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The Majorana field $\psi$ can be thought of as a reality condition $\psi=\psi^c$ (and $\overline{\psi}=\overline{\psi^c}$) on the Dirac field. So how does one write the Lagrangian for the Majorana field?

The way I am going about it consist of first writing down the Dirac field Lagrangian $$\mathcal{L}=i\overline{\psi}\gamma^\mu\partial_\mu\psi-m\overline{\psi}\psi$$ and substitute $\psi=\psi^c$ and $\overline{\psi}=\overline{\psi^c}$ into it. This apparently changes the Lagrangian to $$\mathcal{L}^\prime=i\overline{\psi^c}\gamma^\mu\partial_\mu\psi^c-m\overline{\psi^c}\psi^c.$$ But since $\overline{\psi^c}\psi^c\sim \overline{\psi}\psi$ and $\overline{\psi^c}\gamma^\mu\psi^c\sim \overline{\psi}\gamma^\mu\psi$ (the notation '$\sim$' means equality apart from a sign), $\mathcal{L}^\prime=\mathcal{L}$. This means that Lagrangian of the Dirac field $\mathcal{L}$ and the Majorana field $\mathcal{L}^\prime$ are same. I think this is wrong.

Next, I can try the following. Maybe changing both $\psi$ and $\overline{\psi}$ simultaneously to $\psi^c$ and $\overline{\psi^c}$ respectively was wrong. Only $\overline{\psi}$ to $\overline{\psi^c}$ in which case the correct Lagrangian is either $$\mathcal{L}^{\prime\prime}=i\overline{\psi^c}\gamma^\mu\partial_\mu\psi-m\overline{\psi^c}\psi$$ or $$\mathcal{L}^{\prime\prime\prime}=i\overline{\psi}\gamma^\mu\partial_\mu\psi^c-m\overline{\psi}\psi^c.$$

Response to the comment I have checked that $\overline{\psi^c}\psi=\overline{(\psi_L)^c}\psi_L+\overline{(\psi_R)^c}\psi_R$ and $\psi\overline{\psi^c}=\overline{\psi_L}(\psi_L)^c+\overline{\psi_R}(\psi_R)^c$ which means that they are different. In fact, the terms $\overline{\psi^c}\psi$ and $\psi\overline{\psi^c}$ are hermitian conjugates of each other. I think there is a problem of lack of hermiticity of here which came from the original Dirac Hamiltonian which was non-hermitian. See this.

Please help! Tell me which one is correct and which are wrong and why.

Qmechanic
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Solidification
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3 Answers3

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You can use
$$ S[\psi]= \frac 12 \int d^dx \,\psi^T {\mathcal C}({D\!\! / }+m)\psi. $$ where ${\mathcal C}$ is the charge conjugation matrix that gives $$ {\mathcal C}\gamma^\mu {\mathcal C}^{-1}=- (\gamma^\mu)^T $$ and in terms of which $$ \psi^c= {\mathcal C}^{-1} \bar\psi^T={\mathcal C}^{-1}\gamma^0 \psi^*. $$

The ${\mathcal C}$ matrix must be antisymmetric for this action to be non-zero, and so we must be in $d=$ 2, 3, 4 (mod 8) dimensions. These are the same dimensions in which $(\psi^c)^c=\psi$ making $\psi^c=\psi$ consistent, and allowing Majorana fermions to be possible.

mike stone
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The mass term has to stay invariant under general Lorentz transformations. Since Majorana fermions are their own antiparticles the usual 4-component spinor $\psi = \left( \psi_R, \psi_L \right)$ reduces to $\psi_R$. It is easy to check that $\psi_R^\dagger \psi_R$ is not invariant, since the infinitesimal transformation is: $$ \delta \psi_R = \frac{1}{2} \left( i \theta_j + \beta_j \right) \sigma_j \psi_R. $$

The usual mass term is: $\psi_R^T \sigma_2 \psi_R$, where $\sigma_2$ is a Pauli matrix: $$ \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}. $$

Darkseid
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  • Please note that $\psi$ is a 4-component spinor. A product like $\psi^T\sigma_2\psi$ is meaningless unless you write in terms of the Weyl fields in $\psi$. Also, $\bar{\psi}\psi$ is Lorentz invariant. That is the whole point of not writing a term $\psi^\dagger\psi$ but $\bar{\psi}\psi$. – Solidification Jan 19 '20 at 17:18
  • Oh snap! Well, it is a 2-component spinor for Majorana fermions as far as I know. Let me update the answer to make it more specific. – Darkseid Jan 19 '20 at 17:43
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The standard Lagrangian for a Majorana field is $$\mathcal{L}=\frac{1}{2}\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi.$$ Since $\psi^c=\psi$ for a Majorana field, we first re-express the Dirac Lagrangian $$\mathcal{L}=\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ in terms of $\psi^c$. Remembering, $\overline{\psi^c}\psi^c=\overline{\psi}\psi$ and $\overline{\psi^c}\gamma^\mu\psi^c=\overline{\psi}\gamma^\mu\psi$ the Dirac Lagrangian is transformed into, $$\mathcal{L}=\overline{\psi^c}(i\gamma^\mu\partial_\mu-m)\psi^c.$$ The final task is to make the replacement $\psi^c=\psi$, to get $$\mathcal{L}=\frac{1}{2}\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ where the factor $1/2$ in front is written by hand in accordance with the answer here. The is the required Majorana Lagragian. If $\psi=\psi_L+(\psi_L)^c$, the mass term becomes $$-\frac{m}{2}(\overline{\psi}_L(\psi_L)^c+\overline{(\psi_L)^c}\psi_L,$$ and if $\psi=\psi_R+(\psi_R)^c$, the mass term becomes $$-\frac{m}{2}(\overline{\psi}_R(\psi_R)^c+\overline{(\psi_R)^c}\psi_R.$$

Solidification
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