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1 Selection Rule for Electric Dipole Transition Masatsugu Sei Selection rule for electric dipole transition Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: December 29, 2014) Electric dipole transition is the dominant effect of an interaction of an electron in an atom with the electromagnetic field. The transition dipole moment or transition moment, usually denoted l',m' rˆ l,m for a transition between an initial state, l,m , and a final state, l,m , is the electric dipole moment associated with the transition between the two states. The emission of photon occurs as a result of the transition. The polarization of the photon (linearly polarized photon, the right-hand circularly polarized photon, and the left-hand circularly polarized photon) is dependent on the detail of the transitions. The angular momentum of the photon can be derived from the conservation of the total angular momentum. Here we discuss the selection rule of the electric dipole transition. We need to discuss first the properties of the commutation relations for the angular momentum (the orbital angular momentum Lˆ , the spin angular momentum Sˆ , the total angular momentum Jˆ ) for the electrons in an atom. 1. Orbital angular momentum of electron The orbital angular momentum Lˆ of the electron is defined as Lˆ rˆ pˆ , or ˆ Lz xˆpˆ y yˆpˆ x , ˆ Lx yˆpˆ z zˆpˆ y , ˆ Ly zˆpˆ x xˆpˆ z . The following commutation relations are valid: ˆ ˆ ˆ L L iL , or 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ [Ly , Lz ] iLx , [Lz , Lx ] iLy , [Lx , Ly ] iLz . We also have the commutation relations, ˆ [Lz , zˆ] [xˆpˆ y yˆpˆ x , zˆ] 0 , ˆ [Lz , xˆ] [xˆpˆ y yˆpˆ x , xˆ] [yˆpˆ x , xˆ] yˆ[ pˆ x , xˆ] iyˆ , ˆ [Lz , yˆ] [xˆpˆ y yˆpˆ x , yˆ] [xˆpˆ y , yˆ] ixˆ , or ˆ ˆ ˆ [Lz , xˆ iyˆ] [Lz xˆ] i[Lz , yˆ] iyˆ i(ixˆ) (xˆ iyˆ) , ˆ ˆ ˆ [Lz , xˆ iyˆ] [Lz xˆ] i[Lz , yˆ] iyˆ i(ixˆ) (xˆ iyˆ) . We also note that ˆ2 ˆ2 2 ˆ2 [L ,[L , xˆ]] 2 {xˆ, L }, (1) where {Aˆ, Bˆ} AˆBˆ BˆAˆ , (anti-commutation) and ˆ2 ˆ 2 ˆ 2 ˆ 2 ˆ 2 L Lx Ly Lx Lz . ((Mathematica)) The proof of the above commutation relation of the orbital angular momentum (Eq.(1)) is given by Mathematica. (Proof) 2 Clear "Global`" ; ux 1, 0, 0 ;uy 0, 1, 0 ; uz 0, 0,1 ; r x, y, z ; Lx : — ux. Cross r, Grad , x, y, z & Simplify; Ly : — uy. Cross r, Grad , x, y, z & Simplify; Lz : — uz. Cross r, Grad , x, y, z & Simplify; Lsq : Lx Lx Ly Ly Lz Lz & ; eq2 Lsq Lsq x x, y, z Lsq x Lsq x, y, z Lsq x Lsq x, y, z x Lsq Lsq x, y, z FullSimplify; eq3 2 —2 x Lsq x, y, z Lsq x x, y, z FullSimplify; eq2 eq3 Simplify 0 ______________________________________________________________________________ 2. Eigenkets of orbital angular momentum for electron ˆ2 ˆ l,m is the simultaneous eigenstate of L and Lz , where ˆ2 ˆ [L , Lz ] 0 where ˆ2 2 L l,m l(l 1) l,m , ˆ Lz l,m m l,m , where l is an integer (l = 0, 1, 2, 3…), m = l, l-1, l-2,…, -l. We note that 3 ˆ L l,m (l m)(l m 1) l,m 1 , ˆ L l,m (l m)(l m 1) l,m 1 , where ˆ ˆ ˆ ˆ ˆ ˆ L Lx iLy , L Lx iLy . 3. Selection rule-I Using the relation ˆ Lz l,m m l,m , we have ˆ l',m'[Lz , zˆ] l,m 0, or ˆ ˆ l',m' Lz zˆ zˆLz l,m 0 , or (m'm) l',m' zˆ l,m 0 . l',m' zˆ l,m 0, only if m' = m. 4. Selection rule-II Using the relation ˆ Lz l,m m l,m , we have ˆ l',m' [Lz , xˆ iyˆ] l,m l',m' xˆ iyˆ l,m , 4 or ˆ ˆ l',m' Lz (xˆ iyˆ) (xˆ iyˆ)Lz l,m l',m' xˆ iyˆ l,m , or (m'm 1) l',m' xˆ iyˆ l,m 0 . or l',m' xˆ iyˆ l,m 0 , only if m' m 1. 5. Selection rule-III Using the relation ˆ Lz l,m m l,m , we have ˆ l',m' [Lz , xˆ iyˆ] l,m l',m' xˆ iyˆ l,m , or ˆ ˆ l',m' Lz (xˆ iyˆ) (xˆ iyˆ)Lz l,m l',m' xˆ iyˆ l,m , or (m'm 1) l',m' xˆ iyˆ l,m 0 . or l',m' xˆ iyˆ l,m 0 , only if m' m 1. 6. Selection rule-IV Using the commutation relation ˆ2 ˆ2 2 ˆ2 [L ,[L , xˆ]] 2 {xˆ, L }, 5 we get the following equation, ˆ2 ˆ2 2 ˆ2 l',m' [L ,[L , xˆ]] l,m 2 l',m' {xˆ, L } l,m , or ˆ2 ˆ2 ˆ2 ˆ2 ˆ2 ˆ2 2 ˆ2 ˆ2 l',m' L L xˆ 2L xˆL xˆL L l,m 2 l',m' xˆL L xˆ l,m . Here we use the relation ˆ2 2 ˆ2 2 L l,m l(l 1) l,m , and l,m L l(l 1) l,m . Then we have 4 2 2 2 2 [l' (l'1) 2l'(l'1)l(l 1) l (l 1) 2l'(l'1) 2l(l 1)] l',m' xˆ l,m 0, or (l'l 1)(l'l 1)(l'l)(l'l 2) l',m' xˆ l,m 0 , The last factor yields the selection rule l' l 1. This means that no transition occurs if l' l . ______________________________________________________________________________ ((Mathemtica)) Proof of the identity: [l'2 (l'1)2 2l'(l'1)l(l 1) l 2 (l 1)2 2l'(l'1) 2l(l 1)] (l'l 1)(l'l 1)(l'l)(l'l 2) g1 a2 a 1 2 2 a a 1 b b 1 b 2 b 1 2 2 a a 1 2 b b 1 Factor 1 a b 1 a b a b 2 a b 6 ______________________________________________________________________________ Since l’ and l are both non-negative, the (l'l 2) term cannot vanish, and the (l'l ) term can only vanish for l’ = l = 0. However, this selection rule cannot be satisfied, since the states with l’ = l = 0 are independent of direction, and therefore these matrix elements of xˆ vanish. Formally, one easily shows this l 0,m 0 xˆ l 0,m 0 0 , using the property of the parity operator ˆ . ((Proof)) ˆxˆˆ xˆ , (property of ˆ ) where the parity operator satisfies the relations, ˆ ˆ , ˆ 2 1ˆ , 0,0 ˆxˆˆ 0,0 0,0 xˆ 0,0 , (1) We note that ˆ l,m (1)l l,m , In other words, 0,0 has the even parity, ˆ 0,0 0,0 , 0,0 ˆ 0,0 . Then we get 0,0 ˆxˆˆ 0,0 0,0 xˆ 0,0 . (2) From Eqs.(1) and (2), we get 0,0 xˆ 0,0 0 . 7 ______________________________________________________________________________ 7. Dipole selection rule The dipole radiation is emitted if M f ε rˆ i ε f rˆ i ε Dfi , does not vanish, where ε is the electric field (the polarization vector), and Dfi f rˆ i f xˆ i ex f yˆ i ey f zˆ i ez . We assume that the initial state i l,m and the final state f l',m' . Then we have Dfi l',m' xˆ l,m ex l',m' yˆ l,m ey l',m' zˆ l,m ez . or D fi l',m' xˆex yˆey zˆez l,m l',m' xˆex yˆey l,m l',m' zˆ l,m ez l',m' rˆe rˆe l,m l',m' zˆ l,m ez e l',m' rˆ l,m e l',m' rˆ l,m l',m' zˆ l,m ez where e ie e ie e x y , e x y . (Jones vector notation) 2 2 or 1 1 e (e e ) , e (e e ) . x 2 y 2i and xˆex yˆe y rˆe rˆe . Note that 8 e e 0 . e e 0 , e e 1. e e 1. (i) For m' m , l' l 1 l',m' zˆ l,m 0, l',m' rˆ l,m 0 , l',m' rˆ l,m 0 So we have Dfi l',m' zˆ l,m ez D fi is directed along the z axis. (a) Suppose that the wavevector k of the emitted photon is along the z axis. There is no radiation in the z-direction since the polarization vector is perpendicular to D fi (the z axis). (b) For example, we consider light going in the x direction. It can have two directions of polarization, either in the z or in the y direction. A transition in which m = 0, can produce only light which is linearly polarized in the z direction. 9 z Dfi e k x Fig. m' = m. D fi // z . The light propagating along the x direction. It is a linearly polarized wave (along the z axis). We now consider the matrix element with m' m 1. (ii) For m' m 1, and l' l 1 l',m' rˆ l,m 0 , l',m' rˆ l,m 0 , l',m' zˆ l,m 0. Dfi l',m' xˆex yˆey l,m e l',m' rˆ l,m has the same direction of the left circularly polarization vector (e ).
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