Damn you, Otto Stern and Walther Gerlach

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 2K views
Onamor
Messages
76
Reaction score
0

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

|+>y = 1/[tex]\sqrt{2}[/tex] (|[tex]\alpha[/tex]>y + i|[tex]\beta[/tex]>y) and

|->y = 1/[tex]\sqrt{2}[/tex] (|[tex]\alpha[/tex]>y - i|[tex]\beta[/tex]>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<+|[tex]\alpha[/tex]> and y<-|[tex]\alpha[/tex]>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

 

Attachments

  • IMAG0095.jpg
    IMAG0095.jpg
    40.7 KB · Views: 704
Last edited by a moderator:
Physics news on Phys.org
you need to take each of the possible beams in the previous question which have states [tex]\left| \Psi_i \right\rangle[/tex] say and contract them with the +y eigenstate to get a coefficient. The previous intensities can be discarded
 
Thanks, i think I've done it now with your help. Each beam has 50/50 probability as you would expect. thanks again.