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I was just thinking about set theory and somehow this question just comes up to my head.

The cardinality of the set of all function from $\mathbb{R}^m$ to $\mathbb{R}^n$ is $2^{\mathfrak{c}}$. However, not all functions from $\mathbb{R}^m$ to $\mathbb{R}^n$ are valid wavefunctions, as not all of them are square integrable. So, supposed we only consider square integrable functions, is the cardinality still $2^{\mathfrak{c}}$ ?
Lucenaposition
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Tensor
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2 Answers2

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Actually the wavefunctions are the elements of a $L^2(R^n)$ space. That space is separable, so to fix an element you have to fix a sequence of complex numbers. The fact that the sequence must converge in $\ell^2$ to a non vanishing number and that one has still to take a quotient with respect to common complex factors do not change the final result: the cardinality is $2^{\mathbb{N}}$.

Valter Moretti
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Strictly speaking, L^2-space really consists of equivalence classes of functions. Two functions represent the same L^2-function if the set where they differ has measure zero. It is not hard to see that this makes <f,g> an inner product, because <f,f>=0 if and only if f=0 almost everywhere. A good way to think of an L^2-function is as a density function, so only its integral on sets with positive measure matter.

https://mathworld.wolfram.com/L2-Space.html

The set of equivalence classes has cardinality $2^{\mathbb N}$, but the set of functions has cardinality $2^c$ (each equivalence class has $2^c$ members).

Acccumulation
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