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

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
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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.
 
  • #3
Bill_K
Science Advisor
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center o bass, Did you see this expression in Peskin & Schroeder, or somewhere else?
 

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