- #1
ralqs
- 99
- 1
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:
[tex] {\langle x | s \rangle}_{via 1} = \langle x | 1 \rangle \langle 1 | s \rangle[/tex]"
My question is, why is this true? Obviously, probabilities must multiply, so
[tex]|{\langle x | s \rangle}_{via 1}|^2 = |\langle x | 1 \rangle|^2 \langle 1 | s \rangle|^2[/tex]
meaning
[tex]{\langle x | s \rangle}_{via 1} = \langle x | 1 \rangle \langle 1 | s \rangle e^{i \vartheta}[/tex]
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
[tex]{\langle x | s\rangle}_{via 1} + {\langle x | s\rangle}_{via 2} = \langle x | s\rangle[/tex]
So what allows us to ignore the phase?
"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:
[tex] {\langle x | s \rangle}_{via 1} = \langle x | 1 \rangle \langle 1 | s \rangle[/tex]"
My question is, why is this true? Obviously, probabilities must multiply, so
[tex]|{\langle x | s \rangle}_{via 1}|^2 = |\langle x | 1 \rangle|^2 \langle 1 | s \rangle|^2[/tex]
meaning
[tex]{\langle x | s \rangle}_{via 1} = \langle x | 1 \rangle \langle 1 | s \rangle e^{i \vartheta}[/tex]
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
[tex]{\langle x | s\rangle}_{via 1} + {\langle x | s\rangle}_{via 2} = \langle x | s\rangle[/tex]
So what allows us to ignore the phase?