[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].