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Appendix B  ZGNR Optical Matrix Elements

B.1  Bloch Wave Functions Prefactors

To obtain CA and CB in Eq. 18, one can substitute Eq. 6 and Eq. 19 into the Shrödinger equation H|ψ⟩=E|ψ⟩. Considering an A-type carbon atom at some atomic site n, and its three nearest neighbors, the Hamiltonian can be written as:

H  = t | BNn+1⟩ ⟨ An |  + t | BNn+1⟩ ⟨ An |  + t | BNn⟩ ⟨ An |.     (1)

Using Eq. 1 along with the wave functions obtained in Eq. 12, one obtains:

ECA ei kx xnAsin(n θ) =  tCBei kx xNn+1Bsin((Nn+1)θ) 
 +tCBei kx xNn+1Bsin((Nn+1)θ) 
 +tCBei kx xNnBsin((Nn)θ).
    (2)

Therefore, the relation between CA and CB can be written as:

ECAsin(nθ)
=tCB



e
ikx
xNn+1BxnA
 
+e
i kx
xNn+1BxnA
 


 
  sin
(Nn+1)θ
+e
ikx
xNnBxnA
 
sin
(Nn
e

xB
 


 
= tCB


2 cos


3
2
kx acc


sin((Nn+1)θ)+sin((Nn)θ)


.
    (3)

By employing the relation sin(x)sin(y)=(1/2)[cos(xy)−cos(x+y)] and using Eq. 22,

ECA = −tCB
sin(θ)
sin((N+1)θ)
.     (4)

Analogously, for the Nn+1th B-type carbon atom one can obtain the following relation:

ECB sin

Nn+1
θ
tCA 


2 cos


3
2
 kx acc


sin(nθ) + sin((n−1)θ)


    (5)

which gives

ECB = −tCA
sin(θ)
sin((N+1)θ)
.     (6)

From Eq. 4 and Eq. 6, one can find that CACB.
Also, the dispersion relation can be found by multiplying Eq. 3 by Eq. 5,

E2CACB sin(nθ)sin((Nn+1)θ) 
 
= t2CACB



4cos2


3
2
 kx acc


sin((Nn+1)θ)sin(nθ) 
 
   +2cos


3
2
 kx acc


sin((Nn+1)θ)sin((n−1)θ) 
 
   +2cos


3
2
 kx acc


sin((Nn)θ) sin(nθ) 
 
   +sin((Nn)θ) sin((n−1)θ)



.
    (7)

With the help of trigonometric identities and Eq. 22, this expression can be reformatted as

E = ± t 


1+4cos2


3
2
 kx acc


+4cos


3
2
 kx acc


cos
θ



1/2



 
.     (8)

B.2  Transverse Wave Functions Amplitude

To solve the recursive formula,

φn+1C φnn−1=0,     (9)

one can consider the ansatz φn=tn and follow similar equation,

t2Ct+1=0.     (10)

This equation is the generating polynomial of the recursive formula 9.
The roots of 10 are

t1,2=


C±
C2−4
 

2
.     (11)

The general solution of the difference equation is

φn=α t1n+β t2n,     (12)

since t1 is a root of the equation, the other root t2 can be written as: t2=t1−1.
By substituting those two roots in 12 one obtains

φn=α t1n+β t1n.     (13)

Imposing the initial condition φ0=0 results in

α+β =0,  α =−β,     (14)

and from the 13,

φn=α(t1nt1n).     (15)

We obtain

α=
φ1
C2−4
,   β=−
φ1
C2−4
.     (16)

By substituting 11 and 16 in 15, one obtains

φn=
φ1
C2−4



C+
C2−4
2



n



 
φ1
C2−4



C
C2−4
2



n



 
.     (17)

17 can be rewritten as

φn=



C+
C2−4
2



 n



 



C
C2−4
2



 n



 
 C2−4
φ1.     (18)

B.3  Optical Matrix Elements

Using Eq. 12 and Eq. 13 the matrix elements pθ,θ(kx) ≡ ⟨ +, θ, kx| px | −, θ, kx⟩ for an interband transition from a valence band state | −, θ, kx⟩ to a conduction band state | +, θ, kx⟩ are obtained as

Pθ,θ=(xθxθ)
im0
⟨ θ | H | θ ⟩     (19)
Pθ,θ
im0
ℏ Ω
 
N
n=1
 
N
m=1
 



 
ei k (xmB − xnA) sin(nθ) sin(m θ ) ⟨ An | H | Bm ⟩ (xmB − xnA)   
 
− ei k (xmA − xnB) sin(nθ ) sin(mθ) ⟨ Bn | H | Am ⟩ (xmA − xnB



    (20)

Considering only the nearest neighbors, each atom with some index n has two neighbors with index Nn+1 and one neighbor with index Nn, see Fig. 4.1. Therefore, the index m has only three values with ⟨ An | H | Bm ⟩ = t. So we have

 Pθ,θ=



i m0
ℏΩ






i
3
acct
2



N
n=1






e
i 
3
kx a/2
 
e
i 
3
kx a/2
 


sin(nθ) sin((Nn+1) θ 
 


e
i 
3
kx a/2
 
e
i 
3
kx a/2
 


sin(nθ ) sin((Nn+1)θ)



    (21)

after some algebra and replacing Ω from Eq. 16, the optical matrix elements are

Pθ,θ=
−2
3
m0 acc t
ℏ (2N+1)
 sin


3
2
kacc


N
n=1




sin(nθ)sin((Nn+1) θ )−
sin(nθ )sin((Nn+1)θ)



.
    (22)

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