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Reading through David Tong lecture notes on QFT.

On page 100, he solves the Dirac equation by Pauli four-vector. See below link:

QFT notes by Tong, Chapter 4

In (4.107) he gives the solution in terms of $\sqrt{p.\sigma}$ where $p$ and $\sigma$ are four vectors.

When I follow the proof this equation makes sense:

$$(p.\sigma)(p.\bar{\sigma})=m^2$$ because $$(p.\sigma)(p.\bar{\sigma})= p_{\mu}p^{\mu} I_{2} =m^2I_{2}$$ Both sides are matrices.

But $\sqrt{p.\sigma}$ doesn't make sense. Are we taking the square root of a matrix and treating it like a scalar? In particular to show that:

$$(p.\sigma)\sqrt{p.\bar{\sigma}}=m\sqrt{p.\sigma}$$

How we expand the Pauli and momentum four vectors in $\sqrt{p.\sigma}$?

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    $\begingroup$ Did you try searching for the square root of a matrix? If no, do that first. If yes, what specifically do you need to have explained more carefully about this concept? $\endgroup$ – ACuriousMind Dec 3 '16 at 23:22
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It pays to first inspect the extraordinary matrix you are given. The Pauli 4-vector is $$ p \cdot \sigma= \begin{pmatrix} p_0-p_3&-p_1+ip_2\\-p_1-ip_2&p_0+p_3\end{pmatrix}. $$ It is hermitian and so unitarily diagonalizable.

Can you see its determinant is $p\cdot p=m^2$? Did you now find its eigenvalues $p_0\pm\sqrt{\vec{p}^2}$, and the corresponding eigenvectors, and so the unitary matrix U diagonalizing it?

So, putting math fussing aside until finding one's bearings, that same U matrix also, ipso facto, diagonalizes any power of this Pauli 4-vector matrix, and so any Taylor expandable function thereof, and so, ahem, any function. Damn the torpedoes and assume this holds for the sqrt.

So your answer might be $U\left(\operatorname{diag}(\sqrt{p_0+\sqrt{\vec{p}^2}},~\sqrt{p_0-\sqrt{\vec{p}^2}}~)\right)U^\dagger$.

Now, inspect $p\cdot \bar{\sigma}$. Does it have the same determinant? The same eigenvalues? The same eigenvectors? The same U? But are they in the same position in the diagonalized matrix? You never had to do this to get the formal identity your write above, but can you see it now, trivially, in this explicit form? You never had to evaluate U explicitly, after all.

I hope you get the idea... These formal square roots correspond to something tangible, but you can bypass details with the crucial symmetrization underlying the identity you wrote.

While you are inspecting it, it would be a shame to not look at the celebrated static limit of it, for a nontrivial m, and, by contrast, the even more celebrated m=0 limit: you already isolated the 0 eigenvalue.

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