- #1
Illuminatum
- 8
- 0
Hi guys,
I'm currently struggling to show something my lecturer told us in class. We have that
[tex]\Psi\left(x\right) \rightarrow S\left(L\right)\Psi\left(L^{-1}x\right)[/tex]
under a Lorentz transform defined
[tex]L = exp\left(\frac{1}{2}\Omega_{ij}M^{ij}\right)[/tex]
with
[tex]S\left(L\right) = exp\left(\frac{1}{2}\Omega_{ij}S^{ij}\right)[/tex]
where [tex]M^{ij}[/tex] are the Lorentz generators for "position" based transformations and [tex]S^{ij} = \frac{1}{4}\left[\gamma^{i}, \gamma^{j}\right][/tex] 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
[tex]\Psi\left(x, 0\right) = u\left(0\right)e^{-imt} = \sqrt{2m}\begin{pmatrix} \chi_{s} \\ 0\end{pmatrix}e^{-imt}[/tex] for (the sake of argument take spin-up
[tex]\chi_{s} = \begin{pmatrix}1 \\ 0\end{pmatrix}[/tex]
and apply a Lorentz boost to it, such that in the new frame it would have momentum p, we would get the full spinor
[tex]\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}[/tex]
So to get a boost in the x direction, we want to take only [tex]\Omega_{i0} = -\Omega_{0i} = -\phi[/tex] so that
[tex]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}[/tex]
where I guess I should point out that
[tex]M^{0i} = -M^{i0} = \begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0& 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}[/tex]
Then in the representation (which we have been using all through the course, and I want to use for this question)
[tex]\gamma^{0} = \begin{pmatrix}I & 0 \\ 0 & -I\end{pmatrix}[/tex]
[tex]\gamma^{i} = \begin{pmatrix}0 & \sigma_{i} \\ -\sigma_{i} & 0\end{pmatrix}[/tex]
I find
[tex]S^{i0} = -S^{0i} = \begin{pmatrix}0 & \frac{\sigma_{x}}{2} \\ \frac{ \sigma_{x}}{2} & 0 \end{pmatrix}[/tex]
So taking the exponential expression for S(L) I find
[tex]S(L) = \begin{pmatrix} \cosh{(\phi / 2)} & \sigma_{x}\sinh{(\phi / 2)} \\ \sigma_{x}\sinh{(\phi / 2)} & \cosh{(\phi /2)}\end{pmatrix}[/tex]
Ok, so here are my problems...
Firstly, I'm not sure exactly how to apply the [tex]\Psi(\L^{-1}x)[tex] 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?
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...
Any thoughts would be greatly appreciated.
Thanks,
James
I'm currently struggling to show something my lecturer told us in class. We have that
[tex]\Psi\left(x\right) \rightarrow S\left(L\right)\Psi\left(L^{-1}x\right)[/tex]
under a Lorentz transform defined
[tex]L = exp\left(\frac{1}{2}\Omega_{ij}M^{ij}\right)[/tex]
with
[tex]S\left(L\right) = exp\left(\frac{1}{2}\Omega_{ij}S^{ij}\right)[/tex]
where [tex]M^{ij}[/tex] are the Lorentz generators for "position" based transformations and [tex]S^{ij} = \frac{1}{4}\left[\gamma^{i}, \gamma^{j}\right][/tex] 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
[tex]\Psi\left(x, 0\right) = u\left(0\right)e^{-imt} = \sqrt{2m}\begin{pmatrix} \chi_{s} \\ 0\end{pmatrix}e^{-imt}[/tex] for (the sake of argument take spin-up
[tex]\chi_{s} = \begin{pmatrix}1 \\ 0\end{pmatrix}[/tex]
and apply a Lorentz boost to it, such that in the new frame it would have momentum p, we would get the full spinor
[tex]\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}[/tex]
So to get a boost in the x direction, we want to take only [tex]\Omega_{i0} = -\Omega_{0i} = -\phi[/tex] so that
[tex]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}[/tex]
where I guess I should point out that
[tex]M^{0i} = -M^{i0} = \begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0& 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}[/tex]
Then in the representation (which we have been using all through the course, and I want to use for this question)
[tex]\gamma^{0} = \begin{pmatrix}I & 0 \\ 0 & -I\end{pmatrix}[/tex]
[tex]\gamma^{i} = \begin{pmatrix}0 & \sigma_{i} \\ -\sigma_{i} & 0\end{pmatrix}[/tex]
I find
[tex]S^{i0} = -S^{0i} = \begin{pmatrix}0 & \frac{\sigma_{x}}{2} \\ \frac{ \sigma_{x}}{2} & 0 \end{pmatrix}[/tex]
So taking the exponential expression for S(L) I find
[tex]S(L) = \begin{pmatrix} \cosh{(\phi / 2)} & \sigma_{x}\sinh{(\phi / 2)} \\ \sigma_{x}\sinh{(\phi / 2)} & \cosh{(\phi /2)}\end{pmatrix}[/tex]
Ok, so here are my problems...
Firstly, I'm not sure exactly how to apply the [tex]\Psi(\L^{-1}x)[tex] 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?
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...
Any thoughts would be greatly appreciated.
Thanks,
James