Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Pauli matrices WITHOUT ladder operators

  1. Oct 4, 2004 #1
    Does anyone know of an alternative way of calculating the Pauli spin matrices[tex] \mbox{ \sigma_x}[/tex] and [tex] \mbox{ \sigma_y}[/tex] (already knowing [tex] \mbox { \sigma_z} [/tex] and the (anti)-commutation relations), without using ladder operators [tex] \mbox{ \sigma_+}[/tex] and [tex] \mbox{ \sigma_- }[/tex]?

    Thanks!
     
    Last edited by a moderator: Oct 4, 2004
  2. jcsd
  3. Oct 4, 2004 #2

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How about brute force ? Knowing that you need traceless hermitean 2x2 matrices, you put in the unknowns, write out all the equations anti-comm relations ... and solve ?

    cheers,
    Patrick.
     
  4. Oct 4, 2004 #3
    Brute force was my initial plan :shy: problem is: Only using comm and anti-comm I get a whole bunch of possible solutions (like e.g. \sigma_x'=-\sigma_x=(0 & -1 \\ -1 & 0) and \sigma_y'=-sigma_y=(0 & i \\ -i \\ 0) ) also obeying these commutation relations.

    I would like to restrict these solutions to the 'traditional' Pauli matrices... Am I forgetting some basic equations somewhere that 'll do just that? :cry:
     
  5. Oct 4, 2004 #4
    Any representation is as good as another !
    You find one, and make a rotation to go to the one you want.
     
  6. Oct 5, 2004 #5
    I give here an alternative way to find the Pauli matrices, which seems natural to me. Any [tex]U(2)[/tex] matrix can be parameterized by :
    [tex]
    M_{U(2)} = \left(
    \begin{array}{cc}
    e^{\imath u}\cos(\theta) & e^{\imath v}\sin(\theta)\\
    -e^{\imath w}\sin(\theta) & e^{\imath (w+v-u)}\cos(\theta)
    \end{array}
    \right)
    [/tex]

    and this reduces in the subgroup [tex]SU(2)[/tex] to [tex]w+v=0[/tex] or :

    [tex]
    M_{SU(2)} = \left(
    \begin{array}{cc}
    e^{\imath u}\cos(\theta) & e^{\imath v}\sin(\theta)\\
    -e^{-\imath v}\sin(\theta) & e^{-\imath u}\cos(\theta)
    \end{array}
    \right)
    [/tex]

    Now as usual to find the generators, one differentiate with respect to each parameters, and takes the values near the identity :

    [tex]
    \frac{\partial M}{\partial\theta} = \left(
    \begin{array}{cc}
    -e^{\imath u}\sin(\theta) & e^{\imath v}\cos(\theta)\\
    -e^{-\imath v}\cos(\theta) & -e^{-\imath u}\sin(\theta)
    \end{array}
    \right)_{\theta=0,u=0,v=0}
    =
    \left(
    \begin{array}{cc}
    0 & 1\\
    -1& 0
    \end{array}
    \right)
    [/tex]


    [tex]
    \frac{\partial M}{\partial u} = \left(
    \begin{array}{cc}
    \imath e^{\imath u}\cos(\theta) & 0\\
    0 & -\imath e^{-\imath u}\cos(\theta)
    \end{array}
    \right)_{\theta=0,u=0,v=0}
    =
    \left(
    \begin{array}{cc}
    \imath & 0\\
    0& -\imath
    \end{array}
    \right)
    [/tex]


    [tex]
    \frac{\partial M}{\partial w} = \left(
    \begin{array}{cc}
    0 & \imath e^{\imath v}\sin(\theta)\\
    \imath e^{-\imath v}\sin(\theta) & 0
    \end{array}
    \right)_{\theta=0,u=0,v=0}
    =
    \left(
    \begin{array}{cc}
    0 & 1\\
    1& 0
    \end{array}
    \right)
    [/tex]


    But these are not the Pauli matrices, they differ by a factor [tex]-\imath[/tex]. This is exactly what is done : the Pauli matrices define an arbitrary [tex]SU(2)[/tex] matrix by :
    [tex]
    M_{SU(2)} =e^{\imath \vec{L}\cdot\vec{\sigma}\alpha/2}=\sigma_0\cos(\frac{\alpha}{2})
    -\imath \vec{L}\cdot\vec{\sigma}\sin(\frac{\alpha}{2})
    [/tex] with [tex]\sigma_0[/tex] the identity, [tex]\vec{L}[/tex] a unitary vector directing the rotation axis, and [tex]\alpha[/tex] the rotation angle. By differentiating this near the identity, one recovers the correct [tex]-\imath[/tex] factor w.r.t. the previously calculated matrices.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding Pauli matrices WITHOUT ladder operators
  1. Pauli Matrices (Replies: 7)

  2. Pauli Matrices (Replies: 6)

  3. Pauli matrices? (Replies: 1)

Loading...