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I am having trouble understanding the steps taken in (11.27) equation in Quantum Field Theory and the Standard Model by M.D.Schwartz.

I don't understand how to get the middle diagonal matrix in the second last statement of the equation.

Here is my attempt. I am writing $$ \begin{pmatrix} \sqrt{p \cdot \sigma} \xi_{s'} \\ \sqrt{p \cdot \bar \sigma} \xi_{s'} \end{pmatrix} = \begin{pmatrix} \sqrt{p \cdot \sigma} \quad 0\\ 0 \quad \sqrt{p \cdot \bar \sigma} \end{pmatrix} \begin{pmatrix}\xi_{s'} \\ \xi_{s'} \end{pmatrix} \quad and \quad \begin{pmatrix} \sqrt{p \cdot \sigma} \xi_{s} \\ \sqrt{p \cdot \bar \sigma} \xi_{s} \end{pmatrix}^\dagger = \begin{pmatrix}\xi_{s} \\ \xi_{s} \end{pmatrix}^\dagger \begin{pmatrix} \sqrt{p \cdot \sigma} \quad 0\\ 0 \quad \sqrt{p \cdot \bar \sigma} \end{pmatrix}^\dagger $$

and then proceeding as follows:

$$ \quad \begin{pmatrix} \sqrt{p \cdot \sigma} \xi_s \\ \sqrt{p \cdot \bar \sigma} \xi_s \end{pmatrix}^\dagger \begin{pmatrix} 0 \ \mathbb{I}_2 \\ \mathbb{I}_2 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{p \cdot \sigma} \xi_{s'} \\ \sqrt{p \cdot \bar \sigma} \xi_{s'} \end{pmatrix} \quad \quad (x) $$

$$ = \begin{pmatrix}\xi_{s} \\ \xi_{s} \end{pmatrix}^\dagger \begin{pmatrix} \sqrt{p \cdot \sigma} \quad 0\\ 0 \quad \sqrt{p \cdot \bar \sigma} \end{pmatrix}^\dagger \begin{pmatrix} 0 \ \mathbb{I}_2 \\ \mathbb{I}_2 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{p \cdot \sigma} \quad 0\\ 0 \quad \sqrt{p \cdot \bar \sigma} \end{pmatrix} \begin{pmatrix}\xi_{s'} \\ \xi_{s'} \end{pmatrix} \quad \quad \quad (a)$$

$$ = \begin{pmatrix}\xi_{s} \\ \xi_{s} \end{pmatrix}^\dagger \begin{pmatrix} \sqrt{p \cdot \sigma} \quad 0\\ 0 \quad \sqrt{p \cdot \bar \sigma} \end{pmatrix} \begin{pmatrix} 0 \ \mathbb{I}_2 \\ \mathbb{I}_2 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{p \cdot \sigma} \quad 0\\ 0 \quad \sqrt{p \cdot \bar \sigma} \end{pmatrix} \begin{pmatrix}\xi_{s'} \\ \xi_{s'} \end{pmatrix} \quad \quad \quad (b)$$

$$ = \begin{pmatrix}\xi_{s} \\ \xi_{s} \end{pmatrix}^\dagger \begin{pmatrix} \sqrt{p \cdot \sigma} \quad 0\\ 0 \quad \sqrt{p \cdot \bar \sigma} \end{pmatrix} \begin{pmatrix} 0 \quad \sqrt{p \cdot \bar \sigma} \\ \sqrt{p \cdot \sigma} \quad 0 \end{pmatrix} \begin{pmatrix}\xi_{s'} \\ \xi_{s'} \end{pmatrix} \quad \quad \quad (c)$$

$$ = \begin{pmatrix}\xi_{s} \\ \xi_{s} \end{pmatrix}^\dagger \begin{pmatrix} 0 \quad \sqrt{(p \cdot \sigma )( p \cdot \bar \sigma)} \\ \sqrt{(p \cdot \bar \sigma )( p \cdot \sigma)} \quad 0 \end{pmatrix} \begin{pmatrix}\xi_{s'} \\ \xi_{s'} \end{pmatrix} \quad \quad \quad (d)$$

Clearly, my middle matrix is very different than the correct* one. Can anyone please point out or hint what am I doing wrong? Or how the author obtained the middle diagonal matrix?

*Update: The "correct" expression ( i.e the one given in the book is ):

$$ = \begin{pmatrix}\xi_{s} \\ \xi_{s} \end{pmatrix}^\dagger \begin{pmatrix} \sqrt{(p \cdot \sigma )( p \cdot \bar \sigma)} \quad 0 \\ 0 \quad \sqrt{(p \cdot \sigma )( p \cdot \bar \sigma)} \end{pmatrix} \begin{pmatrix}\xi_{s'} \\ \xi_{s'} \end{pmatrix} \quad \quad \quad (y)$$

To summarize, equation 11.27 says: $$ (x) = (y) = 2m\delta_{ss'}$$

My main concern is how did the author go from (x) to (y)?

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    $\begingroup$ Examine your third to last equation and the second to last one: you didn't compute the matrix product properly $\endgroup$
    – daydreamer
    Oct 16, 2020 at 7:12
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    $\begingroup$ Sorry @daydreamer, I still don't see the mistake. There is just one matrix multiplication that I did in going from (b) to (c). I checked it by multiplying 4x4 matrices manually but got the same result that I have mentioned in (c). Were you referring to the same step? $\endgroup$
    – baba26
    Oct 16, 2020 at 14:31
  • $\begingroup$ Sorry I was sleepy when I wrote that. Your product is all right. But, can you please give what "the true answer" should be? Otherwise, in order to obtain a diagonal matrix I'd suggest trying to diagonalize the matrix in the middle of d) $\endgroup$
    – daydreamer
    Oct 16, 2020 at 14:36

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I got the book and think I got it: he just switched columns 1 and 2. And you can do it without any doubt since your spinor components are "the same"

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