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[QUE/QFT-03005] QFT-PROBLEMNode id: 4013pageShow that conformal transformations consisting of dilations \[x^\mu \to x^\mu = e^{-\rho} x^\mu \] and special conformal transformations (SCT) \[x^\mu \to x^{\prime\,\mu}= \frac{x^\mu + c^\mu x^2}{ 1 + 2c \cdot x + c^2 x^2},\] and usual Poincaré transformations form a group. Find the commutation relations for the generators of infinitesimal transformations of this group.
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24-07-31 08:07:18 |
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[QUE/QFT-03006] QFT-PROBLEMNode id: 4014page
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 \(\hat{n}\).
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22-02-06 18:02:38 |
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[QUE/QFT-03004] QFT-PROBLEMNode id: 4012pageDefine Pauli Lubanski operator \(W_\sigma\) by \[ W_\sigma = -\frac{1}{2} \epsilon_{\mu\nu\lambda\sigma} M^{\mu\nu} P^\sigma\] where \(P^\mu\) is energy momentum four vector and \(M^{\mu\nu}\) angular momentum tensor. Prove the following relations
- \(W^\sigma P^\sigma =0\)
- \(\big[W^\sigma , P^\mu\big] =0 \)
- \(W_\sigma\) is a four vector, {\it i.e.} \(\big[M_{\mu\nu}, W_\sigma\big] = -i(W_\mu g_{\nu\sigma}-W_\nu g_{\mu\sigma})\)
- \( \big[W_\lambda, W_\sigma \big] = i\epsilon_{\lambda\sigma\alpha\beta}W^\alpha P^\beta\)
- \(P_\mu P^\mu\) and \(W^\sigma W_\sigma\) are Cashimir invariants of the Poincare group, they commute with all the ten generators of the Poincare group.
- Prove that \(W^2 = -\frac{1}{2} M_{\mu\nu} M^{\mu\nu} P^2 + M_{\mu\sigma}M^{\nu\sigma} P^\mu P_\sigma \)
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22-02-06 18:02:04 |
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[QUE/QFT-03003] QFT-PROBLEMNode id: 4011pageIt is given that the generators \(M^{\mu\nu}\) of Lorentz group satisfy the commutation relations \begin{eqnarray}\nonumber [M^{\mu\nu} , P^\sigma] &=& i(g^{\sigma\mu} P^\nu - g^{\sigma\nu} P^\mu )\\\nonumber [M^{\mu\nu} , M ^{\rho\sigma}] &=& i(M^{\mu\rho} g^{ \nu\sigma} + M^{\nu\sigma} g^{\mu\rho} - M^{\nu\rho} g^{\mu\sigma} - M^{\mu\sigma} g^{\nu\rho} ) \end{eqnarray}
- Compute the commutator \(\big[ \epsilon^{\mu\nu\lambda\sigma} M_{\mu\nu}M_{\lambda\sigma}, M_{\alpha\beta}\big].\)
- Define \(K_i= M_{i0}\) and \(J_i=-\frac{1}{2}\epsilon_{imn}M^{mn}\) and prove the following commutation relations. \begin{eqnarray}\nonumber \Big[ J_i, P_k\Big]= -\epsilon_{ik\ell} P_ell, &\qquad& \big[J_i,P_0\big] =0 \\{}\nonumber \Big[ K_i, P_k\Big]= - P_0 g_{ik}, &\qquad& \big[K_i,P_0\big] =-iP_0 . \end{eqnarray}
- The commutators of operators \(\vec{J}, \vec{K}\) are given by \begin{eqnarray}\nonumber \big[J_\ell,J_m\big] &=& i\epsilon_{\ell m n }J_n\\\nonumber \big[J_\ell,J_m\big] &=& i\epsilon_{\ell m n }K_n\\\nonumber \big[K_\ell,K_m\big] &=& -i\epsilon_{\ell m n }J_n. \end{eqnarray}
- Show that \(\vec{j}^{\pm}\) defined by \(j^{(\pm)}_m = \frac{1}{2}\big(J_m\pm i K_m\big)\) obey angular momentum commutations relations \begin{eqnarray}\nonumber \big[j^{(\pm)}_\ell, j^{(\pm)}_m\big] &=& i \epsilon_{\ell m n }j^{(\pm)}_n\\\nonumber \big[j^{(+)}_\ell, j^{(-)}_m\big] &=&0. \end{eqnarray}
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22-02-06 18:02:32 |
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[QUE/QFT-03002] QFT-PROBLEMNode id: 4010page
Write an infinitesimal Lorentz transformation as
\begin{equation} \widetilde{X{'}} = \widetilde{\Lambda} \widetilde{X} \end{equation} where \(\widetilde{\Lambda}\) is a four by four matrix. Taking \(Lambda\) to differ infinitesimal transformation of the form \[\Lambda = I + \Omega \] and find restrictions on the matrix \(\Omega\) using the fact that under Lorentz transformations \(x^\mu x_\mu \) remains invariant. Count independent number of parameters in \(\Omega\) and write its most general form.
The most general form of \(\Omega\) is \begin{equation*} \Omega = \begin{pmatrix} a & \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha_1 & b & \alpha_4 & \alpha_5 \\ \alpha_2 & -\alpha_4 & c & \alpha_6\\ \alpha_3 & -\alpha_5 & -\alpha_6 & d \end{pmatrix} \end{equation*} }\end{Answer} \begin{Solution} The Lorentz invariance of \(x^\mu x_\mu= x^\mu g_{\mu\nu} x^\nu\) implies \[ \widetilde{X}^{{'} \,T} \tilde{g} \widetilde{X}{'} = {X}^{{'} \,T} \tilde{g} {X}{'}\] where \(\tilde{g}\) is the diagonal matrix with diagonal entries (1,-1,-1,-1). The requirement of Lorentz invariance of the four scalar product gives the following condition on \(\Lambda\). \begin{equation} \widetilde{\Lambda} \tilde{g} \widetilde{\Lambda} = \tilde{g} \end{equation} Using \(\widetilde{X}^{{'} \,T} g \widetilde{X}{'}\) We rewrite the above equation in the form \begin{equation} (I + \Omega^T) \tilde{g} (1+ \Omega) = \tilde{g} \end{equation} Simplifying and keeping only first order terms in \(\Omega\), gives \begin{equation}\label{EQ04} \Omega^T \tilde{g} + \tilde{g} \Omega = 0. \end{equation}
Writing and multiplying out the matrices gives
\begin{eqnarray} \Omega \tilde{g} &=& \begin{pmatrix} a & -\alpha_1 & -\alpha_2 & -\alpha_3 \\ \alpha_1 & b & \alpha_4 & \alpha_5 \\ \alpha_2 & -\alpha_4 & c & \alpha_6\\ \alpha_3 & -\alpha_5 & -\alpha_6 & d \end{pmatrix}\\
\tilde{g}\Omega &=& \begin{pmatrix} a & \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha_1{'} & b & \alpha_4 & \alpha_5 \\ \alpha_2{'} & -\alpha_4{'} & c & \alpha_6\\ \alpha_3{'} & -\alpha_5{'} & -\alpha_6{'} & d \end{pmatrix} \end{eqnarray}
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22-02-06 18:02:56 |
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[QUE/QFT-03001] QFT-PROBLEMNode id: 4009pageProve the following commutation relations for the infinitesimal generators of the Lorentz group \begin{eqnarray}\nonumber [P^\mu , P^\nu ] &=& 0\\\nonumber [M^{\mu\nu} , P^\sigma] &=& i(g^{\sigma\mu} P^\nu - g^{\sigma\nu} P^\mu )\\\nonumber [M^{\mu\nu} , M ^{\rho\sigma}] &=& i(M^{\mu\rho} g^{ \nu\sigma} + M^{\nu\sigma} g^{\mu\rho} - M^{\nu\rho} g^{\mu\sigma} - M^{\mu\sigma} g^{\nu\rho} ) \end{eqnarray} Our notation is the same as that of Gasiorowicz, {\it Elementary Particle Physics} John Wiley and Sons (1968) |
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22-02-05 22:02:16 |
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