We study the Schrodinger equation for a quantum particle in an external potential $V$. We fix an initial wave function $\psi_0 \in L^2(\mathbb{R}^d)$ with $\Vert \psi_0 \Vert_{L^2} =1$ describing the quantum particle at time $t=0$.
We consider the following initial value problem.
$$i\frac{\partial}{\partial t} \psi(x,t)=-\Delta \psi(x,t)+V(x)\psi(x,t)$$ with $\psi(x,0)=\psi_0(x).$
$V$ is the multiplication operator given by the map $$V: D(V)\subset L^2(\mathbb{R}^d) \to L^2(\mathbb{R}^d)$, \; (V\psi)(x):=V(x)\psi(x).$$ We take $D(V):=\{\psi \in L^2(\mathbb{R}^d), V\psi \in L^2(\mathbb{R}^d)\}.$
Let $\psi$ be a classical solution of the Schrodinger equation that is
(i) For every $x\in \mathbb R^d$ we have $\psi(x,\cdot)\in C^1([0,\infty)).$
(ii) For $s>0,$ there exist $\tau>0, h\in L^2(\mathbb R^d)$ (all of which depend on $s$ such that for all $x\in \mathbb R^d$ and $t\in [s-\tau, s+\tau]$ we have $$|\frac{\partial}{\partial t} \psi(x,t)|+|\psi(x,t)|\le h(x).$$
(iii) For every $t\ge 0,$ we have $\psi(\cdot,t)\in C^2(\mathbb R^d)$ and $\nabla \psi(\cdot,t)\in L^2(\mathbb R^d)$.
(iv) For each fixed $t\ge 0,$ we have $V(\cdot)\psi(\cdot,t)\in L^2(\mathbb{R}^d)$, where $V$ is the external potential.
Now we have the conservation of $L^2$ norm for classical solutions. That is, if $\psi$ is a classical solution as above then the quantity $\int_{\mathbb{R}^d} |\psi(x,t)|^2dx$ is conserved in time, i.e. $\frac{d}{dt} \int_{\mathbb{R}^d} |\psi(x,t)|^2 dx=0$.
The proof shows this holds for the Schwartz functions then concludes that the general case follows by a density argument.
I am not sure how the density argument works here. So the proof proceeds as follows.
$$\frac{d}{dt} \int_{\mathbb{R}^d} |\psi(x,t)|^2 dx = \int_{\mathbb{R}^d} \big[ \frac{\partial}{\partial t} \overline{\psi(x,t)} \psi(x,t)+\overline{\psi(x,t)}\frac{\partial}{\partial t}\psi(x,t)\big] dx. $$
Then using the Schrodinger equation we rewrite the right hand side above as $$=\int_{\mathbb{R}^d} \big[-i\Delta \overline{\psi(x,t)}\psi(x,t)+iV(x)|\psi(x,t)|^2+i\overline{\psi(x,t)}\Delta\psi(x,t)-iV(x)|\psi(x,t)|^2dx = \int_{\mathbb{R}^d} \big[-i\Delta \overline{\psi(x,t)}\psi(x,t)+i\overline{\psi(x,t)}\Delta \psi(x,t)\big]dx.$$
Now we integrate by parts as write the above as
$$=i\int_{\mathbb{R}^d} \big[\overline{\nabla \psi(x,t)}\cdot \nabla \psi(x,t)-\overline{\nabla \psi(x,t)}\cdot \nabla\psi(x,t)\big]dx=0.$$
So all but the last step holds for any classical solution. And now by the properties (ii) and (iii) we know that any classical solution $\psi(\cdot,t)$ is in $H^2$, i.e. $\psi, \nabla \psi, \Delta \psi$ are all in $L^2$.
So it suffices to show that $-\int \overline{\Delta g(x)} g(x) dx= \int |\nabla g(x)|^2 dx$ for all $g$ in the Sobolev space $H^2$ and then take $g(x)=\psi(x,t)$ to deduce the result.
I know that if $g_n \to g$ in $H^2$ then $\int |\nabla g_n|^2 \to \int |\nabla g|^2$ since the $H^2$ norm is $|u|_2 = (\sum_{0\le |\alpha|\le 2}\int |\partial^\alpha u(x)|^2 dx)^{1/2}$.
But how can I show that $\int \overline{\Delta g_n(x)}g_n(x)dx \to \int \overline{\Delta g(x)}g(x)dx$?
I would greatly appreciate an explanation on how the density argument applies for this case.
For the very specific question you asked at the end:
$$\begin{align*} \Big|\int \overline{\Delta g_n} g_n - \overline{\Delta g} g\Big| & = \Big|\int \overline{\Delta g_n} (g_n - g) + (\overline{\Delta g_n} - \overline{\Delta g}) g \Big|\\ & \leq \|\overline{\Delta g_n} \|_{L^2} \|g_n - g\|_{L^2} + \| \overline{\Delta g_n} - \overline{\Delta g}\|_{L^2} \|g\|_{L^2} \end{align*} $$
by Cauchy-Schwarz. Since $g_n$ converges in $H^2$, necessarily $\|\Delta g_n\|_{L^2}$ is uniformly bounded. So there is some constant $K$ such that
$$ \leq K \big( \|g_n - g\|_{L^2} + \|\Delta g_n - \Delta g\|_{L^2} \big) $$
which goes to zero since $g_n \to g$ in $H^2$.