Fourier transformations:
$$\phi(\vec{k}) = \left( \frac{1}{\sqrt{2 \pi}} \right)^3 \int_{r\text{ space}} \psi(\vec{r}) e^{-i \mathbf{k} \cdot \mathbf{r}} d^3r$$
for momentum space and
$$\psi(\vec{r}) = \left( \frac{1}{\sqrt{2 \pi}} \right)^3 \int_{k\text{ space}} \phi(\vec{k}) e^{i \mathbf{k} \cdot \mathbf{r}} d^3k$$
for position space.
How do we know that $\psi$ is not the Fourier transform of $\phi$ but we suppose that its the other way around ($\psi$ would be proportional to $\exp[-ikr]$ and $\phi$ would be proportional to $\exp[ikr]$)? If there was no difference in the signs, wouldn't there be a problem in the integration from minus inf. to plus inf. if the probability is asymmetric around zero?
What is the physical reason that in the integral for momentum space we have $\exp[-ikr]$? I agree about the exponent for position space which can be explained as follows: its the sum of all definite momentum states of the system, but what about the Fourier of the momentum space? How can we explain the integral (not mathematically)?
Best Answer
Let's say $\Phi$ is a delta function, $\Phi(k)=\delta(k-k_0)$. Presumably, you want this to be an eigenstate of the momentum operator with momentum $\hbar k_0$. With the convention you've chosen, we can convert this to a real-space wavefunction (I'm ignoring normalization for convenience):
$$ \Psi(r)= \int dk \delta(k-k_0)e^{ikr}=e^{ik_0 r} $$
We can then find the momentum of the state by applying the momentum operator $-i\hbar \frac{\partial}{\partial r}$ and finding the eigenvalue. We see that this state has momentum $\hbar k_0$, as desired.
Had you defined the Fourier transform with your signs switched, you would find that the state defined by $\Phi(k)=\delta(k-k_0)$ would have momentum $-\hbar k_0$, which would be inconvenient. That's why we define the Fourier transform as above. Without any particular preference as to what we want $\Phi(k)$ to represent, we could have chosen either one as long as we were consistent.