2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Have you written out the terms (in terms of gamma matrices etc.) to see whether $\bar{\psi^c}\psi$ and $\bar{\psi}\psi^c$ are actually different? $\endgroup$
    – ACuriousMind
    Jan 19, 2020 at 17:03
  • $\begingroup$ See the added comment above. $\endgroup$ Jan 19, 2020 at 17:09

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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: $$ \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}. $$

$\endgroup$
2
  • $\begingroup$ 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 $\bar{\psi}\psi$. $\endgroup$ Jan 19, 2020 at 17:18
  • $\begingroup$ 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. $\endgroup$
    – Darkseid
    Jan 19, 2020 at 17:43
1
$\begingroup$

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 $$-\frac{m}{2}(\overline{\psi}_R(\psi_R)^c+\overline{(\psi_R)^c}\psi_R.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.