Sign of the time derivative of the Majorana Lagrangian

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

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