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For coherent EM wave propagating along the +z axis (Figure 07a), the two transversal electric fields can be expressed as : E _{x} = E_{x0} cos(t - kz + _{x}) ---------- (53a)E _{y} = E_{y0} cos(t - kz + _{y}) ---------- (53b)where E _{x0}, E_{y0} are amplitudes, and _{x}, _{y} are the phase angles of the x, y components respectively (Figure 07b). These equations can be simplified somewhat if the view is facing the x-y plane fixed at z=0 (such as the view in Figure 07c), and by using the relative phase angle = _{x} - _{y} : | ||

## Figure 07a Polari-zation |
## Figure 07b Phase Angles |
E_{x} = E_{x0} cos(t) ---------- (53c)E _{y} = E_{y0} cos(t - ) ---------- (53d) |

= tan

the type of polarization can be readily discerned by the values of E

Phase Angle | tan() | Polarization | |
---|---|---|---|

0^{o} |
0 | 0^{o} |
Linear along x-axis |

0^{o} |
90^{o} |
Linear along y-axis | |

0^{o} |
1 | 45^{o} |
Linear at 45^{o} |

45^{o} |
(1/2)^{1/2}(E_{y0}/E_{x0})[1 + tan(t)] |
Right-handed rotation | Elliptical or circular |

90^{o} |
(E_{y0}/E_{x0})[tan(t)] |
Right-handed rotation | Elliptical |

90^{o} |
tan(t) | t | Right-handed Circular for E_{y0}=E_{x0} |

180^{o} |
0 | 0^{o} |
Linear along x-axis |

180^{o} |
- | -90^{o} |
Linear along y-axis |

180^{o} |
-1 | -45^{o} |
Linear at -45^{o} |

-90^{o} |
-(E_{y0}/E_{x0})[tan(t)] |
Left-handed rotation | Elliptical |

-90^{o} |
-tan(t) | -t | Left-handed Circular for E_{y0}=E_{x0} |

Table 01 shows that for linear polarization the angle is constant with E_{x} and E_{y} varying in such a way that its ratio is still a constant. For circular polarization the length of the field |E| is constant, but changes with a frequency determined by . For elliptical polarization even |E| is not a constant. Another way to check out the polarization is to view the animation provided by Amanogawa in Figure 07c. Table 02 shows the types of polarization associated with different sets of parameters in the application. | |

## Figure 07c Polarization of EM Wave [see animation] |

Amplitude x | Phase x | Amplitude y | Phase y | Polarization |
---|---|---|---|---|

1.0 | 0^{o} |
1.0 | 0^{o} |
Linear at 45^{o} |

1.0 | 0^{o} |
0.0 | 0^{o} |
Linear along x-axis |

0.0 | 0^{o} |
1.0 | 0^{o} |
Linear along y-axis |

1.0 | 45^{o} |
0.0 | 0^{o} |
Linear along x-axis |

1.0 | 45^{o} |
1.0 | 0^{o} |
Right-handed elliptical |

0.5 | 0^{o} |
1.0 | 0^{o} |
Linear at 63.4^{o} |

0.5 | 30^{o} |
1.0 | 30^{o} |
Linear at 63.4^{o} |

1.0 | 0^{o} |
1.0 | 90^{o} |
Left-handed circular |

0.2 | 0^{o} |
1.0 | 90^{o} |
Left-handed elliptical |

|e> = cos |e

where |e

For = 90

|e> = (e

where

|e

|e

are the unit vectors for circular polarization to the left and right respectively. Thus, circular polarization can be expressed in terms of the combination of linear polarizations and vice versa as shown below :

|e

|e

Since the circular unit vectors again satisfies the orthogonality relation : < e

|e> = e

The angular momentum density of classical electromagnetic waves is :

where

where U=|E|

Transition to quantum theory is accomplished by expressing U=N/V, where N is the number of photons in volume V, and the energy has been quantized to . Thus, the angular momentum of a photon (N = 1 in unit volume V = 1) can be parallel or anti-parallel to the z-axis :

s

This formula shows that angular momentum of electromagnetic wave is related to the right- or left-handed polarization, which becomes photon spin of + or - respectively when goes over to quantum theory, i.e., the photon is spin 1 particle. It is said to have right-handed helicity if the spin aligns with the direction of motion (+), or left-handed helicity in opposite direction (-) as shown in Figure 07d. As usual in quantum theory, the photon would be in a superposition of both the right- and left-handed helicity states, it is only when a measurement is performed that it assumes a definite right- or left-handed helicity state (see Copenhagen Interpretation). | |

## Figure 07d Photon Helicity [view large image] |

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