The polarization states of the photon
There are a number of other two-state systems which are interesting to study, and the first new one we would like to talk about is the photon. To describe a photon we must first give its vector momentum. For a free photon, the frequency is determined by the momentum, so we don’t have to say also what the frequency is. After that, though, we still have a property called the polarization. Imagine that there is a photon coming at you with a definite monochromatic frequency (which will be kept the same throughout all this discussion so that we don’t have a variety of momentum states). Then there are two directions of polarization. In the classical theory, light can be described as having an electric field which oscillates horizontally or an electric field which oscillates vertically (for instance); these two kinds of light are called $x$-polarized and $y$-polarized light. The light can also be polarized in some other direction, which can be made up from the superposition of a field in the $x$-direction and one in the $y$-direction. Or if you take the $x$- and the $y$-components out of phase by $90^\circ$, you get an electric field that rotates—the light is elliptically polarized. (This is just a quick reminder of the classical theory of polarized light that we studied in Chapter 33, Vol. I.)  Now, however, suppose we have a single photon—just one. There is no electric field that we can discuss in the same way. All we have is one photon. But a photon has to have the analog of the classical phenomena of polarization. There must be at least two different kinds of photons. At first, you might think there should be an infinite variety—after all, the electric vector can point in all sorts of directions. We can, however, describe the polarization of a photon as a two-state system. A photon can be in the state $\ket{x}$ or in the state $\ket{y}$. By $\ket{x}$ we mean the polarization state of each one of the photons in a beam of light which classically is $x$-polarized light. On the other hand, by $\ket{y}$ we mean the polarization state of each of the photons in a $y$-polarized beam. And we can take $\ket{x}$ and $\ket{y}$ as our base states of a photon of given momentum pointing at you—in what we will call the $z$-direction. So there are two base states $\ket{x}$ and $\ket{y}$, and they are all that are needed to describe any photon at all.  For example, if we have a piece of polaroid set with its axis to pass light polarized in what we call the $x$-direction, and we send in a photon which we know is in the state $\ket{y}$, it will be absorbed by the polaroid. If we send in a photon which we know is in the state $\ket{x}$, it will come right through as $\ket{x}$. If we take a piece of calcite which takes a beam of polarized light and splits it into an $\ket{x}$ beam and a $\ket{y}$ beam, that piece of calcite is the complete analog of a Stern-Gerlach apparatus which splits a beam of silver atoms into the two states $\ket{+}$ and $\ket{-}$. So everything we did before with particles and Stern-Gerlach apparatuses, we can do again with light and pieces of calcite. And what about light filtered through a piece of polaroid set at an angle $\theta$? Is that another state? Yes, indeed, it is another state. Let’s call the axis of the polaroid $x'$ to distinguish it from the axes of our base states. See Fig. 11-2. A photon that comes out will be in the state $\ket{x'}$. However, any state can be represented as a linear combination of base states, and the formula for the combination is, here, \begin{equation} \label{Eq:III:11:33} \ket{x'}=\cos\theta\,\ket{x}+\sin\theta\,\ket{y}. \end{equation} That is, if a photon comes through a piece of polaroid set at the angle $\theta$ (with respect to $x$), it can still be resolved into $\ket{x}$ and $\ket{y}$ beams—by a piece of calcite, for example. Or you can, if you wish, just analyze it into $x$- and $y$-components in your imagination. Either way, you will find the amplitude $\cos\theta$ to be in the $\ket{x}$ state and the amplitude $\sin\theta$ to be in the $\ket{y}$ state.  Now we ask this question: Suppose a photon is polarized in the $x'$-direction by a piece of polaroid set at the angle $\theta$ and arrives at a polaroid at the angle zero—as in Fig. 11-3; what will happen? With what probability will it get through? The answer is the following. After it gets through the first polaroid, it is definitely in the state $\ket{x'}$. The second polaroid will let the photon through if it is in the state $\ket{x}$ (but absorb it if it is the state $\ket{y}$). So we are asking with what probability does the photon appear to be in the state $\ket{x}$? We get that probability from the absolute square of amplitude $\braket{x}{x'}$ that a photon in the state $\ket{x'}$ is also in the state $\ket{x}$. What is $\braket{x}{x'}$? Just multiply Eq. (11.33) by $\bra{x}$ to get \begin{equation*} \braket{x}{x'}=\cos\theta\,\braket{x}{x}+\sin\theta\,\braket{x}{y}. \end{equation*} Now $\braket{x}{y}=0$, from the physics—as they must be if $\ket{x}$ and $\ket{y}$ are base states—and $\braket{x}{x}=1$. So we get \begin{equation*} \braket{x}{x'}=\cos\theta, \end{equation*} and the probability is $\cos^2\theta$. For example, if the first polaroid is set at $30^\circ$, a photon will get through $3/4$ of the time, and $1/4$ of the time it will heat the polaroid by being absorbed therein.  Now let us see what happens classically in the same situation. We would have a beam of light with an electric field which is varying in some way or another—say “unpolarized.” After it gets through the first polaroid, the electric field is oscillating in the $x'$-direction with a size $\Efield$; we would draw the field as an oscillating vector with a peak value $\Efield_0$ in a diagram like Fig. 11-4. Now when the light arrives at the second polaroid, only the $x$-component, $\Efield_0\cos\theta$, of the electric field gets through. The intensity is proportional to the square of the field and, therefore, to $\Efield_0^2\cos^2\theta$. So the energy coming through is $\cos^2\theta$ weaker than the energy which was entering the last polaroid.  The classical picture and the quantum picture give similar results. If you were to throw $10$ billion photons at the second polaroid, and the average probability of each one going through is, say, $3/4$, you would expect $3/4$ of $10$ billion would get through. Likewise, the energy that they would carry would be $3/4$ of the energy that you attempted to put through. The classical theory says nothing about the statistics of the thing—it simply says that the energy that comes through will be precisely $3/4$ of the energy which you were sending in. That is, of course, impossible if there is only one photon. There is no such thing as $3/4$ of a photon. It is either all there, or it isn’t there at all. Quantum mechanics tells us it is all there $3/4$ of the time. The relation of the two theories is clear.  What about the other kinds of polarization? For example, right-hand circular polarization? In the classical theory, right-hand circular polarization has equal components in $x$ and $y$ which are $90^\circ$ out of phase. In the quantum theory, a right-hand circularly polarized (RHC) photon has equal amplitudes to be polarized $\ket{x}$ or $\ket{y}$, and the amplitudes are $90^\circ$ out of phase. Calling a RHC photon a state $\ket{R}$ and a LHC photon a state $\ket{L}$, we can write (see Vol. I, Section 33-1) \begin{equation} \begin{aligned} \ket{R}&=\frac{1}{\sqrt{2}}\,(\ket{x}+i\,\ket{y}),\\[4ex] \ket{L}&=\frac{1}{\sqrt{2}}\,(\ket{x}-i\,\ket{y}). \end{aligned} \label{Eq:III:11:34} \end{equation} —the $1/\sqrt{2}$ is put in to get normalized states. With these states you can calculate any filtering or interference effects you want, using the laws of quantum theory. If you want, you can also choose $\ket{R}$ and $\ket{L}$ as base states and represent everything in terms of them. You only need to show first that $\braket{R}{L}=0$—which you can do by taking the conjugate form of the first equation above [see Eq. (8.13)] and multiplying it by the other. You can resolve light into $x$- and $y$-polarizations, or into $x'$- and $y'$-polarizations, or into right and left polarizations as a basis.  Just as an example, let’s try to turn our formulas around. Can we represent the state $\ket{x}$ as a linear combination of right and left?  Yes, here it is: \begin{equation} \begin{aligned} \ket{x}&=\frac{1}{\sqrt{2}}\, (\ket{R}+\ket{L}),\\[4ex] \ket{y}&=-\frac{i}{\sqrt{2}}\, (\ket{R}-\ket{L}). \end{aligned} \label{Eq:III:11:35} \end{equation}  Proof: Add and subtract the two equations in (11.34). It is easy to go from one base to the other.  One curious point has to be made, though. If a photon is right circularly polarized, it shouldn’t have anything to do with the $x$- and $y$-axes. If we were to look at the same thing from a coordinate system turned at some angle about the direction of flight, the light would still be right circularly polarized—and similarly for left. The right and left circularly polarized light are the same for any such rotation; the definition is independent of any choice of the $x$-direction (except that the photon direction is given). Isn’t that nice—it doesn’t take any axes to define it. Much better than $x$ and $y$. On the other hand, isn’t it rather a miracle that when you add the right and left together you can find out which direction $x$ was? If “right” and “left” do not depend on $x$ in any way, how is it that we can put them back together again and get $x$? We can answer that question in part by writing out the state $\ket{R'}$, which represents a photon RHC polarized in the frame $x',y'$. In that frame, you would write \begin{equation*} \ket{R'}=\frac{1}{\sqrt{2}}\,(\ket{x'}+i\,\ket{y'}). \end{equation*} How does such a state look in the frame $x,y$? Just substitute $\ket{x'}$ from Eq. (11.33) and the corresponding $\ket{y'}$—we didn’t write it down, but it is $(-\sin\theta)\,\ket{x}+(\cos\theta)\,\ket{y}$. Then  \begin{align*} \ket{R'}\!&=\!\frac{1}{\sqrt{2}}[ \cos\theta\,\ket{x}\!+\sin\theta\,\ket{y}\!- i\sin\theta\,\ket{x}\!+i\cos\theta\,\ket{y}]\\[2ex] &=\frac{1}{\sqrt{2}}[ (\cos\theta-i\sin\theta)\ket{x}\!+ i(\cos\theta-i\sin\theta)\ket{y}]\\[2ex] &=\frac{1}{\sqrt{2}}(\ket{x}\!+i\ket{y}) (\cos\theta-i\sin\theta). \end{align*}  The first term is just $\ket{R}$, and the second is $e^{-i\theta}$; our result is that \begin{equation} \label{Eq:III:11:36} \ket{R'}=e^{-i\theta}\,\ket{R}. \end{equation} The states $\ket{R'}$ and $\ket{R}$ are the same except for the phase factor $e^{-i\theta}$. If you work out the same thing for $\ket{L'}$, you get that1 \begin{equation} \label{Eq:III:11:37} \ket{L'}=e^{+i\theta}\,\ket{L}. \end{equation}  Now you see what happens. If we add $\ket{R}$ and $\ket{L}$, we get something different from what we get when we add $\ket{R'}$ and $\ket{L'}$. For instance, an $x$-polarized photon is [Eq. (11.35)] the sum of $\ket{R}$ and $\ket{L}$, but a $y$-polarized photon is the sum with the phase of one shifted $90^\circ$ backward and the other $90^\circ$ forward. That is just what we would get from the sum of $\ket{R'}$ and $\ket{L'}$ for the special angle $\theta=90^\circ$, and that’s right. An $x$-polarization in the prime frame is the same as a $y$-polarization in the original frame. So it is not exactly true that a circularly polarized photon looks the same for any set of axes. Its phase (the phase relation of the right and left circularly polarized states) keeps track of the $x$-direction. 