The ladder operators are not axiomatic. They are derived by a quite long chain of conclusions from the commutation relations. $$\begin{align} [L_x,L_y]=i\hbar L_z \\ [L_y,L_z]=i\hbar L_x \\ [L_z,L_x]=i\hbar L_y \end{align} \tag{1}$$ For angular momentum $\mathbf{L}$ these relations (1) can be derived from the definition $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ and the canonic commutators $[r_i,p_j]=i\hbar\delta_{ij}$. But for spin angular momentum $\mathbf{S}$ these relations cannot be derived from anything. So for spin we need to postulate (1) as an axiom. And then we can show that the total angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ also satisfies the relations (1).

From (1) we can prove $[L^2,L_z]=0$. Therefore we know there are simultaneous eigenstates of $L^2$ and $L_z$. We can call these states $|l,m\rangle$ where $l$ labels the $L^2$-eigenvalue and $m$ labels the $L_z$-eigenvalue.

Now we define two operators $L_+$ and $L_-$. $$L_\pm=L_x\pm iL_y \tag{2}$$ These definitions are grabbed from thin air here. So we need to prove that these two operators are actually ladder operators.

It is not hard to calculate the commutators between operators (2) and $L^2$ and $L_z$. Using (1) we find: $$[L^2,L_\pm]=0 \tag{3}$$ $$[L_z,L_\pm]=\pm\hbar L_\pm \tag{4}$$

Now we can look for some properties of the states $L_+|l,m\rangle$ and $L_-|l,m\rangle$.
From (3) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L^2$, then $L_\pm|l,m\rangle$ is an eigenstate of $L^2$ to the same eigenvalue.

And from (4) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L_z$ to eigenvalue $m\hbar$, then $L_\pm|l,m\rangle$ is an eigenstate of $L_z$ to the eigenvalue $(m\pm 1)\hbar$.

It is these last two properties which show that $L_+$ and $L_-$ are ladder operators. $L_+$ steps up the $L_z$ eigenvalue, and $L_-$ steps down the $L_z$ eigenvalue. $$L_\pm|l,m\rangle \propto |l,m\pm 1\rangle \tag{5}$$

Angular momentum is quantized because of how we defined the ladder operators. But where do these operators come from? How do they naturally arise when dealing with angular momentum?

The ladder operators for angular momentum are not arbitrary. They are derived by a quite long chain of conclusions from the commutation relations of angular momentum. $$\begin{align} [L_x,L_y]=i\hbar L_z \\ [L_y,L_z]=i\hbar L_x \\ [L_z,L_x]=i\hbar L_y \end{align} \tag{1}$$ For angular momentum $\mathbf{L}$ these relations (1) can be derived from the definition $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ and the canonic commutators $[r_i,p_j]=i\hbar\delta_{ij}$. But for spin angular momentum $\mathbf{S}$ these relations cannot be derived from anything. So for spin we need to postulate (1) as an axiom. And then we can show that the total angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ also satisfies the relations (1).

From (1) we can prove $[L^2,L_z]=0$. Therefore we know there are simultaneous eigenstates of $L^2$ and $L_z$. We can call these states $|l,m\rangle$ where $l$ labels the $L^2$-eigenvalue and $m$ labels the $L_z$-eigenvalue.

Now we define two operators $L_+$ and $L_-$. $$L_\pm=L_x\pm iL_y \tag{2}$$

what is the logic that made physicists think "hey lets define a very specific operator to change the angular momentum eigenvalue in units of $\hbar$"? How did we think the define it as $L_1±iL_2$?

These definitions (2) seem like arbitrarily grabbed from thin air. But they are carefully chosen based on the commutator relations (1). We actually need to prove that these two operators are ladder operators. This proof (without the details) is sketched below.

It is not hard to calculate the commutators between operators (2) and $L^2$ and $L_z$. Using (1) we find: $$[L^2,L_\pm]=0 \tag{3}$$ $$[L_z,L_\pm]=\pm\hbar L_\pm \tag{4}$$

Now we can look for some properties of the states $L_+|l,m\rangle$ and $L_-|l,m\rangle$.
From (3) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L^2$, then $L_\pm|l,m\rangle$ is an eigenstate of $L^2$ to the same eigenvalue.

And from (4) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L_z$ to eigenvalue $m\hbar$, then $L_\pm|l,m\rangle$ is an eigenstate of $L_z$ to the eigenvalue $(m\pm 1)\hbar$.

It is these last two properties which show that $L_+$ and $L_-$ are ladder operators. $L_+$ steps up the $L_z$ eigenvalue, and $L_-$ steps down the $L_z$ eigenvalue by $\hbar$. $$L_\pm|l,m\rangle \propto |l,m\pm 1\rangle \tag{5}$$

The ladder operators are not axiomatic. They are derived by a quite long chain of conclusions from the commutation relations. $$\begin{align} [L_x,L_y]=i\hbar L_z \\ [L_y,L_z]=i\hbar L_x \\ [L_z,L_x]=i\hbar L_y \end{align} \tag{1}$$ For angular momentum $\mathbf{L}$ these relations (1) can be derived from the definition $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ and the canonic commutators $[r_i,p_j]=i\hbar\delta_{ij}$. But for spin angular momentum $\mathbf{S}$ these relations cannot be derived from anything. So for spin we need to postulate (1) as an axiom. And then we can show that the total angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ also satisfies the relations (1).

From (1) we can prove $[L^2,L_z]=0$. Therefore we know there are simultaneous eigenstates of $L^2$ and $L_z$. We can call these states $|l,m\rangle$ where $l$ labels the $L^2$-eigenvalue and $m$ labels the $L_z$-eigenvalue.

Now we define two operators $L_+$ and $L_-$. $$L_\pm=L_x\pm L_y \tag{2}$$ These definitions are grabbed from thin air here. So we need to prove that these two operators are actually ladder operators.

It is not hard to calculate the commutators between operators (2) and $L^2$ and $L_z$. Using (1) we find: $$[L^2,L_\pm]=0 \tag{3}$$ $$[L_z,L_\pm]=\pm\hbar L_\pm \tag{4}$$

Now we can look for some properties of the states $L_+|l,m\rangle$ and $L_-|l,m\rangle$.
From (3) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L^2$, then $L_\pm|l,m\rangle$ is an eigenstate of $L^2$ to the same eigenvalue.

And from (4) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L_z$ to eigenvalue $m\hbar$, then $L_\pm|l,m\rangle$ is an eigenstate of $L_z$ to the eigenvalue $(m\pm 1)\hbar$.

It is these last two properties which show that $L_+$ and $L_-$ are ladder operators. $L_+$ steps up the $L_z$ eigenvalue, and $L_-$ steps down the $L_z$ eigenvalue. $$L_\pm|l,m\rangle \propto |l,m\pm 1\rangle \tag{5}$$

The ladder operators are not axiomatic. They are derived by a quite long chain of conclusions from the commutation relations. $$\begin{align} [L_x,L_y]=i\hbar L_z \\ [L_y,L_z]=i\hbar L_x \\ [L_z,L_x]=i\hbar L_y \end{align} \tag{1}$$ For angular momentum $\mathbf{L}$ these relations (1) can be derived from the definition $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ and the canonic commutators $[r_i,p_j]=i\hbar\delta_{ij}$. But for spin angular momentum $\mathbf{S}$ these relations cannot be derived from anything. So for spin we need to postulate (1) as an axiom. And then we can show that the total angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ also satisfies the relations (1).

From (1) we can prove $[L^2,L_z]=0$. Therefore we know there are simultaneous eigenstates of $L^2$ and $L_z$. We can call these states $|l,m\rangle$ where $l$ labels the $L^2$-eigenvalue and $m$ labels the $L_z$-eigenvalue.

Now we define two operators $L_+$ and $L_-$. $$L_\pm=L_x\pm iL_y \tag{2}$$ These definitions are grabbed from thin air here. So we need to prove that these two operators are actually ladder operators.

It is not hard to calculate the commutators between operators (2) and $L^2$ and $L_z$. Using (1) we find: $$[L^2,L_\pm]=0 \tag{3}$$ $$[L_z,L_\pm]=\pm\hbar L_\pm \tag{4}$$

Now we can look for some properties of the states $L_+|l,m\rangle$ and $L_-|l,m\rangle$.
From (3) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L^2$, then $L_\pm|l,m\rangle$ is an eigenstate of $L^2$ to the same eigenvalue.

And from (4) we can prove:

• If $|l,m\rangle$ is an eigenstate of $L_z$ to eigenvalue $m\hbar$, then $L_\pm|l,m\rangle$ is an eigenstate of $L_z$ to the eigenvalue $(m\pm 1)\hbar$.

It is these last two properties which show that $L_+$ and $L_-$ are ladder operators. $L_+$ steps up the $L_z$ eigenvalue, and $L_-$ steps down the $L_z$ eigenvalue. $$L_\pm|l,m\rangle \propto |l,m\pm 1\rangle \tag{5}$$