound instead of theinverse addition of longitudinal and tranverse velocitiesbecause in the calculated phonon dispersion relation thetransverse acoustic line is by far the most deviant fromexperimental value In addition it should be noted that

6

FIG 3 Calculated total and individual-mode density ofstates

use of the longitudinal acoustic velocity provides a rea-sonably accurate ΘD of 2200 K (with experimental datasuggesting a ΘD of 2400 K [6]

[1] J Hone Book Series Topics in Applied Physics 80 273(2001)

[2] V K Tewary and B Yang Physical Review B (CondensedMatter and Materials Physics) 79 075442 (pages 9)(2009) URL httplinkapsorgabstractPRBv79

e075442[3] L A Falkovsky Soviet Journal of Experimental

and Theoretical Physics 105 397 (2007) arXivcond-mat0702409

[4] M Mohr J Maultzsch E Dobardzic S ReichI Milosevic M Damnjanovic A Bosak M Krisch andC Thomsen Physical Review B (Condensed Matter andMaterials Physics) 76 035439 (pages 7) (2007) URLhttplinkapsorgabstractPRBv76e035439

[5] L M Woods and G D Mahan Phys Rev B 61 10651(2000)

[6] W DeSorbo and W W Tyler The Journal of Chemi-cal Physics 21 1660 (1953) URL httplinkaiporg

linkJCP2116601

7

FIG 4 Calculated specific heats using calculated density ofstates 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) Kand 5(c) M points [4]

Electronic Band Structure of Graphene

The band structure properties of graphene form thebasis of understanding the electronic spectra of carbon-based allotropes of various geometries such as graphiteor carbon nanotubes We can use an analytical and nu-merical descriptions of carbonrsquos valence electrons to bet-ter understand the conduction and semiconductor prop-erties of these materials

1 BAND STRUCTURE BACKGROUND

QUESTIONS

How many extended orbital basis func-tions will you have for your material Why

A graphene sheet forms a two-dimensional hexagonalcrystal lattice with a primitive cell containing two atoms(A and B) its lattice structure of graphene and its firstBrillouin zone is shown in Figure 1 The electron con-figuration of free carbon atoms is 1s22s22p2 For thetwo atoms in the basis there are four valence orbitalsyielding eight extended orbital basis functions for ourmaterial Because of its planar structure atoms undergobonding with four hybridized sp2-wave functions of theform

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

Conceivably one could also construct theLCAO wave function out of core orbitals aswell as valence orbitals How many extendedorbital basis functions would you have to usefor your material How large would yourHamiltonian matrix be in this case Howdo you expect your results would differ fromthose you would get with just valence orbitalsWhat if we used higher (totally unoccupied)orbitals too How many orbitals per atomwould we have to use to get an exact bandstructure Why

If we were to add the 1s orbital to our LCAO basis setwe would have five orbitals per atom in a two atom basisyielding 10 LCAO wavefunctions and a 10x10 Hamilto-nian matrix The additional of this orbital to our basishas a negligible effect due to the statistically insignificant

interactions between the 1s and 2s2pi orbitals Beingindependent of the valence orbitals the two extra rows ofthe Hamiltonian are block diagonalized from other eightrows

2 CONSTRUCTION OF HAMILTONIAN

MATRIX

What are the atomic configurations of thetwo atoms in your material Which orbitalsdo you expect to play a significant role inbonding

The electron configuration of free carbon atoms is1s22s22p2 The valence electrons in the 2s 2px 2pyand 2pz orbits play a significant role in bonding

Draw all the atoms in the basis and alltheir nearest neighbors with appropriate or-bitals on each atom Label the orbitals accord-ing to their lattice vector basis vector orbitaltype and the type atom they are associatedwith (cation or anion)

The lattice structure of graphene and its first Brillouinzone is shown in Figure 1 The lattice contains two sub-lattices 0 and 1 which differ by their bond orientationsThe first atom A in the primitive cell has three first neigh-bors in the other sublattice 1 with relative unit vectors

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

and six next-nearest neighbors in the same sublattice 0

2

with relative unit vectors

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

The carbon basis atom A at cell vector R has orbitals|2s (R)〉 and |2pi (R)〉 (i = x y z) and orbitals fromthree nearest neighbors shown in Figure 1

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

and the carbon basis atom B at cell vector R has orbitals|2s (R)〉 and |2pi (R)〉 (i = x y z) and orbitals fromthree nearest neighbors

A ∣

∣2sB (R minus eB)rang

and∣

∣2pBi (R minus eB)

rang

(26)

E ∣

∣2sB (R minus eC)rang

and∣

∣2pBi (R minus eC)

rang

(27)

F ∣

∣2sB (R minus eD)rang

and∣

∣2pBi (R minus eD)

rang

(28)

[ ] approximate all the nearest neigh-bor interactions Is this approximation rea-sonable Calculate values for Ess Esp Exxand Exy in terms of Vssσ Vspσ etc

In examining the orbital interactions contributing tothe matrix elements of the Hamiltonian we will referto several figures taken from Saito and Dresselhaus thatdepict carbonrsquos valence orbital overlaps [1] The hy-bridized orbitals contributing to the LCAO approxima-tion are given by the sp2 orbitals

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Figure 2 shows that all nearest neighbor bonding inter-actions are described by each of

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

where all remaining orbital overlaps going to zero

(a)Four sp2 hybridized orbitals in carbon bonding

(b)The rotation of 2px andσ-bond band parameteroverlaps for the 2s and

2pi (i = x y z) orbitals

(c)Sample matrix elements forσ-bonding

FIG 2 Orbital overlaps in graphene Figure 2(b) shows howto project 2px along the σ and π components and the non-vanishing (1-4) and vanishing (5-8) elements of the Hamilto-nian matrix Figure 2(c) shows examples of Hamiltonian ma-

trix elements for σ orbitals˙

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

respectively Figures taken from [1]

The matrix elements for the Bloch orbitals betweenthe A and B atoms of the basis are obtained by tak-ing the components of the 2px and 2py orbitals in the σand π basis From rotating the orbitals we have one ofeight overlap configurations given by Figure 2 For exam-ple the wavefunction |2px〉 is determined by projectingit onto the σ and π basis such that

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

Using these rotation transformations we can determinethe energies given in Equation 210 in terms of Vijk over-lap terms For example the third energy overlap termshown in Figure 2(b) is given by

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

4(Vppσ + Vppπ) eminusikxa2

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

The remaining overlap terms are summarized in Equa-tion 2 in the next section of this paper

Finally in our LCAO approximation note thatlang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

The first is valid because of the uniformity of ourmedium valence orbitals will be the same regardless ofwhere the electrons are located Permutations of orbitalsleave energy overlaps invariant because of the symmetryof the interactions The latter expression is also validas orbital self-interactions will remain the same amongstidentical atoms In the case both basis atoms are carbon

Construct a set of extended atomic or-bitals Be explicit about your choice of phasefactors Write your trial wave function as alinear combination of these extended orbitals

Recall that for the tight-binding solution that we usethe trial wave function

|ψ (r)〉 =sum

α

sum

Rl

cα [Rl] |φα (r minus Rl)〉 (214)

where α denotes both each type of wave function for eachatom and also the type of atom in the basis and Rl

denotes the direct lattice vectors Therefore our LCAOtrial wave function is given by hybridized wave functionsfor the atoms in the primitive basis

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

Find the Hamiltonian matrix for the near-est neighbor approximation

For a systems Hamiltonian operator H the Hamilto-nian matrix in the LCAO method is

H (k) =sum

Rp

H (Rp) eminusikmiddotRp (215)

where the matrix elements in right side of the equalityare

Hβα (RnRm) = |φβ (r minus Rn)〉 H |φα (r minus Rm)〉 (215)

Assuming only nearest neighbor interactions and thatS (k) the Hamiltonian matrix of a two-dimensionalhexagonal crystal of carbon atoms is given by a blockdiagonal matrix

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

FIG 3 The energy dispersion relations for graphene areshown through the whole region of the Brillouin zone Thelower and the upper surfaces denote the valence π and theconduction πlowast energy bands respectively The coordinatesof high symmetry points are Γ = (0 0) K = (0 2π3a) andM = (2π

radic

3a 0) The energy values at the K M and Γpoints are 0 t and 3t respectively (httpwwwiuetuwienacatphdpourfathnode18html)

with

A2times2 = diag (Es Ep Ep Ep) (215)

and

B2times2 =

Vssσg0 Vspσg1 Vspσg2 0Vspσg1 Vppσg3 + Vppπg4 (Vppσ + Vppπ) g5 0Vspσg2 (Vppσ + Vppπ) g5 Vppσg3 + Vppπg4 0

0 0 0 Vppπg0

(215)The phase factors in B2times2 are

g0 = 1 + eminusikmiddotRB + eminusikmiddotC g1 = 1 minus 1

2eminusikmiddotRC minus 1

2eminusikmiddotRC

g2 =

radic3

2

(

eminusikmiddotRC minus eminusikmiddotRC)

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

Conduction in the plane is limited entirely to π bond-ing so the graphene band structure of graphene is limitedto a subset of the matrix given in Equation 2

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

Look up a real energy band diagram foryour material Include a copy in your report

An accepted electronic band energy diagram is shownin Figure 3

Write a matlab program to plot the free-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 bandsWhere is the Fermi level located How doesthe free electron band structure compare to thereal band structure

Our calculated free-electron band structure forgraphene is included in Figure 4 In this model we as-sume that the electron experiences no potential energyand therefore has energies described by

E (k) =~

2k

2

2m (30)

This is applied in the reduced-zone scheme where theevaluation to reciprocal lattice vectors K in the first Bril-louin zone Subsequently the energies are evaluated atK for

E (k) =~

2

2m(kprime + K)

2 (30)

where kprime is restricted to the symmetries of the first Bril-

louin zoneFor the free electron model of graphene using the six

nearest neighbors in reciprocal space there are six de-generacies at the K point two double degeneracies andone triple degeneracy at both the Γ and M points Wesee two degenerate pairs along K to Gamma one along Γto M and three along M to K For this model the Fermienergy is found exactly where the conduction and valencebands touch at the K point It is clear when comparingthis band structure to those found in the literature thatthe free electron model is not a good approximation forelectron transport in graphene We see several band de-generacies that do not match the actual band structureand the general shapes are very different

Algebraically diagonalize the Hamiltonianmatrix at the Γ point What are the differ-ent energies and eigenvectors and what do

FIG 5 Energy dispersion relations for graphene

they correspond to physically Using Har-risons Solid State Table (attached) find nu-merical values for Esa Esc Epa Epc EssEsp Exx and Exy Compare your calculatedenergies at the Γ point with values from theliterature

Our calculated total density of states for the conduc-tion and valence bands is included in Figure 5 Denotingt = Eppπ as the tight-binding energy from the valenceorbitals the band structure depicted in Figures 5 and 6is given analytically by

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

(30)after diagonalizing Equation 2 Dresselhaus cites t =minus3033eV differing from optimized value of t = minus81eV(see Matlab code)[1]

Write a Matlab program to plot the LCAOenergy bands along the same directions asabove along with the approximate location of

5

FIG 6 Graphene energy dispersion from π-bonding

the Fermi level How do your results com-pare (qualitatively) with the band structureyou found in the literature If you wish op-timize the matrix elements for your material

The LCAO energy dispersion is shown in Figure 6Because this figure is based on a known analytic solutionthe results compare almost exactly in qualitative featuresto the known literature [3]

Where are the valence band maximum andthe conduction band minimum located Whatis the energy gap Is your material direct orindirect

The valence band maximum and the conduction bandminimum are degenerate at the K-point yielding a zeroenergy band gap Because the points are coincident in theplane the material is direct Notably because the energydispersion around the K point is linear corresponding todispersion relation of a relativistic Dirac fermion [3]

4 EFFECTIVE MASSES CONSTANT ENERGY

SURFACES AND DENSITY OF STATES

Plot constant energy contours near the va-lence and conduction band edges for appropri-ate planes For example if your minimum isat kmin along ΓminusX you should probably plotan energy contour for the kx minus ky plane andfor the plane parallel to ky minus kz that containskmin

A close up of the conduction and valence band edgesand a contour plot is shown in Figures 7 and 8 respec-tively

Solve for the energy at a number of kpoints near the valence and conduction band

FIG 7 Valence and conduction band edges for graphene

FIG 8 Energy contours for graphene

edges Fit these points using a quadratic poly-nomial (be sure to think about your resultsfrom D1 when you do this) Use your resultsto find the effective masses for both the va-lence bands and the conduction band Howdo your results compare with results from theliterature How would you improve your re-sults

The energy surface near the conduction band edgesform cones that touch each other precisely at the K pointsas shown in Figure 6 making graphene effectively a zero-bandgap material This differs greatly from the form ofcubic semiconductors which have band edges that can beapproximated by parabaloids Since the band structureis resultant from a 2x2 Hamiltonian its scale is com-pletely determined by the energy offset (which we set tozero) and a scaling factor ηppπ The results are as closeas they possibly can be for our near-neighbor approxima-tion as we initially scaled our band structure along itshigh-symmetry points to that of the literature Given thelinear shape of the dispersion relation around the bandedges an electron in graphene for our approximation has

6

FIG 9 Electronic density of states for graphene

an effective mass of zero everywhere around the bandedge except exactly where the bands meet where it hasan infinite effective mass

Plot the total density of states (histogrammethod include all bands) versus energy Useyour calculated effective masses to determinean approximate expression for the density ofstates near the valence and conduction bandedges How does this calculation compare withthe total DOS

Our calculated total density of states for the conduc-tion and valence bands is included in Figure 9 As shownour generated density of states defies simple expressionas unlike a cubic system we cannot represent the den-sity of states as a straightforward

radicE relation As shown

in Figure 9 our calculated DOS matches the shape ofexperimental data for the conduction band well

Using your total DOS calculate theelectronic specific heat of your material as afunction of temperature Compare this withyour calculations for the phonons and com-ment

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 matchingwell with the literature However the electronic specificis significantly smaller than its phonon counterpart Ac-cordingly the specific heat is completely dominated bythe 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 10 Electronic heat capacity for graphene

Use the band structure you found in theliterature to discuss the characteristics ofyour material What electronicoptical appli-cations would your material be goodbad forWhy

Because the expression of the π band in graphene is ex-act our band structure calculation is exactly similar tothe those found in the literature Figure 6 shows the up-per πlowast-energy anti-bonding and lower π-energy bondingbands that are degenerate at the K-point the two elec-trons in the π band are fully occupy the lower π bandFrom the degeneracy at theK-point the density of statesat the Fermi level is zero making graphene a zero bandgap semiconductor [3] The planar π-bonding bestowsgraphene with extremely high room temperature electronmobility allowing a conductivity greater than silver Thelinear shape of the energy dispersion formed the basis offrequency multiplier chip produced by MIT [3]

[1] R Saito G Dresselhaus and M S Dresselhaus Phys-

ical Properties of Carbon Nanotubes (World ScientificPublishing Company 1998) ISBN 1860940935 URLhttpwwwamazoncaexecobidosredirecttag=

citeulike09-20ampamppath=ASIN1860940935[2] J Hone Book Series Topics in Applied Physics 80 273

(2001)[3] A H C Neto F Guinea N M R Peres K S Novoselov

and A K Geim Reviews of Modern Physics 81 109(pages 54) (2009) URL httplinkapsorgabstract

RMPv81p109

Part IV Electronic Band Structure of Graphene Continued

(Dated May 8 2009)

1 A OPTICAL PROPERTIES

Utilizing the (optimized) LCAO band-structure from Part III use the lowest energydirect bandgap (from valence band to conduc-tion band) to estimate the momentum matrixelement

The band structure around graphenersquos bandedge (theK point) is a pair of intersecting cones As such thedouble gradient of the band structure in that region isdiscontinuous yielding an effective mass mlowast of zero at theK point and infinity in the immediate vicinity Lookingat the momentum matrix element calculation outlined inLecture 27

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

Eg= infin at the K point which gives

us intuition that something special is happening at thatlocation If we think about the band structure we real-ize that the matrix element must be zero at the K pointsince the valence and conduction bands are degeneratetherendashthere is no gap across which to absorb a photonHowever if we move away from the K point and no longerhave a linear dispersion relation Eg becomes nonzero asdoes |pcv| but the effective mass becomes zero Clearlythis equation which was formulated for zinc-blende crys-tals cannot help us with determining the momentum ma-trix element aside from suggesting Since 11 is not di-rectly useful in analytically finding |pcv| for our conicbandstructure approximation we instead turned to theliterature

Within the literature [1] [2] the momentum matrixelements are derived explicitly by the dot product of aphoton polarization vector P and the dipole vector D

where

D equiv ltΨf |nabla|Ψigt (12)

and

|pcv| = P middot D = P middot ltΨf |nabla|Ψigt (13)

erjimitedu

this formulation is then broken down into its Blochconstituents and further simplified Orienting the sheetof graphene in the x-y plane it is pointed out that fortwo lattice atoms Rj1 and Rj2

ltφ(r minus Rj1)| part

partz|φ(r minus Rj2)gt = 0 (14)

since the integrand is an odd function of z Thus wecan restrict the atomic dipole vector to the x-y planeie D = (dx dy 0) The matrix element P middot D is furthersimplified using a Taylor expansion around the K point(kx0 ky0) until the arrival of the result

|pcv| = P middotD =3M

2k(py(kx minus kx0) minus px(ky minus ky0)) (15)

where M is the optical matrix element for two nearest-neighbor atoms separated by b1

M = ltφ(R + b1)| part

partx|φ(R)gt (16)

The above can be numerically evaluated for a tight-binding model including 2s and 2p orbitals giving a valueof M = 0206 [1]

Utilize the optimized LCAO bandstructureand momentum matrix element from A1 toestimate the optical absorption Plot the ab-sorption coefficient for your material from thedirect bandedge to 5 kBT above the bandedge

Using the absorption coefficient formula derived in Lec-ture 27 we arrive at

α(ω) =πq2cmicro0

m2

0ωn

|3M

2k|2(py∆kx minus px∆ky)|2pr(hω minus Eg)

(17)where pr(hω minus Eg) is the reduced density of states we

calculated in Part IIAs shown in Figure 1 the absorption is zero directly

at the K points (as our intuition suggested) and growsas k moves away in most directions Notice the line ofzero absorption however for k directions associated with(py∆kx minus px∆ky) = 0

Compare your estimate for the interbandabsorption coefficient with literature valuesWhat are the major sources of discrepancy be-tween your theory and experiment

2

FIG 1 Absorption spectrum for graphene for polarization P= (01) [1] Darker areas correspond to stronger absorption

Considering that our method of derivation was almostidentical to that of the literature it matches very wellIn many of the articles we reviewed we found that the op-tical absorption was solely shaped by a density of statesscaled by constants that appeared to be of approximatelythe same form as those in Equation 17

2 B ELECTRON-PHONON SCATTERING

Utilize the LCAO bandstructure from PartIII to estimate the deformation potential forelectrons near the conduction bandedge andholes at the valence bandedge Compare theestimated deformation potentials with the lit-erature values for the longitudinal acousticdeformation potential

We can calculate the deformation potential directlyfrom the bandstructure since we have it in analytic formfrom Part III as

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

(21)where t = 81eV a = 142A and A = a2 At the

bandedge point K kx = 2πradic3a

and ky = 2π3a

To perform

the calculation we took a small increment and divide itby the change in atomic areal lattice spacing

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

asymp E(a + da) minus E(a)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

the resulting calculation is in very good accordancewith experimental values from graphene devices grownon SiO2 DA(emp) = 18 plusmn 1eV [3]

Utilize the phonon dispersion from PartII to estimate the phonon occupancy for allmodes within the Brillouin zone Assume alattice temperature of 300K

Considering the lattice temperature is so high wecan approximate the phonon occupancy function in theBoltzmann limit ie

N(w) =kBT

w(25)

Using the LCAO density-of-states and theresults from B1 and B2 estimate the longi-tudinal acoustic phonon scattering for yourmaterial Compare your result with literaturevalues

Beginning with Eq (276) from Lundstrom modifiedfor 2D (after doing the integral present in Eq (275) theonly difference is the removal of a factor of β)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

and since wersquore calculating the acoustic deformationpotential

Cβ =πmlowastD2

A

ρvs

pΩ (27)

wersquore operating in the Boltzmann limit as well so

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

We can calculate cl using ρ = 65 middot 10minus7 and using thelongitudinal sound velocity vs = 1312kms from PartIII of the project and TL is 300K The first part of the

3

FIG 2 Density of states histogram calculated in Part III

final equation form yields a constant factor to propor-tion g2D(E) which is the density of states calculated viahistogram method in Part III and shown in Figure 2

cl = v2

s lowast ρ = 851 middot 10minus3kgs (214)

Within the relaxation time approximationin the Boltzmann limit estimate the heavy-hole mobility Compare your result with lit-erature values What are the major sources ofdiscrepancy between your theory and experi-ment

The verdict is still out on the methodology to de-termine the mobility of graphene Measured values ofgraphene mobility show that it has extremely high mo-bilities theoretically infinite at the Dirac points butthat these values are limited by acoustic phonon scat-tering [3] Graphene has been experimentally shown tohave hole mobilities commonly in the range of 3400-4400Vcm2s [4] and as high as 15000 Vcm2s [5] Current re-search suggests that graphene mobility is limited by thedensity of charged impurities within the lattice and sothe key to increasing mobility for use in higher-frequencyelectronics is to reduce the number of impurities whenbuilding graphene components [6]

[1] R Saito A Grneis G Samsonidze G DresselhausM Dresselhaus A Jorio L Canado M Pimenta andA S Filho Applied Physics A Materials Science amp Pro-cessing 78 1099 (2004) URL httpdxdoiorg10

1007s00339-003-2459-z[2] A K Gupta O E Alon and N Moiseyev Physical

PRBv68e205101[3] J Chen C Jang S Xiao M Ishigami and M S Fuhrer

Nat Nano 3 206 (2008) ISSN 1748-3387 URL http

dxdoiorg101038nnano200858[4] Y Q Wu P D Ye M A Capano Y Xuan Y Sui

M Qi J A Cooper T Shen D Pandey G Prakashet al Applied Physics Letters 92 092102 (2008) URLhttplinkaiporglinkAPL920921021

[5] A K Geim and K S Novoselov Nat Mater 6 183(2007) ISSN 1476-1122 URL httpdxdoiorg10

1038nmat1849[6] E H Hwang S Adam and S D Sarma Physical Review

Letters 98 186806 (2007) URL httplinkapsorg

abstractPRLv98e186806

Preliminary Questions
- Lattice Structure
- Reciprocal Lattice Structure
- Atomic form factors
- - X-Ray Diffraction
  - - Planes in the Reciprocal Lattice
    - Structure Factors and X-Ray Diffraction Intensities
    - Crystal Structure of Silicon
    - - References
      - Background for the Bohr Model
      - Parameters of Bohr Model
        
        Nearest Neighbor Couplings
        
        Elastic Properties
        
        Construction of the Dynamical Matrix
        
        Lattice and Reciprocal Space Structures
        
        Born Force Model
        
        Dynamical Matrix
        
        Model Optimization and Comparison to Macroscopic Properties
        
        Comparison with Published Theoretical and Experimental Data
        
        Phonon Dispersion Relations
        
        Density of States
        
        Specific Heat
        
        References
        
        Band Structure Background Questions
        
        Construction of Hamiltonian Matrix
        
        Band Calculations
        
        Effective Masses Constant Energy Surfaces and Density of States
        
        References
        
        A Optical Properties
        
        B Electron-Phonon Scattering
        
        References

Page 2: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

2

FIG 2 Reciprocal lattice planes [001] [110] and [111]

given by

b1 =2π

a

(

1radic3minus1 0

)

|b1| =2π

a

2radic3

b2 =2π

a

(

1radic3 1 0

)

|b2| =2π

a

2radic3

b3 =2π

c(0 0 1) |b3| =

2π

c (14)

The reciprocal lattice plane generated by the b1 and b2

vectors forms the outline of the first Brillouin zone asdepicted in Figure 1 The intersection of the the planeskz = plusmn2πc with the plane forms a hexagonal prism ofheight 4πc

13 Atomic form factors

As carbon is the only element present in graphene andgraphite the atomic form factor is uniform across theentire crystal and thus can be factored out when calcu-lating the structure factor Thus the atomic form factorhas no effect on the relative intensities of x-ray diffrac-tion occuring in different planes of graphite Accordingto the NIST Physics Laboratory the atomic form factorof carbon varies from 600 to 615 eatom with incidentradiation ranging from 2 to 433 KeV [2]

2 X-RAY DIFFRACTION

21 Planes in the Reciprocal Lattice

Provide pictures of the crystal and of thereciprocal lattice in the [100] [110] and [111]planes Indicate the vertical positions ofatoms with respect to the plane

Pictures of the crystal and of the reciprocal lattice inthe [100] [110] and [111] planes are included in Fig-ure 2 In MATLAB the crystal was represented as a set

of points in space using the specified lattice vectors andatom bases Normals generated from lattice vector sumswere used to extract planes and display them Varyingcolors were used to depict vertical spacing between adja-cent planes

22 Structure Factors and X-Ray Diffraction

Intensities

Calculate the structure factor for all thereciprocal lattice vectors Kl lt 16 (2πa)

2

The structure factor is calculated as

Mp (Ki) = fc

nsum

j=1

(Ki) eminusiKimiddotρi

where fc is the structure factor of Carbon and ρi are thebasis vectors of our lattice We find that only four uniquenon-zero values of Mp (Ki) occur in the reciprocal latticeEach of these corresponds to the height of a diffractionintensity peak and their relative values are referenced inTable I

Calculate the ratio of the intensities ex-pected for the following lines of the diffractionpattern with respect to the [111] line [100][200] [220] [311] and [400]

Including the structure factor there are other severalfactors contributing to the intensities of the diffractionpeaks [3]

1 The Lorenz correction is a geometric relation al-tering the intensity of an x-ray beam for differentscattering angles θ

2 The multiplicity factor p is defined as the num-ber of different planes having the same spacingthrough the unit cell Systems with high symme-tries will have different planes contributing to thesame diffraction thereby increasing the measuredintensity

3 Temperature absorption polarization each con-tribute higher-order corrections ultimately ignoredin our calculation These include Doppler broad-ening from thermal vibrations in the material ab-sorption of x-rays and the polarization of initiallyunpolarized x-rays upon elastic scattering

These factors all contribute to the relative intensity of a[hkl] diffraction peak given by

I[hkl] (θ) = p|Ma (Km) |2(

1 + cos2 2θ

sin2 θ cos θ

)

(21)

Using the formula from the previous question to calcu-late the ratios of the structure factors in the given planes

3

Si C (Graphite)

[111] 2847 10000 mdash (3) mdash

[100] mdash mdash 4277 4 (3) 345

[002] mdash mdash 2674 106 (16) 100

[220] 4735 6431 mdash (1) mdash

[311] 3731 3731 mdash (3) mdash

[400] 6921 958 mdash (1) mdash

diffraction

d0 = ~0 d4 =a

4(1 3 3)

d1 =a

4(1 1 1) d5 =

a

4(2 2 0)

d2 =a

4(3 3 1) d6 =

a

4(2 0 2)

d3 =a

4(3 1 3) d7 =

a

4(0 2 2) (22)

Ma (Km) =

nsum

j=1

= f(1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

+(minus1)h+k+l

+ (minusi)3h+k+l

+ (minusi)3h+3k+1

+(minusi)h+3k+1

)

= f(

1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

)

middot(

1 + (minusi)h+k+l

)

(23)

default

=1

+1

2αφ

(

(11)

V (u[Ri t])) =

V0 +sum

n

sum

m

(

part2V

)

eq

un[Ri t]

(12)

n

2

2MC [ 3radic

3a2

radic

MATRIX

(

2πaradic

3 0)

and kK =(

2πaradic

3 2π3a)

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

22 Born Force Model

3

A B C D E F G H I J

xy yx 0 0radic

3αs4 minusradic

3αφ4radic

3αφ4 0radic

3αφ4 minusradic

3αφ4 0

3αs4radic

Vs (R) =1

+1

Vφ

=1

2αφ

(

+1

2αφ

(

+1

2αφ

(

(24)

23 Dynamical Matrix

Dij (k) =sum

Rp

Vprimeprime

(26)

and u =(

u1x u2

y u1x u2

y

D (k) =

A0 B0 C D

B0 A1 D B1

(27)

yx (BA)

A0 = Vprimeprime

xx (AE) eminusi

ldquo radic3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AF ) eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AG) eminusikya

+ Vprimeprime

3

2kxaminus

kya

2

rdquo

+ Vprimeprime

3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AJ) eikya

=3αs

2+ 3αφ minus 3αs

4eminusi

ldquo radic3

2kxa+

kya

2

rdquo

minus 3αs

4eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ 0 minus 3αφ

4eildquo radic

3

2kxaminus

kya

2

rdquo

minus 3αφ

4eildquo radic

3

2kxa+

kya

2

rdquo

+ 0

=3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

(29)

4

Γ M K Γ M K

ωLO 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

vLA 217 1312

vTA 14 621

vZA mdash mdash

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

PROPERTIES

Experimental Data

5

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

34 Specific Heat

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 3: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

3

Si C (Graphite)

[111] 2847 10000 mdash (3) mdash

[100] mdash mdash 4277 4 (3) 345

[002] mdash mdash 2674 106 (16) 100

[220] 4735 6431 mdash (1) mdash

[311] 3731 3731 mdash (3) mdash

[400] 6921 958 mdash (1) mdash

diffraction

d0 = ~0 d4 =a

4(1 3 3)

d1 =a

4(1 1 1) d5 =

a

4(2 2 0)

d2 =a

4(3 3 1) d6 =

a

4(2 0 2)

d3 =a

4(3 1 3) d7 =

a

4(0 2 2) (22)

Ma (Km) =

nsum

j=1

= f(1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

+(minus1)h+k+l

+ (minusi)3h+k+l

+ (minusi)3h+3k+1

+(minusi)h+3k+1

)

= f(

1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

)

middot(

1 + (minusi)h+k+l

)

(23)

default

=1

+1

2αφ

(

(11)

V (u[Ri t])) =

V0 +sum

n

sum

m

(

part2V

)

eq

un[Ri t]

(12)

n

2

2MC [ 3radic

3a2

radic

MATRIX

(

2πaradic

3 0)

and kK =(

2πaradic

3 2π3a)

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

22 Born Force Model

3

A B C D E F G H I J

xy yx 0 0radic

3αs4 minusradic

3αφ4radic

3αφ4 0radic

3αφ4 minusradic

3αφ4 0

3αs4radic

Vs (R) =1

+1

Vφ

=1

2αφ

(

+1

2αφ

(

+1

2αφ

(

(24)

23 Dynamical Matrix

Dij (k) =sum

Rp

Vprimeprime

(26)

and u =(

u1x u2

y u1x u2

y

D (k) =

A0 B0 C D

B0 A1 D B1

(27)

yx (BA)

A0 = Vprimeprime

xx (AE) eminusi

ldquo radic3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AF ) eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AG) eminusikya

+ Vprimeprime

3

2kxaminus

kya

2

rdquo

+ Vprimeprime

3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AJ) eikya

=3αs

2+ 3αφ minus 3αs

4eminusi

ldquo radic3

2kxa+

kya

2

rdquo

minus 3αs

4eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ 0 minus 3αφ

4eildquo radic

3

2kxaminus

kya

2

rdquo

minus 3αφ

4eildquo radic

3

2kxa+

kya

2

rdquo

+ 0

=3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

(29)

4

Γ M K Γ M K

ωLO 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

vLA 217 1312

vTA 14 621

vZA mdash mdash

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

PROPERTIES

Experimental Data

5

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

34 Specific Heat

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 4: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

=1

+1

2αφ

(

(11)

V (u[Ri t])) =

V0 +sum

n

sum

m

(

part2V

)

eq

un[Ri t]

(12)

n

2

2MC [ 3radic

3a2

radic

MATRIX

(

2πaradic

3 0)

and kK =(

2πaradic

3 2π3a)

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

22 Born Force Model

3

A B C D E F G H I J

xy yx 0 0radic

3αs4 minusradic

3αφ4radic

3αφ4 0radic

3αφ4 minusradic

3αφ4 0

3αs4radic

Vs (R) =1

+1

Vφ

=1

2αφ

(

+1

2αφ

(

+1

2αφ

(

(24)

23 Dynamical Matrix

Dij (k) =sum

Rp

Vprimeprime

(26)

and u =(

u1x u2

y u1x u2

y

D (k) =

A0 B0 C D

B0 A1 D B1

(27)

yx (BA)

A0 = Vprimeprime

xx (AE) eminusi

ldquo radic3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AF ) eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AG) eminusikya

+ Vprimeprime

3

2kxaminus

kya

2

rdquo

+ Vprimeprime

3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AJ) eikya

=3αs

2+ 3αφ minus 3αs

4eminusi

ldquo radic3

2kxa+

kya

2

rdquo

minus 3αs

4eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ 0 minus 3αφ

4eildquo radic

3

2kxaminus

kya

2

rdquo

minus 3αφ

4eildquo radic

3

2kxa+

kya

2

rdquo

+ 0

=3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

(29)

4

Γ M K Γ M K

ωLO 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

vLA 217 1312

vTA 14 621

vZA mdash mdash

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

PROPERTIES

Experimental Data

5

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

34 Specific Heat

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 5: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

2

2MC [ 3radic

3a2

radic

MATRIX

(

2πaradic

3 0)

and kK =(

2πaradic

3 2π3a)

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

22 Born Force Model

3

A B C D E F G H I J

xy yx 0 0radic

3αs4 minusradic

3αφ4radic

3αφ4 0radic

3αφ4 minusradic

3αφ4 0

3αs4radic

Vs (R) =1

+1

Vφ

=1

2αφ

(

+1

2αφ

(

+1

2αφ

(

(24)

23 Dynamical Matrix

Dij (k) =sum

Rp

Vprimeprime

(26)

and u =(

u1x u2

y u1x u2

y

D (k) =

A0 B0 C D

B0 A1 D B1

(27)

yx (BA)

A0 = Vprimeprime

xx (AE) eminusi

ldquo radic3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AF ) eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AG) eminusikya

+ Vprimeprime

3

2kxaminus

kya

2

rdquo

+ Vprimeprime

3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AJ) eikya

=3αs

2+ 3αφ minus 3αs

4eminusi

ldquo radic3

2kxa+

kya

2

rdquo

minus 3αs

4eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ 0 minus 3αφ

4eildquo radic

3

2kxaminus

kya

2

rdquo

minus 3αφ

4eildquo radic

3

2kxa+

kya

2

rdquo

+ 0

=3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

(29)

4

Γ M K Γ M K

ωLO 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

vLA 217 1312

vTA 14 621

vZA mdash mdash

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

PROPERTIES

Experimental Data

5

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

34 Specific Heat

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 6: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

3

A B C D E F G H I J

xy yx 0 0radic

3αs4 minusradic

3αφ4radic

3αφ4 0radic

3αφ4 minusradic

3αφ4 0

3αs4radic

Vs (R) =1

+1

Vφ

=1

2αφ

(

+1

2αφ

(

+1

2αφ

(

(24)

23 Dynamical Matrix

Dij (k) =sum

Rp

Vprimeprime

(26)

and u =(

u1x u2

y u1x u2

y

D (k) =

A0 B0 C D

B0 A1 D B1

(27)

yx (BA)

A0 = Vprimeprime

xx (AE) eminusi

ldquo radic3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AF ) eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AG) eminusikya

+ Vprimeprime

3

2kxaminus

kya

2

rdquo

+ Vprimeprime

3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AJ) eikya

=3αs

2+ 3αφ minus 3αs

4eminusi

ldquo radic3

2kxa+

kya

2

rdquo

minus 3αs

4eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ 0 minus 3αφ

4eildquo radic

3

2kxaminus

kya

2

rdquo

minus 3αφ

4eildquo radic

3

2kxa+

kya

2

rdquo

+ 0

=3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

(29)

4

Γ M K Γ M K

ωLO 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

vLA 217 1312

vTA 14 621

vZA mdash mdash

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

PROPERTIES

Experimental Data

5

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

34 Specific Heat

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 7: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

4

Γ M K Γ M K

ωLO 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

vLA 217 1312

vTA 14 621

vZA mdash mdash

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

PROPERTIES

Experimental Data

5

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

34 Specific Heat

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 8: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

5

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

34 Specific Heat

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 9: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

6

linkJCP2116601

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 10: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

7

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 11: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

8

(c)M point

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 12: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

QUESTIONS

1radic3

(

|2s〉 +radic

2 |2pi〉)

(i = x y z) (11)

MATRIX

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 13: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

2

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

B ∣

∣2sA (R + eB)rang

and∣

∣2pAi (R + eB)

rang

(23)

C ∣

∣2sA (R + eC)rang

and∣

∣2pAi (R + eC)

rang

(24)

D ∣

∣2sA (R + eD)rang

and∣

∣2pAi (R + eD)

rang

(25)

A ∣

and∣

rang

(26)

E ∣

and∣

rang

(27)

F ∣

and∣

rang

(28)

∣

∣φAB

1

rang

=1radic3

∣

∣2sABrang

plusmnradic

2

3

∣

∣2pABx

rang

∣

∣φAB

2

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

3

rang

=1radic3

∣

∣2sABrang

∓ 1radic6

∣

∣2pABx

rang

plusmn 1radic2

∣

∣2pABy

rang

∣

∣φAB

4

rang

=∣

∣2pABrang

(29)

Ess =lang

2sA (0)∣

∣ H∣

∣2sB (eB)rang

Esp =lang

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

Exx =lang

2pAx (0)

∣

∣ H∣

∣2pBx (eB)

rang

Exy =lang

2pAx (0)

∣

∣ H∣

∣2pBy (eB)

rang

(210)

2sA˛

˛ H˛

˛2pBx

cedil

and˙

2pAx

˛

˛ H˛

˛2pBy

cedil

|2px〉 = cos(π

3

)

|2pσ〉 + sin(π

3

)

|2pπ〉 (211)

lang

2pAx

∣

∣ H∣

∣2pBy

rang

=3

radic3eikya2

minus 3

radic3eminusikya2

= i

radic3

radic3 sin

kya

2

(212)

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 14: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

3

2sA (0)∣

∣ H∣

∣2pBx (eB)

rang

=lang

2sB (0)∣

∣ H∣

∣2pAx (eB)

rang

lang

2sA (0)∣

∣ H∣

∣2sA (0)rang

=lang

2sB (eB)∣

∣ H∣

∣2sB (eB)rang

(213)

|ψ (r)〉 =sum

α

sum

Rl

|ψ〉 = aA

∣

∣2sA (R)rang

+sum

i=xyz

biA∣

∣2pAi (R)

rang

+ aB

∣

∣2sB (R + eB)rang

+sum

i=xyz

biB∣

∣2pBi (R + eB)

rang

(215)

H (k) =sum

Rp

H3D =

[

A2times2 B2times2

Bdagger2times2

A2times2

]

(215)

radic

with

and

B2times2 =

0 0 0 Vppπg0

2eminusikmiddotRC

g2 =

radic3

2

(

g3 = 1 +1

4eminusikmiddotRC +

1

4eminusikmiddotRC

g4 =3

4

(

g5 =

radic3

4

(

H2D =

[

Ep Vppπg0Vppπg

lowast0

Ep

]

(213)

3 BAND CALCULATIONS

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 15: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

4

E (k) =~

2k

2

2m (30)

E (k) =~

2

2m(kprime + K)

2 (30)

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 16: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

5

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 17: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

6

RMPv81p109

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 18: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

(Dated May 8 2009)

1

mlowast =1

m+

2

m2

|pcv|2Eg

(11)

we see that 1

mlowast= 1

where

and

erjimitedu

partx|φ(R)gt (16)

α(ω) =πq2cmicro0

m2

0ωn

|3M

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 19: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

2

Eplusmn2D = plusmnt

radic

radic1 + 4 cos

(radic3kxa

2

)

cos

(

kya

2

)

+ 4 cos2(

kya

2

)

and ky = 2π3a

To perform

DA =partEn(K)

partA|eqA =

partEn(K)

parta|eq middot

a

2(22)

damiddot a

2=

1

2

a

da∆E (23)

= 1415eV (24)

N(w) =kBT

w(25)

1

τ=

Ω

4π2

int βmax

βmin

(Nβ +1

2∓ 1

2)Cββdβ (26)

Cβ =πmlowastD2

A

ρvs

pΩ (27)

1

τ=

Ω

4π2

int βmax

βmin

NωsCββdβ (28)

Nωs=

kBTL

ωs

=kBTL

βvs

(29)

we find that

1

τ=

mlowastD2

AkBTL

4π2clp

int βmax

βmin

dβ (210)

=mlowastD2

AkBTL

4π2clp

2p

(211)

=D2

AkBTL

2cl

mlowast

π2(212)

=D2

AkBTL

2cl

g2D(E) (213)

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

Page 20: Crystal Structure of Graphite, Graphene and Siliconmiholcomb/graphene.pdf · 2009. 3. 13. · Crystal Structure of Graphite, Graphene and Silicon Dodd Gray, Adam McCaughan, Bhaskar

3

cl = v2

- Lattice Structure
    - - References
        
        
        Elastic Properties
        
        
        
        Born Force Model
        
        Dynamical Matrix
        
        
        
        
        Density of States
        
        Specific Heat
        
        References
        
        
        
        Band Calculations
        
        
        References
        
        
        
        References

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