# On the dimensionality of the Hilbert space for a central potential [duplicate]

So, we have that for particles in a central potential, the wavefunction can be expressed as: $$\psi(\mathbf{x})=R(r)Y^m_l(\theta,\phi)$$

Explanation (not crucial in the question):

$$Y^m_l(\theta,\phi)$$ are spherical harmonics (functions associated to the eigenvalues of the orbital angular momentum $$\mathbf{L}^2$$ and one of the components of $$\mathbf{L}$$). And $$u(r)=rR(r)$$ satisfy the following form Schrödinger equation: $$Eu(r)=-\frac{\hbar^2}{2\mu}\frac{d^2u(r)}{dr^2}+\left[V(r)+\frac{l(l+1)\hbar^2}{2\mu r^2}\right]u(r)$$ Source: Weinberg, S. (2015). Lectures on Quantum Mechanics. Cambridge: Cambridge University Press. doi:10.1017/CBO9781316276105

My question is:

In this case what is the dimensionality of the space $$\mathcal{H}$$ of the solutions?

On the one hand, the solutions in the particular case of the Hydrogen atom are products of Laguerre polynomials and Spherical Harmonics, which can be identified using the quantum numbers $$n, l, m$$, thus the states are linear combinations of $$\varphi^m_{n,l}$$, a countable set:$$\psi=\sum_{n,l,m}C_{n,l,m}\varphi^m_{n,l}$$

On the other hand, as functions, the $$\psi(\mathbf{x})$$ can be thought of as linear combinations of the position basis $$\delta(\mathbf{x})$$, since there are as many $$\delta(\mathbf{x})$$ as there are $$\mathbf{x}\in \mathbb{R}^3$$, this set is uncountable: $$\psi=\int_{\mathbb{R}^3}d\mathbf{x}\:C(\mathbf{x})\delta(\mathbf{x})$$

This baffles me since on the one hand we have that the basis is countable ($$\aleph_0$$-dimensional) and on the other the basis appears to be uncountable ($$\aleph_1$$-dimensional*). Clearly they can't both be right.

*assuming that the continuum hypothesis holds.

• These things are tricky. There are other questions which deal with the (somewhat related) topic of what the position "basis" really means. physics.stackexchange.com/q/441503
– cxx
Commented Nov 29, 2018 at 3:25
• Possible duplicate of Hilbert space of harmonic oscillator: Countable vs uncountable? Commented Nov 29, 2018 at 3:52
• Thank you for your feedback! I did a quick search before asking but it didn't occur to me to search specifically for Harmonic Oscillator related questions.
– user137661
Commented Nov 29, 2018 at 3:57
• @DanYand I was going to write an answer but I found myself almost quoting you verbatim - perhaps you should post your comment as an answer. Commented Nov 29, 2018 at 4:10

As noted in a comment and acknowledged by the OP, the answer has already been explained in this post:

Hilbert space of harmonic oscillator: Countable vs uncountable?

I'll just list a few key points to help put the answer in perspective:

• Quantum field theory, and all of its various useful approximations (like nonrelativistic single-particle quantum mechanics), always uses a separable Hilbert space. By definition, a separable Hilbert space has a countable basis.

• All infinite-dimensional separable Hilbert spaces are isomorphic to each other — meaning that as far as their abstract Hilbert-space structure is concerned, they are all the same. This unique Hilbert space can be constructed in many different-looking ways. Different constructions are useful in different models. One construction, especially useful in nonrelativistic single-particle quantum mechanics, involves representing vectors in the Hilbert space as functions $$\psi:\mathbb{R}^3\mapsto\mathbb{C}$$, and then the inner product of $$\psi$$ with itself is represented by the integral of $$|\psi|^2$$. This is the construction used in the OP's question.

• If $$|\psi\rangle$$ is any vector in a Hilbert space, the inner product of $$|\psi\rangle$$ with itself is a well-defined (finite) real number. Therefore, the function $$\exp(i\mathbf{p}\cdot\mathbf{x})$$ does not represent any vector in the Hilbert space, even though it can be useful as a computational device. The same comment applies to the "function" $$\delta(\mathbf{x})$$. This resolves the paradox that was described in the question. The Hilbert space has a countable basis that suffices for constructing all other vectors in the Hilbert space.

• The inner product of $$|\psi\rangle$$ with itself is not zero unless $$|\psi\rangle$$ itself is zero. In particular, the inner product of the difference $$|a\rangle-|b\rangle$$ is not zero unless $$|a\rangle=|b\rangle$$. Therefore, two functions that differ from each other only on a subset $$\subset\mathbb{R}^3$$ of measure zero both represent the same vector in the Hilbert space. This is nicely explained in tparker's answer to the question linked above.

• Then this makes me wonder, is there such a thing as a "position basis"? Because if $\delta(\mathbf{x})$ is not a state vector, then it obviously can't form a basis for the Hilbert space. Then what do we understand when we use the "$\ket{\mathbf{x}}$"?
– user137661
Commented Nov 30, 2018 at 0:48
• @SalvadorVillarreal Strictly speaking, the answer is no, there is no such thing as a "position basis," and the notation $|x\rangle$ is only a shortcut. It is still convenient, and people often write $\langle x|f\rangle$ as an abbreviation for $f(x)$. Similarly, the position operator $X$ (and momentum operator $P$) doesn't have any eigenvectors, at least not in the Hilbert space. It has a continuous spectrum, but no eigenvalues. When books talk about "eigenvectors of $X$", they are using a shortcut. Good books should at least mention this, like pp 101-102 in Griffiths' Intro to QM. Commented Nov 30, 2018 at 2:35