operator in ${{\mathcal H}}$.

Consider the quadratic form $u \mapsto \|H^{1/2}Pu\|^2$ with form domain $D[H^{1/2}P]$. Note that $H^{1/2}P$ involved here is a closed operator and $HP$ has the same property. Let $H_P :=
(H^{1/2}P)^*(H^{1/2}P)$ be the self-adjoint operator associated with this quadratic form. In general, $H_P$ may not be densely defined in which case it is a self-adjoint operator in a closed subspace of ${{\mathcal H}}$. More specifically, it is obviously defined and acts nontrivially in a closed subspace ${\mathrm{Ran}\,}P$ of the closure of the form domain $D[H^{1/2}P]$, while in the orthogonal complement $({\mathrm{Ran}\,}P)^\perp$ it acts as zero.

The quadratic form $u \mapsto \|H^{1/2}Pu\|^2$ defined on $D[H^{1/2}P]$ is a closed extension of the form $u \mapsto \langle
Pu,HPu \rangle$ defined on $D[HP]$, but the former is not in general the closure of the latter. Indeed, if $H$ is unbounded, $D[H]$ is a proper subspace of $D[H^{1/2}]$. Take $u_0 \in D
[H^{1/2}]\backslash D[H]$ such that the vector $H^{1/2}u_0$ is nonzero, and set $P$ to be the orthogonal projection onto the one-dimensional subspace spanned by $u_0$. Taking into account that $D[HP] = \{u \in {\cal H};\,\, Pu \in D[H]\}$ which $u_0
=Pu_0$ does not belong to, we find $HPu = 0$ for 