0
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – d_b
    Apr 8 at 3:50
  • $\begingroup$ @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! $\endgroup$ Apr 8 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.