I am following the convention of FT for operators, normalized to a length scale $L$, whereby $$ a_{k}=\frac{1}{\sqrt{L}}\int dx\,e^{-ikx}a(x)\,\,\,\,\,\,\,\,\,\,\,(1) \\ a^{\dagger}_{k}=\frac{1}{\sqrt{L}}\int dx\,e^{-ikx}a^{\dagger}(x)\,\,\,\,\,\,\,\,\,\,\,(2) $$ where the sign of the phase factor is not flipped upon conjugation of $a_{k}$ (see https://arxiv.org/abs/0810.4729 Table 1). How can I show that $$ a^{\dagger}(x)\equiv[a^{\dagger}(-x)]\,\,?\,\,\,\,\,\,\,\,\,\,\,(3) $$ Here's my attempt: I take the IFT of the operators such that $$ a(x)=\frac{1}{\sqrt{L}}\int dk\,e^{ikx}a_{k}\,\,\,\,\,\,\,\,\,\,\,(4) \\ a^{\dagger}(x)=\frac{1}{\sqrt{L}}\int dk\,e^{ikx}a_{k}^{\dagger}\,\,\,\,\,\,\,\,\,\,\,(5) $$ If I look at Eq.(4), I first take $x\rightarrow -x$ $$ a(-x)=\frac{1}{\sqrt{L}}\int dk\,e^{-ikx}a_{k}\,\,\,\,\,\,\,\,\,\,\,(6) $$ and taking the conjugate of Eq.(6) $$ [a(-x)]^{\dagger}=\frac{1}{\sqrt{L}}\int dk\,e^{-ikx}a_{k}^{\dagger}\,\,\,\,\,\,\,\,\,\,\,(7) $$ where I've followed the convention that taking the dagger does not change the phase in the integral. Clearly, Eq.(5) is not the same as Eq.(7), but I feel like I'm missing something and that they should be a constitutive relation relating $x\rightarrow-x$ to the dagger of the operators.
Thank you.
You start with some operators $a_k$ and their adjoints $a_k$ (in the usual sense, where conjugation is anti-linear, so flips the phases). Then you define their respective Fourier transforms $a(x)$ and $a^\dagger(x)$ by equations $(1)$ and $(2)$. Here, since you are defining new operators, you have all the freedom to chose the sign of the phases (in other words, you can chose whether you take the Fourier transform or its inverse).
However, with the choice we are making here, the adjoint of equation $(1)$ is not $(2)$, but rather : \begin{align} a_k &= \frac{1}{\sqrt{L}}\int \text dx e^{-ikx}a(x) \tag{1} \\ a_k^\dagger &= \frac{1}{\sqrt{L}}\int \text dx e^{ikx}(a(x))^\dagger \tag{$1^\dagger$} \end{align}
In other words, because of the choices we made, $a^\dagger(x)$ is not the adjoint of $a(x)$. Changing $x\to -x$ in the RHS, we get : $$a_k^\dagger = \frac1{\sqrt L} \int \text dx e^{-ikx} (a(-x))^\dagger$$ and comparing with $(2)$ we see that $a^\dagger(x) = (a(-x))^\dagger$.
As I point out, your calculations go wrong between equations $(6)$ and $(7)$. You are taking a hermitian adjoint. There is only one convention here, and it flips phases.