Question about Schrodinger equation I meet this problem when I am reading an introductary PDE textbook. I wonder if you could give a rigorous proof of it.
The wave function $u(t,\mathbf{x})$ is governed by the $Schr\ddot{o}dinger$ equation:
$$\dfrac{\partial u}{\partial t}-i\Delta u=0 $$
(ignoring the physical constants). Suppose the $u(t,\mathbf{x})$ is a solution for $t\in[0,\infty)$ and $\mathbf{x}\in\mathbb{R}^n$, with initial condition
$$u(0,\mathbf{x})=g(\mathbf{x}).$$
Assume that
$$\int_{\mathbb{R}^n}|g|^2d^n\mathbf{x} < \infty .$$
Show that for all $t\geq 0$,
$$\int_{\mathbb{R}^n}|u(t,\mathbf{x})|^2d^n\mathbf{x}=\int_{\mathbb{R}^n}|g|^2d^n\mathbf{x} $$
 A: This is a sketch of one possible approach. (The usual factors of $\pi,2\pi,(\sqrt{2\pi})^{-n},$ etc. of Fourier analysis are ignored for clarity.)
Consider the spacial Fourier transform $$\hat u(t,\xi):= \int_{\mathbb R^n}u(t,x)e^{-ix\cdot\xi}\,dx$$ of a solution $u(t,x)$ to the Schrödinger equation with initial data $g(x)$ to see that $u(t,x)$ may be written
$$
u(t,x) = \int e^{ix\cdot\xi}e^{it|\xi|^2}\hat g(\xi)\,d\xi.
$$
In other words, for each $t$, $\hat u(t,\xi) = e^{it|\xi|^2}\hat g(\xi)$. Since $|e^{it|\xi|^2}| = 1$, use Plancherel's theorem to conclude $\|u(t,\cdot)\|_{L^2} = \|g\|_{L^2}$.
A: (You are missing a minus sign in your equation. It should read $\frac{\partial u}{\partial t}\color{red}{+}i\Delta u=0$)
A direct approach.
Let $\psi:\Bbb R_{\geq 0}\times\Bbb R^n\to\Bbb C$ be a solution of the dimensionless Schrodinger equation for a free particle,
$$\partial_t\psi=-i\nabla^2\psi\tag{1}$$
Define the probability density
$$P=|\psi|^2=\psi^*\psi$$
We can see that
$$\partial_t P=\psi\partial_t\psi^*+\psi^*\partial_t\psi \\ =\psi(\partial_t \psi)^*+\psi^*\partial_t\psi \\ =\psi(-i\nabla^2\psi)^*+\psi^*(-i\nabla^2\psi) \\ =\psi(-i)^* (\nabla^2\psi)^*-i\psi^*\nabla^2\psi \\ =i\big(\psi\nabla^2\psi^*-\psi^*\nabla^2\psi\big)$$
Now define the probability current,
$$\boldsymbol J=i\big(\psi^*\nabla\psi-\psi\nabla\psi^*\big)$$
See that
$$\nabla\cdot\boldsymbol J=i\big(\nabla\cdot(\psi^*\nabla\psi)-\nabla\cdot(\psi\nabla\psi^*)\big) \\ \frac{1}{i}\nabla\cdot\boldsymbol J=\psi^*\nabla^2\psi+\nabla\psi^*\cdot\nabla\psi-\psi\nabla^2\psi^*-\nabla\psi\cdot\nabla\psi^* \\ =\psi^*\nabla^2\psi-\psi\nabla^2\psi^*$$
In particular, we have
$$\partial_tP+\nabla\cdot\boldsymbol J=0\tag{2}$$
Now, consider
$$F(t)=\int_{\Bbb R^n}P(t,\boldsymbol x)\mathrm d^n \boldsymbol x$$
Via the Leibniz rule,
$$F'(t)=\int_{\Bbb R^n}\partial_t P(t,\boldsymbol x)\mathrm d^n \boldsymbol x$$
Using $(2)$,
$$F'(t)=\int_{\Bbb R^n}\nabla\cdot\boldsymbol J(t,\boldsymbol x)~\mathrm d^n \boldsymbol x$$
We regard the integral over $\Bbb R^n$ as a limit of integrals over balls,
$$\int_{\Bbb R^n}\nabla\cdot\boldsymbol J(t,\boldsymbol x)~\mathrm d^n \boldsymbol x=\lim_{R\to\infty}\int\limits_{\Bbb B(0,R)}\nabla\cdot\boldsymbol J(t,\boldsymbol x)~\mathrm d^n \boldsymbol x$$
Now we can apply the divergence theorem,
$$\int\limits_{\Bbb B(0,R)}\nabla\cdot\boldsymbol J(t,\boldsymbol x)~\mathrm d^n \boldsymbol x=\int\limits_{\partial \Bbb B(0,R)}\boldsymbol J(t,\boldsymbol x)\cdot\boldsymbol n(\boldsymbol x)~\mathrm d^n \boldsymbol x$$
We can estimate this integral as
$$\int\limits_{\partial \Bbb B(0,R)}\boldsymbol J(t,\boldsymbol x)\cdot\boldsymbol n(\boldsymbol x)~\mathrm d^n \boldsymbol x\leq A_{n-1}R^{n-1}\max_{\Vert\boldsymbol x\Vert=R} \Vert \boldsymbol J(t,\boldsymbol x)\Vert\tag{*}$$
Where
$$A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$$
Is the "surface area" of the unit $n-1$ sphere.

YOUR TASK: Show that if $\psi(t,\cdot)\in L^2(\Bbb R^n,\Bbb C)$, that $\Vert \boldsymbol J\Vert\to 0$ as $\Vert \boldsymbol x\Vert\to \infty$. In fact, you need to show that it vanishes faster than $1/\Vert \boldsymbol x\Vert^{n-1}$. (This is not too hard - just remember that $\mathrm d^n\boldsymbol x=r^{n-1}\mathrm dr\cdot \text{angular terms}$.)
This will show that the integral in equation $(*)$ goes to $0$ as $R\to\infty$, which shows $F'(t)=0$ which implies $F(t)=F(0)$ which is your desired result.
