1. Feb 5, 2008

### dx

Hi, I'm just beginning to learn about the quantum world, so excuse me if this is naive.

Consider the double-slit experiment except that the slits are imaginary. call the slits A and B.

you have a probability amplitude that a particles goes to the point x on the screen via A (an imaginary slit) and also an amplitude that it goes to the screen via B :

$$< xA | s >$$
$$< xB | s >$$

these two amplitudes are clearly the amplitudes for indistinguishable final states and therefore should interfere. but since the slits are imaginary, you could also write it as a single amplitude <x|s> and the probability is then just |<x|a>|^2.

2. Feb 5, 2008

### dx

you could object that im not considering all possible paths to the screen, so let me modify it a little.

instead of two imaginary slits, say we have an imaginary point between the screen and the source. then there are two possible classes of paths to the screen : the amplitude for the paths that go above the point and the ones that go below the point. the final states by the two types of paths are indistinguishable and so the amplitudes interfere with eachother.

3. Feb 14, 2008

### joyer2

What's the problem? Everything is fine if one uses the amplitude
$$<x|A|s>+<x|B|s>+...$$ for n paths
which becomes
$$n<x|A|s>$$
when n paths are 'indistinguishable'.