Finding |<+y|-θ>|^2 in terms of θ

[NanoChrisK]

Homework Statement

What is the probability that an electron emerging from the plus channel of the first device will end up in the minus channel of the second device? Express your answer in terms of θ. (Refer to the attached image)

Homework Equations

This uses the outcome probability rule that says that "In an experiment that determines the value of an observable, the probability of any given result (i.e., the probability that the quanton's state will collapse to that result's eigenvector) is the absolute square of the inner product of the quanton's original state and the result's eigenvector" (From "Six Ideas that Shaped Physics: Unit Q" by Thomas A. Moore pp. 103")

The Attempt at a Solution

This is saying that the result will be 50% regardless of θ. This can't be right. Graphing 1/2(sinθ/2-cosθ/2)^2 gives a probability between 0% and 100%, which I would expect, but that's not what I get when I simplify it.

I suspect I've done something wrong when I've said sin^2(1/2*θ)+cos^2(1/2*θ)=1, but if the values inside sin^2 and cos^2 are the same, why can't I just say sin^2(A)+cos^2(A)=1?

On a side note: The second part of the question asks what the quanton's spin state vector would be after leaving the second device. I believe I would use the superposition rule to find this out, but I'm not sure. But I can put that in a different thread if I can't figure it out.

UPDATE:

Ok, I think I've found the solution. I was implying that (A-B)^2=A^2+B^2, which was INCORRECT!. I guess the problem is that I need to re-take basic algebra. Here my new attempt:

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[Matterwave]

Your updated solution looks correct to me, assuming you have constructed the $\left|-\theta\right>$ ket correctly (I can't recall the results well enough to know if you have a minus sign missing or something like that).