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Firstly I am sorry for any awkward English expressions.

Recently I'm reading "Feynman Lectures on Physics - Quantum Mechanics" and come to have a single question.

In the book Feynman explains, "You must never add amplitudes for different and distinct final states".

For example) in double slit experiment with two holes (hole 1, hole 2), if we call the probability amplitude of an electron through the hole 1 to an electron detector as Y1 and hole 2 as Y2,

without any disturbing measurments such as a light, we can say, "the probability of electron to reach the detector is [Y1 + Y2]^2 ." and......

however with some light emitting device to distinguish which hole is taken by the electrons - in this case we can know the choice of electrons by photons - the probability of electron to reach the detector is not just [Y1 + Y2]^2 but [Y1]^2 + [Y2]^2.

I do know that the difference between the results depends on the distinguishability of alternatives, but the problem is why distinguishability makes the results different. Why can't the probability amplitudes interfere if we can know which alternative is chosen?

(Additionally, Feynman says, the probabilities of passing hole 1 and hole 2 are independent each other, but I cannot find any relationship between this mention and my problem.)

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It is easy to understand from Feynman's first rule--- if you measured the location, something is different, namely the stuff that contains the information of the location, so you don't do interference. –  Ron Maimon Jul 9 '12 at 4:29

2 Answers 2

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Why can't the probability amplitudes interfere if we can know which alternative is chosen?

The knowledge of the alternative "chosen" comes with a price; the price of interaction with the electron.

Regardless of the state of the electron before the interaction, the state after the interaction, the interaction with the detection apparatus at either slit, localizes the electron; the wave function "collapses" to a wave packet localized at one slit and unitarily evolving thereafter (until interacting with the detector). Thus there are two distinguishable states and they don't interfere because, due to the localized nature of the wave packet, there is vanishing amplitude at the other slit and so, the probabilities add.

However, if there is no interaction with the electron before it is detected, the state unitarily evolves until detection. The wave function propagates through both slits and so the amplitude is significant at both slits; the electron "takes both paths". The wave function between the slits and detector is essentially proportional to the sum of the two distinguishable states above. When we square this sum of amplitudes to find the probability, there's an interference term.

To summarize, by detecting which path, you "throw away" the other possibility and with it, the interference.

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collapse! it is a clear word for me to understand. Thank you. –  Epsilon Jul 8 '12 at 23:19

In The Feynman Lectures on Physics Vol. III, Section 3-2 The two-slit interference pattern, the reason why "probability amplitudes [do not] interfere if we can know which alternative is chosen" is explained simply and quantitatively for the two-slit experiment, in terms of the laws for combining quantum mechanical amplitudes discussed in the book, and without resorting to philosophical notions, such as "collapse of the wave function."

(Please refer to Figure 3-3. I was not allowed to post it here, unfortunately.)

Let phi_1 = < x | 1 >< 1 | s > be the amplitude for an electron to go from the source to hole 1 and from hole 1 to position x on the backstop, and similarly

Let phi_2 = < x | 2 >< 2 | s > be the amplitude for an electron to go from the source to hole 2 and from hole 2 to position x on the backstop,

Let a = the amplitude for an electron at hole 1 to scatter a photon into detector D1, which we assume by symmetry also equals the amplitude for an electron at hole 2 to scatter a photon into detector D2, and

Let b = the amplitude for an electron at hole 1 to scatter a photon into detector D2, which we assume by symmetry also equals the amplitude for an electron at hole 2 to scatter a photon into detector D1.

Then the amplitude to detect an electron at position x coincident with a photon at D1 is a*phi_1 + b*phi_2, while the amplitude to detect an electron at position x coincident with a photon at D2 is a*phi_2 + b*phi_1. Since these two cases are distinguishable, their amplitudes must each be squared, to find their respective probabilities, before being added together to find the total probability of detecting an electron at position x, which is (See Eq. 3.10)

 |a*phi_1 + b*phi_2|^2 + |a*phi_2 + b*phi_1|^2.

To figure out what this distribution in x looks like, you can imagine different cases of 'a' and 'b'. If, for example, you have arranged things carefully so that electrons at hole 1 are almost certain to scatter photons into detector D1 and not into D2 (and visa versa for electrons at hole 2), so that the magnitude of amplitude a is close to 1 while that of b is close to 0, then you get, for the total probability of detecting an electron at position x, something close to

 |a*phi_1|^2 + |a*phi_2|^2

which is just the distribution |phi_1|^2 + |phi_2|^2, multiplied by a factor |a|^2. In this case, evidently, there is no interference between phi_1 and phi_2. On the other hand, if you have arranged things so that you are completely uncertain as to whether a photon detected at D1 was coincident with an electron at hole 1 or with one at hole 2 (and similarly for photons detected at D2), so that amplitudes a and b are approximately equal in magnitude, then the total probability of detecting an electron at position x is something close to

  2*|a*phi_1 + a*phi_2|^2

which is just the distribution |phi_1 + phi_2|^2, multiplied by a factor 2|a|^2. And in this case there is interference between phi_1 and phi_2.

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