If you have to do a lot of manipulations with creation and annihilation operators, but you can't remember (like me) where the minus signs, daggers and $i$'s go, the following cheat sheet might come in handy.

$[a_p,a_q^\dagger]=\delta_{pq}$ | Rule |

$\quad a_p a_q^\dagger=a_q^\dagger a_p+\delta_{pq}$ | (Consequence) You can swap $a_p$ and $a_q^\dagger$, at the cost of an extra $\delta_{pq}$ |

$\quad a_q^{\dagger}a_p=a_p a_q^\dagger - \delta_{pq}$ | Likewise, but note the minus sign |

$[a_p,a_q]=0$ | Rule |

$\quad a_p a_q = a_q a_p$ | (Consequence) You can swap $a_p$ and $a_q$ at no extra cost |

$\quad a_p^\dagger a_q^\dagger = a_q^\dagger a_p^\dagger$ | Likewise |

$q_k=\frac{1}{\sqrt{2}}(a_k+a_k^\dagger)$ $p_k=\frac{1}{i \sqrt{2}}(a_k-a_k^\dagger)$ |
Definition, with $a_k$, $a_k^\dagger$ bosonic operators |

$a_k=\frac{1}{\sqrt 2}(q_k+ip_k)$ $a_k^\dagger=\frac{1}{\sqrt 2}(q_k-ip_k)$ |
Inverted definition |

$[q_k,p_\ell]=i\, \delta_{k\ell}$ $[q_k,q_\ell]=0$ $[p_k,p_\ell]=0$ |
Rule (also follows from definition and bosonic commutation relations) |

$( q_k = \sqrt{\frac{m \omega}{\hbar}} \tilde q_k )$ $( p_k = \sqrt{\frac{1}{\hbar m \omega}} \tilde p_k )$ |
Substitute this above if your are working with standard quantum harmonic oscillators and you don't want to use the nondimensionalized $q_k$ and $p_k$. The operators $\tilde q_k$ and $\tilde p_k$ have dimension of length and momentum, respectively |

$\{a_p,a_q^\dagger\}=\delta_{pq}$ | Rule |

$\quad a_p a_q^\dagger=-a_q^\dagger a_p+\delta_{pq}$ | (Consequence) You can swap $a_p$ and $a_q^\dagger$ at the cost of a minus sign and a $\delta_{pq}$ |

$\quad a_q^{\dagger}a_p=-a_p a_q^\dagger + \delta_{pq}$ | Likewise |

$\{a_p,a_q\}=0$ | Rule |

$\quad a_pa_q=-a_qa_p$ | (Consequence 1) You can always swap $a_p$ and $a_q$ at the cost of a minus sign, no matter $p$ and $q$ |

$\quad a_p^\dagger a_q^\dagger = - a_q^\dagger a_p^\dagger$ | Likewise |

$\quad \left(a_p\right)^2=\left(a_p^\dagger\right)^2=0$ | (Consequence 2) The Pauli exclusion-principle. There is no such consequence for bosons |

$\gamma_{p1}=\frac{1}{2}(a_p+a_p^\dagger)$ $\gamma_{p2}=\frac{1}{i2}(a_p-a_p^\dagger)$ |
Definition, with $a_p, a_p^\dagger$ fermionic operators |

$\{\gamma_{pa},\gamma_{qb}\}=2\delta_{pq}\delta_{ab}$ | Rule (follows from the rules for fermionic operators) |

$\quad \gamma_{pa}\gamma_{qb}=-\gamma_{qb}\gamma_{pa} \quad (pa \neq qb)$ | (Consequence 1) As long as the two operators are not exactly the same, you can swap them, always at the cost of a minus sign |

$\quad (\gamma_{pa})^2=\mathbb 1$ | (Consequence 2) Majorana fermions are their own anti-particle |

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