quantum mechanics - Reconciling Wikipedia and textbook descriptions of ladder operator method

I'm trying to reconcile the work in my textbook, Quantum Field Theory and the Standard Model by Schwartz, which I'm finding difficult to follow, with the Wikipedia article for the ladder operator method of quantum harmonic oscillators.

The Wikipedia article proceeds as follows:

The commutation property yields

$$\begin{align} Na^{\dagger} \mid n \rangle &= (a^{\dagger} N + [N, a^{\dagger}]) \mid n \rangle \\ &= (a^{\dagger} N + a^{\dagger}) \mid n \rangle \\ &= (n + 1)a^{\dagger} \mid n \rangle\end{align}$$

and similarly,

$N a \mid n \rangle = (n - 1)a \mid n \rangle$

I understood everything, except the first part:

$$Na^{\dagger} \mid n \rangle = (a^{\dagger} N + [N, a^{\dagger}]) \mid n \rangle$$

1. How did they get this? What "commutation property" is being referred to for this?

The textbook proceeds as follows:

$$\begin{align} \hat{N} a^{\dagger} \mid n \rangle &= a^{\dagger} a a^{\dagger} \mid n \rangle \\ &= a^{\dagger} \mid n \rangle + a^{\dagger} a^{\dagger} a \mid n \rangle \\ &= (n + 1) a^{\dagger} \mid n \rangle \end{align}$$

2. Here, I don't understand how the author got

$$a^{\dagger} a a^{\dagger} \mid n \rangle = a^{\dagger} \mid n \rangle + a^{\dagger} a^{\dagger} a \mid n \rangle$$

and

$$a^{\dagger} \mid n \rangle + a^{\dagger} a^{\dagger} a \mid n \rangle = (n + 1) a^{\dagger} \mid n \rangle$$

?

I would greatly appreciate it if people could please take the time to clarify this.

Answer

1 -This is simply the definition of a commutator:

$$[N,a^\dagger] := N a^\dagger - a^\dagger N$$

so:

$$Na^\dagger = a^\dagger N + [N,a^\dagger]$$

2 - Remember the commutation relation for the ladder operators $[a,a^\dagger] = 1$:

$$a^\dagger a a^\dagger = a^\dagger (a^\dagger a + [a,a^\dagger]) = a^\dagger a^\dagger a + a^\dagger 1 = a^\dagger a^\dagger a + a^\dagger$$

$\quad$ Meaning that:

$$a^\dagger a a^\dagger |n \rangle = a^\dagger a^\dagger a |n \rangle + a^\dagger |n \rangle$$

3 - Remember that $a^\dagger a \equiv N$, so:

$$a^\dagger |n \rangle + a^\dagger \underbrace{a^\dagger a}_{N} |n \rangle = a^\dagger |n \rangle + a^\dagger N |n \rangle = (1+n) a^\dagger |n \rangle$$