Lorentz transformations on Spinors

[Illuminatum]

Hi guys,

I'm currently struggling to show something my lecturer told us in class. We have that
$$\Psi\left(x\right) \rightarrow S\left(L\right)\Psi\left(L^{-1}x\right)$$
under a Lorentz transform defined
$$L = exp\left(\frac{1}{2}\Omega_{ij}M^{ij}\right)$$
with
$$S\left(L\right) = exp\left(\frac{1}{2}\Omega_{ij}S^{ij}\right)$$
where $$M^{ij}$$ are the Lorentz generators for "position" based transformations and $$S^{ij} = \frac{1}{4}\left[\gamma^{i}, \gamma^{j}\right]$$ are the spinor based generators for the spinor representation.

I have been told that if we take a solution to the Dirac equation with zero 3-momentum
$$\Psi\left(x, 0\right) = u\left(0\right)e^{-imt} = \sqrt{2m}\begin{pmatrix} \chi_{s} \\ 0\end{pmatrix}e^{-imt}$$ for (the sake of argument take spin-up
$$\chi_{s} = \begin{pmatrix}1 \\ 0\end{pmatrix}$$
and apply a Lorentz boost to it, such that in the new frame it would have momentum p, we would get the full spinor
$$\Psi\left(x, p\right) = u\left(p\right)\sqrt{E_{p} + m}\begin{pmatrix} \chi_{s} \\ \frac{\sigma.p}{E + m}\chi_{s}\end{pmatrix}e^{-p.x}$$

So to get a boost in the x direction, we want to take only $$\Omega_{i0} = -\Omega_{0i} = -\phi$$ so that
$$L = \begin{pmatrix}\cosh{\phi} & -\sinh{\phi} & 0 & 0\\ -\sinh{\phi} & \cosh{\phi} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$
where I guess I should point out that
$$M^{0i} = -M^{i0} = \begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0& 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}$$

Then in the representation (which we have been using all through the course, and I want to use for this question)
$$\gamma^{0} = \begin{pmatrix}I & 0 \\ 0 & -I\end{pmatrix}$$
$$\gamma^{i} = \begin{pmatrix}0 & \sigma_{i} \\ -\sigma_{i} & 0\end{pmatrix}$$
I find
$$S^{i0} = -S^{0i} = \begin{pmatrix}0 & \frac{\sigma_{x}}{2} \\ \frac{ \sigma_{x}}{2} & 0 \end{pmatrix}$$

So taking the exponential expression for S(L) I find
$$S(L) = \begin{pmatrix} \cosh{(\phi / 2)} & \sigma_{x}\sinh{(\phi / 2)} \\ \sigma_{x}\sinh{(\phi / 2)} & \cosh{(\phi /2)}\end{pmatrix}$$

Ok, so here are my problems...

Firstly, I'm not sure exactly how to apply the [tex]\Psi(\L^{-1}x)$$bit. Obviously I can find the inverse Lorentz transform and apply that to the 4-vector x, giving me x'' say, but should that then be plugged into the exponential as (-imt'') or as (-im p'.x'') or what...? And where is the energy contribution to the normalisation going to come in?<br /> <br /> Secondly, the S-matrix I have found must be applied to the spinor components, and it does not seem to give the correct expression for a spinor of momentum p in the x-direction. Though maybe it does after manipulation of the half angle formulae...I can't show it is true...<br /> <br /> Any thoughts would be greatly appreciated.<br /> <br /> Thanks,<br /> James$$[/tex]

 

[Bill_K]

Looks like you're very much on the right track. Here's a hint: working backwards, E + m = m(γ+1), and γ = cosh φ. I think if you apply the half-angle formulas to your expression like you said, you can hook up with this.

 

[Illuminatum]

Ok, that relation looks like it might help, and I'll give the half angle formulae a go...

What is the correct way to handle to transform of the x dependent part of the wavefunction? Do I keep -imt and substitute the new t, or do I go for -ip.x with the new x and p?

Thanks

 

[The_Duck]

Ok, that relation looks like it might help, and I'll give the half angle formulae a go...

What is the correct way to handle to transform of the x dependent part of the wavefunction? Do I keep -imt and substitute the new t, or do I go for -ip.x with the new x and p?

Thanks

Your first equation tells you to swap out x for $$L^{-1} x$$. Now, note that if we write the rest 4-momentum as $$p_r = (m, 0, 0, 0)$$ then $$m t = p_r \cdot x \to p_r \cdot (L^{-1} x) = (p_r L^{-1}) \cdot x = p \cdot x$$ where $$p = p_r L^{-1} = L p_r$$ is the correct boosted 4-momentum.