Ladder operators vs. conjugate variables

Question

In the book Introduction to Many-Body Physics by Piers Coleman, it states on page 12 that

... the particle field and its complex conjugate are conjugate variables.

In other words, the particle field $\psi(x)$ and its complex conjugate $\psi^\dagger(x)$ obey the canonical commutation relation $ [\psi(x), \psi^\dagger(y) ]_\pm = \delta(x-y) $ as given in (2.8) and (2.10), and can therefore be viewed as ladder operators which create and annihilate particles. This does not seem to be a general result, for example

1. Consider the complex Klein-Gordon field $\phi(x)$. From the Lagrangian $\mathcal{L} = \partial_\mu \phi^* \partial^\mu \phi$, the canonically conjugate variables are $\phi(x)$ and $\pi(x) = \partial_0 \phi^*(x) \neq \phi^*(x) $. If we work with a real scalar field, then there isn't even a notion of the complex conjugate field either.
2. Consider Maxwell's equations. From the Lagrangian $\mathcal{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu}$, we have the conjugate variables $A_\mu(x)$ and $\pi^\mu(x) = - F^{0 \mu} \neq A_\mu^*(x)$. Again, this field is real too.

The only fields this seems to be the case for is the Schrodinger and Dirac fields with Lagrangians that contain the term $i\psi^\dagger \partial_0 \psi$.

My question

Does Coleman's statement about position space quantum fields being ladder operators for particles only apply for the Schrodinger and Dirac fields? For other fields that have particle-like excitations, this does not seem to be the case, as one finds the ladder operators appear only in momentum space, i.e., the ladder operators $a^\dagger(p), a(p)$. In particular for real fields, there does not seem to be a notion of a creation/annihilation pair in position space as the operators are Hermitian. It seems that many-body and high energy physics treatments of QFT define fields in different ways.

Answer 1

I imagine that Coleman is restricting himself to non-relatvistic fields. For fields obeying the Schroedinger equation with action $$ S= \int d^dx dt ]\left\{i\hbar \psi^\dagger \partial_t \psi + \frac{\hbar^2}{2m} (\nabla \psi)^\dagger \cdot \nabla\psi +{\rm interactions}\right\} $$ the conjugate field to $\psi$ is $\pi= i\psi^\dagger$ so the commutation relations are $$ [\psi(x,t),\pi(x',t)]_\pm = i\hbar \delta^d(x-x') $$ or $$ [\psi(x,t),\psi^\dagger(x',t)]_\pm = \hbar \delta^d(x-x'). $$ This result is even simpler than the Dirac case because there are no antiparticles. Instead $\psi(x)$ simply annihilates a particle at $x$ and $\psi^\dagger(x)$ creates one. For Dirac $\psi$ can either annihilate a particle or create an antiparticle. So for a Schroedinger field in a static potential $V(x)$ we have a mode expansion $$ \psi(x,t)= \sum_n a_n u_n(x)e^{-iE_nt/\hbar } $$ where the $u_n(x)$ are normalized wavefunctions of enegy $E_n$, and $a_n$ with $[a_n, a_m^\dagger]_\pm = \delta_{nm}$ are the the corresponding annihilation operators.

Answer 2

In second quantization the quantum fields are promoted to operators and defined as integrals over creation and annihilation operators for each momentum $\vec p$. For instance a real scalar field $\phi (\vec x)$ in the Schroedinger picture shows
$\phi (\vec x) = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} \left(a (\vec p) e^{i \vec p \cdot \vec x} + a^\dagger (\vec p) e^{-i \vec p \cdot \vec x}\right)$
where $a (\vec p)$ is the annihilation operator, $a^\dagger (\vec p)$ is the creation operator and $[a_p, a^\dagger_k] = (2 \pi)^3 \delta^3 (\vec p - \vec k)$.
The conjugate operator is defined as $\pi (\vec x) = \frac{\partial \mathcal L}{\partial (\partial_t \phi (\vec x))} = \partial_t \phi (\vec x)$ and $[\phi (\vec x), \pi (\vec y)] = i \delta^3 (\vec x -\vec y)$.

As for a Dirac field $\psi (\vec x)$ the conjugate operator is $\frac{\partial \mathcal L}{\partial (\partial_t \psi (\vec x))} = i \psi^\dagger (\vec x)$ and $[\psi (\vec x), i \psi^\dagger (\vec y)] = i \delta^3 (\vec x - \vec y)$.

The quantum fields and their conjugate are position space ladder operators. In second quantization the Hilbert space is promoted to a Fock space, which is defined as a direct sum of Hilbert spaces of physical $n$-particles states
$\mathcal F = \oplus_n \mathcal H_n$
That is the many-body description, so it is consistent with QFT.