Hamiltonian operator in terms of field operators I'm struggling to derive the second-quantised expression of the Hamiltonian operator of a many-particle system.
Starting from
\begin{equation*}
\renewcommand{\vec}[1]{\mathbf{#1}}
\newcommand{\ket}[1]{\lvert{#1}\rangle}
\newcommand{\bra}[1]{\langle{#1}\rvert}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
H=\sum_{i=1}^n\frac{\vec p_i^2}{2m},
\end{equation*}
I want to derive
\begin{equation*}
H=-\frac{\hbar^2}{2m}\iint\Psi^\dagger(\vec{x})\triangle\Psi(\vec{x})\,\mathrm{d}\vec{x}\,\mathrm{d}\vec{x}'.
\end{equation*}
I know that, by definition, the second quantised form of $H$ is     written as
\begin{equation*}
H=\iint\Psi^\dagger(\vec{x})\bra{\vec{x}}H\ket{\vec{x}'}\Psi(\vec{x}')\,\mathrm{d}\vec{x}\,\mathrm{d}\vec{x}'.
\end{equation*}
I don't know how to tackle the term
\begin{equation*}
\bra{\vec{x}}\vec{p}_i^2\ket{\vec{x}'}
\end{equation*}
that appears in the derivation.
I tried simplifying the problem down to two bodies only, i.e.
\begin{equation*}
H=-\frac1{2m}(\vec{p}_1^2+\vec{p}_2^2)=
-\frac1{2m}(\vec{p}^2\otimes 1+1\otimes\vec{p}^2),
\end{equation*}
for which I have calculated
\begin{equation*}
\begin{split}
\bra{\vec{x}_1,\vec{x}_2}\vec{p}^2\otimes 1+1\otimes\vec{p}^2\ket{\vec{x}'_1,\vec{x}'_2}&=
\bra{\vec{x}_1}\vec{p}^2\ket{\vec{x}'_1}\braket{\vec{x}_2}{\vec{x}'_2}+
\braket{\vec{x}_1}{\vec{x}'_1}\bra{\vec{x}_2}\vec{p}^2\ket{\vec{x}'_2}=\\ &=
-\hbar^2\delta(\vec{x}_1-\vec{x}'_1)\delta(\vec{x}_2-\vec{x}'_2)\triangle_1-\hbar^2\delta(\vec{x}_1-\vec{x}'_1)\delta(\vec{x}_2-\vec{x}'_2)\triangle_2=\\ &=
-\hbar^2\delta(\vec{x}_1-\vec{x}'_1)\delta(\vec{x}_2-\vec{x}'_2)(\triangle_1+\triangle_2),
\end{split}
\end{equation*}
but (assuming I did it right) I don't know how to link it to the result I want to prove.
In this last equation I have $n$ Laplacian operators and $n$ coordinate vectors, while in the previous one I had only one...
 A: Well, I found the solution.
I misinterpreted the equation
\begin{equation*}
\renewcommand{\vec}[1]{\mathbf{#1}}
\newcommand{\ket}[1]{\lvert{#1}\rangle}
\newcommand{\bra}[1]{\langle{#1}\rvert}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
H=\iint\Psi^\dagger(\vec{x})\bra{\vec{x}}H\ket{\vec{x}'}\Psi(\vec{x}')\,\mathrm{d}\vec{x}\,\mathrm{d}\vec{x}'.
\end{equation*}
It should be written more clearly as
\begin{equation*}
H=\iint\Psi^\dagger(\vec{x})\bra{\vec{x}}\frac{\vec p^2}{2m}\ket{\vec{x}'}\Psi(\vec{x}')\,\mathrm{d}\vec{x}\,\mathrm{d}\vec{x}',
\end{equation*}
without the $i$ index, since the operator in the integral is the single particle operator, not the whole $H$ I had defined in the first equation.
So
\begin{equation*}
H=\iint\Psi^\dagger(\vec{x})\frac{-\hbar^2}{2m}\triangle\delta(\vec{x}-\vec{x}')\Psi(\vec{x}')\,\mathrm{d}\vec{x}\,\mathrm{d}\vec{x}',
\end{equation*}
and integrating by parts (twice) I get
\begin{equation*}
\begin{split}
H&=\int\Psi^\dagger(\vec{x})\frac{-\hbar^2}{2m}\delta(\vec{x}-\vec{x}')\triangle'\Psi(\vec{x}')\,\mathrm{d}\vec{x}\,\mathrm{d}\vec{x}'=\\ &=
-\frac{\hbar^2}{2m}\int\Psi^\dagger(\vec{x})\triangle\Psi(\vec{x})\,\mathrm{d}\vec{x}
\end{split}
\end{equation*}
as I should.
