Let $x_{1},...,x_{n} \in \mathbb{R}^{3}$ and $\mathcal{H}_{n}$ be defined by: \begin{eqnarray} H_{n} := \sum_{i=0}^{n}\bigg{(}-\frac{\hbar^{2}}{2m}\nabla_{x_{i}}+W(x_{i})\bigg{)} + \frac{1}{2}\sum_{i\neq j}v(x_{i}-x_{j}) \tag{1}\label{1} \end{eqnarray} where $\nabla_{x_{i}}$ is the Laplacian operator on $\mathbb{R}^{3}$ associated to $x_{i}$. Moreover, let $L^{2}_{sym}(\mathbb{R}^{3n})$ be the set of all $L^{2}$ functions on $\mathbb{R}^{3n}$ which are symmetric with respect to permutations of its entries $x_{1},...,x_{n}$.We define the bosonic Fock space by: $$\mathcal{F}_{bos}:= \bigoplus_{n=0}^{\infty}\mathcal{F}_{n}$$ where $\mathcal{F}_{0} = \mathbb{C}$, $\mathcal{F}_{n} = L^{2}_{sym}(\mathbb{R}^{3n})$, $n \ge 1$. $\mathcal{F}_{bos}$ is equipped with the inner product: $$\langle \psi, \phi\rangle := \sum_{n=0}^{\infty}\int_{\mathbb{R}^{3n}}\overline{\psi_{n}(x_{1},...,x_{n})}\phi_{n}(x_{1},...,x_{n})d^{n}x$$ We can extend $H_{n}$ to an operator $H$ on $\mathcal{F}_{bos}$ by setting $H := \oplus_{n=0}^{\infty}H_{n}$, i.e. $H\psi = (H_{0}\psi_{0},H_{1}\psi_{1},H_{2}\psi_{2},...) \in \mathcal{F}_{bos}$ where $H_{0} \equiv 0$, $H_{1} = -\frac{\hbar^{2}}{2m}\nabla_{x}+W(x)$ and $H_{n}$ is given by (\ref{1}) for $n \ge 2$. I'd like to prove the following result:
Lemma: If $v$ is real and bounded, $H$ is self-adjoint on $\mathcal{F}_{bos}$.
Well, I get the idea of the proof. By definition, we have: $$\langle H\psi, \phi\rangle = \sum_{n=0}^{\infty}\int_{\mathbb{R}^{3n}}\overline{H_{n}\psi_{n}(x_{1},...,x_{n})}\phi_{n}(x_{1},...,x_{n})d^{n}x $$ and the ideal is to write the above as $\langle \psi, H\phi\rangle$ using (1) integration by parts to change the application of each $\nabla_{x_{i}}$ to $\phi_{n}$ and (2) since $v$ is real, $\bar{v} = v$ so $\int \overline{v\psi}\phi = \int \bar{\psi}v\phi$. However, I have two questions:
Q1: To use integration by parts to apply $\nabla_{x_{i}}$ to $\phi_{n}$ instead of $\psi_{n}$, don't we need some rapid decrease property on either $\psi$ or $\phi$? In other words, why does the independent term vanish?
Q2: Why do we need $v$ to be bounded? It seems to me that $v$ real would do the job. Maybe to garantee that $v\psi$ and $v\phi$ is again on $L^{2}$?