Express a product of Majorana Fermions as an exponential of number operators

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.

$\hat{N} = c^{\dagger}c$ 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.
• 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 $-i \gamma_{o} = \exp[i \pi \hat{N}]$? – greenchile455 Jan 31 '18 at 16:17