In my earlier posting on the Shining Cryptographers net I stated that if Eve the eavesdropper made a measurement on a photon in a pure state in the cryptographer's measurement basis, she would disrupt the final result with probability 1/2. This is not correct, so I want to give the right answer here, with a little more physical analysis. (Note, we are assuming "strong" measurements here and not yet trying to analyze the case of weak measurements which only partially couple to the photon.) The case we will consider is a photon that is in a pure state with respect to the final measurement basis, which we will consider to be in the vertical and horizontal orientation: | | -----+----- | | I.e. the photon is either vertical or horizontal. Now Eve comes along and, without knowing the actual orientation of the measurement basis, makes her own measurement. Her basis will be tilted at a random angle with respect to the actual measurement basis, which we will call x. x ranges from 0 to 90 degrees. H' V V' -__ | / -.|/ H -----+_---- /| '-_ / | ' I have labeled the cryptographer's measurement axes V and H, and Eve's V' and H'. The angle from V to V' is x, as is the angle from H to H'. Supposing the photon is initially in state V, the probability of it being measured by Eve as V' is cos^2(x). The probability of her measuring it as H' is sin^2(x). This is from the basic physics of polarization. (Note: Eve can also use a circular polarization basis, but that only adds an imaginary factor to the coefficients, and since we are taking the square of the norm of the coefficient to get our probabilities, the imaginary term is irrelevant for these calculations.) Once Eve has measured the photon, it is then measured again in the crytographer's basis. If she had measured it as V', the chance that it will then be measured as V is again cos^2(x). If she had measured it as H' the chance that it will be measured as V is sin^2(x). These are mutually exclusive alternatives so we can add the probabilities. Therefore, the chance that a photon incoming as a V will be measured as a V is cos^2(x)*cos^2(x) + sin^2(x)*sin^2(x), or cos^4(x)+sin^4(x). This is the chance for a specific angle x. We must average this for x from 0 to 90 degrees in order to find the final probability that the measurement will be unchanged. So we integrate, from 0 to 90 degrees, (cos^4(x) + sin^4(x)) dx. The answer is 3/4. The probability that Eve's measurement will leave the result unchanged is 3/4, and therefore the probability that she will perturb the result is 1/4. Of course the same calculation applies as well to a photon entering in the H state and remaining unchanged. This corrected result has two implications for my earlier conclusion. First, the chance of Eve being caught is only half as great with a single measurement, 1/4 rather than 1/2 as I said earlier. And second, it is no longer true that multiple measurements are no worse than a single measurement. Because a single measurement only has a 1/4 chance of perturbing things, it is still possible that multiple measurements might make things worse. To analyze the multiple measurement case, consider first two measurements. The idea is again that the photon is in a pure state for the cryptographers' measurement basis. Eve makes a measurement at random angle x with respect to that basis, and now she makes another measurement at random angle y with respect to her earlier measurement. This represents the situation where she is making measurements at two points in the measurement chain and there is an unknown rotation occuring between those two points. There are now three axis pairs: V,H; V',H'; and V'',H''. The second is rotated with respect to the first by x, and the third with respect to the second by y. Again starting with a photon in the V state, there are four possibilities for it to end up as a V. The first measurement can go into the V' or H' direction, and then second measurement can go into the V'' or H'' direction. Two possibilities for two measurements makes four possibilities total. Probabilities for transitions among the V states are based on cos^2 of the relevant angle, and for transitions between V and H states are based on sin^2 of the angle. The actual probability for specific angle x and y works out to be: cos^2(x) cos^2(y) cos^2(x+y) + [For V->V'->V''->V] cos^2(x) sin^2(y) sin^2(x+y) + [For V->V'->H''->V] sin^2(x) cos^2(y) sin^2(x+y) + [For V->H'->H''->V] sin^2(x) sin^2(y) cos^2(x+y) [For V->H'->V''->V] Integrating this for x and y going from 0 to 90 degrees and taking the average, the probability for leaving the state unchanged comes out to be 5/8. Eve's chances of perturbing the measurement have increased from 1/4 to 3/8 by doing two measurements rather than one. Increasing the number of measurements to three reduces the chance of success to 9/16, with a 7/16 chance of perturbation. The pattern is that each additional measurement brings us closer to a 50-50 chance of detection. In practice, then, if Eve needs to make more than a single measurement, her chance of being caught is close to 1/2. It does not appear that a single measurement can tell her anything useful, since that can only give information about the orientation of the photon at one point. Without knowing the measurement basis, that tells her nothing about the information carried by the photon. Given that she has to do at least two measurements her probability of being detected is already close to 1/2, at 3/8. Doing additional measurements will increase the detection probability only slightly further, up to a limit of 1/2. Therefore the ultimate conclusion from the earlier presentation is essentially correct, that Eve might as well do as many measurements as she can in order to get as much information as possible on a single photon, and that she must accept that her chance of being caught is about 1/2. Hal

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