Polarization and Quantum Theory -- confused This is what I'm getting hung up on: If you take a piece of vertically polarized glass and place it in front of a piece of horizontally polorized galss then NO light passes through the second glass because the light is at 90 degrees to the horizontal polarizer once it it passes the vertical polarizer. If we place a third polarizer between the two and set it at 45 degrees to the other two then a quarter of the light entering the vertical polarizer will emerge from the horizontal polarizer (as opposed to none as before). 50% of the light that passed the verticle polarizer passes the middle polarizer and then 50% of this light (25% of the original light) passes through the horizontal polarizer. The explaination given (as I understand it anyway) is that the light can get through the polarizer that is set at 45 degreed because of the light's "probability wave" and the inherent uncertainty about exactly what polarization the light has. I guess that's the part I don't get, how can there be ANY non zero probability that light that ISN'T vertically polarized exists on the opposite side of the vertical polarizer. If the light got through the polarizer then it MUST be vertically polarized (otherwise it simply would not have passed, no?). So doesn't the vertical polarizer act as "detector" of sorts (since we know that only ONE polarity can pass)? If so shouldn't there be a 100% probability that the light traveling between the vertical polarizer and the one set at 45 degrees is vertically polarized? I guess I'm not understanding how a vertical polarizer can trasmit light that could then pass through a polarizer at 45 degrees to the vertical one. Confused... help! Thanks *CA*
I don't think this has anything to do with uncertainty. Take an unpolarized light wave. It's electric field is E, and it can be decomposed into equal parts of x-polarized light (E_{x}) and y-polarized light (E_{y}). The unpolarized state is: E=(1/2)^{1/2}(E_{x}+E_{y}) When you pass it through a y-polarizer, the E_{x} component is blocked out, leaving E'=(1/2)^{1/2}E_{y} When this is passed through an x-polarizer, it is blocked out compelely, so: E''=0 Now if you put a 45^{o} polarizer in there, you have to convert the components of E to the basis states corresponding to that measurement apparatus. This is equivalent to rotating the axes 45^{o} to x',y' as follows: E'=(1/2)^{1/2}E_{y} E'=(1/2)(E_{x'}+E_{y'}) Now if you have an x'-polarizer, you have: E''=(1/2)E_{x'} Now when this wave is incident on the x-polarizer, we have to re-transform back to the original basis to get: E''=(1/2)E_{x'} E''=(2^{1/2}/4)(E_{x}+E_{y}) When this is incident on an x-polarizer, the E_{y} part is blocked out, leaving: E'''=(2^{1/2}/4)E_{x}
So what, if anything, does quantum theory have to do with the passage of some light when the 45 degree polarizer is used? Anything at all? I only ask because the book I'm reading is on the subject of quantum theory, and has spent a large portion of the chapter explaining this effect as one of the "weird wonders" of quantum theory and the wave/particle nature of light. I'm a bit of a layman when it comes to the mathematical end but I like to think I can grasp the basic "ideas" ok. The notion I get from reading the book is that vertical polarizers ONLY passed vertically polarized light, and nothing else AT ALL (and conversely, that horizontal polarizers ONLY pass horizontally polarized light, and nothing else AT ALL). From what you posted I get the impression that vertical polarizers past MOSTLY vetically polarized light, but some "other" polarities get through as well -- and that the more perpendicular the light's polarization is the less the chance it gets through (till, at 90 degrees, it is zero)? If that's the case, what is the "amazing" quantum theory part? Is the author throwing out BS? Am I totally misunderstanding the setup? Thanks again, *CA*
Yes, this has to do with the quantum theory of measurement. In classical mechanics, the observer does not affect the outcome of experiments. Such is not the case in quantum mechanics. Making a measurement on a quantum state necessarily affects the state. Specifically, making a measurement of an observable forces the system to be found in one of the eigenstates of that observable. That is why I had to transform bases when dealing with the 45^{o} polarizer, because I knew that it would let the E_{x'} component through and block the rest out. I characterized "the rest" by E_{y'}, which is orthogonal to E_{x'}.
Tom said it better than I could, but I will add a cent or two... When I was first learning about QM, I actually tried the experiment you described. I couldn't believe that putting a dark lens between the crossed polarizers would increase the light that passed through. (I bought some cheap polarized sunglasses for the experiment.) But actually, I think this experiment is consistent with classical optics as well. Where the difference comes in is that for this to make sense in classical optics, light must be a wave. But in QM, light sometimes behaves as a wave and sometimes behaves as a point particle. This changeover from the wave to the particle theory operates within the Heisenberg Uncertainty Relations. There is no analog to this in classical physics. So when you switch over to testing the particle nature of light, you realize that something really strange is happening. Because a single particle of light (for example an exposure on a photographic plate) can be made to interfere with itself, something that only waves are supposed to be able to do. A point particle cannot interfere with itself in the classical world.