Commutation relations cheat sheet

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

Contents:
Bosons
Fermions
Majorana fermions
Spin / Angular momentum / Pauli operators

Bosons

$[a_p,a_q^\dagger]=\delta_{pq}$ 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
$[a^\dagger a^{\phantom\dagger},a]=-a$
$[a^\dagger a^{\phantom\dagger},a^\dagger]=a^\dagger$
(Consequence) Say you have a Hamiltonian of the form $H= \omega\, a^\dagger a^{\phantom\dagger}$, with eigenstates $|n\rangle$ and corresponding eigenvales $n\omega$. In this case it follows from the relations on the left (together with a normalization condition) that $a|n\rangle=\sqrt{n}|n-1\rangle$ and $a^\dagger_p|n\rangle=\sqrt{n+1}|n+1\rangle$.

Quadratures

$q_k=\frac{1}{\sqrt{2}}(a_k+a_k^\dagger)$
$p_k=\frac{1}{i \sqrt{2}}(a_k-a_k^\dagger)$
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

Fermions

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

Majorana fermions

$\gamma_{p1}=\frac{1}{2}(a_p+a_p^\dagger)$
$\gamma_{p2}=\frac{1}{i2}(a_p-a_p^\dagger)$
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

Spin / Angular momentum / Pauli operators

One spin

Notation:
$\sigma_1=\sigma_x=X$,
$\sigma_2=\sigma_y=Y$,
$\sigma_3=\sigma_z=Z$.

$[\sigma_a,\sigma_b]=2i\varepsilon_{abc}\sigma_c$
Rule, with $\varepsilon$ the Levi-Civita symbol.
$\quad [X,Y]=2iZ$ (Example) By cyclic permutation of the symbols we obtain the other commutators; $[Y,Z]=2iX$ and $[Z,X]=2iY$
$\sigma_a\sigma_b=i\varepsilon_{abc}\sigma_c\quad(a\neq b)$
$\sigma_a^2=I$
(Consequence 1)
$\quad XY=iZ$ (Example) Again, by cyclic permutation, $YZ=iX$ and $ZX=iY$
$\sigma_a\sigma_b=-\sigma_b\sigma_a$ (Consequence) The Pauli matrices anti-commute
$\quad XY=-YX$ (Example)
$\sigma_{\pm}=X\pm iY$ Definition
$[Z,\sigma_{\pm}]= \pm 2 \sigma_\pm$ Consequence

Multiple spins

Notation:
$S_{a}=\frac{1}{2}\sum_j \sigma_{a,j}$, with $\sigma$ the Pauli-$a$ operator that acts on spin $j$ only.

$S^2 \equiv (S_x)^2+(S_y)^2+(S_z)^2$ Definition of the total spin operator. With $|\psi\rangle$ an eigenstate of $S^2$, the quantum number $S$ is defined by $S^2|\psi\rangle=S(S+1)|\psi\rangle$.
$[S^2,S_a]=0$ Consequence of the commutation relations of the Pauli operators
$S_\pm=S_x\pm iS_y$ Definition
$[S_z,S_\pm]=\pm S_\pm$
$[S^2,S_\pm]=0$
Consequence of the commutation relations of the Pauli-operators

Updated