I was under the impression that position and momentum spaces were two different vector spaces; however in the Wikipedia article it is stated that

This quantum state can be represented as a superposition (i.e. a linear combination as a weighted sum) of basis states. In principle one is free to choose the set of basis states, as long as they span the space. If one chooses the eigenfunctions of the position operator as a set of basis functions, one speaks of a state as a wave function $\psi(r)$ in position space...By choosing the eigenfunctions of a different operator as a set of basis functions, one can arrive at a number of different representations of the same state. If one picks the eigenfunctions of the momentum operator as a set of basis functions, the resulting wave function $\phi(k)$ is said to be the wave function in momentum space.[3]

Now to me this seems to imply that the position and momentum eigenstates are bases for the same vector space. But then you must be able to write a position state as a linear combination of momentum stats and vice versa, which I don't see as being the case.

I understand this in terms of the position wavefunction and momentum wavefunction containing the same information since they are related simply via a Fourier transform, but I do not see how the position and momentum wavefunctions exist in the same vector space, or positions and momenta are bases for the same vector space.

  • $\begingroup$ Actually you can write one basis in terms of the other. Have you found expressions like $\langle p|x\rangle$? Because you can write $|x\rangle = \int \mathrm{d}p \, \, |p\rangle \langle p | x \rangle $ and viceversa. Look here link. $\endgroup$ – secavara Feb 20 '18 at 20:11

The Fourier transform is a representation of the position eigenstates as linear superpositions of plane-wave momentum eigenstates, $$ |x\rangle = \int |p\rangle \langle p|x\rangle \mathrm dp = \frac{1}{\sqrt{2\pi\hbar}}\int e^{-ipx/\hbar} |p\rangle \mathrm dp, $$ and vice versa, $$ |p\rangle = \int |x\rangle \langle x|p\rangle \mathrm dx = \frac{1}{\sqrt{2\pi\hbar}}\int e^{ipx/\hbar} |x\rangle \mathrm dx. $$ The position and momentum eigenstates are indeed bases of the same vector space, and the Fourier transformation is the unitary transformation that takes one to the other.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.