# Sesquilinear Forms: Hamiltonian (I)

Given a Hilbert space $\mathcal{H}$.

Consider a dense positive form: $$s:\mathcal{D}\times\mathcal{D}\to\mathbb{C}:\quad s(\varphi,\varphi)\geq0\quad(\overline{\mathcal{D}}=\mathcal{H})$$

Construct its form space: $$\mathcal{H}_s:=\mathcal{D}:\quad\langle\varphi,\psi\rangle_s:=s(\varphi,\psi)+\langle\varphi,\psi\rangle$$

Suppose it is complete: $$s=\hat{s}:\iff\mathcal{H}_s=\hat{\mathcal{H}}_s$$

Then it represents as: $$A=A^*:\quad s(\varphi,\psi)=\langle A\varphi,\psi\rangle\quad(\overline{\mathcal{D}(A)}=\mathcal{H}_s)$$

How can I prove this from scratch?

Scaled Spaces

Key property: $$\varphi\in\mathcal{H}_s:\quad\|\varphi\|^2\leq s(\varphi,\varphi)+\|\varphi\|^2=\|\varphi\|_s^2$$

Regard the embedding: $$\tau:\mathcal{H}\hookrightarrow(\mathcal{H}_s)^*:\quad\tau(\varphi)\chi:=\langle\varphi,\chi\rangle$$

It is well-defined since: $$|\tau(\varphi)\chi|=|\langle\varphi,\chi\rangle|\leq\|\varphi\|\cdot\|\chi\|\leq\|\varphi\|\cdot\|\chi\|_s$$

It is invertible since: $$0=\tau(\varphi)\chi=\langle\varphi,\chi\rangle\quad(\chi\in\mathcal{H}_s)\implies\varphi=0$$

That gives the chain: $$\mathcal{H}\stackrel{\tau}{\hookrightarrow}(\mathcal{H}_s)^*\stackrel{\Phi}{\leftrightarrow}\mathcal{H}_s\stackrel{\iota}{\hookrightarrow}\mathcal{H}$$

For the identification: $$\iota:\mathcal{H}_s\hookrightarrow\mathcal{H}:\quad\iota(\varphi):=\varphi$$

And Riesz representation: $$\Phi:\mathcal{H}_s\leftrightarrow(\mathcal{H}_s)^*:\quad\Phi(\varphi)\chi:=\langle\varphi,\chi\rangle_s$$

Investigate the operator: $$B:=\tau^{-1}\circ\Phi\circ\iota^{-1}\quad B^{-1}:=\iota\circ\Phi^{-1}\circ\tau$$

It acts as desired: $$\langle B\varphi,\chi\rangle=\tau(B\varphi)\chi=\Phi(\varphi)\chi=\langle\varphi,\chi\rangle_s$$

It is symmetric since: $$\langle B\varphi,\psi\rangle=\langle\varphi,\psi\rangle_s=\overline{\langle\psi,\varphi\rangle_s}=\overline{\langle B\psi,\varphi\rangle}$$

Its domain is a form core: $$0=\langle B^{-1}(\psi),\chi\rangle_s=\langle\psi,\chi\rangle\quad(\psi\in\mathcal{H})\implies\chi=0$$

It is densely-defined since: $$\mathcal{Z}\subseteq\mathcal{H}_s:\quad\iota(\overline{\mathcal{Z}})\subseteq\overline{\iota(\mathcal{Z})}$$

Clearly its inverse: $$B^{-1}\in\mathcal{B}(\mathcal{H}):\quad B^{-1}=\left(B^{-1}\right)^*$$

So it is selfadjoint: $$B^*=\left(\left(B^{-1}\right)^*\right)^{-1}=\left(B^{-1}\right)^{-1}=B$$