Lecture 17 Intensity-dependent refractive index Optical Kerr effect. 1 1 Third-order nonlinear optics 2 2 1 NL polarization density, time domain NL polarization. Scalar version; no dispersioion of � " " " " � � = �&� � (�) Time-domain, sum Time-domain NL of all fields polarization density Consider � " to be containing constant for now components at different frequencies 3 3 Third harmonic generation 4 4 2 Third harmonic generation (THG) Fourier component 1 Assume that the input field has only one frequency component � �(�) = � �012 + �. �. 2 . And assume a scalar version of the 3-wave interaction (the input field is �.=�.(�) along x-axis) " (") � = �>>>> NL polarization (scalar version) � " � = � � " �"(�) in time domain: & � " � = � � " (. � �012 + �. �. )"= & 6 . 1 = � � " � �012 + �∗�9012 � �012 + �∗�9012 � �012 + �∗�9012 = 8 & . = . � � " [�"�"012 + 3(� �∗)� �012 +�. �. ] = : & . (17.1) Third Self phase harmonic modulation (SPM) term 5 5 Third harmonic generation (THG) pick only components with ±3� → . � � " [�"�"012 +�. �. ] : & . As usual, we now look " 1 0"12 for a polarization density � � @ = � 3� � + �. �. 3� 2 component at a specific frequency 3� Hence we get: 1 � 3� = � � " �"(�) (17.2) 4 & . - NL polarization at third harmonic 3� - will come back to this topic later. 6 6 3 Intensity- dependent refraction or Self phase modulation (SPM) 7 7 Intensity- dependent refraction Fourier component 1 Assume that the input field has only one frequency component � �(�) = � �012 + �. �. 2 . And assume a scalar version of the 3-wave interaction (all fields are �.=�.(�) along x-axis) NL polarization (scalar version) � " � = � � " �"(�) " (") in time domain: & � = �>>>> 1 � " � = � � " ( � �012 + �. �. )"= & 2 . 1 = � � " � �012 + �∗�9012 � �012 + �∗�9012 � �012 + �∗�9012 = 8 & . = . � � " [�"�"012 + 3(� �∗)� �012 +�. �. ] : & . " " 6 012 pick only components with ±� → � � (|� | � � +�. �. ) : & . 8 8 4 Intensity- dependent refraction Fourier component 3 " 6 � � = � � |� � | � � (16.6) à 4 & . What does this new � � = �EF(�) term mean ? Racall relation between polarization density P and the electric field : � = �&�� � � is a term that can be described by �(�) = ∆� = ∆(�& ��.) = �&∆��. 3 " " 6 " 6 Hence � � |� | � = ∆� = ∆(� �� ) = � ∆�� à ∆� = �&� |�.| K & . & . & . 4 6 on the other hand, ∆� = ∆ �& − 1 = 2�&∆� �& - linear ref. index 3 " 6 refractive index change induced by high intensity à ∆� = � |�.| 8�& . Recall that the intensity � = �� � |� |6 6 & & . " Hence 3 " 2� 3� (17.3) ∆� = � = 6 � 8�& ��&�& 4�&�&� 9 9 Intensity- dependent refraction The ususal way of defining the intensity-dependent refractive index is by means of the equation: linear ref. index intensity dependent ref. index (17.4) � = �& + �6� ∆� (3) Hence we find that �6 is related to χ by 3� " (17.5) �6 = 6 4�&�&� 10 10 5 Cross phase modulation (XPM): two waves with || polarization 11 11 Intensity- dependent refraction: two waves with || polarization Co-polarized beams Now we use two separate beams, where the presence of the strong beam of 1 1 amplitude E (ω) leads to a modification of the refractive index experienced by � � = � �01L2 + � �012 + �. �. 2 2 . 2 6 a weak probe wave of amplitude E1(ω′). As before, we assume a scalar version of the 3-wave interaction (all fields are along x-axis) � " sum of all fields " (") �′ NL polarization: " " " � = �>>>> � � = �&� � (�) � now there are two fields 1 L 1 � " � = � � " ( � �01 2 + � �012 + �. �. )"= & 2 . 2 6 1 " 01M2 ∗ 901M2 012 ∗ 9012 = � � � � + � � +� � + � � total 64 terms 8 & . 6 6 01M2 ∗ 901M2 012 ∗ 9012 ×(�.� + �. � +�6 � + �6� ) 01M2 ∗ 901M2 012 ∗ 9012 ×(�.� + �. � +�6 � + �6� ) pick only components with ±�′ 12 12 6 Intensity- dependent refraction: two waves with || polarization . L " L " . L � � " 6 (� �∗ � �01 2 +�. �. } = � � " (|� |6� �01 2 + �. �. ) = � � " (|� |6� �01 2 +�. �. ) à : & 6 6 . K & 6 . 6 & 6 6 . 1 Look for the nonlinear polarization component at the angular frequency ±�′ in the form � " (�) = �(�′)�012 + �. �. 2 the Fourier component 3 " 6 " (") �(�′) = � � |� | � (17.6) � = �>>>> à 2 & 6 . Hence we find that � in � = � + � � representation, is given by 6 & 6 � " 3� " �′ (17.7) �6 = 6 � 2�&�&� Hence, a strong wave affects the refractive index of a weak wave twice as much as it affects its own refractive index. In fact, a strong wave can be of the same frequency, as soon it is distinguishable from the weak wave (say by angle) 13 13 Cross phase modulation (XPM): two waves with ⊥ polarization 14 14 7 Cross-nonlinear refraction: two waves with ⊥ polarization This is a case commonly encountered in fibers. Now we use two separate beams with diffferent frequencies and orthogonal polarizations: a weak probe wave of amplitude E1(ω′) alond x-axis and a strong beam of amplitude E2(ω) alond y-axis. Now, we us a tensor version of the 3-wave interaction. y Look at cross-nonlinear refraction: how a strong beam with the orthogonal polarization E2(ω) changes ref. index of the ‘probe’ beam x E1(ω′) use tensor form now: (") �0 = �& S �0TVW�T�V�W 1 T,V,W � = � �01L2 + �. �. > 2 . only cross components 1 � = � �012 + �. �. X 2 6 (") 6 6 6 �> = �&{�>>XX�>�X + �>XX> �>�X + �>X>X �>�X } = 1 M 1 1 = � {� + � + � } � �01 2 + �. �. � �012 + �. �. � �012 + �. �. & >>XX >XX> >X>X 2 . 2 6 2 6 pick only components with ±�′ 15 15 Cross-nonlinear refraction: two waves with ⊥ polarization 1 L à � {� + � + � } 2 (� �∗ � �01 2 + �. �. } = 8 & >>XX >XX> >X>X 6 6 . = � {� + � + � }(|� |6� �012 +�. �. ) K & >>XX >XX> >X>X 6 . = � � (|� |6� �012 +�. �. ) K & >>>> 6 . for Kleinman symmetry: �>>XX + �>XX> + �>X>X= �>>>> In the Kleinman limit all the cross-polarization terms are equal in isotropic media Finally we get the Fourier component: 1 à �(�′) = � � " |� |6� & 6 . (17.8) " (") 2 � = �>>>> � " (17.9) And the main result: �6 = 6 2�&�&� This is how a strong beam with the orthogonal polarization changes ref. index of the ‘probe’ beam 16 16 8 Cross-nonlinear refraction: summary � = �& + �6� Relative NL ref. index n 2 NL ref. index Self phase modulation (SPM) 3� " 1 �6 = 6 4�&�&� Cross phase modulation (XPM): 3� " two waves with || polarization �6 = 6 2 2�&�&� � " Cross phase modulation (XPM): � = 2 6 2�6� � two waves with ⊥ polarization & & 3 17 17 Cross-nonlinear refraction: summary Hence for the probe beam: ∆� = �6(�. + 2�6) (17.10a) probe beam pump beam, co-polarized 2 (17.10b) ∆� = � (� + � ) 6 . 3 6 probe beam pump beam, orthogonally polarized 3� " where �6 = 6 - self phase modulation ceff. 4�&�&� 18 18 9 Third-order nonlinear optical coefficients of various materials 19 19 Third-order nonlinear optical coefficients of various materials What is the intensity needed to get π phase shift for L=1m fiber at λ=1 µm? Optical fiber: Fused silica (SiO2) −16 2 n2=3.2×10 cm /W 6] ∆� = Δ �� = �Δ� = � ^ . ^ Δ� = = 5 109b 6 F ∆� = ��6 I=5x10-7 / 3.2×10−16=1.56 GW/cm2 for (10µm)2 core, this corresponds to the power of 1.56 kW - can be achieved with pulsed lasers Optical fiber: As2S3 glass −13 2 n2=2.0×10 cm /W - need the power of 2.5 W only 20 20 10 DC Kerr effect 21 21 Kerr electro-optic effect with a DC field Fourier component 1 Assume that we have an optical field ��{� �012} in the presence of a �(�) = � �012 + �. �. 2 . DC field �ef such that |�.|<<|�ef| �.=�.(�) NL polarization (scalar version) " " " " (") in time domain: � � = �&� � (�) � = �>>>> frequency set: � + � + � " " 1 012 1 ∗ 9012 " h i j � � = �&� ( �.� + �. � + �ef) = 2 2 = -ω, ω, 0 1 �, �, � = � � " 3 �6 � �012 + �. � + ⋯ 2 & ef . ω+ω+ω=3ω 1 -ω-ω-ω=-3ω 1 ω+ω+0=2ω 3 -ω-ω+0=-2ω 3 ω+0+0=ω 3 -ω+0+0=-ω 3 components with ω-ω+ω=ω 3 -ω+ω-ω=-ω 3 components with ±� 0, ±2�, ±3� ω-ω+0=0 3 ω-ω+0=0 3 0+0+0=0 1 total 27 22 22 11 Kerr electro-optic effect with a DC field pick only components with ±� 3 � " � = � � " �6 � �012 + �. � 2 & ef . " 6 → � � = 3�&� �ef�. � Liquid e.g. +V " 6 nitrobenzene ∆� = 3�&� |�.| C6H5NO2 6 ∆� = ∆ �& − 1 = 2�&∆� 3 " 6 (17.11) ∆� = � �ef 2�& pol. Kerr effect – the quadratic electro-optic effect – a change in the refractive index of a material in response to an applied electric field. - not the same as linear EO effect in crystals! 23 23 12.