If the wave function is normalized in the position basis:

$$\int \Psi^*(x)\Psi(x) dx =1.$$

This means the probability of finding the particle somewhere is one.

Similarly, in the momentum basis:

$$\int \Psi^*(p)\Psi(p) dp =1$$

means the probability of measuring a particle's momentum to be whatever value is one.

Both of these make sense to be conditions for a wave function and I would suspect that a wave function should be normalizable in both basis.

The thing is, position eigenstates are not normalizable in the momentum basis and vice versa.

A 'normalized' momentum eigenstate is:



$$\int \frac{e^{-ipx}e^{ipx}}{2π} dx ≠1$$

This means that the probability of finding a particle with known momentum $p$ is not 1. How can I interpret this?

  • $\begingroup$ Can you think of a way to express the wavefunction in position basis with the wavefunction in momentum basis? $\endgroup$
    – Samuel
    Jul 22, 2022 at 15:09
  • $\begingroup$ I can fourier transform it $\endgroup$
    – Habouz
    Jul 22, 2022 at 15:13

1 Answer 1


For a particle on a line (with Hilbert space $L^2(\mathbb R)$), both position and momentum eigenstates are non-normalizable and therefore unphysical. When we talk of a "normalized" momentum eigenstate $\phi_p(x) = e^{ipx}/\sqrt{2\pi}$, we are referring to the so-called delta function normalization $$\int \mathrm dx \ \overline{\phi_p(x)} \phi_q(x) = \delta(p-q)$$ $$\int \mathrm dp \ \overline{\phi_p(x)}\phi_p(x') = \delta(x-x')$$

If we adopt this convention, then the answer to your titular question is yes. Given that $\Psi$ is a normalized position-space wavefunction and that $\Phi(p) = \int \mathrm dx\ \Psi(x) \phi_p(x)$ is the corresponding momentum-space wavefunction, we have that

$$\Vert\Phi\Vert^2 = \int\mathrm dp \ \overline{\Phi(p)} \Phi(p) = \int \mathrm dp \int \mathrm dx \int \mathrm dx' \ \overline{\Psi(x)}\Psi(x') \overline{\phi_p(x)}\phi_p(x')$$ $$= \int \mathrm dx \int \mathrm dx' \overline{\Psi(x)}\Psi(x') \underbrace{\int \mathrm dp \ \overline{\phi_p(x)}\phi_p(x')}_{=\delta(x-x')}$$ $$= \int \mathrm dx \ \overline{\Psi(x)}\Psi(x) = \Vert \Psi\Vert^2$$

For the momentum eigenstates $\phi_p(x)=e^{ipx}/\sqrt{2\pi}$, this takes the form of the familiar Fourier transform. For more general self-adjoint operators with continuous spectra (say, the Hamiltonian for a particle incident on a potential barrier), the generalized eigenstates will be more complex than simple plane waves. However, we are guaranteed by the spectral theorem that we can still choose a delta-function ortho"normalized" basis and expand our wavefunctions as above.


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