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Sign of the time derivative of the Majorana Lagrangian

  1. Mar 31, 2013 #1
    Let [itex]\gamma^{\rho} \in M_{4}(\mathbb{R})[/itex] be the Majorana representation of the Dirac algebra (in spacetime signature [itex]\eta_{00} = -1[/itex]), and consider the Majorana Lagrangian [tex]\mathcal{L} = \mathrm{i} \theta^{\mathrm{T}} \gamma^{0} (\gamma^{\rho} \partial_{\rho} - m) \theta,[/tex] where [itex]\theta[/itex] is a Grassmann-valued four-spinor. The associated gravitational energy-density, the 00-component of the Belinfante energy-momentum tensor, I calculate to be [tex]\Theta^{00} = \frac{\mathrm{i}}{2} [ \theta^{\mathrm{T}} (\partial^{0} \theta) - (\partial^{0} \theta)^{\mathrm{T}} \theta].[/tex] Inserting into it the stationary plane, wave solution [itex]\theta = \mathrm{exp}(\gamma^{0}Et)\eta[/itex], where [itex]\eta[/itex] is some spacetime-independent, Grassmann-valued four-spinor, yields [itex]\Theta^{00} = \mathrm{i} E \eta^{\mathrm{T}} \gamma^{0} \eta[/itex]. Due to [itex](xy)^{*} \equiv y^{*}x^{*}[/itex] for Grassmann-valued quantities, this expression for [itex]\Theta^{00}[/itex] is complex self-conjugate (and nonvanishing), as it should be, but it is not real-valued.

    In comparison, for the Dirac Lagrangian, [tex]\mathcal{L}_{D} = -\mathrm{i} \psi^{\dagger} \gamma^{0} (\gamma^{\rho} \partial_{\rho} - m) \psi,[/tex] also in spacetime signature [itex]\eta_{00} = -1[/itex], a similar calculation of the gravitational energy-density yields for a plane wave solution [itex]\psi = \mathrm{exp}(-\mathrm{i}Et)\psi_{0}[/itex] the real-valued expression [itex]\Theta^{00} = E \psi_{0}^{\dagger} \psi_{0}[/itex]. The exact sign of [itex] -\mathrm{i} \psi^{\dagger} \gamma^{0} \gamma^{0} \partial_{0} \psi = +\mathrm{i} \psi^{\dagger} \partial_{0} \psi [/itex] in [itex]\mathcal{L}_{D}[/itex] is essential for this energy-density to be positive-definite.

    And now to my question: Is it nonsensical to analogously contemplate what the sign of the time derivative should be in the case of the Majorana Lagrangian? And if not, what is it?
     
    Last edited: Mar 31, 2013
  2. jcsd
  3. Apr 1, 2013 #2
    Extracting observables

    Perhaps more concretely the following general question is what I am asking: How are any observables - ordinary, real numbers - to be extracted from a classical (i.e., non-quantum) theory that uses Grassmann numbers?

    Using complex self-conjugate quantities like [itex]\mathrm{i}\theta_{1}\theta_{2}[/itex], say, where [itex]\theta_{1},\theta_{2}[/itex] are Grassmann numbers, is not a solution, for even though such a product does commute with everything, it is not an ordinary, real number, because it squares to zero.
     
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