Thanks again Prof. Hwang. Let me restate my question. From our physical observation, a linear polarizer performs a definite action on a stream of unpolarized light (which we will designate as lúr> state) by realigning the orientation of the photons along its polarizing axis. Hence, a source of non-polarized photons when operated upon by a vertical or v-state polarizer (Mv) will produce photons polarized along this polarizing axis or in the lv> eigen state. Quantum Mechanics express this operation and its projection probabilities as follows: l<v l Mv l úr>l 2= 1/2. If we pass this polarized light in the lv> state through another v-state polarizer (Mv), we will get 100%: l<vl Mv lv>l 2= 1. However, if we rotate the second polarizer by 90 degree and turn it into a horizontal or h-state operator (Mh), the projection probability will be zero: l<hl Mh lv>l 2= 0. The question I am raising is whether there is a distinct Right circularly polarized state or lR> eigen state and a Left circularly polarized state or lL> eigen state when light passes through a Right Circular Polarizer and a Left Circular polarizer respectively. Quantum Mechanics appears to treat the lR> state and the lL> state as somewhat similar to the lv> state and the lh> state as it expresses the projection probabilities when light in the lR> state passes through a Right Circular polarizer as: l<R l MR l R>l 2= 1 and through a Left circular polarizer as: l<Ll ML lR>l 2= 0. Is it correct for Quantum Mechanics to Make this prediction? I thank you in anticipation of your clarification.