An axiomatic definition of normal order can be found in the book "Solitons: Differential Equations, Symmetries and Infinite Dimensional Algebra" by T. Miwa, M. Jimbo and E. Date at page 44-46. This is the best definition I have found so far and goes as follow for products of bosonic operators. Call $\mathcal{A}$ the set of linear combinations of formal finite products of bosonic operators $b_i,\,b_i^\dagger$. The normal order ${:} a {:}$ of $a \in\mathcal{A}$ is a notation defined inductively by the properties
- Linearity, $${:} z_1a_1+z_2a_2 {:}= z_1{:} a_1 {:} + z_2 {:} a_2 {:}\quad \text{for} \quad z_1,\,z_2\in \mathbb{C}\quad \text{and}\quad a_1,\,a_2 \in \mathcal{A}$$
- ${:} 1 {:} = 1$, with $1$ the identity operator in $\mathcal{A}$
- within the dots all the operators $b_i,\,b_i^\dagger$ commute among themselves
- the annihilation operators can be taken out of the columns on the right $${:}a b_i{:} = {:} a{:}\,b_i$$
- the creation operators can be taken out of the columns on the left $${:}b_i^\dagger a{:} = b_i^\dagger\,{:} a{:}$$
The definition for fermionic operators is the same with the only difference that the fermionic operators anticommute among themselves (Property 3). It is important to stress that usually annihilation operators are those operators that annihilate a specified vacuum state, therefore the normal ordering depends on the choice of the vacuum.
The normal order is a notation and not a function that acts on operators (i.e. a superoperator). This means that, whereas $b_ib_i^\dagger$ and $b_i^\dagger b_i +1$ are the same operators according to the canonical commutations relations, they are represented as different elements of $\mathcal{A}$ which is the set of linear combinations of strings of symbols generated by $b_i,b_i^\dagger$. In mathematical terms the normal order is a function defined on the elements of of the free algebra generated by $b_i,b_i^\dagger$, but is not a well defined function on the CCR-algebra (Canonical Commutation Relations algebra). The wrong step that leads to the paradoxical result
$$b_ib_i^\dagger = b_i^\dagger b_i +1 \Rightarrow {:}b_ib_i^\dagger {:} = {:}b_i^\dagger b_i +1{:} = {:}b_i^\dagger b_i{:} + 1 = {:}b_ib_i^\dagger{:} + 1 \Rightarrow 0 = 1$$
is actually the first equality since $b_ib_i^\dagger \neq b_i^\dagger b_i +1$ in $\mathcal{A}$. In another post it was suggested that the normal product is undefined when acting on linear combinations. However this would seriously limit the usefulness of the normal order. For example it is common to take the normal order of infinite series such as exponentials. The definition as a function acting on the free algebra $\mathcal{A}$ is in fact the one employed in practice.
This is a good example of how being mathematically rigor should not be regarded as a nuisance in the physics community, but rather as an important tool to avoid misunderstanding and confusion.
To start with, if you are considering the operators which are linear combination of $S_x$, $S_y$, $S_z$, the answer is yes. In addition, with this condition, note that $S_{\pm}=S_x \pm iS_y$ are the only possible raising/lowering operators for particles of any spin, not just spin-${1 \over 2}$ ones.
The key idea to derive the exact form of $S_{\pm}$ lies in the properties of $S_x$, $S_y$, $S_z$. For simplicity, we let $\hbar=1$ and have the commutators
$$[S_x,S_y]=iS_z, \ [S_y,S_z]=iS_x, \ [S_z,S_x]=iS_y \tag{1}$$
The raising/lowering operators $S_{\pm}=\alpha_{\pm}S_x+\beta_{\pm}S_y+\gamma_{\pm}S_z$ where $\alpha_{\pm}$, $\beta_{\pm}$, $\gamma_{\pm}$ are some constants are the operators which satisfy
$$[S_z,S_{\pm}]=\rho_{\pm}S_{\pm} \tag{2}$$
where $\rho_{\pm}$ are some constants. Eq. (2) comes from the fact that for any $|\psi\rangle$ being a eigenstate of $S_z$, we require $S_{\pm}|\psi\rangle$ still be eigenstates of $S_z$ with the eigenvalues varying from that of $|\psi\rangle$ by some constants $\rho_{\pm}$. The constraint may sound quite harsh, but as we know, $S_{\pm}$ exist, so we do not have to worry about their existence.
Coming to the uniqueness of $S_{\pm}$, we define the set of all linear combinations of spin operators $H=\{\alpha S_x+\beta S_y+\gamma S_z|\alpha,\beta,\gamma \in \mathbb{C}\}$, and it is useful to consider the linear map $M:H \rightarrow H$
$$M(h)=[S_z,h] \tag{3}$$
And we can express the linear map $M$ in the basis of $S_x$, $S_y$, $S_z$, which is
$$\begin{pmatrix}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} \tag{4}$$
which means for any $h=\alpha S_x+\beta S_y+\gamma S_z$,
$$M(h)= \begin{pmatrix}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
\alpha \\
\beta \\
\gamma
\end{pmatrix}=-i\beta S_x+i\alpha S_y$$
If we compare Eq. (2) with Eq. (4), we can find $S_{\pm}=\alpha_{\pm}S_x+\beta_{\pm}S_y+\gamma_{\pm}S_z$ are nothing but the eigenvectors of the matrix in Eq. (4)! Solving them and comparing the solution to the raising/lowering operators, we have $S_{\pm}=S_x\pm iS_y$ as the unique solution to Eq. (2) with $\rho_{\pm}=\pm 1$. Another takeaway message here is that the similar procedure (although they may not be recognized in the same way as in the $\text{su}(2)$ system) occurs in classification of semisimple Lie algebra, which has important applications in particle physics.
Best Answer
Linear field equations are usually reduced to a system of oscillators, each oscillator corresponding to a particular field mode (essentially an eigenstate.) We then refer to a particle as the number of excitations in each mode, i.e., the quantum number of the corresponding oscillator. This interpretation is quite straightforward for the electromagnetic field, since every mode can be characterized by a momentum and polarization, so we can speak of a number of photons having this momentum and polarization.
For those particles that call "particles" in classical sense, the same thing happens in the second quantization formalism - i.e., once we rewrite the Schrödinger equation (for an arbitrary number of particles) in terms of single-particle eigenstates.
See also: How does quantization arise in quantum mechanics?
Appendix:
I will try here to cook up a simple example. Suppose we have a many-particle Hamiltonian: $$ \hat{H} = \sum_{i=1}^{+\infty}\hat{h_i},$$ which is just a sum of one-particle Hamiltonians $$ h_i = \frac{p_i^2}{2m}+v(x_i). $$ We consider only the ground state of the single-particle Hamiltonian, $$h_i\phi_0(x_i)=\epsilon_0\phi_0(x_i) $$ (e.g., we could assume that all the other states are much higher in energy and can be disregarded.) The eigenstates of the full Hamiltonian $H$ are then $$ \psi_1(x_1)=\phi_0(x_1), \text{(one particle)},\\ \psi_2(x_1, x_2)=\phi_0(x_1)\phi_0(x_2), \text{(two particles)},\\ \psi_3(x_1, x_2, x_3)=\phi_0(x_1)\phi_0(x_2)\phi_0(x_3), \text{(three particles)},\\ ...\\\ \psi_3(x_1, x_2, x_3,..., x_n)=\prod_{i=1}^n\phi_0(x_i), \text{(n particles)}, $$ corresponding to (a ladder of) energies $$ \epsilon_0, 2\epsilon_0, 3\epsilon_0,...,n\epsilon_0. $$ (For simplicity I neglect the symmetrization and Pauli exclusion principle, obligatory in the case of fermions.)
We can now introduce operators (in the Hilbert space of the multiparticle Hamiltonian $H$) that move us from a single-particle state, to two-particle state, to three-particle state, and so on, every time increasing the energy by $\epsilon_0$. This is quite similar to what ladder operators do in an oscillator, and we could have push this analogy even closer, by showing that the sum and difference of the creation and annihilation operators thus obtained do behave as the position and momentum of an abstract oscillator. This however requires quite a bit of math, and is covered in many books.