1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to find this spin component

  1. Sep 16, 2007 #1
    We know that [tex] S_x = \frac{\hbar}{2} \left( |+ \rangle \langle - | + | - \rangle \langle+| \right)[/tex]

    But what is [tex]|S_x ; + \rangle[/tex]?

    I think my text says [tex]|S_x ; + \rangle = \frac{1}{\sqrt{2}} \left( |+ \rangle + | - \rangle \right)[/tex] but i dont know how they got this.

    I feel like this is a trivial question but I'm not sure how one finds [tex]|S_x ; + \rangle[/tex]
     
  2. jcsd
  3. Sep 17, 2007 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    But WHAT is [itex] |S_{x},+\rangle [/itex] ?? I've never seen this notation before...And it's not that i've looked into one book...:rolleyes: I haven't looked in your book, apparently, you might share with us the title and the author...
     
  4. Sep 17, 2007 #3
    I imagine that it's the positive spin direction for S_x.

    OP: it's just an eigenvector -- so you find it in the same way that you find any eigenvectors. If it helps, write S_x as a matrix, in the |+>, |-> basis that you've got things in.
     
  5. Sep 17, 2007 #4
    I just figured it out.

    I am using * as the dot product

    [tex]S * \hat n | S * \hat n ; + \rangle = \frac{ \hbar}{2} | S * \hat n ; + \rangle [/tex]
    [tex]| S * \hat n ; + \rangle = \cos \frac{\beta}{2} |+ \rangle + \sin \frac{\beta}{2} e^{i \alpha} | - \rangle [/tex]

    where beta is the polar angle and alpha is the azimuthal angle.

    therefore, an S_x measurement would be where beta = pi/2 and alpha =0

    since the S_x measurement would yield +hbar/2, we get:

    [tex]| S_x; + \rangle = \cos \frac{\pi/2}{2} |+ \rangle + \sin \frac{\pi/2}{2} e^{0} | - \rangle [/tex]

    therefore:
    [tex]|S_x ; + \rangle = \frac{1}{\sqrt{2}} \left( |+ \rangle + | - \rangle \right)[/tex]
     
    Last edited: Sep 17, 2007
  6. Sep 17, 2007 #5

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    They want the state which is an eigenstate of Sx with the eigenvalue +hbar/2.

    So you could write [tex]|S_x ; + \rangle = \alpha
    |+ \rangle + \beta |-
    \rangle [/tex]

    and apply S_x, imposing [itex] S_x |S_x; + > = \frac{\hbar}{2} |S_x;+> [/itex] and then solve for alpha and beta (and normalize at the end)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How to find this spin component
  1. Finding spin state. (Replies: 3)

Loading...