# A Spinor Lorentz Transform via Vectors - Cross Product Issue

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1. Nov 19, 2016

### bolbteppa

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 get the answer?

Last edited: Nov 19, 2016
2. Nov 20, 2016

### vanhees71

Start with the rotations. What's done here is to use the fact that the "fundamental representation" for $\mathrm{SO}(3)$ on 3D real vectors is just the adjoint representation of the group $\mathrm{SU}(2)$. The rotation in the representation $s=1/2$ is given by
$$\hat{R}(\vec{\varphi})=\exp(-\mathrm{i} \vec{\varphi} \cdot \hat{\vec{\sigma}}/2).$$
Now you can evaluate this by using the series expansion of the matrix exponential. Then it's very easy to show that with
$$\vec{x}' \cdot \hat{\vec{\sigma}}=\hat{R} \vec{x} \cdot \hat{\vec{\sigma}} \hat{R}^{\dagger}.$$
$\vec{x}'$ is indeed just the rotated $\vec{x}$ with angle $|\vec{\varphi}|$ around the direction of $\vec{\varphi}$.

The same can be done for the boost with the extended definition, $x^{\mu} \cdot \hat{\sigma}_{\mu}$, where $\sigma_0=\hat{1}$.

For infinitesimal transformations it's even simpler. You just use
$$\exp(-\mathrm{i} \delta \vec{\varphi} \cdot \hat{\vec{\sigma}}/2) = \hat{1}-\mathrm{i} \delta \vec{\varphi} \cdot \hat{\vec{\sigma}}/2$$
for rotations or
$$\exp(-\delta \vec{\eta} \cdot \hat{\vec{\sigma}}/2) = \hat{1}-\mathrm{i} \delta \vec{\eta} \cdot \hat{\vec{\sigma}}/2.$$

3. Nov 20, 2016

### bolbteppa

Landau indeed says we don't actually need to do the calculation and can do what you just said.

However he mentions that it's also possible to directly work the answer out, in the way I have indicated, and that is my question: directly working the answer out.

How in the world does

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

reduce to

$$\lambda = \frac{1}{2}i \sigma \cdot \delta \mathbf{\theta}?$$

4. Nov 20, 2016

### vanhees71

For the rotation you have
$$X'=R X R^{\dagger},$$
and for an infinitesimal $\delta \vec{\varphi}$ you get commutators of $\sigma$ matrices, but these fullfil the commutator relations for angular momenta (up to a factor of $2$),
$$[\sigma_j,\sigma_k]=2\mathrm{i} \epsilon_{jkl} \sigma_l.$$
Tinkering everything together you must get
$$\delta \vec{x} = -\delta \vec{\varphi} \times \vec{x},$$
or in Cartesian Ricci notation
$$\delta x_j=-\epsilon_{jkl} \delta v_k x_l.$$

5. Nov 20, 2016

### bolbteppa

I don't really get what you mean, but I think I found the answer. Using the relation for the cross product given here

https://en.wikipedia.org/wiki/Spinors_in_three_dimensions#Formulation

\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!!!

6. Nov 21, 2016

### vanhees71

What I meant is the following. Starting from the infinitesimal rotation (and not distinguishing upper and lower indices since these are purely Cartesian anyway)
$$x_j'\sigma_j=(1-\mathrm{i} \delta \vec{\phi} \cdot \vec{\sigma}/2)x_j \sigma_{j} (1-\mathrm{i} \delta \vec{\phi} \cdot \vec{\sigma}/2)=x_j \sigma_j -\mathrm{i} \delta \phi_k/2 x_j [\sigma_k,\sigma_j]=x_j \sigma_j +\epsilon_{kjl} \delta \phi_k x^j \sigma_l = [x_j -(\delta \vec{\phi} \times \vec{x})_j]\sigma_j,$$
directly proves that you induce an infinitesimal rotation on the spatial components $x_j$ of the four-position vector (of course you don't change the time component, because it figures in the spinor representation as $x^0 \hat{1}$, and $[\hat{1},\sigma_j]=0$).

7. Nov 21, 2016

Ah, nice!