[Click here for a PDF of this post with nicer formatting]

With the help of Mathematica, a fairly compact form was found for the root of \( p \cdot \sigma \)

\begin{equation}\label{eqn:DiracUVmatricesExplicit:121}

\sqrt{ p \cdot \sigma }

=

\inv{

\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }

}

\begin{bmatrix}

\omega_\Bp- p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 } & – p^1 + i p^2 \\

– p^1 – i p^2 & \omega_\Bp+ p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 }

\end{bmatrix}.

\end{equation}

A bit of examination shows that we can do much better. The leading scalar term can be simplified by squaring it

\begin{equation}\label{eqn:squarerootpsigma:140}

\begin{aligned}

\lr{ \sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} } }^2

&=

\omega_\Bp- \Norm{\Bp} + \omega_\Bp+ \Norm{\Bp} + 2 \sqrt{ \omega_\Bp^2 – \Bp^2 } \\

&=

2 \omega_\Bp + 2 m,

\end{aligned}

\end{equation}

where the on-shell value of the energy \( \omega_\Bp^2 = m^2 + \Bp^2 \) has been inserted. Using that again in the matrix, we have

\begin{equation}\label{eqn:squarerootpsigma:160}

\begin{aligned}

\sqrt{ p \cdot \sigma }

&=

\inv{\sqrt{ 2 \omega_\Bp + 2 m }}

\begin{bmatrix}

\omega_\Bp- p^3 + m & – p^1 + i p^2 \\

– p^1 – i p^2 & \omega_\Bp+ p^3 + m

\end{bmatrix} \\

&=

\inv{\sqrt{ 2 \omega_\Bp + 2 m }}

\lr{

(\omega_\Bp + m) \sigma^0

-p^1 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}

-p^2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}

-p^3 \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}

} \\

&=

\inv{\sqrt{ 2 \omega_\Bp + 2 m }}

\lr{

(\omega_\Bp + m) \sigma^0

-p^1 \sigma^1

-p^2 \sigma^2

-p^3 \sigma^3

} \\

&=

\inv{\sqrt{ 2 \omega_\Bp + 2 m }}

\lr{

(\omega_\Bp + m) \sigma^0 – \Bsigma \cdot \Bp

}.

\end{aligned}

\end{equation}

We’ve now found a nice algebraic form for these matrix roots

\begin{equation}\label{eqn:squarerootpsigma:180}

\boxed{

\begin{aligned}

\sqrt{p \cdot \sigma} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \sigma } \\

\sqrt{p \cdot \overline{\sigma}} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \overline{\sigma}}.

\end{aligned}}

\end{equation}

As a check, let’s square one of these explicitly

\begin{equation}\label{eqn:squarerootpsigma:101}

\begin{aligned}

\lr{ \sqrt{p \cdot \sigma} }^2

&= \inv{2 \omega_\Bp + 2 m }

\lr{ m^2 + (p \cdot \sigma)^2 + 2 m (p \cdot \sigma) } \\

&= \inv{2 \omega_\Bp + 2 m }

\lr{ m^2 + (\omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + \Bp^2) + 2 m (p \cdot \sigma) } \\

&= \inv{2 \omega_\Bp + 2 m }

\lr{ 2 \omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + 2 m (\omega_\Bp – \Bsigma \cdot \Bp) } \\

&= \inv{2 \omega_\Bp + 2 m }

\lr{ 2 \omega_\Bp \lr{ \omega_\Bp + m } – (2 \omega_\Bp + 2 m) \Bsigma \cdot \Bp } \\

&=

\omega_\Bp – \Bsigma \cdot \Bp \\

&=

p \cdot \sigma,

\end{aligned}

\end{equation}

which validates the result.

texttexttext`code`

more code

~~~~