Environmental Engineering Reference
In-Depth Information
when
G
(0)
0,0
(
ω
)
=
g
0,0
(
ω
) is introduced, we can generalize the above
relations for iterationsas
G
(
n
)
n
,
n
(
ω
)
=
g
n
,
n
(
ω
)
+
g
n
,
n
(
ω
)
D
n
,
n
−
1
G
(
n
)
1,
n
(
ω
),
(2.22)
−
n
G
(
n
−
1)
G
(
n
)
)
D
n
−
1,
n
G
(
n
)
n
−
1,
n
(
ω
)
=
n
−
1,
n
−
1
(
ω
n
,
n
(
ω
),
(2.23)
)
D
n
,
n
−
1
G
(
n
−
1)
G
(
n
)
G
(
n
)
n
,0
(
ω
)
=
n
,
n
(
ω
n
−
1,0
(
ω
).
(2.24)
Here, Eqs. 2.22 and 2.23 are also solved self-consistently. At the
end of this iteration process, we can obtain
G
(
N
+
1)
N
+
1,0
(
ω
), which must
be the block of the Green's function 2.12,
G
(
N
+
1)
N
+
1,0
(
ω
)
=
G
N
+
1,0
(
ω
)
because the system generated after (
N
+1)th iteration is equivalent
totheoriginalone.ThisequivalenceisguaranteedsincetheDyson's
Eqs. 2.22, 2.23, and 2.24 are based on the exact perturbative
treatments of the off-diagonal blocks of the dynamical matrix to the
infinite order Now, the level width function is also constructed only
by the intra-buffer-layer block and thereby
G
N
+
1,0
(
ω
) is the block
of the Green's function required for the calculation of the phonon
transmission.
Next, we explain the method to compute the left/right surface
Green's function
G
L/R
(
ω
) [17], which is related to the self energy
L/R
(
ω
)through
L/R
(
ω
)
=
D
LL /RR
G
L/R
(
ω
)
D
LL/ RR
.
(2.25)
Here,
D
LL / RR
denotes the inter-layer dynamical matrix of the
left/right semi-infinite lead. In addition, we denote the intra-layer
matrix for a layer in the semi-infinite lead as
D
L/R
. Note that the
intra-layer dynamical matrix at the surface layer may be different
from the ones in the bulk lead. So, we carry out the same procedure
of the recursive technique depicted in Fig. 2.3b, that is, one layer
is added to the semi-infinite lead in each iteration step. The semi-
infinite system obtained after the iteration should be equivalent
to the semi-infinite subsystem before the iteration. Therefore, by
replacing
G
(
n
)
)and
G
(
n
−
1)
n
,
n
(
ω
n
−
1,
n
−
1
(
ω
)inEqs.2.22and2.23with
ω
G
L/R
(
), wecan derive
)
D
LL / RR
G
L/R
(
G
L/R
(
ω
)
=
g
L/R
(
ω
)
+
g
L/R
(
ω
ω
)
D
LL / RR
G
L/R
(
)
(2.26)
ω
