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

In [1], Prof Osmond explicitly boosts a \( u^s(p_0) \) Dirac spinor from the rest frame with rest frame energy \( p_0 \).
After doing so he claims the identification
\begin{equation}\label{eqn:squarerootpsigma:20}
\begin{aligned}
\sqrt{m} e^{-\inv{2} \eta \sigma^3} &= \sqrt{ p \cdot \sigma } \\
\sqrt{m} e^{\inv{2} \eta \sigma^3} &= \sqrt{ p \cdot \overline{\sigma} },
\end{aligned}
\end{equation}
for the components of \( u^s(\Lambda p_0) \).

Let’s verify this by squaring. First
\begin{equation}\label{eqn:squarerootpsigma:40}
e^{\pm \inv{2} \eta \sigma^3 }
=
\cosh\lr{ \inv{2} \eta \sigma^3 }
\pm
\sinh\lr{ \inv{2} \eta \sigma^3 } \sigma^3,
\end{equation}
which squares to (FIXME: link to uvspinor.nb)
\begin{equation}\label{eqn:squarerootpsigma:60}
\lr{ e^{\pm \inv{2} \eta \sigma^3 } }^2
=
\begin{bmatrix}
e^{\pm \eta} & 0 \\
0 & e^{\mp \eta}
\end{bmatrix}.
\end{equation}

Explicitly boosting the rest energy \( p_0 \) gives
\begin{equation}\label{eqn:squarerootpsigma:80}
\begin{bmatrix}
p_0 \\
0
\end{bmatrix}
\rightarrow
\begin{bmatrix}
\cosh\eta & \sinh\eta \\
\sinh\eta & \cosh\eta \\
\end{bmatrix}
\begin{bmatrix}
p_0 \\
0
\end{bmatrix}
=
p_0
\begin{bmatrix}
\cosh\eta \\
\sinh\eta
\end{bmatrix},
\end{equation}
so after the boost
\begin{equation}\label{eqn:squarerootpsigma:100}
\begin{aligned}
p \cdot \sigma
&\rightarrow
p_0 \lr{ \cosh \eta – \sinh \eta \sigma^3 } \\
&= p_0
\begin{bmatrix}
\cosh\eta – \sinh\eta & 0 \\
0 & \cosh\eta + \sinh\eta
\end{bmatrix} \\
&=
p_0
\begin{bmatrix}
e^{-\eta} & 0 \\
0 & e^{\eta}
\end{bmatrix},
\end{aligned}
\end{equation}
where \( p_0 = m \) is still the rest frame energy. However, according to \ref{eqn:squarerootpsigma:60} this is exactly
\begin{equation}\label{eqn:squarerootpsigma:120}
\lr{\sqrt{m} e^{-\inv{2} \eta \sigma^3 }}^2
\end{equation}

Since \( p \cdot \overline{\sigma} \) flips the signs of the spatial momentum, we have shown that
\begin{equation}\label{eqn:squarerootpsigma:140}
\begin{aligned}
\lr{\sqrt{m} e^{-\inv{2} \eta \sigma^3 }}^2 &= p \cdot \sigma \\
\lr{\sqrt{m} e^{\inv{2} \eta \sigma^3 }}^2 &= p \cdot \overline{\sigma},
\end{aligned}
\end{equation}
which isn’t a full proof of the claimed result (i.e. the most general orientation isn’t considered), but at least validates the claim.

References

[1] Dr. Tobias Osborne. Qft lecture 15, dirac equation, boost from stationary frame. Youtube. URL https://youtu.be/J2lV8uNx0LU?list=PLDfPUNusx1EpRs-wku83aqYSKfR5fFmfS&t=4328. [Online; accessed 18-December-2018].