In the context of many-body quantum systems, the calculation is more efficient if we use the second quantization formulation. Instead of the standard many-body wave function, we use the so-called bosonic and fermionic operators. This post does not discuss the motivation on the second quantization, though. Here, I simply note some important identities.

Bosonic Operators

The bosonic creation and annihilation operators are usually written as \(a^{\dagger}\) and \(a\), respectively. They follow the following commutation relation, \begin{equation} \boxed{\left[a_i, a_j \right] = [a_i^{\dagger}, a_j^{\dagger} ] = 0, \qquad [a_i, a_j^{\dagger} ] = \delta_{ij}} \end{equation} where \([A, B] = AB - BA\).

The commmutator has the following identity:

\[[A, B] = AB - BA = - (BA - AB) = - [B, A]\]

Fermionic Operators

For the fermionic case, people usually denote the creation and annihilation by \(c^{\dagger}\) and \(c\). They follow the anti-commutation relation

\[\boxed{\{ c_i, c_j \} = \{ c_i^{\dagger}, c_j^{\dagger} \} = 0, \quad \{ c_i, c_j^{\dagger} \} = \delta_{ij}}\]

where \(\{ A, B \} = AB + BA\). It is called an anti-commutator. Because addition is commutative, so

\[\{ A, B \} = AB + BA = BA + AB = \{ B, A \}\]

From this identity, we can calculate the following:

\[\{ c_i, c_i \} = c_i c_i + c_i c_i = 2 c_i^2 = 0 \\ \Rightarrow \boxed{c_i^2 = 0}\]

Also,

\[\{ c_i^{\dagger}, c_i^{\dagger} \} = c_i^{\dagger} c_i^{\dagger} + c_i^{\dagger} c_i^{\dagger} = 2 (c_i^{\dagger})^2 = 0 \\ \Rightarrow \boxed{(c_i^{\dagger})^2 = 0}\]

This identity has an interesting physical interpretation. Basically it shows us that we can’t put two fermions in the same mode \(i\) (or the same quantum mechanical state, the term mode is more general). So, it captures the Pauli exclusion principle. Nice.

Number Operator

Let us now define the number operator \(n = a^{\dagger} a\) or \(n = c^{\dagger} c\) which counts the number of bosons or fermions. Note that for the fermion case, \(n\) can only be 0 or 1 whereas for the bosonic case, \(n\) can be any non-negative numbers.

Let’s now consider the following:

\[[ab, c] = abc - cab = abc + (acb - acb) - cab = (abc - acb) + (acb - cab)\]

or

\[[ab, c] = (abc + acb) + (-acb - cab)\]

so now we have this identity

\[\boxed{[ab,c] = a[b, c] + [a, c] b = a\{b, c\} - \{a, c \}b}\]

The, considering the bosonic commutation relation, we get the following

\[\begin{split} [n_j, a_j] &= [a^{\dagger}_j a_j, a_j] \\ &= a_j^{\dagger}[a_j, a_j] + [a_j^{\dagger}, a_j]a_j \\ &= -a_j \end{split}\]

because \([a_j^{\dagger}, a_j] = - [a_j, a_j^{\dagger}]\). Also,

\[\begin{split} [n_j, a_j^{\dagger}] &= [a^{\dagger}_j a_j, a_j^{\dagger}] \\ &= a_j^{\dagger}[a_j, a_j^{\dagger}] + a_j [a_j^{\dagger}, a_j^{\dagger}]\\ &= a_j^{\dagger} \end{split}\]

Okay, now let’s look at the fermionic case and calculate the commutator of \(n_j\) with fermionic operators

\[\begin{split} [ n_j, c_j ] &= [ c_j^{\dagger}c_j, c_j ] \\ &= c_j^{\dagger}\{ c_j, c_j \} - \{ c_j^{\dagger}, c_j \} c_j \\ &= -c_j \end{split}\]

and

\[\begin{split} [ n_j, c_j^{\dagger}] &= [ c_j^{\dagger} c_j, c_j^{\dagger} ]\\ &= c_j^{\dagger} \{ c_j, c_j^{\dagger}\} - \{ c_j^{\dagger}, c_j^{\dagger} \} c_j \\ &= c_j^{\dagger} \end{split}\]

We can also calculate the following

\[n_j^2 = (c_j^{\dagger} c_j)^2 = c_j^{\dagger} (c_j c_j^{\dagger}) c_j = c_j^{\dagger} (1 - c_j^{\dagger} c_j) c_j = c_j^{\dagger} c_j = n_j\]

In summary, we have the following identities

\[\begin{split} \boxed{[n_j, a_j] = -a_j, \quad [n_j, a_j^{\dagger}] = a_j^{\dagger}, \\ [ n_j, c_j ] = -c_j, \quad [ n_j, c_j^{\dagger}] = c_j^{\dagger}, \\ n_j^2 = n_j } \end{split}\]