Do I run into trouble if I interpret the fermionic field operator as a linear combination of a real and an imaginary part?

Question

As some other questions on this website suggest, I have a really hard time with the fermionic field operator $\psi(x)$. I'd like to come to terms with this blockade.

It serves as the smallest building block for a whole bunch of observables, and thus I can't just abandon it and only think about "bigger" operator-terms like $\psi^{\dagger} \psi$, yet $\psi$ itself is not an observable, because it is neither a hermitian operator, nor a normal operator. Even for complex eigenvalues, $\psi$'s eigenvectors won't form a complete orthonormal set.

On the other hand (and this is what is giving me a hard time), $\psi$ appears in the lagrangian. In the classical version (before quantization), we give the time evolution for $\psi$, by solving the dirac equation. Essentially, the whole dynamics of the field (including the hamiltonian) are tied to $\psi$. This is behaviour that I'm used to from observables. Be it $x$, $p$, $A^{\mu}$, $\phi$ (and so on).

It feels completely unnatural to me that an object whose dynamics I describe by giving the hamiltonian and solving the Dirac equation is not an observable.

To overcome this mental limitation of mine, I'd like to view the field $\Psi$ as a linear combination of two fields $\psi_{real} = \frac{1}{2}(\psi + \psi^{\dagger})$ and $\psi_{im} = \frac{1}{2i}(\psi - \psi^{\dagger})$. Each of those is hermitian, and I can find a complete set of orthonormal eigenstates for those (although they are not simultaneously diagonalizable). By that I have a (at least kind of) classical outlook / interpretation on the topic.

Now the question is: Will I run into problems with this view I have towards dirac fields, and should I thus abandon it again?

Answer 1

For what it's worth, physical fermion fields need first and foremost to be Grassmann-valued fields, cf. e.g. this & this Phys.SE posts and the above comment by knzhou. Whether they are real$^1$ or complex-valued depends on the specific theory at hand.

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$^1$ A Grassmann-valued field can be real, i.e. equal to its complex conjugate field.

Answer 2

(Currently taking the QFT course and tried to answer as a practice so don't take it as granted, someone more professionals might have a better explanation.)

For complex scalar field, $\psi(x) $ and $\psi^\dagger (x)$ could be treated as linear independent fields, and could be rewrite as the linear independent fields of $\psi_{R}(x)$ and $\psi_{Im}(x)$ as easily shown from the equation of motion.

For fermionic fields it's more complicated because the anti commutation relation meant the grassman number "$\psi_R(x) \psi_R(x)=0$"[In actual case it's a bunch of linear combination so it's not zero, it was used only to show the trouble in the justifications], which brought the headaches to the lagrangian. Moreover the equation of motion for $\bar \psi $ was not independent of $\psi$, it's best to think there's a fermionic field $\psi$ and one created an operator $\bar \psi$ for the purpose of computation. Also, fermionic fields were vector fields, so keep track of the things like $\gamma^\mu p_\mu$ were a bit complicated.

*Second thought, in path integral formalism there wasn't apparent issue with equation of motion or the Lagrangian, but it wasn't sure if the equation of the motion could be solved. Also, notice that the gamma matrix was sort of complex in nature, and in the canonical formalism the completeness relation of the $\sum_s u^s(p) \bar u^s(p)=\gamma^\mu p_\mu+m$ meant you would not implement the spin indices if you treated $\psi_R$ and $\psi_{Im}$ separately.