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A How spin projector got included in inverse of Matrix?

  1. Dec 10, 2017 #1
    The following matrix A is,
    \begin{equation}
    A=
    \begin{bmatrix}
    a+b-\sigma\cdot p & -x_1 \\
    x_2 & a-b-\sigma\cdot p
    \end{bmatrix}
    \end{equation}
    The inversion of matrix A is,
    \begin{equation}
    A^{-1}=
    \frac{\begin{bmatrix}
    a-b-\sigma\cdot p & x_1 \\
    -x_2 & a+b-\sigma\cdot p
    \end{bmatrix}}{a^2-b^2+p^2-2\sigma\cdot p-x_1x_2}
    \end{equation}
    The textbook shows the formula in different form,
    \begin{equation}
    A^{-1}=
    \frac{(1/2)\begin{bmatrix}
    (\sigma_o+\sigma\cdot \hat{p})(a-b+p) & (\sigma_o+\sigma\cdot \hat{p}) x_1 \\
    -x_2(\sigma_o+\sigma\cdot \hat{p}) & \sigma_y(\sigma_o+\sigma\cdot \hat{p})\sigma_y(a+b-p)
    \end{bmatrix}}{a^2-b^2-p^2+2bp-x_1x_2}
    \end{equation}
    and p hat is p/b and sigma's are 2x2 pauli matrices. The sigmas inside brackets are projectors. How this projector was derived? How projector is used to get the inversion of matrix A that looks different from conventional method used to calculate A^(-1)?
     
  2. jcsd
  3. Dec 10, 2017 #2
    They are taking a 4x4 matrix as if it where a 2x2 with some generalization of the complex numbers. They use that to make a formula similar to 2x2 inversión formula. The sigmas aren't proyectors, but their anticonmutation relations are a dirac delta dot identity of 2x2. If you look at how to construct the inverse using a similar construction, you'll understand, i think.
     
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