# In quantum mechanics, why position and momentum are related by Fourier Transformation (only)? [duplicate]

We know that if we take Fourier transform of momentum we go to position space. But why Fourier transform only.

• Possible duplicate: physics.stackexchange.com/q/39442
– user7757
Commented Jun 11, 2014 at 10:30
• – user7757
Commented Jun 11, 2014 at 10:31

The answer to this question begins and ends with the canonical commutation relationship (CCR), the ubiquitous $[\hat{X},\,\hat{P}]=i\,\hbar\,\mathrm{id}$ where of course $\hat{X}$ and $\hat{P}$ are the position and momentum observables. The CCR alone, as I show in this answer here implies there is an orthogonal co-ordinate system (which we call position co-ordinates) for the quantum state space wherein:
$$\begin{array}{lcl}\hat{X} f(x) &=& x\,f(x)\\ \hat{P} f(x) &=& -i\,\hbar\,\mathrm{d}_x\,f(x)\end{array}$$
This being so, if we want to transform to a co-ordinate system wherein the momentum, and thus the derivative $\mathrm{d}_x$ becomes a multiplication operator, then the basis states (strictly speaking, as tempered distributions as discussed here) have to be the eigenvectors of the derivative operator. These are, of course, distributions of the form $e^{i\,k\,x}$, and to find their relative weights in a quantum state, i.e. superposition, we decompose that superposition with the Fourier transform. So there you are!