How to transform free field Hamiltonian from position to momentum space? I'm reading Srednicki's Quantum Field Theory.
The equation (3.1) says
$$
H=\int\mathrm{d}^3xa^\dagger(\boldsymbol{x})\left(-\frac{1}{2m}\nabla^2\right)a(\boldsymbol{x})
$$
will be transformed
$$
H=\int\mathrm{d}^3p\frac{1}{2m}\boldsymbol{p}^2\tilde{a}^\dagger(\boldsymbol{p})\tilde{a}(\boldsymbol{p})
$$
when eq. (3.2)
$$\tilde{a}(\boldsymbol{p})=\int\frac{\mathrm{d}^3x}{(2\pi)^{3/2}}\mathrm{e}^{-\mathrm{i}\boldsymbol{p}\cdot\boldsymbol{x}}a(\boldsymbol{x})$$
I want to show this.
I tried to transform lower equation by substituting (3.2) and $\tilde{a}^\dagger(\boldsymbol{p})=\int\frac{\mathrm{d}^3x}{(2\pi)^{3/2}}\mathrm{e}^{-\mathrm{i}\boldsymbol{p}\cdot\boldsymbol{x}}a^\dagger(\boldsymbol{x})$, and canonical quantization $\boldsymbol{p}\to-\mathrm{i}\nabla,(\hbar\equiv1)$. I got equation
$$
H=\frac{-1}{2m\cdot(2\pi)^3}\int\mathrm{d}^3p\ \nabla^2\int\mathrm{d}^3x\mathrm{e}^{-\mathrm{i}\boldsymbol{p}\cdot\boldsymbol{x}}a^\dagger(\boldsymbol{x})\int\mathrm{d}^3x^\prime\mathrm{e}^{-\mathrm{i}\boldsymbol{p}\cdot\boldsymbol{x}^\prime}a(\boldsymbol{x}^\prime).
$$
I predicted Dirac's delta function and Leibniz's law will be needed for transformation. Though, I couldn't find what to do specifically.
What is the specific method should I do next?
 A: Start with the inverse Fourier transform:
$$
a(\mathbf x)=\int\frac{\mathrm d^3 p}{(2\pi)^{3/2}}e^{-i\mathbf{p\cdot x}}\ \tilde a(\mathbf p)
\\H=\int\mathrm{d}^3x\ a^\dagger(\mathbf{x})\left(-\frac{1}{2m}\nabla^2\right)a(\mathbf{x})
\\=\int\mathrm{d}^3x\ \int\frac{\mathrm d^3 p}{(2\pi)^{3/2}}e^{i\mathbf{p\cdot x}}\ \tilde a^\dagger(\mathbf p)\left(-\frac{1}{2m}\nabla^2\right)\int\frac{\mathrm d^3 q}{(2\pi)^{3/2}}e^{-i\mathbf{q\cdot x}}\ \tilde a(\mathbf q)
\\=\int\mathrm{d}^3x\ \int\frac{\mathrm d^3 p}{(2\pi)^{3/2}}\int\frac{\mathrm d^3 q}{(2\pi)^{3/2}}e^{i\mathbf{p\cdot x}}\ \tilde a^\dagger(\mathbf p)\left(-\frac{1}{2m}\nabla^2\right)e^{-i\mathbf{q\cdot x}}\ \tilde a(\mathbf q)
\\=\int\mathrm{d}^3x\ \int\frac{\mathrm d^3 p}{(2\pi)^{3/2}}\int\frac{\mathrm d^3 q}{(2\pi)^{3/2}}e^{i\mathbf{p\cdot x}}\ \tilde a^\dagger(\mathbf p)\left(\frac{\mathbf q^2}{2m}\right)e^{-i\mathbf{q\cdot x}}\ \tilde a(\mathbf q)
\\=\int\mathrm{d}^3x\ \int\frac{\mathrm d^3 p}{(2\pi)^{3/2}}\int\frac{\mathrm d^3 q}{(2\pi)^{3/2}}\ \tilde a^\dagger(\mathbf p)\left(\frac{\mathbf q^2}{2m}\right)e^{i\mathbf{(p-q)\cdot x}}\ \tilde a(\mathbf q)
\\=\int\mathrm d^3 p\int\mathrm d^3 q\ \tilde a^\dagger(\mathbf p)\left(\frac{\mathbf q^2}{2m}\right)\delta(\mathbf{p-q})\ \tilde a(\mathbf q)
\\=\int\mathrm d^3 p\ \tilde a^\dagger(\mathbf p)\left(\frac{\mathbf p^2}{2m}\right)\ \tilde a(\mathbf p)
$$
The key step is the use of the delta function identity
$$
\delta(\mathbf k)=\int\frac{\mathrm d^3 x}{(2\pi)^3}\ e^{i\mathbf{k\cdot x}}
$$
