Hydrogen atom in partial differential equations For the hydrogen atom, if
$$\int |u|^2 ~dx = 1,$$
at $t = 0$,
I am trying to show that this is true at all later times.
What I need help is with differentiating the integral with respect to $t$, and taking care about the solution being complex valued. Except that my notation is getting me mixed up. I think this might get me there. 
Following Ben's hint, here is what I have:


*

*Change $|u|^ 2$   into $u^* u$.

*Bring the derivative inside the integral. 

*Apply the product rule. 

*Apply the Schrödinger equation and try to show that the result is zero. 


$$\int u^* u ~dx = 1 $$
and to bring the derivative inside the integral, isn't $dx$ already inside? 
From Schrödinger equation I have:
$$-i\hslash u_t = \sum_{i=1}^n \frac{\hslash^2}{2m_i}(u_{x_i x_i} + u_{y_i y_i} + u_{z_i z_i}) + V(x_1,\ldots,z_n)u$$
for $n$ particles and the potential would $V$ depend on all of the $3n$ coordinates.
I'm not sure how to extend it to even 2 dimensions with the notation below

For a beginning, define $3n$ orthonormal basis vectors with respect to an inner product "$\cdot$", $\hat x_j, \hat y_j, \hat z_j$ in a real vector space, and define 
$$ \nabla = \sum_j \frac{\hbar}{\sqrt{2 m_j}}\left(\hat x_j \frac{\partial}{\partial x_j}+ \hat y_j \frac{\partial}{\partial y_j}+ \hat z_j\frac{\partial}{\partial z_j} \right)
$$
Then $$\nabla \cdot \nabla u = \sum_j \frac{\hbar^2}{2 m_j}\left(u_{x_i x_i} + u_{y_i y_i} + u_{z_i z_i} \right) $$
The Schrodinger equation and its complex conjugate, multiplied with $u^*$ and $u$ respectively gives
$$-i \hbar u_t u^* = (\nabla \cdot \nabla u ) u^* + V u u^* $$  and $$ i\hbar u_t^* u  = (\nabla \cdot \nabla u^*)u + V u^* u $$. From this,
$$u_t u^* + u_t^* u = \frac{1}{i \hbar} \left( -(\nabla \cdot \nabla u)u^* + (\nabla \cdot \nabla u^*)u \right). $$ 
Right hand side  can be written 
$$\frac{1}{i \hbar} \nabla \cdot ((\nabla u)u^* + (\nabla u^*)u) =\nabla \cdot F(u).$$ 
When $W$ is a volume in $3n$-dimensional space independent of time, then 
$$ \frac{\partial}{\partial t}\int_Wu^* u \;\; d w = \int_W u_t u^* + u u_t^* \;\; dw = \int_W \nabla \cdot F(u) \;\; dw .$$
From  the divergence theorem, assuming that the surface $A$ of the volume $W$ is so far out that the wavefunction $u$ is zero there, and then also $F(u),$  the last expression equals 
$$ \int_A F(u) \cdot da = 0.$$ 
That is, $ \int_W |u|^2 dw$ is independent of time. 
