- #1

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## Homework Statement

This isn't exactly a homework question, but rather from the lecturer's slides. I'm not sure if I understand what's going on.

## Homework Equations

## The Attempt at a Solution

__Branch a is blocked__

1. Atoms all with +ve spin enter

2. all the atoms can approach the terminal a (+)

3. Just before crossing over the gap, half tries to go through a(+), half tries to go through b(-).

**factor = 1/2**

4. atoms passes through b, +ve spin gets converted into -ve spin

5. second analyser at 90

^{o}, so splits into cos

^{2}(90/2) = 1/2 with +ve spin, splits into 1 - cos

^{2}(90/2) = sin

^{2}(90/2) 1/2 with -ve spin

**factor = 1/2**

6. Result: 1/4 with -ve spin, 1/4 with positive spin. Ones with +ve spin approaches and crashes at blue barrier. Ones with -ve spin approaches below. Result = 1/4 -ve spin.

__Neither Branch is blocked__

1. atoms approach a(+)

2. Just before crossing over, splits into a(+) and (b-);

**factor = 1/2**

3. Recombines.

**factor becomes 1.**

4. second analyser at 90

^{o}, so splits into cos

^{2}(90/2) = 1/2 with +ve spin, splits into 1 - cos

^{2}(90/2) = sin

^{2}(90/2) 1/2 with -ve spin

**factor = 1/2**

5. Result: 1/2 with -ve spin, 1/2 with positive spin. Ones with +ve spin approaches and crashes at blue barrier. Ones with -ve spin approaches below. Result = 1/2 -ve spin.

But, from the slides I can infer that going to an inferometer that's 90 degrees does not change the particle's spin at all..if it's all +ve heading towards a perpendicular inferometer, it remains +ve.

However, this seems to contradict what i'm currently reading, which is also notes provided by the same lecturer. It suggests that passing from a z-plane to a x-plane stern-gerlach apparatus appears to 'reset' the spins back to having both +ve and -ve.