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For coherent EM wave propagating along the +z axis (Figure 07a), the two transversal electric fields can be expressed as : Ex = Ex0 cos( t - kz + x) ---------- (53a)Ey = Ey0 cos( t - kz + y) ---------- (53b)where Ex0, Ey0 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 |
Ex = Ex0 cos( t) ---------- (53c)Ey = Ey0 cos( t - ) ---------- (53d) |
= tan-1(Ey/Ex) = tan-1{(Ey0/Ex0)[cos(
) + tan(
t)sin(
)]} ---------- (53e)
as shown in Table 01 below.Phase Angle ![]() |
tan( ) |
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Polarization |
|---|---|---|---|
| 0o | 0 | 0o | Linear along x-axis |
| 0o | ![]() |
90o | Linear along y-axis |
| 0o | 1 | 45o | Linear at 45o |
| 45o | (1/2)1/2(Ey0/Ex0)[1 + tan( t)] |
Right-handed rotation | Elliptical or circular |
| 90o | (Ey0/Ex0)[tan( t)] |
Right-handed rotation | Elliptical |
| 90o | tan( t) |
t |
Right-handed Circular for Ey0=Ex0 |
| 180o | 0 | 0o | Linear along x-axis |
| 180o | -![]() |
-90o | Linear along y-axis |
| 180o | -1 | -45o | Linear at -45o |
| -90o | -(Ey0/Ex0)[tan( t)] |
Left-handed rotation | Elliptical |
| -90o | -tan( t) |
- t |
Left-handed Circular for Ey0=Ex0 |
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Table 01 shows that for linear polarization the angle is constant with Ex and Ey 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 | 0o | 1.0 | 0o | Linear at 45o |
| 1.0 | 0o | 0.0 | 0o | Linear along x-axis |
| 0.0 | 0o | 1.0 | 0o | Linear along y-axis |
| 1.0 | 45o | 0.0 | 0o | Linear along x-axis |
| 1.0 | 45o | 1.0 | 0o | Right-handed elliptical |
| 0.5 | 0o | 1.0 | 0o | Linear at 63.4o |
| 0.5 | 30o | 1.0 | 30o | Linear at 63.4o |
| 1.0 | 0o | 1.0 | 90o | Left-handed circular |
| 0.2 | 0o | 1.0 | 90o | Left-handed elliptical |
= 0, the direction of oscillation
depends on the ratio Ey0/Ex0. Let's take Ey0=Ex0 to simplify the formulas, then the polarization vector can be expressed as :
|ex> + sin
|ey> ---------- (53f)
and |ey> =
are the unit vector in the x (upper component) and y (lower component) directions respectively. It is similar to the Jones vector in classical electrodynamics and the polarization vectors in Eq.(52). Since the unit vectors satisfies the orthogonality relation : < ei|ej> =
(i-j) for (i, j) = (x or y), < e|e> = 1. If it is interpreted as the total probability, then cos2
and sin2
can be interpreted as the probability in polarization state |ex> or |ey> respectively.
= 90o or -90o,
= +
t (right-handed circular polarization) or -
t (left-handed circular polarization). The polarization vector in Eq.(53f) can be expressed in the form :
t/
) |eL> + (ei
t/
) |eR> ---------- (53g)
=
/
---------- (53h)
=
/
---------- (53i)
---------- (53j)
---------- (53k)
(i-j) for (i, j) = (L or R), < e|e> = 1. If it is interpreted as the total probability, then
there are 50% probability for each of the circular polarization state in Eq.(53g). In order to obtain a definite circular polarization, we have to pick either +
t or -
t but not both. A linear polarization will be produced if we keep both +
t and -
t in Eq.(53g). For example, the purely right-handed rotating polarization vector is described by :
t |eR> ---------- (53l)
[E(r,t)
B(r,t)]/(4
c) ---------- (53m)
U/|
| ---------- (53n)
) is energy of the wave per unit volume, the + sign is for right-handed circular polarization, the - sign for the left-handed one. An experiment back in 1936 had demonstrated conclusively that circular electromagnetic wave does carry angular momentum.
/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 :
---------- (53o)![]() |
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] |