Free field rigorous quantization - possibly a misunderstanding?

Question

I'm sorry if this is not the right place to ask this question but I've been struggling with this for days now (and I think this is too technical/specific for math stack).

Notation: A conjugation $C$ on a Hilbert space $\mathscr{H}$ is a anti-linear isometry satisfying $C^{2} = I$, the identity map. Let $\mathscr{H}_{C} := \{f \in \mathscr{H}: \hspace{0.1cm} Cf = f\}$. Finally, $\Phi_{s}(\cdot)$ denotes the Segal quantization operator.

On section X.7 on Reed & Simon, vol 2. there is a nice discussion on the rigorous quantization of the free Klein-Gordon field. At some point, the authors introduce the following definition: if $f \in \mathscr{H}_{C}$ then $f \mapsto \varphi(f) := \Phi_{s}(f)$ is called the canonical free field over $(\mathscr{H},C)$ and $f \mapsto \pi(f) := \Phi_{s}(if)$ is called the canonical conjugate momentum.

Then, they especialize to the case where $\mathscr{H} := L^{2}(H_{m},d\Omega_{m})$ where $H_{m}$ is the mass shell $H_{m}$ and $\Omega_{m}$ is associate Lorentz invariant measure. A conjugation $C$ on $\mathscr{S}(\mathbb{R}^{4})$ is defined by: $$(Cf)(p_{0},{\bf{p}}) := \overline{f(p_{0},-{\bf{p}})}$$

In order to define the time-zero field, the authors then introduce the following maps. For each $f \in \mathscr{S}(\mathbb{R}^{4})$, let $Ef := \sqrt{2\pi}\hat{f}|_{H_{m}}$ where $\hat{f}$ is the Fourier transform: $$\hat{f}(p) = \frac{1}{(2\pi)^{2}}\int e^{ip\cdot \tilde{x}}f(x) dx $$ with $p\cdot x := p_{0}x_{0}-p_{1}x_{1}-p_{2}x_{2}-p_{3}x_{3}$ is the Lorentz inner product. Then if $f\in \mathscr{S}(\mathbb{R}^{4})$ is real-valued, one defines: $$\varphi_{m}(f) := \varphi(Ef) \quad \mbox{and} \quad \pi_{m}(f) := \pi(\mu Ef) \quad \mu:= \sqrt{m^{2}+|{\bf{p}}|^{2}} $$ and extends to all $\mathscr{S}(\mathbb{R}^{4})$ by linearity.

According to the authors, $\varphi_{m}(\cdot)$ and $\pi_{m}(\cdot)$ are given, in terms of the creation and annihilation operators, by: \begin{eqnarray} \varphi_{m}(f) = \frac{1}{\sqrt{2}}\{ (a^{\dagger}(Ef) + a(CEf)\} \quad \mbox{and} \quad \pi_{m}(f) = \frac{i}{\sqrt{2}}\{a^{\dagger}(\mu Ef) - a(C\mu E f)\} \tag{1}\label{1} \end{eqnarray}

Here are my questions concerning the above. First, if $f \in \mathscr{S}(\mathbb{R}^{4})$, it does not follow that $Ef \in \mathscr{H}_{C}$, so I don't really get the defintions of $\varphi_{m}(\cdot)$ and $\pi_{m}(\cdot)$. One can argue that this is just notation for defining $\varphi_{m}(f) = \Phi_{s}(Ef)$ and $\pi_{m}(f) = \Phi_{s}(i\mu Ef)$, but then how to explain the $C$ factor inside each $a(\cdot)$ in (\ref{1})? It seems to me that these factors are there because we should account for the anti-linearity of $a(\cdot)$ by taking the conjugation $C$, but (as far as I understand) this expression only holds if $Ef \in \mathscr{H}_{C}$, so I'm getting really confused with these definitions. I know these are probably trivial questions, but I'm really stuck there for days and what comes next in the book depends on all these.

Answer 1

I think I figured it out. Let $f \in \mathscr{H}$ be arbitrary, where $\mathscr{H}$ here is a complex Hilbert space. Then $f$ can be written uniquely as: $$f = f_{1} + if_{2} $$ where $f_{1},f_{2} \in \mathscr{H}_{C}$ are given by: $$f_{1} = \frac{1}{2}(f+Cf) \quad \mbox{and} \quad f_{2} = \frac{1}{2i}(f-Cf)$$ Now, $\varphi$ and $\pi$ are, in principle, only defined for $f\in \mathscr{H}_{C}$ but we can extend these definitions using linearity and the above decomposition. Thus, we define, for arbitrary $f \in \mathscr{H}$: $$\varphi(f) := \varphi(f_{1}) + i\varphi(f_{2}) \quad \mbox{and} \quad \pi(f) := \pi(f_{1}) + i\pi(f_{2})$$

Let's now take $\mathscr{H} = L^{2}(H_{m},d\Omega_{m})$. If $f \in \mathscr{S}(\mathbb{R}^{4})$ then $Ef \in L^{2}(H_{m},d\Omega_{m})$, so we can write: $$Ef = (Ef)_{1} +i(Ef)_{2}$$ using the above decomposition. Setting $\varphi_{m}(f) := \varphi(Ef)$, we now have: \begin{align} \varphi_{m}(f) &= \frac{1}{\sqrt{2}}(a^{\dagger}((Ef)_{1})+a((Ef)_{1})) + \frac{i}{\sqrt{2}}(a^{\dagger}((Ef)_{2})+a((Ef)_{2}))) \\ &=\frac{1}{\sqrt{2}}a^{\dagger}((Ef)_{1}+i(Ef)_{2}) + \frac{1}{\sqrt{2}}a((Ef)_{1}-i(Ef)_{2}) \\ &= \frac{1}{\sqrt{2}}a^{\dagger}(Ef) + \frac{1}{\sqrt{2}}a(C((Ef)_{1}+i(Ef)_{2})) \\ &= \frac{1}{\sqrt{2}}(a^{\dagger}(Ef)+a(CEf)) \end{align} as claimed.