lly, one refers to the ccw case as representing right-circular polarization and the CW case as representing left-circular polarization.16 As in the linear-polarization case, we may view the two circular polarizations as representing two orthogonal states (orthogonality to be shown 16 This convention is easy to remember when looking at one's right or left hands, the thumb pointing up and the fingers bent. With the right hand, the fingers indicate a ccw direction, while with the left hand, they indicate a cw direction. 25.8 Electromagnetic waves, polarization states, photons, and quantum measurements hereafter). Occasionally, I shall designate the right- and left-circular polarization states by the symbols U and n, respectively. Back to complex notation, we may represent the circularly polarized E-fields according to the decomposition: (25.13) where Q = 84/2 is an arbitrary phase factor. For circularly polarized E-fields we have (25.14) The decomposition in Eq. (25.13) is immediately reminiscent of a quantum state super- position! Indeed, if we define the second term in the decomposition as + |1), the super- position is equal to |0) + |1) = 21+) in the upper case (+, or cw, or left-circular polarization, or n), and to |0) - in the lower case (-, or ccw, or right- circular polarization, or U). Within a normalization factor of 1/2, we see that the two circularly polarized states n, U are equivalent to the quantum states or qubits |+), |->. We can now check the states orthogonality by effecting the (Hilbert-space) scalar product: E+ )-(i]o (25.15) which is the expected result, just as |+), |-> are orthogonal states. It is left as an easy exercise to show that the superposition of two circularly polarized E-fields having oppo- site directions (i.e., E++E-) yield linearly polarized E-fields. This result is expected, since within a 1/2 normalization factor, we know for a fact that |+) )+|-> = |0) and |+)-|-- = |1). Thus, circular polarization states can be viewed as the superposition of two linear polarization states (os