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G βc. γ 2 βa & # ) 8πω 4 σ = 4 (βc + γ 2 + βa). c
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23000 22000 21000
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18000 rg -
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7
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n
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E',F+ E',F' E(,+E E(,'* E(,(( E(,1+ E(,1F EF,01
ωis
:!= ? −1
FE((F FE(*F FE*** FE/F/ FE/+' FE01* FE0(' FE001
∆ω = ω0 − ωis
E)1) E*+) E(+( EFEE EFEE E1(( E11' E11*
:!= ? J G 2 :%!2 52? FF0(+ F),0 HIH E0,E HIH E),( HIH (1,E FF00* J FF0(+ (1,E HIH E11,1 FF00* 0,F*
ω0
−1
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n
EF,+1 EF,*) EF,'' EF,(* EF,F1 EF,1+ EE,0+ EE,+) EE,)E EE,(1
:!= ? −1
∆ω = ω0 − ωis
FF1(+ FFE*F FFFE1 FFF*1 FF(E* FF(/E FF'E/ FF'/E FF*0F FF/FE
:!= ? J G 2 :%!2 52? F(1(0 /,( FF0(+ F+,+ FF00* +,) F(1(0 *,) FF0(+ E*,* FF00* ),0 FF0(+ FE,* FF00* /,( FF0(+ ),0 HIH F0,1
ω0
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−1
( ( 3 1F( n
E1,'1 0,+* 0,'F 0,(* 0,E1 +,00 +,0E +,'0 +,F' /,/E /,'F
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−1
0
is
ω0
:!= ? G 2 F'/1* HKH HKH HKH HKH HKH HKH HKH HKH HKH HKH −1
(1( 3 J2 NNQ4( G!% F( n ω :!= ? ∆ω = ω − ω ω :!= ? G 2 0,0+ F(F)+ E'(/ F'/1* 0,+* F((FE E(+' HKH 0,)/ F((0( E(EF HKH 0,'F F('0( EFEF HKH 0,(' F(*F' EE+E HKH 0,10 F()F) E1/0 HKH +,00 F())0 E1() HKH +,0E F(/1E E11' HKH +,)' F(+E1 +0* HKH +,'0 F(+)/ +(+ HKH +,(/ F(0E0 /+) HKH +,E( F'1E' )0E HKH /,0) F'1+( )FF HKH /,/E F'E+' *FE HKH /,*+ F'F(+ ')/ HKH is
−1
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−1
(F( 3 H ? " &
RSR2( G!% C( n ω :!= ? ∆ω = ω − ω ω :!= ? G 2 0,+* F((FE E(+' F'/1* 0,)/ F((0( E(EF HKH 0,'F F('0( EFEF HKH 0,F+ F(**1 EE** HKH +,00 F())0 E1() HKH +,0E F(/1E E11' HKH +,'0 F(+)/ +(+ HKH +,F( F(0/* /(1 HKH is
−1
0
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:!= ? G 2 HKH HKH HKH −1
(C( 8! # " #!% ( " !==
A1
B1
G!% B( ν :!= ? J % 2 −1
*FE /+) E11' E1(1 EE+E EFEE E(+1 E*1' E)1* F)F1 (1*' (1)/ (1)/ (') )F( 0/1 E1+E EE** EF/+
1,F' E,1/ F,*' 1,/E1 J E,1*1 J J E,E01 J J J J J 1,FE* J J J J
J
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" !==
ν
B1
A2
B2
:!= ?
J
E(EE E'(0 E')1 E*+) F0+1 (1(F (1/' '1* +'1 0/1 FE/ ')* )01 '(1 /(1 +0*
J J J 1,*F1' J J J 2 =2 J J 1,1/ J J J J J
−1
% 2
J
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(B( ! ( G!% :( 8 # " C σ ## C2ν σ z 2
yz
r
H
r
r
xz
rH
C2ν
rC
r
r
r
r r
A1 A2 B1 B2 N X
E
E E E E E* (0
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E E
σ(xz)
σ(yz)
−1
−1
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−1
E
−1
−1
* (
E( E(
E
−1
E * *
r
89
3
X(E) = (15 − 2)(1 + 2 cos 0) = 39, X(C2) = (5 − 2)(1 + 2 cos π/2) = 3, X(σxz ) = 13(−1 + 2 cos 0) = 13, X(σyz ) = 5(−1 + 2 cos 0) = 5. 1 nA1 = (39 + 3 + 13 + 5) = 15, 4 1 nA2 = (39 + 3 − 13 − 5) = 6, 4 1 nB1 = (39 − 3 + 13 − 5) = 11, 4 1 nB2 = (39 − 3 − 13 + 5) = 7, 4 Γ = 15A1 + 6A2 + 11B1 + 7B2.
T !" 15 + 6 + 11 + 7 = 39( T ! @ " ! 3N − 6 = 39( N H # . ?@ # "& # ?" % ## ( (:( G # αxy !" ## A1. A2. B1. B2( G #!% βxy (A1). βxy (A2). βxy (B1). βxy (B2)( ,9G57G*7 2( PP 4 . 2=F=( B. 1:B( ( PP *D8. 2=CF( :C. (
