- #1
AdrianMay
- 121
- 4
Hi folks,
I just read the Feynman Lectures Vol 3, Chapter 6 about spin 1/2 particles going through various arrangements of Stern-Gerlach filters. I think I get it, but I'm not sure why the same arguments don't apply to the spin 1 particles in the preceding chapter.
The chapter goes something like this (skip to ##### if you already read it): first define the axes: Z is upwards and is the direction of the non-uniform magnetic field that splits the beam. Y is along the splitter, presumably from your left hand to your right hand, and X points at you. The splitters split the beam in two so you can see how bright they are and block them as you will, and then recombine them.
Suppose we start with a beam of particles polarised along X or Y, then watch them split and recombine and pop out the other end still knowing which way they were polarised. The only way the two beams could remember that is in the relative phase of the two beams. We can define +X polarisation as the two being in phase, -X as 180' out of phase, and +/-Y as 90/270' out of phase (any other convention will also do.) Then he considers putting another splitter at various funny angles after the first and deduces that:
1) a rotation around the z axis must leave the magnitudes of +Z and -Z alone but should rotate X in and out of Y so
|+Z> -> |+Z> . exp (i@/2)
|-Z> -> |-Z> . exp (-i@/2)
does the trick. E.g. with a 90' rotation, the beams would have gone 45' in opposite directions so their relative phase has swapped X for +/-Y as required.
2) other rotations may not turn X into Y or vice versa so they must leave the relative phase alone. But they should swap Z for X or Y. So if you start with all |Z+> and rotate 90', you should get equal magnitudes of |Z+> and |Z-> but in phase for a rotation about Y:
|+Z> -> |+Z>.cos(@/2) + |-Z>.sin(@/2)
|-Z> -> |+Z>.-sin(@/2) + |-Z>.cos(@/2)
3) and out of phase for a rotation about X:
|+Z> -> |+Z>.cos(@/2) + |-Z>.i.sin(@/2)
|-Z> -> |+Z>.i.sin(@/2) + |-Z>.cos(@/2)
#####
Now then, my question is how that would all look look if we do similar stuff with the spin-1 particles of the previous chapter. First I'd like to know how the experiments would turn out:
First split the beam from the oven into |+X>, |0X> and |-X>, filter it down to |+X>, shove it through an unfiltered Z splitter/recombiner just to admire the three beams (are they all the same or is the middle one brighter?) then put another splitter after that at various orientations (turned around Z, Y or X by e.g 90' or 180') and measure everything you can.
What happens? I think it should remember being |+X> at least up until the rotation, but what comes after that I have no idea. Can anybody explain what the experimental results would be?
Thanks in advance,
Adrian.
I just read the Feynman Lectures Vol 3, Chapter 6 about spin 1/2 particles going through various arrangements of Stern-Gerlach filters. I think I get it, but I'm not sure why the same arguments don't apply to the spin 1 particles in the preceding chapter.
The chapter goes something like this (skip to ##### if you already read it): first define the axes: Z is upwards and is the direction of the non-uniform magnetic field that splits the beam. Y is along the splitter, presumably from your left hand to your right hand, and X points at you. The splitters split the beam in two so you can see how bright they are and block them as you will, and then recombine them.
Suppose we start with a beam of particles polarised along X or Y, then watch them split and recombine and pop out the other end still knowing which way they were polarised. The only way the two beams could remember that is in the relative phase of the two beams. We can define +X polarisation as the two being in phase, -X as 180' out of phase, and +/-Y as 90/270' out of phase (any other convention will also do.) Then he considers putting another splitter at various funny angles after the first and deduces that:
1) a rotation around the z axis must leave the magnitudes of +Z and -Z alone but should rotate X in and out of Y so
|+Z> -> |+Z> . exp (i@/2)
|-Z> -> |-Z> . exp (-i@/2)
does the trick. E.g. with a 90' rotation, the beams would have gone 45' in opposite directions so their relative phase has swapped X for +/-Y as required.
2) other rotations may not turn X into Y or vice versa so they must leave the relative phase alone. But they should swap Z for X or Y. So if you start with all |Z+> and rotate 90', you should get equal magnitudes of |Z+> and |Z-> but in phase for a rotation about Y:
|+Z> -> |+Z>.cos(@/2) + |-Z>.sin(@/2)
|-Z> -> |+Z>.-sin(@/2) + |-Z>.cos(@/2)
3) and out of phase for a rotation about X:
|+Z> -> |+Z>.cos(@/2) + |-Z>.i.sin(@/2)
|-Z> -> |+Z>.i.sin(@/2) + |-Z>.cos(@/2)
#####
Now then, my question is how that would all look look if we do similar stuff with the spin-1 particles of the previous chapter. First I'd like to know how the experiments would turn out:
First split the beam from the oven into |+X>, |0X> and |-X>, filter it down to |+X>, shove it through an unfiltered Z splitter/recombiner just to admire the three beams (are they all the same or is the middle one brighter?) then put another splitter after that at various orientations (turned around Z, Y or X by e.g 90' or 180') and measure everything you can.
What happens? I think it should remember being |+X> at least up until the rotation, but what comes after that I have no idea. Can anybody explain what the experimental results would be?
Thanks in advance,
Adrian.
