How is Gauss's theorem used in this quantum mechanics problem? I'm following my teacher's Quantum Mechanics notes, where he solves the free particle Shrodinger equation by means of a Fourier transform. The free particle equation (no potential energy) is
$$i\partial_t\Psi=\frac{-\hbar}{2m}\nabla^2\Psi$$
He uses Gauss's theorem twice, though he only mentions one of those times. After some work we arrive at the equation
$$i\partial_t\mathcal{F}\Psi=\frac{-\hbar}{2m}\int_{\mathbb{R}^3}\left(\nabla\boldsymbol{\cdot}\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\nabla\Psi\right)+ie^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\mathbf{k}\boldsymbol{\cdot}\nabla\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$
The calligraphic F denotes Fourier transform, and $\mathrm{d}\mu^3(\mathbf{r})$ is the typical three dimensional measure, i.e, $\mathrm{d}x\mathrm{d}y\mathrm{d}z$, just expressed in coordinate-free notation. Now if we had some finite integration region $\Omega\subset\mathbb{R}^3$ we could use Gauss's divergence theorem and write the first term as
$$\int_\Omega \nabla\boldsymbol{\cdot}\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\nabla\Psi\right)\mathrm{d}V=\int_{\partial\Omega}e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\nabla\Psi\boldsymbol{\cdot}\mathbf{n}\mathrm{d}S$$
He claims (without proof) that as we take $\Omega\to\mathbb{R}^3$, that $\nabla\Psi\boldsymbol{\cdot}\mathbf{n}$ vanishes, but offers no proof of this. I kind of understand why this might be true because $|\psi(\mathbf{r},t)|\to 0$ as $\Vert\mathbf{r}\Vert\to\infty$ in order to be square integrable, but who's to say $\Vert\mathbf{n}\Vert$ doesn't shoot off to infinity?
Now onto the second usage. We reach the equation
$$\partial_t\mathcal{F}\Psi=\frac{-\hbar}{2m}\int_{\mathbb{R}^3}\mathbf{k}\boldsymbol{\cdot}\left(\nabla\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)+i\mathbf{k}e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$
He again claims, without proof

The first term,
$$\int_{\Bbb{R}^3}\mathbf{k}\boldsymbol{\cdot}\nabla\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$
Vanishes due to Gauss's theorem.

This one is even more mysterious to me. How do we even apply Gauss's theorem in this case?
Any justification of this, even hand wavy justification, would be much appreciated. Thanks.
 A: $\mathbf{n}$ is a unit normal, i.e. $||\mathbf{n}||=1$, so that's not really an issue. However, usually in order to be able to calculate the Fourier transform of a function, the function has to be a rapidly decreasing function (see e.g. here), and in particular the first derivative must vanish faster than any polynomial grows. It seems this is being assumed in the assertion that $\int_{\partial \Omega}e^{-i\mathbf{k}\cdot\mathbf{r}}\nabla\Psi\cdot\mathbf{n}\,\mathrm{d}S\to 0$ as $\Omega\to\mathbb{R}^3$; the normal derivative vanishes faster than the surface area grows (as the square of the radius).
As to your second question, you can simply pull the $\mathbf{k}$ inside the derivative,
$$
\int_{\Bbb{R}^3}\mathbf{k}\boldsymbol{\cdot}\nabla\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r}) = \int_{\Bbb{R}^3}\nabla\cdot\left(\mathbf{k} e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$
and then apply Gauss' Theorem to the second form. It will vanish for similar reasons to above, i.e. $\Psi$ vanishes faster than any polynomial grows.
