In general, when one talks about the spectrum of an self-adjoint operator, it is naturally considered in a complex Hilbert space (say $L^2(\mathbb{R}^N,\mathbb{C})$). Moreover, the spectral projections associated with the self-adjoint operator is also defined on the complex Hilbert space.
I have read some contexts about the spectral properties of Schrodinger operators/elliptic operators in the form $L:=-\Delta+V(x)$ ($V(x)$ is a real function on $\mathbb{R}^N$), where the spectrum is considered in $L^2(\mathbb{R}^N,\mathbb{R})$. In different assumptions on $V(x)$ the spectrum of $L$ varies: for example, $\sigma(L)$ is the union of disjoint closed intervals of $\mathbb{R}$ when $V(x)$ is periodic in $x_i$ and $\sigma(L)$ has only discrete point spectrum when $V(x)\to\infty$ as $|x|\to\infty$.
When the periodic case is considered ($V(x)\neq\text{constant}$), a special case is when $0$ lies in a spectral gape (that is there exists $a>0$ such that $\sigma(L)\subset\mathbb{R}\setminus(-a,a)$ and $\sigma(L)\cap(-\infty,-a]\neq\emptyset$). In this case, $L$ induce a dicomposition of $L^2(\mathbb{R}^N,\mathbb{R})$ into positive and negative parts $$ L^2(\mathbb{R}^N,\mathbb{R})=L^+\oplus L^- $$ where $L$ is positive (resp. negative) defined on $L^+$ (resp. $L^-$).
My question is that why it is correct to consider $\sigma(L)$ on the real space $L^2(\mathbb{R}^N,\mathbb{R})$ ? And if I first consider the spectrum and spectral projections in the complex space $L^2(\mathbb{R}^N,\mathbb{C})$, can I say that all the spectral properties in $L^2(\mathbb{R}^N,\mathbb{R})$ can be inherited from the complex case?
If you consider $L=-\Delta +V$ on the linear space of complex functions, then the involution operator of complex conjugation on this space commutes with $L$. If you start with some domain $\mathcal{D}(L)$ which is invariant under this involution on which $L$ is symmetric, then $L^{\star}$ commutes with conjugation.
I cannot conceive of a reason that you would not want the domain of a selfadjoint restriction $L_{e}$ of $L^{\star}$ to be closed under complex conjugation, especially because it is always possible to choose an extension $L_{e}$ to have such a domain. That said, if you do choose a domain which is closed under conjugion, then the spectral measure $E$ for $L_{e}$ will also be invariant under complex conjugation. So the restriction of $L_{e}$ to real functions has the same spectral resolution as $L_{e}$, and $E$ can be recovered on the full complex space by setting $Ef=Ef_{r}+iEf_{i}$ where $f_{r}$, $f_{i}$ are the real and imaginary parts of $f$.