Let me first describe how I got to that problem. We know that Majorana Lagrangian (here I choose left-handed but for right-handed problem is analogue) $${\cal L}=\psi_{L}^{\dagger}i\bar{\sigma}^{\mu}\partial_{\mu}\psi_{L}-\frac{m}{2}\left[\psi_{L}^{T}\epsilon\psi_{L}+\psi_{L}^{\dagger}\epsilon\psi_{L}^{*}\right]\tag{1}$$ (where $\epsilon =\scriptscriptstyle\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$) doesn't obey $U(1)$ symmetry, therefore Majorana Fermion doesn't cary a charge understood in a sense of electric like charge.

Nevertheless, it obeys $SO(2)$ symmetry (acting on $\psi$ by 2x2 representation of rotation), thus by the Noether theorem, there should be conserved current and charge of that symmetry. This is how I calculate it:

Noether current is given by the formula $$j^{\mu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi_{L,\alpha})}\epsilon_{\alpha\beta}\psi_{L,\beta}-K^{\mu},\tag{2}$$ where $K^{\mu}$ is $0$ as given symmetry is not only a symmetry of action but also the symmetry of Lagrangian and $\epsilon$ is an infinitesimal form of a rotation. Now we could be tricky. In the case of Majorana fermions, $\psi$ and $\psi^{\dagger}$ are not independent (relation given by the Majorana equation) therefore we could differentiate over $\psi^{\dagger}$ to calculate the current. Alternatively, we could move the derivative to $\psi^{\dagger}$ by integration by parts of the kinetic term of our Lagrangian.

All in all, we got the Noether current equal to $0$ and therefore conserved charge equal to $0$. My first question is did I calculate that charge correctly? And then if yes, what does it mean that a Noether current and charge are equal to 0? In my opinion, it doesn't give us any information about the dynamics of a system. Does it give any information about that symmetry? What does it mean in that particular example? And finally, are we able to predict that a given symmetry gives $0$ charge?

  • $\begingroup$ How do Lorentz transformations act on your Majorana spinor? $\endgroup$
    – octonion
    Commented Apr 7, 2022 at 14:23
  • $\begingroup$ @octonion By $\exp\left(-\frac{i}{2}\omega_{\mu\nu}\sigma^{\mu\nu}\right)$ as it's $\left(0,\frac{1}{2}\right)$ representation of $SO(1,3)$, but I don't see how does it refer to the problem. $\endgroup$ Commented Apr 7, 2022 at 14:59
  • $\begingroup$ It's another transformation that has $U(1)$ subgroups that act on the spinor index. If the $SO(2)$ you are considering is a genuine symmetry I doubt it is distinct from this. $\endgroup$
    – octonion
    Commented Apr 7, 2022 at 18:02
  • $\begingroup$ @octonion sorry, but I still didn't get your point we have the Coleman-Mandula theorem that states that a group of all symmetries of a theory is a direct product of $ISO(1,3)$ and internal symmetries. So why do we bother about Lorentz's rep acting on spinor? By simple calculation, we see that $U(1)$ is not a symmetry of the Lagrangian (if we want it to hold, we need to have $\psi$ contracted with its hermitian conjugate which is not the case here) and $SO(2)$ is as rotation is orthogonal and we have contraction of $\psi$ with its transpose in each term of the Lagrangian. $\endgroup$ Commented Apr 7, 2022 at 18:56
  • 1
    $\begingroup$ Oh wait a second @octonion indeed you are right $U(1)$ representation is a subgroup of Lorentz representation acting on spinor. But the question is then why is the Majorana spinor neutral as its obey $U(1)$ symmetry? $\endgroup$ Commented Apr 7, 2022 at 19:12

1 Answer 1

  1. Counterexample to the title question (v2): Consider zeroth-order Lagrangian density ${\cal L}=-\frac{m^2}{2}\phi^2$. The EOM is $\phi\approx 0$. Take (infinitesimal) spacetime translation symmetry $\delta \phi =\epsilon^{\mu} \partial_{\mu}\phi$. The corresponding Noether current is the canonical stress-energy-momentum tensor $T^{\mu}{}_{\nu}\approx 0$, which vanishes on-shell.

  2. One may show that the Noether current in OP's example (1) with $SO(2)$ symmetry is non-vanishing.

  3. Be aware that the order of factors (and whether one uses left or right derivatives) become important in the Noether current formula (2) when dealing with Grassmann-odd variables.


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