1. Limited time only! Sign up for a free 30min personal 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!

Elementary quantum spin in Sakurai

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data

    I am currently working on a seemingly straightforward eigenvalue problem appearing as problem 1.8 in Sakurai's Modern QM. He asks us to find an eigenket [itex]\vert\vec S\cdot\hat n;+\rangle[/itex] with [tex]\vec S\cdot\hat n\vert\vec S\cdot\hat n;+\rangle = \frac\hbar 2\vert\vec S\cdot\hat n;+\rangle[/tex] where the unit vector n is defined by polar angle alpha and azimuthal angle beta.

    2. Relevant equations

    The definitions of the Pauli sigma matrices, along with the formula [tex]\hat n = (\sin\alpha\cos\beta,\sin\alpha\sin\beta,\cos\alpha)[/tex] for conversion to Cartesian coordinates.

    3. The attempt at a solution

    [tex]\vec S\cdot\hat n = \frac\hbar 2\sin\alpha\cos\beta\begin{pmatrix}0&1\\1&0\end{pmatrix}+\sin\alpha\sin\beta\begin{pmatrix}0&-i\\i&0\end{pmatrix}+\cos\alpha\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \frac\hbar 2\begin{pmatrix}\cos\alpha&\sin\alpha e^{-i\beta}\\\sin\alpha e^{i\beta}&-\cos\alpha\end{pmatrix}.[/tex] Thus we have the [itex]\lambda=1[/itex] eigenvalue problem for the matrix [tex]\begin{pmatrix}\cos\alpha&e^{-i\beta}\sin\alpha\\e^{i\beta}\sin\alpha&-\cos\alpha\end{pmatrix}.[/tex] This becomes the system of equations [tex]\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\cos\alpha+ye^{-i\beta}\sin\alpha-x\\-y\cos\alpha+xe^{i\beta}\sin\alpha-y\end{pmatrix}.[/tex] Thus beta appears to be completely arbitrary, and alpha = n*pi, however this appears to have no solutions as we get cosines equaling 2 which is nonsense. The characteristic polynomial predicts a lambda=1 eigenvalue - where am I going wrong?
     
  2. jcsd
  3. Apr 4, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    This looks good to me.

    I don't understand your matrix on the right hand side. How did you get the ##-x## and the ##-y## terms on the far right of the matrix?

    ##\alpha## and ##\beta## are fixed by the choice of ##\hat{n}##. You need to find ##x## and ##y##.
     
    Last edited: Apr 4, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Elementary quantum spin in Sakurai
  1. Quantum spin (Replies: 1)

Loading...