[QUE/QFT-03006] QFT-PROBLEM

The matrix for an infinitesimal Lorentz boost along direction \(\hat{n}\) transformation can be written as \begin{equation} \Lambda = I + \Delta v (\hat{n}\cdot \vec{Y}), \end{equation} where the matrices \(\vec{Y}\) are given by \begin{equation} Y_1=\begin{pmatrix} 0 &1 & 0 & 0\\1 & 0 & 0& 0\\0 & 0&0&0\\0&0&0& 0\\ \end{pmatrix}\qquad; Y_2=\begin{pmatrix} 0 &0 & 1 & 0\\0 & 0 & 0& 0\\1& 0&0&0\\0&0&0& 0\\ \end{pmatrix}\qquad; Y_3=\begin{pmatrix} 0 &0 & 0 &1 \\0 & 0 & 0&0 \\0 & 0&0&0\\1&0&0& 0\\ \end{pmatrix}. \end{equation} Work out the transformation matrix \(\Lambda\) for a finite boost by velocity \(v\) by taking it as \(N\) successive transformations and considering the limit \(N\to \infty\). Hence show that \begin{eqnarray} x^{{'}\,0} &=& \cosh \alpha x^0 +\sinh\alpha (\hat{n}\cdot\vec{x})\\ \vec{x}\,^{'} &=& [\vec{x}- (\hat{n}\cdot \vec{x})\hat{n}] + \hat{n} [\cosh \alpha (\hat{n}\cdot\vec{x}) +\sinh \alpha x^0] \end{eqnarray} where \(\tanh\alpha=v\).

We first compute powers of \(\hat{n}\cdot\vec{Y}\), where \(\hat{n}=(n_1,n_2,n_3)\) is a unit vector.
\begin{eqnarray} \hat{n}\cdot\vec{Y} &=& \begin{pmatrix} 0 & n_1 & n_2 & n_3 \\ n_1 & 0 & 0 & 0\\ n_2 & 0 & 0 & 0\\n_3 & 0 & 0 & 0 \end{pmatrix}\\ (\hat{n}\cdot\vec{Y})^2 &=& \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & n_1^2 & n_1 n_2 & n_1 n_3 \\ 0 & n_2 n_1 & n_2^2 & n_2 n_3 \\ 0 & n_3n_1 & n_3 n_2 & n_3 n_1 \end{pmatrix} (\hat{n}\cdot\vec{Y})^3 &=& \hat{n}\cdot\vec{Y} \end{eqnarray} Hence we get \begin{eqnarray} \exp\big(-\omega \hat{n}\cdot \vec{Y} \big) &=& 1- \omega (\hat{n}\cdot\vec{Y}) + \frac{\omega^2}{2!}(\hat{n}\cdot\vec{Y})^2 - \frac{\omega^3}{3!} (\hat{n}\cdot\vec{Y})^3 + \frac{\omega^4}{4!}(\hat{n}\cdot\vec{Y}) -\frac{\omega^5}{5!}(\hat{n}\cdot\vec{Y}) + \ldots. \end{eqnarray} Using \eqRef{EQ07} we get \begin{eqnarray}\nonumber \exp(-\omega \hat{n}\cdot \vec{Y}) &=& I- (\hat{n}\cdot\vec{Y}) \Big[ (\omega +\frac{\omega^3}{3!}+ \frac{\omega^5}{5!} +\ldots\Big] + (\hat{n}\cdot\vec{Y})^2 \Big[ \frac{\omega^2}{2!} + \frac{\omega^4}{4!} + \frac{\omega^6}{6!}+ \ldots \Big]\\ &=& I - (\hat{n}\cdot\vec{Y})\sinh \omega + (\hat{n}\cdot\vec{Y})^2\, (\cosh \omega-1) \end{eqnarray} Using \EqRef{EQ05} and \eqRef{EQ06} we get \begin{equation} \exp(-\omega \hat{n}\cdot\vec{Y}) = \begin{pmatrix} \cosh \omega &-n_1 \sinh \omega & -n_2\sinh\omega &-n_3\sinh \omega\\ -n_1 \sinh\omega &1+n_1^2(\cosh\omega-1) &n_1n_2(\cosh \omega -1) &n_1n_3(\cosh \omega -1) \\ -n_2 \sinh\omega &n_2n_1(\cosh \omega -1) &1+n_2^2(\cosh\omega-1) & n_2n_3(\cosh \omega -1)\\ -n_3 \sinh \omega& n_3n_1(\cosh \omega -1)& n_3n_2(\cosh \omega -1)& 1+ n_3^2(\cosh\omega-1) \end{pmatrix} \end{equation} Writing \begin{equation} \begin{pmatrix} x^{0\,{'}} \\x^{1\,{'}}\\x^{2\,{'}}\\x^{3\, {'}} \end{pmatrix} = \Lambda \begin{pmatrix} x^0 \\x^1\\x^2\\x^3 \end{pmatrix} \end{equation} Performing the matrix multiplication in the right hand side we get \begin{eqnarray}\label{EQ12} x^{{'}\,0} &=& \cosh \omega - \sinh \omega (\hat{n}\cdot\vec{x})\\ \vec{x}\,{'} &=& -\sinh \omega\, x^0 + (\cosh \omega -1) (\hat{n}\cdot\vec{x}) \hat{n} + \vec{x}.\label{EQ13} \end{eqnarray} Taking dot product of both sides in \eqRef{EQ13}, we get
\begin{eqnarray} x^{{'}\,0} &=& \cosh \omega x^0 - \sinh \omega (\hat{n}\cdot\vec{x})\\ (\hat{n}\cdot\vec{x}\,{'}) &=& -\sinh \omega\, x^0 + \cosh \omega (\hat{n}\cdot\vec{x}) \end{eqnarray}
Identifying \(\tanh \omega = (v/c)\), the above form give the standard Lorentz boost along the direction \(\hat{n}\).