One-body operator for fermions - equality

Question

I am reading Modern Many-Particle Physics by Lipparini. In chapter 1.5 he talks about the matrix elements of the one-body operator:
$$F_1 = sum_{i=1}^{N}f(x_i)$$
He mentions that the matrix elements are given by:
$$ langle phi | F_1|psi rangle=int dx_1...dx_N phi^*(x_1,...,x_N)sum_{i=1}^{N}f(x_i) psi(x_1,...,x_N)=Nint dx_1...dx_N phi^*(x_1,...,x_N)f(x_1) psi(x_1,...,x_N)$$
and that this is true for both fermions and bosons.
My doubt is in the equality between the two integrals.
I understand how they are equal in the case for bosons, but why the equality holds for fermions?
For example, if we have $N=2$ then $psi(x_1,x_2)=phi_1(x_1)phi_2(x_2)-phi_1(x_2)phi_2(x_1)$. Then, we would have:
$$
sum_{i=1}^{N}f(x_i)
=[f(x_1)+f(x_2)] psi(x_1,x_2)
=big( [f(x_1)phi_1(x_1)]phi_2(x_2)-phi_1(x_2)[f(x_1)phi_2(x_1)] big) 
+big( phi_1(x_1)[f(x_2)phi_2(x_2)]-[f(x_2)phi_1(x_2)]f(x_1)phi_2(x_1) big) 
$$
inside the first integral. But this is not equal to twice the first term (only applying $f(x_1)$.
Of course, am assuming I am doing something wrong, but what is it?

Jakob · Answer

I think that you do not have to write the many-body wave function explicitly in terms of the single-particle wave functions (although you're of course allowed to do so). As already mentioned in the comments by ragnar, you did not calculate a matrix element.
Consider the following integral
$$int mathrm{d}x_1,mathrm{d}x_2 , phi^*(x_1,x_2), (f(x_1)+f(x_2)) , psi(x_1,x_2) quad ,$$
where $phi(x_1,x_2)$ and $psi(x_1,x_2)$ are some wave functions of two identical fermions and hence anti-symmetrized. We can write the above integral as
$$ int mathrm{d}x_1,mathrm{d}x_2 , phi^*(x_1,x_2), f(x_1), psi(x_1,x_2) +int mathrm{d}x_1,mathrm{d}x_2 , phi^*(x_1,x_2), f(x_2), psi(x_1,x_2)  $$
and due to the properties of the wave functions, this equals to
$$2 , int mathrm{d}x_1,mathrm{d}x_2 , phi^*(x_1,x_2), f(x_1), psi(x_1,x_2)  quad , $$
which is exactly the special case $N=2$ of the equation you gave for $langle phi|F_1|psirangle$. Note that the same equation holds for wave functions of bosons (i.e. symmetrized wave functions), as you correctly stated.

One-body operator for fermions - equality

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