Question

Prove that position and momentum representations are fourier trnasform quantum

Answer 1

Momentum is not the Fourier transform of position.

In the position representation, position is the operator of multiplication by x, whereas momentum is a multiple of differentiation with respect to x. These observables (operators) are not Fourier transforms of each other.

In the momentum representation, momentum is the operator of multiplication by p, whereas position is a multiple of differentiation with respect to p. These observables (operators) are not Fourier transforms of each other.

The reason why these representations are appropriate for position and momentum is the fact that in both representations, the commutators satisfy the canonical commutation relations, the quantum analogue of the Poisson bracket relation {p,q}=1.

The Fourier transform comes in only as the means to switch from the position representation to the momentum representation or conversely. The reason is that apart from a factor, differentiation of the Fourier transform of a function ψ is equivalent to multiplication of ψ, and differentiation of ψ is equivalent to multiplication of the Fourier transform of ψ.

Hope this helps.