 We've seen that a photon with z-component of spin angular momentum, h-bar, corresponds to classical right-handed circular polarization. And the opposite component of spin corresponds to left-handed circular polarization. So it seems we've got the photon figured out. The photon is a spin-1 particle, and that spin manifests itself as circular polarization. But what about linear polarization? We use that all the time in radio systems. How do we get linear polarization if photons are circularly polarized? Well, we've already seen that a vertically polarized field can be formed by the superposition of left and right-hand circularly polarized fields. But what does this mean in terms of photons? Before we go any further, let's put things on an empirical basis by considering how we can measure polarization. Suppose we have a thin wire. Electrons are free to move within the wire. If we introduce a perpendicular electric field, the electron's motion is constrained and the wire has essentially no effect. But in response to a field parallel to the wire, the electrons move freely. In doing so, they generate a field which cancels out the applied field. Effectively, the applied field is shorted out and gets reflected. Therefore, a grid of wires will, ideally, pass the electric field component perpendicular to the wires and reflect the component parallel. At microwave frequencies, this is precisely how a polarization filter can be constructed. To illustrate that this really works, this picture shows two microwave antennas based on this principle. One has vertically oriented bars and will reflect and receive only vertically polarized microwaves. The other has horizontally oriented bars and will reflect and receive only horizontally polarized microwaves. This allows two independent channels to operate over the same range of frequencies. Optical frequencies are roughly a million times microwave frequencies, so the corresponding structures in an optical polarization filter are microscopic, but the principle is the same. Now, real polarization filters are imperfect. They don't transmit or block fully 100% of any polarization. In practice, you have to take this into account in analyzing experimental results. But for simplicity going forward, we're going to assume our filters are ideal. Placing a polarization filter in front of a lamp and looking closely, we notice that the light intensity is decreased. In fact, it's reduced to one half its original value. Adding another filter doesn't seem to change anything. Apparently, it's redundant. Unless we rotate the filters relative to each other, then the transmitted field gets weaker, and when the two filters are oriented at 90 degrees, the field is completely blocked. When a field with any combination of v and h polarization illuminates a vertically oriented filter, only the v component passes through. The h component is blocked or filtered out. Likewise, if the filter is horizontally oriented, only the h component passes through. In direct notation, we can represent this by introducing a new symbol, the bra, which is a mirror image of the ket. Get it? Bra, ket, bracket. A bra allows us to represent the projection of one state on another. The projection of the v-state on the v-state is one. All of a v-polarized wave will pass through a v-oriented filter. The projection of the h-state on the v-state is zero. None of an h-polarized wave will pass through a v-filter. Likewise, the projection of the v-state on the h-state is zero, and the h-state on itself is one. We can now represent the v-oriented filter as an operator composed of a v-ket followed by a v-bra, and the h-oriented filter as an operator composed of an h-ket followed by an h-bra. The top experiment is mathematically described as applying the v-ket v-bra operator to the quantum state of the incident field. Applying the v-bra to the two terms of the quantum superposition and substituting the appropriate one and zero projection values, we obtain the output quantum state as the v-component of the input state. So now that we have a consistent formalism for representing polarization measurements, let's go back to our original issue. Photons have spin. Spin implies circular polarization. So what is linear polarization? Yes, we can say that vertical polarization is a quantum superposition of right and left circular polarization, but what does this mean in terms of photons? First, as a check, let's see if the projection of the superposition onto the v-state is one, as it should be. This projection is 1 over square root 2 times the projection of the r-state onto the v-state, plus a similar term for the l-state. The r and l-states in turn can be expressed as superpositions of the v in delayed h-state. Remember that the imaginary unit i represents a time delay of one-quarter period of the oscillation. Putting r or l-polarized light through a vertical filter removes the h-term, leaving 1 over square root 2 times the v-state. The projection of the r and l-states onto the v-state is 1 over square root 2. It follows that v-polarized light represented as a superposition of r and l-polarized light when put through a v-filter is completely transmitted as it must be. So our bookkeeping works, and we can indeed say that the v-state is a superposition of the r and l-states. And again, this makes complete sense mathematically and classically. But again, what about photons? In a v-polarized wave, are half the photons r-polarized and the other half l-polarized? Should we think of a vertically polarized wave as consisting of lots of photons, half in each of the two circular polarizations? If so, what happens if we just have a single photon? How would that correspond to a v-polarized field? Suppose the photon is r-polarized. If it encounters a vertical filter, what comes out is 1 over square root 2 times the v-state. Now the projection of the input state onto itself is 1, which we can interpret as the intensity of the input state. The projection of the output state onto itself is only one-half. We can interpret this classically as 100% of the light illuminated the filter, but only 50% made it through. 50% of the energy in an r-polarized wave will pass through a v-filter. You can have 50% of a wave, but you can't have 50% of a photon. It's all or nothing. Wave particle duality forces us to interpret this result as a probability for the photon to pass through the filter. So if a vertically polarized wave consisted of equal numbers of r and l-circularly polarized photons, only half of them would make it through a v-filter. But all of the photons in a v-polarized wave have to make it through. We're forced to conclude that in the v-quantum superposition, each individual photon is simultaneously in both the r and l states. Now, this means that the photon does not have a well-defined z-component of angular momentum. But if we measure this, we know we'll find either h-bar or minus h-bar. And this is very strange. We've seen that this does not mean that before measurement, the angular momentum is either plus or minus h-bar, right or left polarization, and the wave function simply represents our ignorance of which. No, the photon is in a very real sense simultaneously in both states, but a measurement will find it in only one state. Does this mean that the angular momentum is somehow not real until it's measured? Is it somehow blurred before that? Yes, very strange. Indeed, it's the same kind of strangeness we encountered in the two-pinhole experiment. That made perfect sense in terms of classical waves. Part of the wave went through the upper-pinhole and part through the lower-pinhole. Those two parts then spread out and created an interference pattern. But it made no sense when we thought of a single photon somehow going through both pinholes and interfering with itself. We essentially had to think of the photon as being in a superposition of the state in which it went through the upper-pinhole and the state in which it went through the lower-pinhole. One more bit about polarization measurement. A so-called birefringent crystal, calcite is one example, has the property that the speed of light in the crystal is different for different linear polarizations. If we arrange for v-polarization to be the fast wave and h-polarization to be the slow wave and we choose an appropriate crystal thickness, we can cause the h-polarization to emerge from the crystal delayed relative to the v-polarization by one-quarter wavelength, hence by one-quarter oscillation period. We represent that delay by the imaginary unit I. So if we illuminate the crystal with equal amounts of v and h-polarization, which we earlier saw produces a linear polarization at 45 degrees to the vertical, the output will be the superposition of v and delayed h-polarizations that corresponds to a right-hand circularly polarized wave. Rotating the crystal, we can get left-hand circular polarization. And reversing the arrangement, we can convert circular polarization into linear polarization. We can use this trick to make circular polarization filters. Moreover, we can make so-called polarizing beam splitters for either linear or circular polarization, in which one polarization continues in a straight line and the other is reflected 90 degrees. Passing a single vertically polarized photon through a circular polarization beam splitter with detectors for the two polarizations, the photon will end up at one or the other of the detectors, each with 50% probability. So we go from a superposition of the R and L states to a measurement of a single one of them. And of course, after the measurement, we know which state it is with 100% certainty. This is an example of wave function collapse. It's not described by the Schrodinger equation. It's a special peculiar formalism we have to use to describe quantum mechanical measurement. As weird as these ideas are, they provide us with an incredibly precise description of the real world. To quote physicist W. H. Zurich, quantum mechanics works exceedingly well in all practical applications. No example of conflict between its predictions and experiment is known. The only failure of quantum theory is its inability to provide a natural framework for our prejudices about the workings of the universe. Given this, it should be no surprise that the interpretation of quantum mechanics has always been a source of controversy, even though there is general agreement about how to apply the theory to physical problems.