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Solution to the dirac equation and the square root of a matrix?

  1. Sep 15, 2012 #1
    Hi. I'm currently reading about (negative frequency) solutions to the Dirac equations which can be written on the form

    [tex] \Psi = ( \sqrt{p \cdot \sigma} \chi, \sqrt{p \cdot \bar{\sigma}} \chi)^T e^{-i p \cdot x}[/tex]


    For any two component spinor Chi. But the dot product with the four vector p and the sigma vector is a matrix, so here one is taking the square root of a matrix. What do we mean by that? Or am I interpreting this wrong?
     
  2. jcsd
  3. Sep 16, 2012 #2
    The square root of a matrix A is another matrix B such that BB = A. Note that, as with the square roots of numbers, there is more than one square root of a matrix. Off the top of my head, I think that a general NxN matrix could have up to 2^N distinct square roots.

    If you can diagonalize A, so that A = M^-1 D M, where D is diagonal, then D lists the eigenvalues of A. Then you can verify that M^-1 sqrt(D) M is a square root of A, where sqrt(D) is a diagonal matrix whose entries are the square roots of the entries of D. So the square root of a matrix A has the same eigenvectors as A, but with eigenvalues that are the square roots of the eigenvalues of A.

    Are you reading Peskin & Schroeder? I seem to recall that they make a comment to the effect that by convention they take the positive square root. So when they take the square roots of the eigenvalues, they choose positive signs.
     
  4. Sep 17, 2012 #3

    Bill_K

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    center o bass, Did you see this expression in Peskin & Schroeder, or somewhere else?
     
  5. Sep 19, 2012 #4
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