Let
$$\hat{T} = \frac{\hbar^2}{2m} \int d^dr d^d r' \psi^\dagger (\vec r ) (\int \frac{d^dk}{(2 \pi)^d} |\vec k|^2 e^{i \vec k \cdot(\vec r - \vec r')})\psi(\vec r')$$
The transform in the middle is $\nabla^2$ at $\vec r - \vec r'$.
$\psi$ has the commutation relation $[\psi(\vec r),\psi^{\dagger}(\vec r')]=\delta^d(\vec r - \vec r')$
I still can't show that $$\hat{T} = \frac{\hbar^2}{2m} \int d^dr \nabla\psi^\dagger (\vec r ) \cdot \nabla\psi (\vec r )$$
Any help would be appreciated.
Doing the fourier transform in the middle (Remembering that $\lambda=\frac{h}{p} \to p=\hbar k $ and so $k$ correspond to $\frac{\partial}{\partial r_i}$) evaluating the fourier transform in the middle gives: $$\int \frac{d^dk}{(2 \pi)^d} |\vec k|^2 e^{i \vec k \cdot(\vec r - \vec r')}=\nabla^2$$ But as a function of the delta $\vec r-\vec r'$, and I don't know what that means for $\nabla^2$.
In order to get rid of the $d^dr'$ in the integral one can try integration by parts (There is probably a mistake here):
$$\hat{T} = \frac{\hbar^2}{2m} \int d^dr d^d r' \psi^\dagger (\vec r ) \nabla^2\psi(\vec r') = \frac{\hbar^2}{2m} [\int d^dr \psi^\dagger (\vec r ) \cdot \nabla \psi(\vec r') - \int d^dr d^d r' \nabla\psi^\dagger (\vec r ) \cdot\nabla\psi(\vec r')]$$
If that is even correct it still does not help.
I'm asking this here because I can't ask on the physics stack exchange site.