Feynman's third general principle of QM

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Feynman's third general principle of quantum mechanics states that the amplitude for a particle's route can be expressed as the product of the amplitude for each segment of the journey. Specifically, in the double slit experiment, the amplitude from the electron source (s) to a point on the screen (x) via hole 1 is given by the equation: {\langle x | s \rangle}_{via 1} = \langle x | 1 \rangle \langle 1 | s \rangle. The discussion highlights the significance of ignoring the phase term (e^{i \vartheta}) in the total amplitude calculation, emphasizing that this is an axiom of quantum mechanics rather than a question of derivation. Feynman’s principles serve as a foundational framework for understanding quantum behavior, though they are not intended as a comprehensive axiomatic system.

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ralqs
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I'm reading vol. 3 of the Feynman lectures on physics, and in chapter three he describes his third general principle of quantum mechanics as follows:

"The third general principle: When a particle goes by some particular route the amplitude for that route can be written as the product of the amplitude to go part way with the amplitude to go the rest of the way. For the [double slit experiment] the amplitude to go from the [electron source, s] to [some point on the screen, x] by way of hole 1 is equal to the amplitude to go from s to 1, multiplied by the amplitude to go from 1 to x:
{\langle x | s \rangle}_{via 1} = \langle x | 1 \rangle \langle 1 | s \rangle"

My question is, why is this true? Obviously, probabilities must multiply, so
|{\langle x | s \rangle}_{via 1}|^2 = |\langle x | 1 \rangle|^2 \langle 1 | s \rangle|^2
meaning
{\langle x | s \rangle}_{via 1} = \langle x | 1 \rangle \langle 1 | s \rangle e^{i \vartheta}

But why can we drop the theta term? It's clearly important, because if we want to total amplitude to go to x via s we will need to add
{\langle x | s\rangle}_{via 1} + {\langle x | s\rangle}_{via 2} = \langle x | s\rangle

So what allows us to ignore the phase?
 
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My question is, why is this true?
It's an axiom. You can't ask why an axiom is true. The only possible answer is that it leads to the desired results.

Feynman is trying to explain the difference between classical and quantum mechanics in terms of a small set of principles. The fact that wavefunctions are complex and so we deal with complex probability amplitudes rather than probabilities themselves is what gives quantum mechanics its wavelike features.
 
Bill_K said:
It's an axiom.

My impression is that Feynman wasn't treating this as an axiom. In the first chapter of the volume, he seems to give all the axioms of quantum behavior. As I understand it, amplitudes are usually determined up to a phase constant...is it not possible that the phase here got absorbed into either \langle x|1 \rangle or \langle 1|s \rangle?
 
ralqs said:
s it not possible that the phase here got absorbed into either \langle x|1 \rangle or \langle 1|s \rangle?
Yes, but you may then take an example with two points on the way: \langle x|s \rangle =\langle x|1 \rangle\langle 1|s \rangle=\langle x|1 \rangle\langle 1|a \rangle\langle a|s \rangle
Probably Feynman should make it as an axiom, but don't forget that this argumentation was created by him just to be used in a series of lectures, not as a new formal approach to QM, so (as it is too late to discuss it with Him) - forgive him a small inconsistency.
 
What are his complete set of principles of QM? Thanks.
 
Oh, his principles are not a quite serious axiomatic approach to QM... They was made just for the purpose of series of introductory lectures. So don't demand them to be complete and fully consistent.
Have a great reading with his book! That was the one I learned my basic QM from. And the one I still recommend to all students..
 

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