Sign of the time derivative of the Majorana Lagrangian

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SUMMARY

The discussion focuses on the Majorana Lagrangian, specifically its gravitational energy-density and the implications of its time derivative. The Majorana Lagrangian is defined as \(\mathcal{L} = \mathrm{i} \theta^{\mathrm{T}} \gamma^{0} (\gamma^{\rho} \partial_{\rho} - m) \theta\), leading to a complex self-conjugate energy-density \(\Theta^{00} = \mathrm{i} E \eta^{\mathrm{T}} \gamma^{0} \eta\). In contrast, the Dirac Lagrangian yields a positive-definite energy-density \(\Theta^{00} = E \psi_{0}^{\dagger} \psi_{0}\). The discussion raises the question of the sensibility of determining the sign of the time derivative in the Majorana case and how to extract real observables from a classical theory involving Grassmann numbers.

PREREQUISITES
  • Understanding of Majorana representation of the Dirac algebra
  • Familiarity with Grassmann-valued four-spinors
  • Knowledge of the Belinfante energy-momentum tensor
  • Concepts of classical field theory and observables
NEXT STEPS
  • Research the properties of Grassmann numbers and their applications in field theory
  • Study the implications of complex self-conjugate quantities in quantum mechanics
  • Explore the differences between Majorana and Dirac Lagrangians in detail
  • Investigate methods for extracting observables from classical theories using Grassmann variables
USEFUL FOR

The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory, as well as researchers exploring the implications of Grassmann numbers in classical and quantum contexts.

JustMeDK
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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?
 
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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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