**The field theory is fully analogous for Hermitian and non-Hermitian fields**

The Hermitian operator $\varphi$ still creates and/or annihilates particles and the number of these particles $N$ is still well-defined (at least if we ignore interactions and problems with loops and divergences).

The only difference from the non-Hermitian field is that the particles and the antiparticles are the same things, so instead of two different "counts" of a particle species – one for an electron and one for a positron, for example – we only have one species (e.g. the Higgs boson) and one operator $N$.

If the operator $\varphi$ acts on an $N$ eigenstate with the eigenvalue $n$, it produces a superposition of two states, one of which is an $N$ eigenstate with the eigenvalue $n+1$ and the other one has the eigenvalue $n-1$.

In the charged, non-Hermitian case, there is usually a simple quantity that is conserved, the "charge" – it's the difference of the number of positrons minus the number of electrons, for example. In the case of the Hermitian field, there is no analogous operator (except for the discrete $(-1)^N$) that is conserved. But otherwise there is no difference between the two cases.

**It is not true that the modes with opposite momenta are linked**

Your conclusion that $a_k=a^\dagger_{-k}$ is simply incorrect. If you rewrite the sum you wrote as the sum of two sums, then the second sum is self-evidently the Hermitian conjugate of the first sum (I didn't want to talk about the terms for a fixed value of $\vec k$, to avoid discussions about the difference between $\vec k$ and $-\vec k$), so the total expression for $\varphi(x)$ is Hermitian without any additional conditions just like $z+\bar z$ is real without any additional conditions on a complex $z$.

Obviously, it can't be true that a creation operator is equal to an annihilation operator. They are qualitatively different (each annihilation operator annihilates the vacuum, while no creation operator does so) so they cannot be equal.

With $\hat \phi(x_A)$ a complex scalar field with mode decomposition as you wrote, the correlation functions 1. and 2. in your post vanish identically, as suggestive of the physical interpretation you gave. So, the only independent correlator one can then build is 3. which indeed coincides with the Feynman propagator for a real scalar field.

The two time ordering interpretations you describe are correct. Both orderings are encapsulated in a lorentz covariant manner through the explicit time ordering V.E.V and this is all that we ever work with in perturbation theory.

## Best Answer

As the formula clearly shows, $\phi(x)$ cannot be interpreted as a pure creation operator of any type. It is a combination of creation and annihilation operators. Creation operators are those called $a(k)^\dagger$ and annihilation operators are called $a(k)$.

So yes, if $\phi(x)$ acts on a generic state with a well-defined number of particles $N$, it produces a linear superposition of states that have $N+1$ and $N-1$ particles, respectively. When it acts on the vacuum, for example, however, the annihilation operator piece drops out and it creates a 1-particle state.

It's somewhat hard to understand what you mean by "interpretation". The only right interpretation is the right calculation. It is an operator that gives something if it acts on a state, and all these answers may be calculated. They shouldn't be interpreted, they should be calculated.