Convention mostly used for Fourier transform I know that mathematically it doesn't matter what sign of $i$ we use to Fourier-transform a wavefunction from real- to momentum-space and vice versa, as long as we consistently change the sign when transforming it back to its original space. But I've seen many text-books (lecture notes) use $-i$, i.e.
$$ \phi(\vec{k}) = N^{-1} \int_{V_r} \psi(\vec{r})\ \ \exp(-i\vec{k}\cdot\vec{r})\ \ d^3r\tag{1}$$
to FT from the real- to momentum-space, and $+i$, i.e. 
$$ \psi(\vec{r}) = N^{-1} \int_{V_k} \phi(\vec{k})\ \ \exp(+i\vec{k}\cdot\vec{r})\ \ d^3k\tag{2}$$
to FT from the momentum-space back to real-space. I might have missed something from my quantum mechanics course, but is there some physical reason(s) behind this convention?
 A: A wave that propagates in the $\vec{k}$ direction is given by
$$\left<\vec{r}\Big|\vec{k}\right>=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}}e^{+i\vec{k}\cdot\vec{r}}$$
and thus it is common to decompose a function as
$$\psi\left(\vec{r}\right)=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}}\int\tilde{\psi}\left(\vec{k}\right)e^{+i\vec{k}\cdot\vec{r}}{\rm d}^{3}k$$
If you add a minus sign to the exponent, then $\tilde{\psi}\left(\vec{k}\right)$ is the amplitude of a wave propagating in the $-\vec{k}$ direction.
EDIT 1: As @ZeroTheHero has pointed out, I've implicitly assumed that the time dependent component is
$$T\left(t\right)=e^{-i\omega t}$$
such that a wave traveling in the positive direction is given by
$$\psi\left(\vec{r},t\right)=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}}e^{i\left(\vec{k}\cdot\vec{r}-\omega t\right)}$$
A: The correct way to pin this down is to look at the full, time-dependent exponential.  You can choose to describe a wave moving towards $+x$ either as 
\begin{align}
\Psi(x,t)&=e^{i(kx-\omega t)}\, ,\tag{1} \\
\hbox{ or  }\quad  \Psi(x,t)&= e^{i(\omega t-kx)}\, .\tag{2}
\end{align}
Whichever you choose then determines the sign of the Fourier transform and its inverse.  Assuming (1) for the purpose of the example, then the spatial part of a plane wave moving in the $+x$ direction would be
$$
\psi(x)=\langle x\vert p\rangle=N e^{ipx/\hbar}\, . \tag{3}
$$
with $N$ a normalization factor.  The rest is basically using the completeness relation.  If, for symmetry reason, one sets
$$
\hat I=\int dx\,\vert x\rangle \langle x\vert = 
\int dp\,\vert p\rangle \langle p\vert \tag{4}
$$
(which I think is easier to remember) then
\begin{align}
\langle p\vert \psi\rangle = \psi(p)&=
\int dx\,\langle p\vert x\rangle \langle x\vert\psi\rangle\, ,\\
&= \int dx N^* e^{-ipx/\hbar}\psi(x)\, , \tag{5}
\end{align}
Note that in (5) I've explicitly used $\langle p\vert x\rangle = N^*e^{-ipx/\hbar}$, which will pin down the $N$ through
\begin{align}
\langle x\vert \psi\rangle = \psi(x)&=
\int dp\,\langle x\vert p\rangle \langle p\vert\psi\rangle\, ,\\
&= \int dp N e^{ipx/\hbar}\psi(p)\, ,\\
&= \int dp dx N N^* \psi(x)\, ,\\
\end{align}
With this you find $NN^*=1/(2\pi\hbar)$.  The choice of $N=1/\sqrt{2\pi \hbar}$ makes the transformation between the direct and inverse transformation very symmetric since
$$
\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}
=\langle p\vert x\rangle^*\, .
$$
