Feynman's third general principle of QM

1. Aug 18, 2011

ralqs

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?

2. Aug 19, 2011

Bill_K

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.

3. Aug 19, 2011

ralqs

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 $$\langle 1|s \rangle$$?

4. Aug 19, 2011

xts

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.

5. Aug 22, 2011

edpell

What are his complete set of principles of QM? Thanks.

6. Aug 22, 2011

xts

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 learnt my basic QM from. And the one I still recommend to all students..