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Sping Matrices and Commutation Relations

  1. Dec 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Check that the spin matrices (Equations 4.145 and 4.147) obey the fundamental commutation relations for angular momentum, Equation 4.134.


    2. Relevant equations
    Eq. 4.147a --> [tex]S_{x} = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}[/tex]

    Eq. 4.147b --> [tex]S_{y} = \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}[/tex]

    Eq. 4.145 --> [tex]S_{z} = \frac{\hbar}{2}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}[/tex]

    [tex]\left[ S_{x}, S_{y}\right] = i\hbar S_{z}[/tex]

    [tex]\left[ S_{y}, S_{z}\right] = i\hbar S_{x}[/tex]

    [tex]\left[ S_{z}, S_{x}\right] = i\hbar S_{y}[/tex]


    3. The attempt at a solution

    Well, from Eq. 4.147 I have [tex]S_{x} = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}[/tex] and also [tex]S_{y} = \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}[/tex]

    So [tex]\left[ S_{x}, S_{y} \right] = i\hbar S_{z} \Rightarrow[/tex]

    [tex]\left[ \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}, \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \right] \Rightarrow[/tex]

    [tex]\frac{\hbar^{2}}{4}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} - \frac{\hbar^{2}}{4}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \Rightarrow[/tex]

    [tex]\frac{\hbar^{2}}{4}\begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix} - \frac{\hbar^{2}}{4}\begin{pmatrix}-i & 0 \\ 0 & i \end{pmatrix} = \frac{\hbar^{2}}{4}\begin{pmatrix}2i & 0 \\ 0 & -2i \end{pmatrix}[/tex]

    It doesn't quite seem right to me because the answer shows:

    [tex]\left[ S_{x}, S_{y}\right] = i\hbar S_{z}[/tex]

    But I can't see how I'd get that from where I'm going.
     
    Last edited: Dec 1, 2007
  2. jcsd
  3. Dec 1, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Your answer IS i*hbar*S_z.
     
  4. Dec 3, 2007 #3
    Err... hey, you're right. :D

    This idiot thanks you.
     
    Last edited: Dec 3, 2007
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