What is the correct way of looking at the Dirac field? All quantum fields are operators in QFT. However, the Dirac field operator $\hat{\psi}$ has the following difference with the scalar field operator $\hat{\phi}$:

For the $\hat{\psi}$, it makes sense to define the operation $$M \cdot \hat{\psi}$$ where $M$ is any $4\times 4$ matrix and "$\cdot$" denotes matrix multiplication. Such an operation is not defined for a scalar field $\hat{\phi}$. It makes sense for the Dirac field $\hat{\psi}$ because, in the Fourier mode expansion of $\hat{\psi}$, each creation and annihilation operator is multiplied by $4\times 1$ column matrix called $u$-spinor and $v$-spinor (which are not operators).

This brings me to the following question. How should we think of the mathematical object $\hat{\psi}$ as opposed to $\hat{\phi}$? I agree that both are operators but $\hat{\psi}$ is more than an operator. It is also a matrix. What is the correct way to think about $\hat{\psi}$ so as to distinguish it from $\hat{\phi}$?
 A: 
I agree that both are operators but $\hat{\psi}$ is more than an operator. It is also a matrix. What is the correct way to think about $\hat{\psi}$ so as to distinguish it from $\hat{\phi}$?

For starters, you can think of $\hat \psi$ as a set of operator fields instead of just one operator field (like $\hat \phi$). This is analogous to how a vector is different from a scalar. The set of $\hat\psi$ operators rotate into one another under transformation (like rotations), whereas the $\hat\phi$ just stays as it is.
In addition, the $\hat \phi$ and $\hat \psi$ usually describe particles of different statistics (bosons vs fermions, respectively). This is reflected in the commutation vs anti-commutation properties of the partial creation/annihilation operators.
In terms of path integrals, the "classical" $\psi$ fields are anti-commuting number fields, whereas the classical $\phi$ fields are regular numbers.
A: To add to what hft said, $\hat \psi$ is valued in a tensor product of operator densities and the spinor bundle. If you want to get an operator, you need to pair it with a smooth section of the dual bundle and integrate. In flat space, that can be a constant section times a bump function, which corresponds to creating or destroying a particle with a polarized spin at the bump.
