Classical Fermion and Grassmann number In the theory of relativistic wave equations, we derive the Dirac equation and Klein-Gordon equation by using representation theory of Poincare algebra. 
For example, in this paper 
http://arxiv.org/abs/0809.4942
the Dirac equation in momentum space (equation [52], [57] and [58]) can be derived from the 1-particle state of irreducible unitary representation of the Poincare algebra (equation [18] and [19]). The ordinary wave function in position space is its Fourier transform (equation [53], [62] and [65]).  
Note at this stage, this Dirac equation is simply a classical wave equation. i.e. its solutions are classical Dirac 4-spinors, which take values in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$.
If we regard the Dirac waves $\psi(x)$ and $\bar{\psi}(x)$ as a 'classical fields', then the quantized Dirac fields are obtained by promoting them into fermionic harmonic oscillators. 
What I do not understand is that when we are doing the path-integral quantization of Dirac fields, we are, in fact, treating $\psi$ and $\bar{\psi}$ as Grassmann numbers, which are counter-intuitive for me. As far as I understand, we do path-integral by summing over all 'classical fields'. While the 'classical Dirac wave $\psi(x)$' we derived in the beginning are simply 4-spinors living in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$. How can they be treated as Grassmann numbers instead? 
As I see it, physicists are trying to construct a 'classical analogue' of Fermions that are purely quantum objects. For instance, if we start from a quantum anti-commutators
$$[\psi,\psi^{\dagger}]_{+}=i\hbar1
\quad\text{and}\quad
[\psi,\psi]_{+}=[\psi^{\dagger},\psi^{\dagger}]_{+}=0, $$
then we can obtain the Grassmann numbers in the classical limit $\hbar\rightarrow0$. This is how I used to understand the Grassmann numbers. The problem is that if the Grassmann numbers are indeed a sort of classical limit of anticommuting operators in Hilbert space, then the limit $\hbar\rightarrow0$ itself does not make any sense from a physical point of view since in this limit $\hbar\rightarrow0$, the spin observables vanish totally and what we obtain then would be a $0$, which is a trivial theory.
Please tell me how exactly the quantum Fermions are related to Grassmann numbers.  
 A: $\require{cancel}$


*

*First of all, recall that a super-Lie bracket $[\cdot,\cdot]_{LB}$ (such as, e.g., a super-Poisson bracket $\{\cdot,\cdot\}$ & the super-commutator $[\cdot,\cdot]$), satisfies super-antisymmetry
$$ [f,g]_{LB} ~=~ -(-1)^{|f||g|}[g,f]_{LB},\tag{1} $$
and the super-Jacobi identity
$$\sum_{\text{cycl. }f,g,h} (-1)^{|f||h|}[[f,g]_{LB},h]_{LB}~=~0.\tag{2}$$
Here $|f|$ denotes the Grassmann-parity of the super-Lie algebra element $f$. Concerning supernumbers, see also e.g. this Phys.SE post and links therein.

*In order to ensure that the Hilbert space has no negative norm states and that the vacuum state has no negative-energy excitations, the Dirac field should be quantized with anticommutation relations 
$$  [\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)]_{+}
~=~ \hbar\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1}
~=~[\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)]_{+},  $$
$$  [\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)]_{+}
~=~ 0, \qquad [\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)]_{+}~=~ 0, \tag{3} $$
rather than with commutation relations, cf. e.g. Ref. 1 and this Phys.SE post.

*According to the correspondence principle between quantum and classical physics, the supercommutator is $i\hbar$ times the super-Poisson bracket (up to possible higher $\hbar$-corrections), cf. e.g. this Phys.SE post. Therefore the corresponding fundamental super-Poisson brackets read$^1$
$$  \{\psi_{\alpha}({\bf x},t), \psi^{\ast}_{\beta}({\bf y},t)\} 
~=~ -i\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})
~=~\{\psi^{\ast}_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\},  $$
$$  \{\psi_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\} 
~=~ 0, \qquad \{\psi^{\ast}_{\alpha}({\bf x},t), \psi^{\ast}_{\beta}({\bf y},t)\}~=~ 0. \tag{4} $$

*Comparing eqs. (1), (3) & (4), it becomes clear that the Dirac field is Grassmann-odd, both as an operator-valued quantum field $\hat{\psi}_{\alpha}$ and as a supernumber-valued classical field $\psi_{\alpha}$. 

*It is interesting that the free Dirac Lagrangian density$^2$
$$ {\cal L}~=~\bar{\psi}(\frac{i}{2}\stackrel{\leftrightarrow}{\cancel{\partial}} -m)\psi \tag{5} $$
is (i) real, and (ii) its Euler-Lagrange (EL) equation is the Dirac equation$^3$
$$(i\cancel{\partial} -m)\psi~\approx~0,\tag{6}$$
irrespectively of the Grassmann-parity of $\psi$!  

*The Dirac equation (6) itself is linear in $\psi$, and hence agnostic to the Grassmann-parity of $\psi$.
References: 


*

*M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 3.5. 

*H. Arodz & L. Hadasz, Lectures on Classical and Quantum Theory of Fields, Section 6.2.
--
$^1$ In this answer, we are for simplicity just considering dequantization, i.e. going from a quantum system to a classical system. Normally in physics, one is faced with the opposite problem: quantization. Given the Lagrangian density (5), one could (as a first step in quantization) find the Hamiltonian formulation via the Dirac-Bergmann recipe or the Faddeev-Jackiw method. The Dirac-Bergmann procedure leads to second class constraints. The resulting Dirac bracket becomes eq. (4). The Faddeev-Jackiw method leads to the same result (4). For more details, see also this Phys.SE post and links therein.
$^2$ The variables $\psi^{\ast}_{\alpha}$ and $\bar{\psi}_{\alpha}$ are not independent of $\psi_{\alpha}$, cf. this Phys.SE post and links therein. We disagree with the sentence "Let us stress that $\psi_{\alpha}$, $\bar{\psi}_{\alpha}$ are independent generating elements of a complex Grassmann algebra" in Ref. 2 on p. 130. 
$^3$ Conventions. In this answer, we will use $(+,-,-,-)$ Minkowski sign convention, and Clifford algebra 
$$\{\gamma^{\mu}, \gamma^{\nu}\}_{+}~=~2\eta^{\mu\nu}{\bf 1}_{4\times 4}.\tag{7}$$
Moreover,
$$\bar{\psi}~=~\psi^{\dagger}\gamma^0, \qquad (\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad  (\gamma^0)^2~=~{\bf 1}.\tag{8} $$
The Hermitian adjoint of a product $\hat{A}\hat{B}$ of two operators $\hat{A}$ and $\hat{B}$ reverses the order, i.e.
$$(\hat{A}\hat{B})^{\dagger}~=~\hat{B}^{\dagger}\hat{A}^{\dagger}.\tag{9} $$
The complex conjugation of a product $zw$ of two supernumbers $z$ and $w$ reverses the order, i.e. $$(zw)^{\ast}~=~w^{\ast}z^{\ast}.\tag{10} $$
