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
I will do for you the case of one particle in one dimension. See if you can do the many particles in three dimension case yourself. 

It suffices to show $$ \frac{d}{dt} \int u^*u \mathrm{d}x = 0 $$

Bringing the derivative inside the integral sign you need to compute
$$ \int \partial_t(u^*) u + u^* \partial_t u \mathrm{d}x \tag{1}$$
Now we plug in Schroedinger's equation
$$ \partial_t(u) = \frac{i \hbar}{2m} u_{xx} + \frac{i}{\hbar} V u $$
which implies, by taking the complex conjugate (we assume that $V$ is a real valued potential; otherwise the conservation law may not be true)
$$ \partial_t (u^*) = - \frac{i \hbar}{2m} u^*_{xx} - \frac{i}{\hbar} V u^* $$
So expression (1) can be rewritten as
$$ \int \frac{i}{\hbar}\left( V |u|^2 - V |u^2|\right) + \frac{i \hbar}{2m} \left(u^* u_{xx} - u^*_{xx} u\right) \mathrm{d}x \tag{2}$$
The terms inside the first set of parentheses clearly cancel. We treat the second set of parentheses by integration by parts. This shows that the terms inside the second parentheses cancel each other. From which we conclude that
$$ (1) = (2) = 0 $$
as desired. 



Now, one quick word on the idea behind the computation above: the magic all sits behind the $i$ factor in the $i\partial_t$ of Schroedinger's equation. Observe the right hand side is given by some real valued function times the wave-function $u$ itself (okay, there is also the Laplacian term; we need that the Laplacian is "self-adjoint", which means that as an operator it is "real-valued"). If we divide the $i$ over the the right hand side, the equation is schematically 
$$ \partial_t u = i \times \text{something real} \times u $$
Now using our elementary complex analysis, we know that "$i$ times a complex number $z$ is orthogonal to $z$ itself", and this is what, when all is said and done, allows you to say that the $L^2$ norm of $u$ is conserved under the Schroedinger flow.
Wave function and time: With a potential (V is the potential) that does not depend on time, a general technique is to see what one gets when one writes the wave function as a product of a function only dependent on time, and another function only dependent on place. The Schrodinger equation then gives the result that the complex conjugate of the time-dependent part (up to a constant) equals the reciprocal. One needs to know what the derivative of the exponential function is.