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
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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)
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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
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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 ,
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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 },
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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
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______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 .
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______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.
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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 ). Then the emitted photon which is right circularly polarized (e+), can propagate along the z axis, since M ε Dfi and
e e 1. A photon with right-hand circular polarization carries a spin +ħ in the z direction (the propagation direction).
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k +Ñ
e
D
Fig. The case of m' m 1 (right circularly polarization, ). A right circularly
polarized photon (e+) propagates with a wavevector k in the z direction. Note that
the electric field is denoted by cos(kz t)ex sin(kz t)e y . This electric field rotates in clock-wise sense with time t, and rotates in counter clock-wise sense with z (as the wave propagates forward). The corresponding spin of the photon is directed in the positive z direction (ħ). D ( e ) . ε( e ) . ( e e 1,e e 0 ). f i
(iii) 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 ,
is parallel to the right circularly polarization vector e . The emitted photon with left circularly polarization ( e ) can propagate along the z axis, since M ε Dfi and e e 1. A photon with the left-hand polarization carries a spin (-ħ), that is, a spin direction opposite to the z direction.
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k -Ñ
e
D
Fig. The case of m' m 1 (left circularly polarization, _ ). A left circularly
polarized photon (e-) propagates with a wavevector k in the z direction. Note that
the electric field is given by cos(kz t)ex sin(kz t)e y . This electric field rotates in counter clock-wise sense with time t, and rotates in clock-wise sense with z (as the wave propagates forward). The corresponding spin of the photon is directed in the negative z direction, as -ħ. D ( e ) . ε( e ) . ( e e 1 , f i
e e 0 ).
The rules on m can be understood by realizing that and _ circularly polarized photons carry angular momenta of + ans - , respectively, along the z axis, and hence m must change by one unit to conserve angular momentum. For linearly polarized light along the z axis, the photons carry no z-component of momentum, implying m 0 , while x or y-polarized light can be considered as a equal combination of and _ photons, giving m 1.
______REFERENCES G. Baym, Lectures on Quantum Mechanics (Westview Press, 1990). D. Bohm, Quantum Theory (Prentice Hall, New York, 1951). H. Lüth, Quantum Physics in the Nanoworld (Springer, 2013). D. Park, Introduction to the Quantum Theory, 2nd-edition (McGraw-Hill, New York, 1974). E. Hecht Optics, 4-th edition (Addison Wesley, 2002).
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G. Grynberg, A. Aspect, and C. Fabre, Introduction to Quantum Optics (Cambridge University Press, 2010). J. Schwinger, Quantum Mechanics: Symbolism of Atomic Measurements (Springer 2001). F. Yang and J.H. Hamilton, Modern Atomic and Nuclear Physics (McGraw-Hill, 1996). J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics, second edition (Addison-Wesley, 2011). D.L. Andrews and M. Babiker edited, The Angular Momentum of Light (Cambridge University Press, 2013).
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