Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook