I'm a math student, the only physics education I had was in high school, so I'm sorry if I don't use the correct jargon. I've been somewhat interested in quantum mechanics recently because of its usage of linear algebra.

However, I've been having a bit of a hard time understanding what a wavefunction is exactly. The Wikipedia page for it says: "The wave function is a function of the degrees of freedom corresponding to some maximal set of commuting observables". I'm not sure what "degrees of freedom" means, my guess is that it's complex valued function on some n-dimension manifold.

I'd also like to know what kind of structure a vector space of wavefunctions is exactly, what kind of topology one would impose on it and how one would describe it.

  • 1
    $\begingroup$ The way you formulated the question suggests to me that you may find that interesting to look at: physics.stackexchange.com/q/187841 $\endgroup$
    – Amit
    Commented Feb 3, 2023 at 22:20

2 Answers 2


I'm sure you'll get rigorous answers on this site as well, but you wouldn't be here if the highly rigorous definition you did not appreciate hadn't defined itself out of unambiguous meaning. I'll work by example which might or might not satisfy you, but that's the point: define something so it makes some sense to most bona-fide seat-of-the-pants users.

Start with the most trivial case of one degree of freedom, in this case x, the real line. The wavefunction $\psi(x)$ is a complex normalized ($\int\!\!dx ~ \psi^*(x) \psi(x)=1$) function which serves to produce a probability measure, among other things, but no matter. x is the location on that line, the "observable", and there is no other observable operator commuting with $\hat x$, the operator associated with it. (E.g. the momentum operator $\hat p$ does not commute with it.) The expectation of $\hat x$ for that state is $\int\!\!dx ~ \psi^*(x) \psi(x) ~x=\langle \hat x\rangle$. You may think of $\psi(x)$ as a complex infinite-dimensional vector. In the spectacularly more meaningful language of rigged Hilbert space, it is but the expansion coefficients in the unnormalizable $|x\rangle$ basis, of the abstract state $|\psi\rangle$, $$ |\psi\rangle= \int\!\!dx~ \psi(x) |x\rangle, ~~\leadsto\\ \psi(x)=\langle x|\psi\rangle, ~~\leadsto\\ \langle x|\hat x|\psi\rangle= x\psi(x). $$

You may proceed to 3-dimensional space, $\mathbb R ^3$, where the observables $\hat x, \hat y, \hat z$ further commute among themselves. The variables are now 3-vectors, x, the integrals are 3-dimensional, etc., 3 degrees of freedom. Hardly any topology involved.

Now here comes a twist, all but guaranteed to drive mathematicians mad. You may choose to focus on further observables, e.g., $\psi ({\mathbf x}; E, \ell, m)$, commuting among themselves, but not with x, separated by the semicolon in the wavefunction. They are the eigenvalues E of $\hat H$, $\hbar^2 \ell (\ell +1)$ of $ L^2$, and ℏm of $L_z$ in atoms. That is, $$ \langle {\mathbf x}|\hat H |\psi\rangle = E~\psi ({\mathbf x}; E, \ell, m),\\ \langle {\mathbf x}|\hat L_z/\hbar |\psi\rangle = m~\psi ({\mathbf x}; E, \ell, m),\\ \langle {\mathbf x}|\hat L^2/\hbar^2 |\psi\rangle = \ell (\ell+1)~\psi ({\mathbf x}; E, \ell, m). $$

This is to say the wavefunction corresponds to the state vector $|\psi\rangle$ which is an eigenstate of the mutually commuting $\hat H$, $\hat L^2$, and $\hat L_z$, but, even as these do not commute with $\hat {\mathbf x}$, the wavefunction corresponding to that state is the unambiguous expression above; a bit of a centaur, based on the projection of $|\psi\rangle$ on $|x\rangle$. $|\psi\rangle$ is not an eigenstate of $\hat x$, but only of the three observables after the semicolon, but the wavefunction is! A visitor from another field is justified to be puzzled, and you might improve the wikipedia presentation when you find your footing. You might sort out any topological issues, if you believed there are any.


Considering the first question: To describe a wave function mathematically, consider what Wikipedia further states: „Once such a representation is chosen, the wave function can be derived from the quantum state.“ There is a difference between the quantum state $|\psi\rangle$, a ket vector in a certain Hilbert space called the position space (also explained by me here for example), whose basis vectors are ket vectors $|x\rangle$ for every $x\in\mathbb R^3$, and the wave function $\psi$, a function $\mathbb R^3\rightarrow\mathbb C$ (or $\mathbb R^{1+3}\rightarrow\mathbb C$ to include time-dependence). They shall not be confused, but are related: $$\psi(x) =\langle x|\psi\rangle.$$ The wave function is therefore just a different way to express the quantum state. A function is easier to handle than a vector, as you can use a lot of tools from analysis like differential calculus, for example in the Schrödinger equation. It is important to note, that no information got lost in the upper transition from quantum state to wave function, as you can recover everything using the completeness relation (See for example here, equation 22, 23 and 24): $$\int_{\mathbb R^3}|x\rangle\langle x|\mathrm d^3x=1,$$ so applying $|\psi\rangle$ on both sides and use the upper equation to obtain: $$|\psi\rangle =\int_{\mathbb R^3}\psi(x)|x\rangle\mathrm d^3x.$$ Concerning the degrees of freedom: When solving the Schrödinger equation, the solution does not have to be unique (for example for the hydrogen atom) and the different solutions are classified by quantum numbers on which the wave functions are dependent on.

Considering the second question: A wavefunction $\psi\colon\mathbb R^3\rightarrow\mathbb C$ has to fulfill: $$\int_{\mathbb R^3}|\psi(x)|^2\mathrm d^3x=1$$ as the probability of finding the particle in the entire space $\mathbb R^3$ is one. The space of wave functions is therefore given by the unit ball in the Hilbert space $L^2(\mathbb R^3,\mathbb C)$, from which you can deduce the topology (induced by the scalar product or the induced norm respectivly).

In case you are not familiar with Lebesgue spaces: You can find useful informations about them on Wikipedia or on nLab. An $L^p$ space basically contains the $p$-integrable functions (meaning the $p$-th potence is integrable) on a space (here $\mathbb R^3$) and is a normed vector space with the norm being given by that integral. For quantum mechanics we are interested in the square-integrable functions as seen above, so the space $L^2$. This is a special one among the $L^p$ spaces as is it a Hilbert space, whose scalar product induces its norm and is well-known in quantum mechanics as it is given by: $$\langle\phi|\psi\rangle =\int_{\mathbb R^3}\phi(x)^*\psi(x)\mathrm d^3x.$$ Compare this with the integral above to get $\langle\psi|\psi\rangle=1$.

  • $\begingroup$ But then how is $\psi(x) = \langle x|\psi\rangle$ well defined in $L^2$? Edit: I'm not sure proper formatting for bra's and ket's so I used langle and rangle. $\endgroup$
    – TC159
    Commented Feb 3, 2023 at 22:24
  • $\begingroup$ What do you mean with the function not being well-defined? Here on the right, you have the scalar product of the position space, not the scalar product of the L^2 space. Put they are related using the completeness relation and using $\phi(x)^*=\langle x|\phi\rangle^*=\langle\phi|x\rangle$: $\langle\phi(x),\psi(x)\rangle_{L^2}=\int_{\mathbb R^3}\phi(x)^*\psi(x)\mathrm d^3x=\langle\phi|\int_{\mathbb R^3}|x\rangle\langle x|\mathrm d^3x|\psi\rangle=\langle\phi|\psi\rangle$. (\langle and \rangle are correct for ket and bra vectors, < and > don't look that good.) $\endgroup$ Commented Feb 3, 2023 at 22:37
  • $\begingroup$ I mean, $L^2$ vectors are equivalence classes of square-integrable functions that differ in a set with measure $0$. My question then is, how is $\psi(x) = \langle x| \psi \rangle$ well defined at discontinuities? $\endgroup$
    – TC159
    Commented Feb 3, 2023 at 22:57
  • $\begingroup$ Integrable functions can have discontinuities. After all, Lebesgue integration theory is based on integrating characteristic functions of measurable sets and then simple functions, so discontinuous functions in particular. $\endgroup$ Commented Feb 3, 2023 at 23:12
  • $\begingroup$ So what you're saying is that $\psi(x)$ is not an evaluation of $\psi$ at point $x$ as it suggests (and would be incorrect to do), and rather is just an abuse of notation to denote $\langle x| \psi \rangle$? $\endgroup$
    – TC159
    Commented Feb 4, 2023 at 17:51

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