Dirac bracket for the Majorana Lagrangian Note: See update below.
Consider the Majorana Lagrangian 
$$\mathcal{L}=-\psi ^{\mathrm{T}}\mathrm{i}%
\gamma ^{0}\left( \gamma ^{\rho }\partial _{\rho }+m\right) \psi ,\tag{1}$$ 
where $%
\psi \in \mathbb{R}^{4}$ is Grassmann-valued, and where $\gamma ^{\rho }\in \mathrm{M}_{4}\left( 
\mathbb{R}\right) $ is the irreducible, real-valued
representation of $\mathrm{Cliff}\left( 3,1\right) $. The generalized
coordinates $\psi ^{a}=\psi ^{a}\left( \mathbf{x}\right) $ and their conjugate
momenta $\pi _{a}\left( \mathbf{x}\right) \equiv \pi _{a}^{L}\left( \mathbf{x%
}\right) =-\pi _{a}^{R}\left( \mathbf{x}\right) $, where
\begin{eqnarray*}
\pi _{a}^{L} &\equiv &\frac{\partial _{L}\mathcal{L}}{\partial \dot{\psi}^{a}%
}\equiv \frac{\overrightarrow{\partial }}{\partial \dot{\psi}^{a}}\mathcal{L}%
=+\mathrm{i}\left( \psi ^{\mathrm{T}}\right) _{a}=+\mathrm{i}\psi _{a}, \\
\pi _{a}^{R} &\equiv &\frac{\partial _{R}\mathcal{L}}{\partial \dot{\psi}^{a}%
}\equiv \mathcal{L}\frac{\overleftarrow{\partial }}{\partial \dot{\psi}^{a}}%
=-\mathrm{i}\left( \psi ^{\mathrm{T}}\right) _{a}=-\mathrm{i}\psi _{a},\tag{2}
\end{eqnarray*}
following answer by Qmechanic, are obviously not independent, but constrained by
$$
0=\chi _{a \mathbf{x}} \equiv \psi _{a}\left( \mathbf{x}\right)
+\mathrm{i}\pi _{a}\left( \mathbf{x}\right) .\tag{3}
$$
The Poisson brackets of these constraints with themselves
are given by
\begin{eqnarray*}
C_{a \mathbf{x},b \mathbf{y}} &\equiv &\left\{ \chi _{a \mathbf{x}},\chi _{b \mathbf{y}} \right\} _{\text{P}} \\
&=&\mathrm{i}\left\{ \psi _{a}\left( \mathbf{x}\right) ,\pi _{b}\left( 
\mathbf{y}\right) \right\} _{\text{P}}+\mathrm{i}\left\{ \pi _{a}\left( 
\mathbf{x}\right) ,\psi _{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\
&=&-2\mathrm{i}\delta _{ab}\delta ^{\left( 3\right) }\left( \mathbf{x}-%
\mathbf{y}\right) ,\tag{4}
\end{eqnarray*}
using
\begin{eqnarray*}
\left\{ F,G\right\} _{\text{P}} &=&\int d^{3}x\left[ \frac{\delta _{R}F}{
\delta \psi ^{a}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \pi
_{a}^{R}\left( \mathbf{x}\right) }-\frac{\delta _{R}F}{\delta \pi
_{a}^{L}\left( \mathbf{x}\right) }\frac{\delta _{L}G}{\delta \psi ^{a}\left( 
\mathbf{x}\right) }\right]  \\
&\equiv &-\int d^{3}x\left[ \frac{\delta _{R}F}{\delta \psi ^{a}\left( 
\mathbf{x}\right) }\frac{\delta _{L}G}{\delta \pi _{a}\left( \mathbf{x}
\right) }+\frac{\delta _{R}F}{\delta \pi _{a}\left( \mathbf{x}\right) }\frac{
\delta _{L}G}{\delta \psi ^{a}\left( \mathbf{x}\right) }\right] ,\tag{5}
\end{eqnarray*}
following again answer by Qmechanic. The matrix $C_{a \mathbf{x},b \mathbf{y}}$ being invertible,
$$
\left( C^{-1}\right) _{a\mathbf{x},b\mathbf{y}}=\frac{\mathrm{i}}{2}\delta
_{ab}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{y}\right),\tag{6}
$$
implies, in the terminology of Dirac, that these constraints are second class. The Dirac bracket is thus given by
\begin{eqnarray*}
\left\{ F,G\right\} _{\text{D}} &\equiv &\left\{ F,G\right\} _{\text{P}
}-\int d^{3}x\int d^{3}y\left\{ F,\chi ^{a\mathbf{x}}\right\} _{\text{P}
}\left( C^{-1}\right) _{a\mathbf{x},b\mathbf{y}}\left\{ \chi ^{b\mathbf{y}
},G\right\} _{\text{P}} \\
&=&\left\{ F,G\right\} _{\text{P}}-\frac{\mathrm{i}}{2}\int d^{3}x\left\{
F,\chi ^{a\mathbf{x}}\right\} _{\text{P}}\left\{ \chi _{a\mathbf{x}
},G\right\} _{\text{P}},\tag{7}
\end{eqnarray*}
so that
\begin{eqnarray*}
\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right)
\right\} _{\text{D}} &=&\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi
^{b}\left( \mathbf{y}\right) \right\} _{\text{P}}-\frac{\mathrm{i}}{2}\int
d^{3}z\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\chi ^{c\mathbf{z}}\right\}
_{\text{P}}\left\{ \chi _{c\mathbf{z}},\psi ^{b}\left( \mathbf{y}\right)
\right\} _{\text{P}} \\
&=&\frac{\mathrm{i}}{2}\int d^{3}z\left\{ \psi ^{a}\left( \mathbf{x}\right)
,\pi ^{c}\left( \mathbf{z}\right) \right\} _{\text{P}}\left\{ \pi _{c}\left( 
\mathbf{z}\right) ,\psi ^{b}\left( \mathbf{y}\right) \right\} _{\text{P}} \\
&=&\frac{\mathrm{i}}{2}\int d^{3}z\delta ^{ac}\delta ^{\left( 3\right)
}\left( \mathbf{x}-\mathbf{z}\right) \delta _{c}^{b}\delta ^{\left( 3\right)
}\left( \mathbf{z}-\mathbf{y}\right)  \\
&=&\frac{\mathrm{i}}{2}\delta ^{ab}\delta ^{\left( 3\right) }\left( \mathbf{x
}-\mathbf{y}\right) .\tag{8}
\end{eqnarray*}
Following the 'quantum bracket = $\mathrm{i} \times$ Poisson
bracket'-rule, the quantum bracket is thus supposedly given by
$$
\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi ^{b}\left( \mathbf{y}\right)
\right\} =-\frac{1}{2}\delta ^{ab}\delta ^{\left( 3\right) }\left( \mathbf{x}
-\mathbf{y}\right) .\tag{9}
$$
Can that really be correct? The minus sign looks wrong. And what about the factor of 1/2? But before turning attention to these details, perhaps I should start by asking whether I have made some conceptual errors above. Have I simply misunderstood how to go about quantizing constrained systems?
Update: Now the penny finally dropped. Almost embarrassingly, the wrong sign
boils down to not having taken into account the minus sign in $\gamma
_{0}^{2}=-1_{4}$ in the calculation of the conjugate momenta. Taking proper
care of this, the sign issue evaporates. (Note that in order to preserve the history of this posting, the material above has not been edited accordingly.) Thus to me only remains now the puzzlement over the factor of 1/2, see my comment below.
 A: Comments to the question (v9): 


*

*If we ignore the overall normalization, then OP correctly applies the Dirac-Bergmann$^1$ method, which leads to second-class constraints.$^2$

*Normally the Majorana Lagrangian (1) is defined with a factor $\frac{1}{2}$ in front. Then there will be no factor $\frac{1}{2}$ in the anti-commutator relation (9), see e.g. Ref. 2.

*As for the overall sign of the Lagrangian (1), one should chose consistent sign conventions. In particular, the Hamiltonian should be bounded from below. See e.g. Ref. 2 for a consistent choice.
References:


*

*S. Weinberg, Quantum Theory of Fields, Vol. 1; Section 7.6.

*M. Srednicki, QFT, Chapter 37. A prepublication draft PDF file is available here.
--
$^1$ OP mentions in a comment that he (partially) follows Ref. 1. Note that Ref. 1 does not explain sign conventions for Grassmann-odd fields.
$^2$ For a similar calculation with fermionic second-class constraints, see e.g. my Phys.SE answer here (NB: not Majorana). Instead of the Dirac-Bergmann method, one can use the Faddeev-Jackiw method, cf. e.g. my Phys.SE answer here (NB: not Majorana).
