MIT非线性光学讲义Nonlinear Optics VI

Define L ≡
−1 1 (0) Lρ(1) μ ngEe −i ωt ρ(0) ng = −(ih ) gg − ρnn 2 1 (0) (0) − (ih )−1 μ ng E*eiω t ρgg − ρnn 2 . ⎧ 1 (1) ρng = L−1 ⎨− (ih )−1 μ ng Ee−i ωt ρ(0) − ρ(0) gg nn ⎩ 2 1 (0) (0) ⎫ ⎬ − (ih )−1 μ ng E*eiω t ρgg − ρnn ⎭. 2
149
⎛ 1⎞ ⎜ ⎜ Δω + i T ⎟ ⎟ ⎝ 2⎠ 1 ⎛ I⎞ 1 + Δω 2 + 2 ⎜ ⎟ ⎟ T2 ⎜ I ⎝ s⎠ .
(
(0) ) (ρ(0) gg − ρnn )
−1
.
2 / T2
χ" ω χ'
ω ng
population inversion
145
Nonlinear Optics
Consider
'
χ = χ' + iχ " ,
2 ω ng − ω Nμ (0) (0) χ = ρgg − ρnn 2 2 ε oh ω ng − ω + (1 / T2 )
II. For
χij tensor,
Pi = ε oχij E j
i Pi = N ρng μ ign + c. c.
(
)
)(
-1 N i j 1 (0) − χij = μ gnμ ng ω ng − ω − iT 2 ρ(0) gg − ρnn ε oh .
(
)
III. For isotropic medium, Due to symmetry consideration,
146
Nonlinear Optics
⇒ ⇒
P is parallel to E,
χ is scalar.
Assume 1/3 atoms will have their dipole transition moments parallel to E field,
χ=
1 N −1 -1 (0) μ gnμ ng ω ng − ω − iT2 ρgg − ρ(0) nn 3 ε oh -1 1 N 2 −1 ≅ μ ω ng − ω − iT2 3 εo h
0 det 0 0
− No / T1
Δ ω + i / T2 2iC
C No * C = (Δω + i / T2 )C = n T1 −1 / T1
∴
ρng
No (Δω + i / T2 )C n T1 = =− ⎡⎛ 2 1 ⎞ 1 d 4 2⎤ ⎢⎜ ⎜ Δω + T 2 ⎟ ⎟T +T C ⎥ ⎥ ⎢ ⎝ ⎣ ⎦ 2⎠ 1 2 ⎛ 1⎞ 1 ⎟ Δ ω + i μ ng Ee−iω t No ⎜ ⎜ ⎟ T2 ⎠ 2 h ⎝ =− 2 ⎞ 1 ⎛ 1 Δω 2 + 2 ⎜1 + 2 μ ng E T1T2 ⎟ ⎠ T2 ⎝ h .
(
)(
)
(
)
with
(0) ρ(0) gg = 1, ρnn = 0 . ( Eng >> kT )
Oscillator Strength for g → n transition,
f ng ≡
2 mω ng μ ng 3he2
2
with dimensionless unit.
(with summation over 3 dimensions
Ωng − ω = ωng − ω − i / T2 = Δω − i / T2
−Ω gn − ω = ωng − ω + i / T2 = Δω + i / T2
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Nonlinear Optics
⇒
Δ ω − i / T2 0 C det 0 Δω + i / T2 C* −2 iC* 2iC −1 / T1 ⎡⎛ ⎤ 1⎞ 1 4 2⎥ 2 ⎢ ⎜ ⎟ = − ⎜ Δω + 2 ⎟ + C =d T T T ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ 2 1 2
With I =
1 2 1 Is = 2
P = N μ = N ρngμ gn + c. c .
2 NN o χ=− μ ng hε o
ε 2 E , μ ε μ −2 1 μ h T1T2 ,
(
)
* If I << Is , we are back to the results obtained from perturbation theory.
with •
1 ' Hng = − μ ng Ee −iωt + E*ei ωt 2 1 ' Hgn = − μ gn Ee−iω t + E*eiωt 2
neglect antiresonant terms.
[ [
] ]
•
ρng ~ e−iω t ρgn ~ e iω t
steady state
•
Assume initial population
There is no initial polarization,
(0) ρng = ρ(0) gn = 0 .
II.
Solve for
(1) ρng to first order for induced polarization.
P = N μ = N ρngμ gn + ρgn μ ng = N ρngμ gn + c. c. .
Nonlinear Optics
§ Linear Polarization (with density matrix)
• • I. solved by perturbation theory solved for
(1) ρng (for induced polarization) (0) (0) ρnn , ρgg .
∂ + iΩng . ∂t
(
)
( ( (
) ) )
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Nonlinear Optics
(1) ρng = − (ih )−1(−iω + iΩ ng )
−1
− (ih )−1 iω + i Ωng
2 − 1 μ ng E − iω t (0) = Ω ng − ω e ρgg − ρ(0) nn 2h * −1 μ ng E iω t (0) (0) + Ω ng + ω e ρgg − ρnn 2h .
(
) (
)
,
1 / T2 Nμ (0) χ = ρ(0) − ρ nn 2 2 ε oh gg ω ng − ω + (1 / T2 )
"
2
(
) (). NhomakorabeaIn the limit of
χ << 1 ,
1/ 2
n = (1 + χ )
≅ 1+
= n' + in" .
eikz = e
i nω z c
1 1 χ = 1 + χ + iχ " 2 2
III. Consider steady state response,
(
) (
)
(1) = hΩ ρ + H ′ ρ (0) − ρ (0) H ′ & ng ih ρ nn ng ng ng ng gg
(1) + iΩ ρ = (ih )−1 H ′ ⎛ 0) (0)⎞ & ng ρ ⎜ρ( gg − ρ nn ⎟ ng ng ng
(
)
−1 1
1 (0) (0) μ ng Ee−iω t ρgg − ρnn 2
(0) μ ng E*e iωt ρ(0) gg − ρnn
(
)
(
)
(
)
(
)
.
(
)
(
)
Neglect anti-resonant term, recall Ωng ≡
(1) −1 ρng =( ω ng − ω − iT2 )
∑ f ng = 1)
n
χ=
Nf nge 2 2mω ng εo
(ω ng − ω − iT2−1) .
−1
Saturation in Two Level Atoms
If population disturbance is too large, we can not apply perturbation theory , ⇒ consider saturation.
⎝ ⎠
(1) + iΩ ρ = −(ih )−1 μ ⎡ Ee − iωt + E*eiωt ⎤⎛ 0) − ρ (0)⎞ & ng ρ ⎜ρ( gg nn ⎟ ⎢ ⎥ ng ng ng ⎝ ⎠ ⎣ ⎦ 2
1
(1) ρng oscillates at both ω and -ω frequencies.
−1 ω ng − iT2 ,
(0) (0) − ρnn (ρgg ).
−1 μ ng E − iω t
2h
⎡
e
Recall P = N
(
(1) ρng μ gn + c. c . = ⎢ ε oχ Ee−iω t + c. c.⎥ . ⎣ ⎦
)
1 2
⎤
χ=
N −1 μ ngμ gn ω ng − ω − iT2 εo h
N ≡ ρnn − ρgg