Polarized light
First of all we would like to check on one idea. In Section 11-4 we showed that when RHC polarized light is viewed in a frame rotated by the angle $\phi$ about the $z$-axis7 it gets multiplied by $e^{i\phi}$. Does that mean then that the photons of light that are right circularly polarized carry an angular momentum of one unit8 along the $z$-axis?  Indeed it does. It also means that if we have a beam of light containing a large number of photons all circularly polarized the same way—as we would have in a classical beam—it will carry angular momentum. If the total energy carried by the beam in a certain time is $W$, then there are $N=W/\hbar\omega$ photons. Each one carries the angular momentum $\hbar$, so there is a total angular momentum of \begin{equation} \label{Eq:III:17:30} J_z=N\hbar=\frac{W}{\omega}. \end{equation}  Can we prove classically that light which is right circularly polarized carries an energy and angular momentum in proportion to $W/\omega$?  That should be a classical proposition if everything is right. Here we have a case where we can go from the quantum thing to the classical thing. We should see if the classical physics checks. It will give us an idea whether we have a right to call $m$ the angular momentum. Remember what right circularly polarized light is, classically. It’s described by an electric field with an oscillating $x$-component and an oscillating $y$-component $90^\circ$ out of phase so that the resultant electric vector $\Efieldvec$ goes in a circle—as drawn in Fig. 17–5(a). Now suppose that such light shines on a wall which is going to absorb it—or at least some of it—and consider an atom in the wall according to the classical physics. We have often described the motion of the electron in the atom as a harmonic oscillator which can be driven into oscillation by an external electric field. We’ll suppose that the atom is isotropic, so that it can oscillate equally well in the $x$- or $y$-directions. Then in the circularly polarized light, the $x$-displacement and the $y$-displacement are the same, but one is $90^\circ$ behind the other. The net result is that the electron moves in a circle, as shown in Fig. 17–5(b). The electron is displaced at some displacement $\FLPr$ from its equilibrium position at the origin and goes around with some phase lag with respect to the vector $\Efieldvec$. The relation between $\Efieldvec$ and $\FLPr$ might be as shown in Fig. 17–5(b). As time goes on, the electric field rotates and the displacement rotates with the same frequency, so their relative orientation stays the same. Now let’s look at the work being done on this electron. The rate that energy is being put into this electron is $v$, its velocity, times the component of $q\Efieldvec$ parallel to the velocity: \begin{equation} \label{Eq:III:17:31} \ddt{W}{t}=q\Efield_tv. \end{equation} But look, there is angular momentum being poured into this electron, because there is always a torque about the origin. The torque is $q\Efield_tr$, which must be equal to the rate of change of angular momentum $dJ_z/dt$: \begin{equation} \label{Eq:III:17:32} \ddt{J_z}{t}=q\Efield_tr. \end{equation} Remembering that $v=\omega r$, we have that \begin{equation*} \ddt{J_z}{W}=\frac{1}{\omega}. \end{equation*} Therefore, if we integrate the total angular momentum which is absorbed, it is proportional to the total energy—the constant of proportionality being $1/\omega$, which agrees with Eq. (17.30). Light does carry angular momentum—$1$ unit (times $\hbar$) if it is right circularly polarized along the $z$-axis, and $-1$ unit along the $z$-axis if it is left circularly polarized.  Now let’s ask the following question: If light is linearly polarized in the $x$-direction, what is its angular momentum? Light polarized in the $x$-direction can be represented as the superposition of RHC and LHC polarized light. Therefore, there is a certain amplitude that the angular momentum is $+\hbar$ and another amplitude that the angular momentum is $-\hbar$, so it doesn’t have a definite angular momentum. It has an amplitude to appear with $+\hbar$ and an equal amplitude to appear with $-\hbar$. The interference of these two amplitudes produces the linear polarization, but it has equal probabilities to appear with plus or minus one unit of angular momentum. Macroscopic measurements made on a beam of linearly polarized light will show that it carries zero angular momentum, because in a large number of photons there are nearly equal numbers of RHC and LHC photons contributing opposite amounts of angular momentum—the average angular momentum is zero. And in the classical theory you don’t find the angular momentum unless there is some circular polarization.  We have said that any spin-one particle can have three values of $J_z$, namely $+1,0,-1$ (the three states we saw in the Stern-Gerlach experiment). But light is screwy; it has only two states. It does not have the zero case. This strange lack is related to the fact that light cannot stand still. For a particle of spin $j$ which is standing still, there must be the $2j+1$ possible states with values of $j_z$ going in steps of $1$ from $-j$ to $+j$. But it turns out that for something of spin $j$ with zero mass only the states with the components $+j$ and $-j$ along the direction of motion exist. For example, light does not have three states, but only two—although a photon is still an object of spin one. How is this consistent with our earlier proofs—based on what happens under rotations in space—that for spin-one particles three states are necessary? For a particle at rest, rotations can be made about any axis without changing the momentum state. Particles with zero rest mass (like photons and neutrinos) cannot be at rest; only rotations about the axis along the direction of motion do not change the momentum state. Arguments about rotations around one axis only are insufficient to prove that three states are required, given that one of them varies as $e^{i\phi}$ under rotations by the angle $\phi$.9
 One further side remark. For a zero rest mass particle, in general, only one of the two spin states with respect to the line of motion ($+j$, $-j$) is really necessary. For neutrinos—which are spin one-half particles—only the states with the component of angular momentum opposite to the direction of motion $(-\hbar/2)$ exist in nature [and only along the motion $(+\hbar/2)$ for antineutrinos]. When a system has inversion symmetry (so that parity is conserved, as it is for light) both components ($+j$, and $-j$) are required. 