The Lorentz transformation operator acting on an undotted, i.e. right-handed, spinor can be expressed as $$e^{-\frac{1}{2} \sigma \cdot \mathbf{\phi} + i\frac{1}{2} \sigma \cdot \mathbf{\theta}}.$$

There is a very cool, almost childlike, derivation of this expression in Landau Vol. 4 S. 18, I've never seen anywhere else, deriving $e^{-\frac{1}{2} \sigma \cdot \mathbf{\phi}}$ first, then $e^{ i\frac{1}{2} \sigma \cdot \mathbf{\theta}}$.

When deriving the second term, a cross product arises in the calculation, and I can't make sense of what to do with it. To properly explain the calculation, I have derived $e^{-\frac{1}{2} \sigma \cdot \mathbf{\phi}}$ first to set the notation, and hopefully pique your interest, and then tried to derive the second term. My question will be: can you finish the calculation, and explain the cross product issue?

Given a position vector $$\mathbf{r} = (x,y,z) = (x^1,y^1,z^1),$$ define $$\sigma \cdot \mathbf{r} = \begin{bmatrix} z & x - iy \\ x + iy & - z \end{bmatrix} = x^i \sigma_i = x \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} + y \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} + x \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$

so that

$$x^i = \frac{1}{2} \mathrm{tr}[(\sigma \cdot \mathbf{r}) \sigma_i],$$

abbreviated as

$$\mathrm{r} = \frac{1}{2} \mathrm{tr}[(\sigma \cdot \mathbf{r}) \sigma].$$

Adding $$tI = x^0 \sigma_0$$ to this gives $$T(t,\mathbf{r}) = T(t,x^1,x^2,x^3) = T(t,x,y,z) = T = x^i \sigma_i = \begin{bmatrix} t + z & x - iy \\ x + iy & t - z \end{bmatrix}$$

so that

$$t = \frac{1}{2} \mathrm{tr}(T)$$ and $$x^i = \frac{1}{2} \mathrm{tr}(T \sigma_i),$$

i.e.

$$\mathbf{r} = \frac{1}{2} \mathrm{tr}(T \sigma).$$

If we perform an infinitesimal Lorentz boost on $(t,\mathbf{r})$ with infinitesimal velocity $ \delta \mathbf{V}$ the Lorentz transformation of $(t,\mathbf{r})$ becomes

\begin{align}

t' &= t - \mathbf{r} \cdot \delta \mathbf{V}, \\

\mathbf{r}' &= \mathbf{r} - t \delta \mathbf{V}

\end{align}

and that of $T$ becomes $$T' = BTB^+ = (I + \lambda)T(I + \lambda^+) = T + \lambda T + T \lambda^+$$

so that our goal is to find $B$, i.e. $\lambda$, via

\begin{align}

t' &= t - \mathbf{r} \cdot \delta \mathbf{V} = t - \frac{1}{2} \mathrm{tr}(T \sigma \cdot \delta \mathbf{V}) \\

&= \frac{1}{2}\mathrm{tr}(T') = \frac{1}{2}\mathrm{tr}(BTB^+ ) \frac{1}{2}\mathrm{tr}[(I + \lambda)T(I + \lambda^+)) = t + \frac{1}{2}\mathrm{tr}[T (\lambda + \lambda^+)]

\end{align}

so that $$\lambda + \lambda^+ = - \sigma \cdot \delta \mathbf{V}$$ implies $$\lambda = \lambda^+ = -\frac{1}{2}\sigma \cdot \delta \mathbf{V}.$$

*(Can justify this fully by expanding $\mathbf{r}'$ in the same way and solving both equations for $\lambda, \lambda^+$)*

giving, for $\delta \mathbf{V} = \mathbf{\phi}$

\begin{align}

B &= I + \lambda = I - \frac{1}{2}\sigma \cdot \mathbf{\phi} \\

&= e^{- \frac{1}{2}\sigma \cdot \mathbf{\phi}},

\end{align}

the first part of our Lorentz transformation operator. Calculating the second part is the issue, hence my question. The cross product complicates things.

Under an infinitesimal rotation $\delta \theta$ we see

\begin{align}

\mathbf{r}' &= \mathbf{r} - \delta \theta \times \mathbf{r} \\

&= \frac{1}{2} \mathrm{tr}(T \sigma) - \delta \theta \times \frac{1}{2} \mathrm{tr}(T \sigma) \\

&= \frac{1}{2}\mathrm{tr}(T'\sigma) = \frac{1}{2}\mathrm{tr}(BTB^+\sigma) \frac{1}{2}\mathrm{tr}[(I + \lambda)T(I + \lambda^+)\sigma) = \frac{1}{2} \mathrm{tr}(T \sigma) + \frac{1}{2}\mathrm{tr}[ \lambda T \sigma + T \lambda^+ \sigma ]

\end{align}

and solving for $\lambda$ in the equality

$$- \delta \theta \times \frac{1}{2} \mathrm{tr}(T \sigma) = \frac{1}{2}\mathrm{tr}[ \lambda T \sigma + T \lambda^+ \sigma ]$$

is unmanageable to me, but the answer is $$\lambda = \frac{1}{2}i \sigma \cdot \delta \mathbf{\theta}.$$

How do you deal with this cross product and finish the calculation, to get the answer?

**Edit:** I think I found the answer. Using the relation for the cross product given here:

\begin{align}

X &= \mathbf{\sigma} \cdot \mathbf{x}, \\

Y &= \mathbf{\sigma} \cdot \mathbf{y}, \\

i \mathbf{\sigma} \cdot ( \mathbf{x} \times \mathbf{y}) &= \frac{1}{2}(XY - YX)

\end{align}

I should have written

\begin{align}

\mathbf{r}' &= \mathbf{r} - \delta \theta \times \mathbf{r} \\

&= \frac{1}{2} \mathrm{tr}(T \sigma) - \frac{1}{2} \mathrm{tr}( \delta \theta \times \mathbf{r}) \\

&= \frac{1}{2}\mathrm{tr}(T'\sigma) = \frac{1}{2}\mathrm{tr}(BTB^+\sigma) \frac{1}{2}\mathrm{tr}[(I + \lambda)T(I + \lambda^+)\sigma) = \frac{1}{2} \mathrm{tr}(T \sigma) + \frac{1}{2}\mathrm{tr}[ \lambda T \sigma + T \lambda^+ \sigma ]

\end{align}

and so I think that settles it, phew!!!