Question

Conceptually, why does a Fourier transform connect the wave function in a position representation to its representation in the momentum representation?

I am referring to what is shown here:

`psi(x) = 1/sqrt(2piℏ) int_(-oo)^(oo) phi(p) e^(ipx//ℏ)dp`

`phi(p) = 1/sqrt(2piℏ) int_(-oo)^(oo) psi(x) e^(-ipx//ℏ)dx`

Using bra-ket notation, a state with definite position `x` is:

`|x>>` `= 1/sqrt(2piℏ)e^(ipx//ℏ)`

and a state with definite momentum `p` is:

`|p>>` `= 1/sqrt(2piℏ)e^(-ipx'//ℏ)`

And so,

`<< x | x' >>`

`= 1/(2piℏ)int_(-oo)^(oo) e^(ip(x-x')//ℏ)dp`

`= delta(x - x')`

and

`<< p | p' >>`

`= 1/(2piℏ)int_(-oo)^(oo) e^(ik(p - p')//ℏ)dk`

`= delta(p - p')`

Answer 1

The Heisenberg Uncertainty Principle.

Explanation:

One of the many definitions of momentum is that `P=kℏ`, k being the wave number, `k=(2pin)/lambda`.

What the Fourier transform relating momentum and position tells us is conducive of the Heisenberg uncertainty principle.
Take for example the wave function for some reason being a sin wave, as a crappy example. Ignoring that the function is not normalized and has no imaginary component, the position, being told by the square magnitude of the wave function at each point, means that we know nothing about its position. The wave function fluctuates over all of space.

HOWEVER, given this relation of the wave function of position and that of momentum, we know the wave function for momentum is the Fourier transform of the position wave function.

This would state that for a wave function that is a sin wave, there is only one possible momentum, as the position wave function has only one frequency of wave associated with it.

The same holds backward for the momentum as well, if there are many many options for the momentum, then the position wave function is the sum of many many weighted waves with the frequencies and weights that the momentum wave function describes.

Someone please check my answer, as I am not really a QM person :)