Two-particle operators in QFT and the factor 1/2 I am learning about QFT through the book Quantum Field Theory for the Gifted Amateur and I am having trouble understanding the factor 1/2 in the definition of two particle field operators. In the book they first introduce an arbitrary second-quantized two-particle operator $\hat{A}$,
$$ \hat{A} = \sum_{\alpha \beta \gamma \delta} \mathcal{A}_{\alpha \beta \gamma \delta} \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta},$$
where
$\mathcal{A}_{\alpha \beta \gamma \delta} = 
\langle \alpha,\beta | \gamma,\delta \rangle.$ Then they state that a two-particle operator written as a function of spatial coordinates is given by
$$ \hat{V} = \frac{1}{2} \int d^3x\, d^3y \hat{\psi}^{\dagger}(\textbf{x}) \hat{\psi}^{\dagger}(\textbf{y}) V(\textbf{x},\textbf{y}) \hat{\psi}(\textbf{y}) \hat{\psi}(\textbf{x}),$$
where the factor $1/2$ is needed so that we don't double count the interactions. Now my question is, why don't we need to include this factor in the definition of $\hat{A}$ where we defined the operator in terms of the momentum-basis?
 A: It basically comes down to the fact that in the first expression, we sum over two-particle states, but in the second expression, we sum over one-particle states.
In my opinion it should be written as
$$\hat{A} = \sum_{(\alpha \beta), (\gamma \delta)} \mathcal{A}_{(\alpha \beta), (\gamma \delta)} \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta}.$$
where $(\alpha \beta)$ and $(\gamma \delta)$ label two-particle states. In terms of one-particle states $|\alpha\rangle$, these can be written as
$$|\alpha \beta\rangle=\frac{1}{\sqrt{2}}\left(|\alpha\rangle|\beta\rangle\pm|\beta\rangle|\alpha\rangle\right)$$
with the $+$ for bosons and $-$ for fermions. We also have that $\mathcal{A}_{(\alpha \beta), (\gamma \delta)}=\langle\alpha \beta|\hat{A} |\delta \gamma\rangle$ (note that $\gamma$ and $\delta$ are backwards in the state, different to what is written in the book).
Now consider e.g.
$$\sum_{(\gamma \delta)} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}.$$
The sum over $(\gamma \delta)$ is half the sum over $\gamma$ and $\delta$ separately because the two-particle state with $\gamma \leftrightarrow \delta$ is the same, i.e.  $|\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}= |\gamma\delta\rangle\hat{a}_{\delta} \hat{a}_{\gamma}$. Thus
$$\sum_{(\gamma \delta)} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}=\frac{1}{2}\sum_{\gamma \delta} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}.$$
We can then write $|\delta\gamma\rangle$ in terms of $|\gamma\rangle$ and $|\delta\rangle$ to give
$$\sum_{\gamma \delta} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}=\frac{1}{\sqrt{2}}\sum_{\gamma \delta} \left(|\delta\rangle|\gamma\rangle\pm|\gamma\rangle|\delta\rangle\right) \hat{a}_{\gamma} \hat{a}_{\delta}=\sqrt{2}\sum_{\gamma \delta} |\delta\rangle|\gamma\rangle \hat{a}_{\gamma} \hat{a}_{\delta}$$
using that $\hat{a}_{\gamma} \hat{a}_{\delta}=\pm\hat{a}_{\delta} \hat{a}_{\gamma}$. Hence overall we find that
$$\sum_{(\gamma \delta)} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}=\frac{1}{\sqrt{2}}\sum_{\gamma \delta} |\delta\rangle|\gamma\rangle \hat{a}_{\gamma} \hat{a}_{\delta}.$$
The same argument shows that
$$\sum_{(\alpha \beta)} \langle\alpha\beta| \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger}  =\frac{1}{\sqrt{2}}\sum_{\alpha \beta} \langle\alpha|\langle\beta| \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger}.$$
Putting these together with $\mathcal{A}_{(\alpha \beta), (\gamma \delta)}=\langle\alpha \beta|\hat{A} |\delta \gamma\rangle$, we finally get
$$\hat{A} = \sum_{(\alpha \beta), (\gamma \delta)} \mathcal{A}_{(\alpha \beta), (\gamma \delta)} \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta}= \frac{1}{2}\sum_{\alpha \beta \gamma \delta} \left(\langle\alpha|\langle \beta|\right)\hat{A} \left(|\delta\rangle |\gamma\rangle\right)  \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta}.$$
This second expression is the form that $\hat{V}$ is in: we have
$$\left(\langle\boldsymbol{x}|\langle \boldsymbol{y}|\right)\hat{V} \left(|\boldsymbol{z}\rangle |\boldsymbol{w}\rangle\right)=V(\boldsymbol{x},\boldsymbol{y})\delta(\boldsymbol{x}-\boldsymbol{z})\delta(\boldsymbol{y}-\boldsymbol{w})$$
and replacing the sum with an integral gives
$$\hat{V} = \frac{1}{2} \int \mathrm{d}^3x \int \mathrm{d}^3y\, \hat{\psi}^{\dagger}(\boldsymbol{x}) \hat{\psi}^{\dagger}(\boldsymbol{y}) V(\boldsymbol{x},\boldsymbol{y}) \hat{\psi}(\boldsymbol{y}) \hat{\psi}(\boldsymbol{x}).$$
