I am a computational chemist who is moving into, what I understand is, condensed matter physics. I have started reading Blaizot and Ripka "Quantum Theory of Finite Systems" and I am having trouble with their problems.

They define two particles which I believe are called Majorana Fermions. If there are n fermion creation and annihilation operators $a^{\dagger}_{j}$ and $a_{j}$ then the Majorana Fermions are $\gamma_{j} = a_{j} + a^{\dagger}_{j}$ and $\gamma_{n+j} = (a_{j} - a^{\dagger}_{j})/i$.

They ask to show that the product of these operators $\gamma_{o} = \Pi_{j=1}^{2n}\gamma_{j} = \exp[i\pi\hat{N}]$ where $\hat{N} = \sum_{j}a^{\dagger}_{j}a_{j}$.

If we assume that there is a single fermion, n=1, I'm already troubled by the relationship since $\gamma_{2}$ is imaginary but $\exp[i \pi N]$ will alternate between -1 and 1 depending on the number of fermions. If we have a single fermion $\gamma_{1} \gamma_{2}$ is imaginary even though $\exp[i \pi]$ is real.

When I multiply out the expression I get $\gamma_{1} \gamma_{2} = i(1 - 2a^{\dagger}_{1}a_{1})$.

Thank you for any information on this or any recommendations other than Blaizot and Ripka that might help me figure this out.


1 Answer 1


If you consider the idempotency of the number operator you should just be able to use the Maclaurin series of e to connect the two. So here are the relevant equations

$\hat{N} = c^{\dagger}c$ and $\hat{N}\hat{N} = \hat{N}$ and $\exp[i\pi\hat{N}] = \sum_{n=0}^{\infty}\frac{(i\pi\hat{N})^{n}}{n!}$

This leads to the result $\exp[i\pi \hat{N}] = 1 + \left [\sum_{n=1}^{\infty}\frac{(i\pi)^{n}}{n!} \right ]\hat{N}$.

Do a quick check on the summation to show that $\exp[i\pi \hat{N}] = 1 -2\hat{N}$ which is relationship you found.

  • $\begingroup$ Thank you. After reading through your explanation I looked back at the book. There is still the factor of $i$ to account for though. Is it possible there is an error in the book? Should it read $-i \gamma_{o} = \exp[i \pi \hat{N}]$? $\endgroup$ Jan 31, 2018 at 16:17

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.