Silviu said:
Hello! I read that the for the lie algebra of the Lorentz group we can parametrize the generators as an antisymmetric tensor ##J^{\mu \nu}## and the parameters as an another antisymmetric tensor ##\omega_{\mu \nu}## and a general transformation would be ##\Lambda = exp(-\frac{i}{2} \omega_{\mu \nu} J^{\mu \nu})##. So based on Einstein notation, this would mean ##\Lambda = exp(-\frac{i}{2} (\omega_{00} J^{00} + \omega_{01} J^{01} + ... ))##. But for example ##\omega_{00}## and ##J^{00}## are numbers, so in the end we will end up with a complex number in the exponential and now with a (4x4 most probably) matrix. Can someone explain to me what I do wrong in reading this notation? Thank you!
We do not ‘
parameterize’ the generators. The generators of any Lie algebra are
independent of the parameters of the corresponding Lie group. This \omega_{\mu\nu}
is not a tensor. It is a set of
six real numbers, \{ \omega_{a}, \ a = 1, ..., 6 \}, (3 rotation angles \theta_{i} \equiv \frac{1}{2}\epsilon_{ijk}\omega_{jk}, and 3 boosts \beta_{i} \equiv \omega_{0i}) grouped together under the condition \omega_{\mu\nu} = - \omega_{\nu\mu}. So, when you use the symbol \omega_{\mu\nu} to represent the (six) parameters, you should keep in mind that \omega_{00} = \cdots = \omega_{33} = 0. This means that there are only six generators (M^{a}, \ a = 1, ... , 6) which (also) can be grouped to satisfy M^{\mu \nu} = - M^{\nu\mu}. Since Lorentz group is a
connected Lie group, then (almost all) its elements can be represented by the exponential map g = e^{\vec{\omega}} = e^{\omega_{a}M^{a}} \equiv e^{\frac{1}{2}\omega_{\mu\nu}M^{\mu\nu}} , where \vec{\omega} is an element of 6-dimensional vector space (the Lorentz algebra) and \omega_{\mu\nu} are its components (local coordinates on the group manifold) “along” the basis vectors M^{\mu\nu}. This should remind you with the school-days vector equation \vec{x} = x_{i}\hat{e}^{i}.
Now, the question whether the generators are Lorentz tensors or not is a different story. To answer it, we need to study the representation of the Lorentz group. Basically, in the representation theory of Lorentz group, you encounter three objects. These are: the coordinates transformation matrix \Lambda, with matrix elements denoted by \Lambda^{\mu}{}_{\nu}; the
finite-dimensional representation matrix D(\Lambda) = \exp \left(\frac{1}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}\right) which acts on the
index space of geometrical objects (classical fields) on Minkowski space-time, i.e., under a Lorentz transformation \Lambda, the components of a given field \varphi_{a}(x) get
mixed by the matrix D(\Lambda); and the third object is the
infinite-dimensional unitary representation operator U(\Lambda) = \exp \left( \frac{i}{2} \omega_{\mu\nu}J^{\mu\nu}\right) which acts on states and operators in the Hilbert space. In QFT, the above three objects appear in one fundamental (transformation) equation, that is U^{\dagger}(\Lambda) \varphi_{a}(x) U(\Lambda) = D_{a}{}^{c}(\Lambda) \ \varphi_{c}(\Lambda^{-1}x) . From this equation we understand that on the one hand the field transforms under the SO(1,3) transformation like a
classical geometrical object in the Minkowski space-time, \varphi_{a}(x) \to \bar{\varphi}_{a}(x) = D_{a}{}^{c}(\Lambda) \ \varphi_{c}(\Lambda^{-1}x) , on the other as an
operator-valued distribution in the Hilbert space, \varphi_{a}(x) \to \bar{\varphi}_{a}(x) = U^{\dagger}(\Lambda) \varphi_{a}(x) U(\Lambda) . Also, the generators \Sigma^{\mu\nu} of the D(\Lambda) transformation are given by
constant numerical (spin)
matrices (i.e.,
not a tensor). To see that consider the transformation law in the case of a vector field A^{\mu}(x), \bar{A}^{\mu}(x) = D^{\mu}{}_{\nu}(\Lambda) \ A^{\nu}(\Lambda^{-1}x) = \Lambda^{\mu}{}_{\nu} A^{\nu}(\Lambda^{-1}x) . This tells you that for a vector field, the finite-dimensional representation D(\Lambda) is given by the Lorentz matrix \Lambda: D^{\mu}{}_{\nu}(\Lambda) = \Lambda^{\mu}{}_{\nu} . To first order in the parameters, this becomes \delta^{\mu}_{\nu} + \frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}{}_{\nu} . This leads to \frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \omega^{\mu}{}_{\nu} , which can be rewritten as \frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \eta^{\mu\rho} \delta^{\sigma}_{\nu} \omega_{\rho\sigma} = \frac{1}{2} \omega_{\rho\sigma} \left( \eta^{\mu\rho} \ \delta^{\sigma}_{\nu} - \eta^{\mu\sigma} \ \delta^{\rho}_{\nu} \right) . Thus
\left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \eta^{\mu\rho} \ \delta^{\sigma}_{\nu} - \eta^{\mu\sigma} \ \delta^{\rho}_{\nu} . This tells you that the generator \Sigma^{\rho\sigma} represents a set of six antisymmetric, 4 \times 4 constant
matrices with matrix elements given by (\Sigma^{\rho\sigma})^{\mu}{}_{\nu}. Of course, you may regard the object \Sigma^{\rho\sigma\mu}{}_{\nu} as a rank-4
invariant tensor, but this does not mean that \Sigma^{\rho\sigma} itself is a rank-2 tensor.
However, Hermitian generator J^{\mu\nu} of the U(\Lambda) transformation
is a rank-2 Lorentz tensor operator. Indeed, under Lorentz transformation, it is easy to show that J^{\mu\nu} \to U^{\dagger}J^{\mu\nu}U = \Lambda^{\mu}{}_{\rho} \ \Lambda^{\nu}{}_{\sigma} \ J^{\rho \sigma} , which is the correct transformation law for rank-2 tensor operator.