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I Lie Algebra of Lorentz group

  1. May 17, 2017 #1
    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!
     
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  3. May 17, 2017 #2

    Orodruin

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    The ##J^{\mu\nu}## are not tensors, they are the basis of the Lie algebra.
     
  4. May 17, 2017 #3
    So they are in the case of Minkowski space 4x4 matrices?
     
  5. May 17, 2017 #4

    Orodruin

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    If you use the (usual) matrix representation of the Lorentz group, yes.
     
  6. May 18, 2017 #5

    vanhees71

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  7. May 18, 2017 #6

    dextercioby

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    They are tensor operators with respect to the Lorentz group.
     
  8. May 18, 2017 #7

    Orodruin

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    They can be represented by tensor operators. This still does not make them tensors in the sense implied by the OP.
     
  9. May 20, 2017 #8

    samalkhaiat

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    We do not ‘parameterize’ the generators. The generators of any Lie algebra are independent of the parameters of the corresponding Lie group. This [itex]\omega_{\mu\nu}[/itex] is not a tensor. It is a set of six real numbers, [itex]\{ \omega_{a}, \ a = 1, ..., 6 \}[/itex], (3 rotation angles [itex]\theta_{i} \equiv \frac{1}{2}\epsilon_{ijk}\omega_{jk}[/itex], and 3 boosts [itex]\beta_{i} \equiv \omega_{0i}[/itex]) grouped together under the condition [itex]\omega_{\mu\nu} = - \omega_{\nu\mu}[/itex]. So, when you use the symbol [itex]\omega_{\mu\nu}[/itex] to represent the (six) parameters, you should keep in mind that [itex]\omega_{00} = \cdots = \omega_{33} = 0[/itex]. This means that there are only six generators ([itex]M^{a}, \ a = 1, ... , 6[/itex]) which (also) can be grouped to satisfy [itex]M^{\mu \nu} = - M^{\nu\mu}[/itex]. Since Lorentz group is a connected Lie group, then (almost all) its elements can be represented by the exponential map [tex]g = e^{\vec{\omega}} = e^{\omega_{a}M^{a}} \equiv e^{\frac{1}{2}\omega_{\mu\nu}M^{\mu\nu}} ,[/tex] where [itex]\vec{\omega}[/itex] is an element of 6-dimensional vector space (the Lorentz algebra) and [itex]\omega_{\mu\nu}[/itex] are its components (local coordinates on the group manifold) “along” the basis vectors [itex]M^{\mu\nu}[/itex]. This should remind you with the school-days vector equation [itex]\vec{x} = x_{i}\hat{e}^{i}[/itex].
    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 [itex]\Lambda[/itex], with matrix elements denoted by [itex]\Lambda^{\mu}{}_{\nu}[/itex]; the finite-dimensional representation matrix [itex]D(\Lambda) = \exp \left(\frac{1}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}\right)[/itex] which acts on the index space of geometrical objects (classical fields) on Minkowski space-time, i.e., under a Lorentz transformation [itex]\Lambda[/itex], the components of a given field [itex]\varphi_{a}(x)[/itex] get mixed by the matrix [itex]D(\Lambda)[/itex]; and the third object is the infinite-dimensional unitary representation operator [itex]U(\Lambda) = \exp \left( \frac{i}{2} \omega_{\mu\nu}J^{\mu\nu}\right)[/itex] which acts on states and operators in the Hilbert space. In QFT, the above three objects appear in one fundamental (transformation) equation, that is [tex]U^{\dagger}(\Lambda) \varphi_{a}(x) U(\Lambda) = D_{a}{}^{c}(\Lambda) \ \varphi_{c}(\Lambda^{-1}x) .[/tex] From this equation we understand that on the one hand the field transforms under the [itex]SO(1,3)[/itex] transformation like a classical geometrical object in the Minkowski space-time, [tex]\varphi_{a}(x) \to \bar{\varphi}_{a}(x) = D_{a}{}^{c}(\Lambda) \ \varphi_{c}(\Lambda^{-1}x) ,[/tex] on the other as an operator-valued distribution in the Hilbert space, [tex]\varphi_{a}(x) \to \bar{\varphi}_{a}(x) = U^{\dagger}(\Lambda) \varphi_{a}(x) U(\Lambda) .[/tex] Also, the generators [itex]\Sigma^{\mu\nu}[/itex] of the [itex]D(\Lambda)[/itex] 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 [itex]A^{\mu}(x)[/itex], [tex]\bar{A}^{\mu}(x) = D^{\mu}{}_{\nu}(\Lambda) \ A^{\nu}(\Lambda^{-1}x) = \Lambda^{\mu}{}_{\nu} A^{\nu}(\Lambda^{-1}x) .[/tex] This tells you that for a vector field, the finite-dimensional representation [itex]D(\Lambda)[/itex] is given by the Lorentz matrix [itex]\Lambda[/itex]: [tex]D^{\mu}{}_{\nu}(\Lambda) = \Lambda^{\mu}{}_{\nu} .[/tex] To first order in the parameters, this becomes [tex]\delta^{\mu}_{\nu} + \frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}{}_{\nu} .[/tex] This leads to [tex]\frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \omega^{\mu}{}_{\nu} ,[/tex] which can be rewritten as [tex]\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) .[/tex] Thus
    [tex]\left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \eta^{\mu\rho} \ \delta^{\sigma}_{\nu} - \eta^{\mu\sigma} \ \delta^{\rho}_{\nu} .[/tex] This tells you that the generator [itex]\Sigma^{\rho\sigma}[/itex] represents a set of six antisymmetric, [itex]4 \times 4[/itex] constant matrices with matrix elements given by [itex](\Sigma^{\rho\sigma})^{\mu}{}_{\nu}[/itex]. Of course, you may regard the object [itex]\Sigma^{\rho\sigma\mu}{}_{\nu}[/itex] as a rank-4 invariant tensor, but this does not mean that [itex]\Sigma^{\rho\sigma}[/itex] itself is a rank-2 tensor.
    However, Hermitian generator [itex]J^{\mu\nu}[/itex] of the [itex]U(\Lambda)[/itex] transformation is a rank-2 Lorentz tensor operator. Indeed, under Lorentz transformation, it is easy to show that [tex]J^{\mu\nu} \to U^{\dagger}J^{\mu\nu}U = \Lambda^{\mu}{}_{\rho} \ \Lambda^{\nu}{}_{\sigma} \ J^{\rho \sigma} ,[/tex] which is the correct transformation law for rank-2 tensor operator.
     
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