Damn you, Otto Stern and Walther Gerlach

[Onamor]

Homework Statement

The last (8 mark) part of this (also in attachment):
[URL]http://img189.imageshack.us/i/imag0095l.jpg/[/URL]

(very sorry for having to post an image, I spent a good hour trying to tex it in this, but there's something wrong with my parsing in the preview post. anyway...)

Homework Equations

the answer to the penultimate part (the 4 marker) is the "up" spin has 0.82 and the "down" has 0.18 relative intensity.

The Attempt at a Solution

Using \hat{S}_{y}=hbar/2\sigma_{y} you can solve the eigenequation \sigma_{y}|\chi> = \lambda|\chi> to find the eigenstates of \hat{S}_{y} in terms of \alpha and \beta:

|+>_y = 1/\sqrt{2} (|\alpha>_y + i|\beta>_y) and

|->_y = 1/\sqrt{2} (|\alpha>_y - i|\beta>_y)
But these are pretty standard results...
They are the states that spin in the +y and -y directions (please correct me if I am wrong on that). Do I just take _y<+|\alpha> and _y<-|\alpha>and get the coefficients for the two beams?

Whether or not i then need to multiple by the intensities found in the part beforehand is another question... (I know intensity != probability but it seems sensible?..)

Ultra thanks to anyone who can help -Im revising for my finals :)

Homework Statement

Homework Equations

The Attempt at a Solution

 

[sgd37]

you need to take each of the possible beams in the previous question which have states \left| \Psi_i \right\rangle say and contract them with the +y eigenstate to get a coefficient. The previous intensities can be discarded

 

[Onamor]

Thanks, i think I've done it now with your help. Each beam has 50/50 probability as you would expect. thanks again.