# LOG#094. Group theory(XIV).

Group theory and the issue of mass: Majorana fermions in 2D spacetime

We have studied in the previous posts that a mass term is “forbidden” in the bivector/sixtor approach and the Dirac-like equation due to the gauge invariance. In fact,

$-i\overline{\Gamma}^\mu\partial_\mu$

as an operator has an “issue of mass” over the pure Dirac equation $i\Gamma^\mu\Psi=0$ of fermionic fields. This pure Dirac equation provides

$\overline{\Gamma_\mu}\Gamma_\nu\partial_\nu\partial_\mu\Psi=\partial_0^2\Psi+\nabla\times (\nabla\times \Psi)=\partial_0^2-\Delta\Psi+\nabla (\nabla\cdot \Psi)=0$

Therefore, $\Psi$ satisfies the wave equation

$\square^2\Psi=\partial^\mu\partial_\mu\Psi=0$

where $\square^2=\Delta-\partial_0^2$ if there are no charges or currents! If we introduce a mass by hand for the $\Psi$ field, we obtain

$(i\Gamma^\mu\partial_\mu-m )\Psi=0$

and we observe that it would spoil the relativistic invariance of the field equation!!!!!!! That is a crucial observation!!!!

A more general ansatz involves the (anti)linear operator V:

$i\Gamma^\mu\partial_\mu\Psi-mV\Psi=0$

A plane wave solution is something like an exponential function $\sim e^{\pm ipx}$ and it obeys:

$p^2=p_\mu p^\mu=-m^2$

If we square the Dirac-like equation in the next way

$i\overline{\Psi}^\nu\partial_\nu (i\Gamma^\mu\partial_\mu\Psi)=-\square \Psi=-m^2\Psi=i\overline{\Psi}^\nu\partial_\nu (mV\Psi)$

and a bivector transformation

$i\overline{\Gamma}^\mu\partial_\mu (V\Psi)-m\Psi=0$

$V(i\overline{\Gamma}^\mu\partial_\mu (v\Psi))-mV\Psi=0$

$Vi\overline{\Gamma}^\mu\partial_\mu (V\Psi)=mV\Psi=i\Gamma^\mu \partial_\mu \Psi$

from linearity we get

$Vi\overline{\Gamma}^\mu=i\Gamma^\mu$

$V^2=I_3$

$V\tilde{S}_aV=-S_a$

$V\tilde{S}_aV^{-1}=-\tilde{S}_a$

if $a=1,2,3$. But this is impossible! Why? The Lie structure constants are “stable” (or invariant) under similarity transformations. You can not change the structure constants with similarity transformations.

In fact, if V is an antilinear operator

$V=\tilde{V}\kappa=iV$ where $\kappa$ is a complex conjugation of the multiplication by the imaginary unit. Then, we would obtain

$\tilde{V}\tilde{V}^*=-I_3$

and

$\tilde{V}\tilde{S}_a^*\tilde{V}^*=\tilde{S}_a$

or equivalently

$\tilde{V}\tilde{S}_a^*\tilde{V}^{-1}=-\tilde{S}_a$

And again, this is impossible since we would obtain then

$\det (V\tilde{V}^*)=\det (V)\det (\tilde{V}^*)=\det \tilde{V}\det \tilde{V}^*>0$

and this contradicts the fact that $\det (-I_3)=-1$!!!

Remark: In 2d, with Pauli matrices defined by

$\sigma=(\sigma_1,\sigma_2,\sigma_3)$

and $\epsilon\tilde{\sigma}^*\epsilon^{-1}=-\tilde{\sigma}$

where

$\epsilon=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}$

and we have

$\epsilon^2=(\eta\epsilon)(\eta\epsilon)^*=-I_2$

with

$\det (\epsilon)=\det (-I_2)$ such that the so-called Majorana equation(s) (as a generalization of the Weyl equation(s) and the Lorentz invariance in 4d) provides a 2-component field equation describing a massive particle DOES exist:

$i\sigma^\mu\partial_\mu\Psi-m\eta\epsilon\Psi^*=0$

In fact, the Majorana fermions can exist only in certain spacetime dimensions beyond the 1+1=2d example above. In 2D (or 2d) spacetime you write

$\boxed{i\sigma^\mu\partial_\mu\Psi-m\eta\epsilon\Psi^*=0}$

and it is called the Majorana equation. It describes a massive fermion that is its own antiparticle, an option that can not be possible for electrons or quarks (or their heavy “cousins” and their antiparticles). The only Standard Model particle that could be a Majorana particle are the neutrinos. There, in the above equations,

$\sigma^\mu=(I_2,\vec{\sigma})$ and $\eta$ is a “pure phase” often referred as the “Majorana phase”.

Gauge formalism and mass terms for field equations

Introduce some gauge potential like

$A=A^R+iA^I=\begin{pmatrix}A_1^R+iA_1^I\\ A_2^R+iA_2^I\\ A_3^R+iA_3^I\end{pmatrix}$

It is related to the massive bivector/sixtor field with the aid of the next equation

$\Psi=i\overline{\Gamma}^\nu\partial_\nu A=(i\partial_0+\nabla\times)(A^R+iA^I)=-\dot{A}^O+\nabla\times A^R+i(\dot{A}^R+\nabla\times A^I)$

It satisfies a massive wave equation

$\square^2A+m^2A=0$

This would mean that

$i\Gamma^\mu\partial_\mu (i\overline{\Gamma}^\nu \partial_\nu A)=(-\partial_0^2-\nabla\times\nabla\times)A=(-\partial_0^2+\Delta-\nabla (\nabla\cdot))A=- m^2A$

and then $\nabla (\nabla\cdot A)=0$. However, it would NOT BE a Lorentz invariant anymore!

Current couplings

From the Ampère’s law

$\partial_t E=\nabla\times B-j$

and where we have absorbed the multiplicative constant into the definition of the current $j$, we observe that $i\Gamma^\mu\partial_\mu\Psi$ can NOT be interpreted as the Dirac form of the Maxwell equations since $j=(j_1,j_2,j_3)$ have 3 spatial components of a charge current 4d density $J=j^\mu=(j^0,\mathbf{j})=(j^0,j^1,j^2,j^3)$ so that

$\partial_t\Psi=-i\nabla\times \Psi-\mathbf{j}$ and

$\nabla\cdot (\partial_t \Psi)=\nabla\cdot (-i\nabla\times \Psi-\mathbf{j})$

or

$\nabla\cdot \dot{\Psi}=-i\nabla\cdot (\nabla\times \Psi)-\mbox{div}\mathbf{j}=\dot{\rho}$

If the continuity equation $\dot{\rho}+\mbox{div}\mathbf{j}=0$ holds. In the absence of magnetic charges, this last equation is equivalent to $\mbox{div} (\dot{E})=\dot{\rho}$ or $\nabla\cdot E=\rho$.

Remark: Even though the bivector/sixtor field couples to the spatial part of the 4D electromagnetic current, the charge distribution is encoded in the divergence of the field $\Psi$ itself and it is NOT and independent dynamical variable as the current density (in 4D spacetime) is linked to the initial conditions for the charge distribution and it fixes the actual charge density (also known as divergence of $\Psi$ at any time; $\Psi$ is a bispinor/bivector and it is NOT a true spinor/vector).

Dirac spinors under Lorentz transformations

A Lorentz transformation is a map

$X'=\Lambda X$

A Dirac spinor field is certain “function” transforming under certain representation of the Lorentz group. In particular,

$\Psi'(x')=Q_D(\Lambda)\Psi (x)$ for every Lorentz transformation belonging to the group $SO(1,3)^+$. Moreover,

$x'_\mu=\Lambda^\mu_{\;\;\; \nu}$

and Dirac spinor fields obey the so-called Dirac equation (no interactions are assumed in this point, only “free” fields):

$i\gamma^\mu\partial_\mu\Psi-m\Psi=0$

This Dirac equation is Lorentz invariant, and it means that it also hold in the “primed”/transformed coordinates since we have

$i\gamma^\mu (\partial_{\mu '}\Psi' (x'))=i\gamma^\mu\partial_{\mu '}(Q_D\Psi (x))=mQ_D\Psi$

and

$i\gamma^\mu\Lambda^{\;\;\; \nu}_\mu\partial_\nu Q_D\Psi=mQ_D\Psi$

Using that $\Lambda^\alpha_{\;\;\; \nu}\Lambda^{\nu}_{\;\;\; \mu}=\delta^\alpha_{\;\;\; \mu}$

we get the transformation law

$\boxed{Q_D^{-1}\gamma^\alpha Q_D=\Lambda^\alpha_{\;\;\; \nu}\gamma^\nu}$

Covariant Dirac equations are invariant under Lorentz transformations IFF the transformation of the spinor components is performed with suitable chosen operators $Q_D$. In fact,

$Q^{-1}\Gamma^\alpha Q=\Lambda^\alpha_{\;\;\; \nu}\Gamma^\nu$

$Q^T\Gamma^\alpha Q=\Lambda^\alpha_{\;\;\; \nu}\Gamma\nu$

$Q^*\Gamma^\alpha Q=\Lambda^\alpha_{\;\;\; \nu}\Gamma\nu$

DOES NOT hold for $\Psi$ bispinors/bivectors. For bivector fields, you obtain

$i\Gamma^\mu\partial_\mu\Psi=-i\mathbf{j}$

and

$i\Gamma^\mu_{ab}\partial_{\mu '}\Psi '(x')=-iJ'_a (x')$

This last equation implies that

$i\Gamma_{ab}^\mu\Lambda_{\mu}^{\;\;\; \nu}\partial_\nu Q_{bc}\Psi_c(x)=-i\Lambda^a_{\;\;\; \nu}j^\nu (x)=-i\Lambda^a_{0}\mbox{div} E(x)-i\Lambda^a_ cJ^c(x)$

$j^\mu=(\mbox{div}E,j^1,j^2,j^3)=(j^0,j^1,j^2,j^3)$

with

$\mbox{div} E=\delta^\nu_c\delta_\nu\Psi_c$ since $\mbox{div} B=\nabla\cdot B=0$ because there are no magnetic monopoles.

If $\tilde{\Lambda}$ is the inverse of the 3d matric $\Lambda^a_{\;\;\; b}$, then we have

$\tilde{\Lambda}^d_a\Lambda^a_b=\delta^d_b$

In this case, we obtain that

$i (\tilde{\Lambda}^d_a\Gamma^\mu_{ab}\Lambda^\nu_\mu Q_{bc}+\tilde{\Lambda}^d_a\Lambda^a_0\delta^\nu_c)\partial_\nu\Psi_c (x)=-i\tilde{\Lambda}^d_a\Lambda^a_cJ^c=-ij^d$

so

$\Gamma^\nu_{dc}=\tilde{\Lambda}^d_a\Gamma^\mu_{ab}\Lambda_{\mu}^\nu Q_{bc}+\tilde{\Lambda}^d_a\Lambda^a_0\delta^\nu_c$

That is, for rotations we obtain that

$\Lambda^a_{\;\;\; b}=Q_{ab}$ $\tilde{\Lambda}^{-1}=Q^{-1}$ $\Lambda^a_{\;\;\; 0}\;\;\forall a=1,2,3$

and so

$\boxed{\Gamma^\nu=\Lambda_{\mu}^{\;\;\; \nu}Q^{-1}\Gamma^\mu Q}$

This means that for the case of pure rotations both bivector/bispinors and current densities transform as vectors under the group SO(3)!!!!

Conclusions of this blog post:

1st. A mass term analogue to the Marjorana or Dirac spinor equation does NOT arise in 4d electromagnetism due to the interplay of relativistic invariance and gauge symmetries. That is, bivector/bispinor fields in D=4 can NOT contain mass terms for group theoretical reasons: Lorentz invariance plus gauge symmetry.

2nd. The Dirac-like equation $i\Gamma^\mu \partial_\mu \Psi=0$ can NOT be interpreted as a Dirac equation in D=4 due to relativistic symmetry, but you can write that equation at formal level. However, you must be careful with the meaning of this equation!

3rd. In D=2 and other higher dimensions, Majorana “mass” terms arise and you can write a “Majorana mass” term without spoiling relativistic or gauge symmetries. Majorana fermions are particles that are their own antiparticles! Then, only neutrinos can be Majorana fermions in the SM (charged fermions can not be Majorana particles for technical reasons).

4th. The sixtor/bivector/bispinor formalism with $F=E+iB$ has certain uses. For instance, it is used in the so-called Ungar’s formalism of special relativity, it helps to remember the electromagnetic main invariants and the most simple transformations between electric and magnetic fields, even with the most general non-parallel Lorentz transformation.