Fourier Transforms of position and momentum space in Quantum Mechanics

Question

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)?

Answer 1

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.

Answer 2

The identification of one transform as the Fourier transform and the other as the inverse transform is a matter of definition. The Fourier transform predates quantum mechanics so the reason for the assignment has nothing to do with QM and everything to do with mathematics history.

In 1807 Fourier submitted a manuscript to the Institut de France containing, among other things, what we now call the Fourier cosine transform and its inverse. These are his transforms:

$$F_c(u)=\frac{2}{\pi}\int_0^\infty f(x)\cos(ux)dx $$ $$f(x)=\int_0^\infty F_c(u)\cos(ux)du. $$

Cauchy's 1827 generalization of Fourier's relations entailed complex-valued functions, and an ineluctable sign asymmetry in form of the transforms. Trying to preserve symmetry does not help. As the article below notes, it may be shown that if the same sign is taken for both the forward and inverse formulae, "one formula is not exactly the inverse of the other one."

It is a long and helpful exercise to verify that $\hat{f}$ and $f$ inhabit dual spaces with a high degree of symmetry. For example, a function contains the same "energy" as its FT (Plancherel). Whether physics would be equally well served had a different convention been chosen is moot, even if we find particular instances that seem to point to the road not taken.

Much of this material can be found in an article the Jan/Feb 2016 issue of IEEE's Pulse, which in turn draws from a sequence of notes by Deakin listed in the references.

Answer 3

The de Broglie hypothesis essentially says that the states of definite momentum $p$ are of the form $\psi(r) = e^{ipr}$. These are orthogonal w.r.t. the inner product

$$\langle f|g\rangle := \int f(r)\overline g(r)dr$$

(for any scaling these would obviously still be orthogonal). By the postulates of quantum mechanics, they generate the full state space.

States of definite position $r_0$ are of the form $\phi(r) = \delta(r_0 - r)$.

In a finite dimensional vector space $V$ with an inner product and a basis $\psi_p$, every element $v\in V$ is of the form

$$v = \sum_pa_p\psi_p.$$

If the $\psi_p$ are orthonormal, it is immediately clear that

$$a_p = \langle v|\psi_p\rangle.\tag1$$

We could view $a$ as a function of $p$: $a(p) := a_p$, giving the coefficient of $v$ in the basis $\psi_p$.

If we have another orthonormal basis $\phi_r$, in which the index has been suggestively denoted by a different symbol, we also have

$$v = \sum_rb_r\phi_r,$$

and

$$b(r) := b_r = \langle v|\phi_r\rangle.$$

For the bases described before $a$ (the coefficient function) would be the momentum representation of $v$ and $b$ as the position representation. We have $\langle\psi_p|\phi_r\rangle = e^{ipr}$, and your first expression is the infinite dimensional analogue of (1), whereas your second expression is the analogue of

$$b(r) \equiv \langle v|\phi_r\rangle = \sum a_p\langle\psi_p|\phi_r\rangle$$