Copyright 2004 by Marcel Dekker, Inc. All Rights Reserved.

Using a similar procedure, one can obtain relative transition probabilitiesto/from the exciton state with F = 1:P U;L1¼ N U;L1sin 2 ðh lp Þ ð28ÞwhereN U 1 ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffipf 2 þ d f þffiffiffiffiffi !3dpffiffiffiffiffiffiffiffiffiffiffiffiffiKP 2 ;6 f 2 þ dN L 1 ¼ ð29Þpffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffi!f 2 þ d þ f 3dpffiffiffiffiffiffiffiffiffiffiffiffiffiKP 26 f 2 þ dThe excitation probability of the F = 1 state is equal to that of the F =1 state. As a result, the total transition probability to the doubly-degeneratejFj = 1 exciton states is equal to 2P 1 U,L .Equations (26) and (28) show that the F = 0 and jFj = 1 state excitationprobabilities for the linear polarized light differ in their dependence on theangle between the light polarization vector and the hexagonal axis of thecrystal. If the crystal hexagonal axis is aligned perpendicular to the lightdirection, only the active F = 0 state can be excited. Alternatively, when thecrystals are aligned along the light propagation direction, only the upper andlower jFj = 1 states will participate in the absorption. In the case of randomlyoriented NCs, polarized excitation resonant with one of these exciton statesselectively excites suitably oriented crystals, leading to polarized luminescence(polarization memory effect) [12]. This effect was experimentally observed inseveral studies [20,21]. Furthermore, large energy splitting between the F = 0and jFj = 1 states can lead to different Stokes shifts in the polarizedluminescence.The selection rules and the relative transition probabilities for circularlypolarized light are determined by the matrix element of the operator e F pˆb,where thepolarizationvector,e F = e x Fie y ,and themomentum, pˆF=pˆxFipˆy,ipˆy, lie in the plane perpendicular to the light propagation direction. In vectorrepresentation, this operator can be written ase F pˆb ¼ epFie ˆV pˆð30Þwhere e ? c, c is the unit vector parallel to the light propagation direction, andeV = (e c); as a result of the eV definition, the scalar product (eeV) = 0. Tocalculate the matrix element in Eq. (24), we expand the operator of Eq. (30) incoordinates that are connected with the direction of the hexagonal axis of theNCs (z direction):e F pˆb ¼ e F pˆ¼ e F zpˆz þ 1 2 ½eF þ pˆþ eF pˆþ Šð31ÞCopyright 2004 by Marcel Dekker, Inc. All Rights Reserved.