# L7 Second-order NLO materials; crystal classes ... ... Second-order خ§(2)nonlinear optical...

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Second-order Χ(2) nonlinear optical materials; crystal classes

and their symmetries and nonlinear optical tensors;

electrooptic effect ;

Lecture 7

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Anisotropic linear media

tensor vector vector

or

In nonmagnetic and transparent materials, 𝜒𝑖𝑗 = 𝜒𝑗𝑖, i.e. the 𝜒 tensor is real and symmetric.

It is possible to diagonalise the tensor by choosing the appropriate coordinate axes, leaving only 𝜒%%, 𝜒'' and 𝜒((. This gives :

Linear susceptibility is a tensor

In an anisotropic medium, such as a crystal, the polarisation field P is not necessarily aligned with the electric field of the light E. In a physical picture, this can be thought of as the dipoles induced in the medium by the electric field having certain preferred directions, related to the physical structure of the crystal.

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Anisotropic linear media

The refractive index is: 𝑛 = 1 + 𝜒

hence 𝑛%% = 1 + 𝜒%%

𝑛'' = 1 + 𝜒''

𝑛(( = 1 + 𝜒((

Waves with different polarizations will see different refractive indices and travel at different speeds.

This phenomenon is known as birefringence and occurs in some crystals such as calcite and quartz.

If 𝜒%% = 𝜒'' ≠ 𝜒((, the crystal is known as uniaxial.

If 𝜒%% ≠ 𝜒'' ≠ 𝜒(( the crystal is called biaxial.

or 𝜀 =

will come back to this in Lecture 8

z-axis or c-axis

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Anisotropic linear media n1=n2 n3 n1 n2 n3

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Nonlinear Susceptibility Tensor

Nonlinear susceptibility is a 3-rd rank tensor 𝜒./0 (2)

(3 x 3 x 3 tensor)

the nonlinear susceptibility tensor 𝜒./0 (2)

When all of the optical frequencies are detuned from the resonance frequencies of the optical medium

has full permutation symmetry:

𝜒./0 = 𝜒.0/= 𝜒/.0= 𝜒/0.= 𝜒0./= 𝜒0/.

Full permutation symmetry can be deduced from a consideration of the field energy density within a nonlinear medium (see Boyd or Stegeman books).

And also from quantum mechanics !

... and does not actually depend of frequencies that participate, so instead of writing we write simply 𝜒./0

(2)

Kleinman’s symmetry condition

𝜒./0 (2)(𝜔5 = 𝜔6+𝜔2, 𝜔6, 𝜔2)

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Nonlinear Susceptibility Tensor Consider mutual interaction of three waves at 𝜔1, 𝜔2, and 𝜔3 = 𝜔1 + 𝜔2

Assume E is the total field vector 𝑬 = 𝑬=6 + 𝑬=2 + 𝑬=5 Or in vector components 𝐸. = 𝐸.,=6 + 𝐸.,=2 + 𝐸.,=5 𝑖 = 1,2,3 (= 𝑥, 𝑦, 𝑧)

For example, 𝑥 −component of the nonlinear polarization P is:

𝑃6 = 𝜖EF /,0

𝜒6/0 𝐸/𝐸0 = 𝜖E { 𝜒666𝐸6𝐸6 + 𝜒662𝐸6𝐸2 + 𝜒665𝐸6𝐸5 +

+ 𝜒626𝐸2𝐸6 + 𝜒622𝐸2𝐸2 + 𝜒625𝐸2𝐸5 +

+ 𝜒656𝐸5𝐸6 + 𝜒652𝐸5𝐸2 + 𝜒655𝐸5𝐸5 }

𝑃2 = 𝜖EF /,0

𝜒2/0 𝐸/𝐸0 = …

𝑃5 = 𝜖EF /,0

𝜒5/0 𝐸/𝐸0 = …

x

y

z

𝑃. = 𝜖EF /,0

𝜒./0 𝐸/𝐸0 𝑖, 𝑗, 𝑘 = 1,2,3 (= 𝑥, 𝑦, 𝑧)then

E are real fields ~ 𝐸 cos(𝜔𝑡)

note that the terms underlined by the same color are equal

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Nonlinear Susceptibility Tensor

Now inroduce the tensor: 𝑑./0 = 1 2𝜒./0

(2) (historical convention !)

and write:

𝑃6 = 2𝜖E ( 𝑑666𝐸6𝐸6 + 𝑑622𝐸2𝐸2 + 𝑑655𝐸5𝐸5 + 2𝑑625𝐸2𝐸5 + 2𝑑665𝐸6𝐸5 + 2𝑑662𝐸6𝐸2)

𝑃2 = 2𝜖E ( 𝑑266𝐸6𝐸6 + 𝑑222𝐸2𝐸2 + 𝑑255𝐸5𝐸5 + 2𝑑225𝐸2𝐸5 + 2𝑑265𝐸6𝐸5 + 2𝑑262𝐸6𝐸2)

𝑃5 = 2𝜖E ( 𝑑566𝐸6𝐸6 + 𝑑522𝐸2𝐸2 + 𝑑555𝐸5𝐸5 + 2𝑑525𝐸2𝐸5 + 2𝑑565𝐸6𝐸5 + 2𝑑562𝐸6𝐸2)

now reduced to only 18 componnents

Reduce 3D tensor to 2D matrix

𝑗𝑘 → 𝑙

𝑗𝑘 = 11 22 33 23,32 31,13 12,21 𝑙 = 1 2 3 4 5 6

Second-order susceptibility described using contracted notation

Matrix with 6x3 components

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Nonlinear Susceptibility Tensor

𝑃% 𝑃' 𝑃(

=

𝐸%2

𝐸'2

𝐸(2 2𝐸'𝐸( 2𝐸%𝐸( 2𝐸%𝐸'

the way to find nonlinear polarizations in 3D case using contracted notation

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Nonlinear Susceptibility Tensor

By further applying Kleinman symmetry, we find that 𝑑.R matrix has only 10 independent elements

for example: d14 is d123 but d25 is d213 =d123 and d36 is d312 =d123

d14=d25=d36à etc.

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Nonlinear Susceptibility Tensor

Only 10 independent elements

à

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Nonlinear Susceptibility Tensor

Spatial symmetries of crystals further reduce the amound of independednt tensor elements

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Nonlinear Susceptibility Tensor

‘Cubic’ crystals e.g. GaAs, GaP, InAs, ZnS, ZnSe – are optically isotropic but nonlinear !

KTP

LiNbO3

GaSe

KDP

GaAszero nonlinearity

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Nonlinear Susceptibility Tensor

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Nonlinear Susceptibility Tensor

andKTP (KTiO2PO4) crystal

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Nonlinear Susceptibility Tensor

KTP (KTiO2PO4) crystal

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Nonlinear Susceptibility Tensor

Crystal of class 3m (e.g. Lithium Niobate, LiNbO3)

only d22 , d31 and d33

Class: 3m à other crystals of this class

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Physical origin of off-diagonal elements in dijk tensor after Stegeman NLO book

x

y z

oxygen oxygen

oxygen

oxygen

phosphorus

active electron

bondsbonds

E-field applied in xy

electron motion

KDP crystal

e-

motion in in xy

motion in z

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Physical origin of off-diagonal elements in dijk tensor

x

y z

As

Ga

active electron

bondsbonds

E-field applied in xy

electron motion

GaAs crystal

e-

motion in in xy

motion in z

As

As

As

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Nonlinear Susceptibility Tensor

Crystal of class 43𝑚 (e.g. Gallium Arsenide, GaAs)

only d14

GaAs

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Nonlinear Susceptibility Tensor

GaAs

=2𝜖E

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Nonlinear Susceptibility Tensor

𝑃% 𝑃' 𝑃(

=

𝐸%2

𝐸'2

𝐸(2 2𝐸'𝐸( 2𝐸%𝐸( 2𝐸%𝐸'

How to calculate effective nonlinearity ?

1) 𝐿𝑖𝑁𝑏𝑂3 𝑑55 ≠0

Simple case: SFG 𝜔3 = 𝜔1 + 𝜔2 all waves (𝜔1𝜔2𝜔3) are polarized along z-axis

𝐸( 𝑡 = 𝐸6,(𝑒.=Z[ + 𝐸2,(𝑒.=\[input fields

Need to find z-component of 𝑃(,=5 (]^) = 𝑃(𝜔5)𝑒.=_[

𝑃( ]^ (𝑡) = 2𝜖E𝑑55𝐸(2 = 2𝜖E𝑑55

6 ` (𝐸6,(𝑒

.=Z[ + 𝐸2,(𝑒.=\[ + c.c.)2= ...

2𝜖E𝑑bcc𝐸(2 𝑑bcc=𝑑55

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Nonlinear Susceptibility Tensor

2𝜖E𝑑55 6 `

(𝐸6,(𝑒.=Z[ + 𝐸2,(𝑒.=\[ + c.c.)2= ...

= 2𝜖E𝑑55 6 ` (𝐸6,(2 𝑒2.=Z[ + 𝐸2,(2 𝑒2.=\[ + 2𝐸6,(𝐸2,(𝑒.(=Zd=\)[ + 2𝐸6,(𝐸2,(∗ 𝑒.(=Zf=\)[ + c. c. ) + 2𝜖E𝑑55

6 2

(𝐸6,(𝐸6,(∗ +𝐸2,( 𝐸2,(∗ ) =...

𝑃( ]^ 𝑡 =

for SFG term: 𝑃 𝜔5 =𝑃 𝜔6 + 𝜔2 – pick only components with ± 𝜔6 + 𝜔2

𝑃( ]^ 𝑡 = 2𝜖E𝑑55

6 ` (2𝐸6,(𝐸2,(𝑒.(=Zd=\)[ +𝑐. 𝑐. ) = 𝜖E𝑑55(𝐸6,(𝐸2,(𝑒.(=Zd=\)[ +𝑐. 𝑐. )

1 2 (𝑃( 𝜔5 + 𝑐. 𝑐)

at 𝜔5

𝑃( 𝜔5 =𝜖E2𝑑55(𝐸6,(𝐸2,()

= 𝜖E𝑑55(𝐸6,(𝐸2,(𝑒.(=Zd=\)[ +𝑐. 𝑐)

real Fourier componnent 𝑃 = 𝑃( 𝜔5 cos(𝜔5t)

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Nonlinear Susceptibility Tensor

same as (5.2)–(5.3) with 𝑑 = 6 2 𝜒(2)

𝑃 𝜔6 + 𝜔2 = 2𝜖E𝑑55𝐸6𝐸2

𝑃 2𝜔6 = 𝜖E𝑑55𝐸62

𝑃 2𝜔2 = 𝜖E𝑑55𝐸22

once you know 𝑑bcc, you can treat fields as scalars

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Nonlinear Susceptibility Tensor

2) Effective nonlinearity, GaAs

focus on this component:

SFG 𝜔3 = 𝜔1 + 𝜔2

𝑬 𝑡 = 𝑬𝟏𝑒.=Z[ + 𝑬𝟐𝑒.=\[for input fields

Need to find 𝑃(𝜔5)

𝐸6 𝐸2 xy

𝑃=5 (]^) 𝐸5z

y

x

GaAs

beam k-vector: goes into the page (so-called 110 direction)

𝐸6 → 1 2 (𝐸6%+𝐸6')𝑒.=Z[

𝐸2 → 1 2 (𝐸2%+𝐸2')𝑒.=\[

𝑃𝑁𝐿(𝑡) = 𝑃((𝑡) = 𝜖E2𝑑6` 2𝐸%𝐸' = 4𝜖E𝑑6` 6 2 6 ` (𝐸6𝑒.=Z[ + 𝐸2𝑒.=\[ + 𝑐. 𝑐)2

𝐸% = 1 2 (𝐸6𝑒.=Z[ + 𝐸2𝑒.=\[)

𝐸' = 𝐸%

𝑃𝑁𝐿 𝑡, 𝜔5 = 𝜖E𝑑6` 6 2 (2𝐸6𝐸2𝑒.(=Zd=\)[ + 𝑐. 𝑐. ) = 𝜖E𝑑6`(𝐸6𝐸2𝑒.(=Zd=\)[ +

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