As I understand, $|\psi\rangle\in\mathcal{H}$, where $\mathcal{H}$ is the Hilbert state space, is a general representation of the wave function of a system. It is a vector that, in itself, is independent of the basis you choose to represent your wave function in. The choice of basis leads to the different representations of state space.

So far, the two representations that have been explained to me in class are:

  • the position representation, that you obtain like this: $\langle \vec{r} |\psi\rangle = \psi(\vec{r})$.
  • the momentum representation, that you obtain like this: $\langle \vec{p} |\phi\rangle = \phi(\vec{p})$.

Assuming I got all that right, are there any other representations that are used in quantum mechanics?

  • $\begingroup$ This site is strict about asking one question at a time. Your three questions are actually quite separate from eachother, so you should pick your favorite and delete the other two. Otherwise I'm sure the question will be quickly closed. Also $\langle p|\psi\rangle=\phi(p)$, not $\psi(p)$. Writing $\psi(p)$ would seem to imply that the position-space wavefunction is for some reason exactly the same function as $\psi(x)$. Like for some reason the wavefunction's value at $x=1$ m is the same as the position-wavefunctions value at $p=1$ kg m/s. (of course the units being different makes this silly) $\endgroup$
    – AXensen
    Sep 15, 2023 at 17:51
  • $\begingroup$ Thanks for the suggestions @AXensen, I've updated the question. $\endgroup$
    – Clerni
    Sep 15, 2023 at 18:17
  • $\begingroup$ @Axensen You need to have some way of denoting that the position-space and momentum-space wave functions are representation of the same state. There are multiple ways to do this. I sometimes use $\langle x |\psi\rangle = \psi(x)$ and $\langle p |\psi\rangle = \tilde{\psi}(p)$. Sometimes I'll use a hat (but that gets mixed up with operator notation), and sometimes I'll use a superscript, like $\psi^{\textrm{mom}}$ or something, but it seems reasonable to want $\psi$ to appear in both wave function names, since they both represent $|\psi\rangle$. $\endgroup$
    – march
    Sep 15, 2023 at 19:47
  • $\begingroup$ @AXensen I'll follow from march to say you can even overload functions if you are sure about units. $\psi(x)$ and $\psi(p)$ could be different functions if you ensure that the function of position uses one definition and the function of momentum uses another. Best to avoid confusion though, I agree $\endgroup$ Sep 15, 2023 at 20:55

1 Answer 1


There are many different representations, which often have to do with physical properties of the system.

Most useful bases come as eigenstates of some pertinent operator. The bases you have mentioned deal with the position and momentum operators, but you can think of many others. For example, the eigenstates of the Hamiltonian for some physical system are often used, especially when solving the Schrödinger equation. These might also be infinite dimensional but can be discrete, as opposed to the continuous bases of position and momentum.

As an example, take quantum mechanics on the real line for a particle with a harmonic oscillator potential. You can define states in the position or momentum eigenbasis, but it is most convenient to represent states as an infinite sum over the harmonic oscillator eigenstates. Each eigenstate can itself be written in a position or momentum representation if you want, with the energy eigenstates being given in the position representation by Hermite polynomials.

Or, take a hyrogen-like atom in three dimensions. The basis of atomic levels and orbitals is much more useful than the position or momentum bases, and the former also comes from the eigenstates of the Hamiltonian.

Finally, you can make useful quantities out of the position and momentum eigenstates even for Hamiltonians that have no potential energy term. For example, you can use wave packets as basis states to more naturally talk about how things move through space over time. Position and momentum are great starting points but they are certainly not all that quantum theory has to offer.


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