# 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}{\mathbf{#1}} \newcommand{\ket}{\lvert{#1}\rangle} \newcommand{\bra}{\langle{#1}\rvert} \newcommand{\braket}{\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...

Well, I found the solution. I misinterpreted the equation $$\begin{equation*} \renewcommand{\vec}{\mathbf{#1}} \newcommand{\ket}{\lvert{#1}\rangle} \newcommand{\bra}{\langle{#1}\rvert} \newcommand{\braket}{\langle{#1}\vert{#2}\rangle} H=\iint\Psi^\dagger(\vec{x})\bra{\vec{x}}H\ket{\vec{x}'}\Psi(\vec{x}')\,\mathrm{d}\vec{x}\,\mathrm{d}\vec{x}'. \end{equation*}$$ 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.