# Majorana fermions algebra confusion

This is a quiet embarrasing question. Consider the Majorana fermion fields $$\psi_i(x)$$ and $$\psi_j(x)$$, where $$i$$ and $$j$$ denote lattice sites and $$x$$ is a spatial coordinate, which satisfy the anticommutation algebra \begin{align} \left\{\psi_i(x),\psi_j(x')\right\}=2\delta_{ij}\delta(x-x'). \end{align} Following this article (in particular, appendix A.1), define the real fermions \begin{align} \psi_i(x)=\frac{1}{\sqrt{2}}\left(c(x)+c(x)^\dagger\right),\quad \psi_j(x)=\frac{1}{i\sqrt{2}}\left(c(x)^\dagger-c(x)\right). \end{align} It is possible to invert this relation to obtain \begin{align} c(x)=\frac{1}{\sqrt{2}}\left(\psi_i(x)+i\psi_j(x)\right),\quad c^\dagger=\frac{1}{\sqrt{2}}\left(\psi_i(x)-i\psi_j(x)\right), \end{align} where $$c$$ and $$c^\dagger$$ are annihilation and creation operators, respectively. I am trying to obtain the algebra of creation and annihilation operators, which is $$\left\{c(x),c^\dagger(x')\right\}\propto \delta(x-x')$$. For this purpose, first I am trying to compute $$c(x)c^\dagger(x')$$ and $$c^\dagger(x)c(x')$$ and then I will try to add them, however, I am not being able to obtain a closed expression for the first product \begin{align} c(x) c^\dagger(x')&=\frac{1}{2}\left(\psi_i(x)+i\psi_j(x)\right)\left(\psi_i(x')-i\psi_j(x')\right)\\ &=\frac{1}{2}\left( \psi_i(x)\psi_i(x')-i\psi_i(x)\psi_j(x')\right.\\ &\hspace{2cm}\left.+i\psi_j(x)\psi_i(x')+\psi_j(x)\psi_j(x')\right), \end{align} and then I do not know how should I proceed (I tried to create a commutator, anticommutator, use $$\psi_i(x)\psi_i(x')=2\delta(x-x')-\psi_i(x')\psi_i(x)$$ but nothing worked).

• It seems like you can just write out $c^{\dagger}(x')c(x)$ like you have here, then add to the first product and use the anticommutator of the $\psi_i$ and $\psi_j$ to get the result, no?
– d_b
Apr 8 at 3:50
• @d_b, yes you are right. Originally I was trying to obtain a number operator $c^\dagger(x)c(x')=i\psi_i(x)\psi_j(x')$, according to equation (A.9) of the article, because that expression will be useful in the future. However, with your help I obtained directly $\left\{c(x),c^\dagger(x')\right\}=2\delta(x-x')$. I was really confused. Thank you very much! Apr 8 at 14:46