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One of the postulates of Quantum Mechanics involves the so-called Born's Rule:

Formulated by Max Born in 1926, it gives the probability that a measurement of a quantum system will yield a given result. In its simplest form, it states that the probability density of finding a particle at a given point, when measured, is proportional to the square of the magnitude of the particle's wave function at that point.

The wave function is a rapidly-decreasing function, $\psi(\mathbf{x}, t) \in \mathscr{S}\left(\mathbb {R} ^{3},\mathbb {C} \right) \times \mathbb{R}$, where

$$ \mathscr{S}\left(\mathbb {R} ^{n},\mathbb {C} \right):=\left\{f\in C^{\infty }(\mathbb {R} ^{n},\mathbb {C} )\mid \forall \alpha ,\beta \in \mathbb {N} ^{n},\|f\|_{\alpha ,\beta }:= \sup _{x\in \mathbb {R} ^{n}}\left|x^{\alpha }(D^{\beta }f)(x)\right|<\infty \right\}$$ is the Schwartz space.

Problem. $\mathscr{S}$ is a Fréchet space! Therefore, it's not possible to define a "norm" ($\|f\|_{\alpha, \beta}$ is just a seminorm). And yet, it's needed in order $\|\psi(\mathbf{x},t)\|^2 d\mathbf{x}$ to be seen as a probability density.

Remark. $\mathscr{S}(\mathbb{R}^n, \mathbb{C})\hookrightarrow L^2(\mathbb{R}^n, \mathbb{C})$, but I don't think this is of any help.

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    $\begingroup$ Are you asking a question? If so, what question are you asking? $\endgroup$
    – s.harp
    Sep 17, 2021 at 21:32
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    $\begingroup$ Where did you get the claim that wavefunctions must be Schwartz functions from? $\endgroup$
    – ACuriousMind
    Sep 18, 2021 at 1:20
  • $\begingroup$ @ACuriousMind are there wave functions $\psi \in L^2\setminus \mathscr{S}$? $\endgroup$
    – ric.san
    Sep 18, 2021 at 6:10
  • $\begingroup$ Different people use "wavefunctions" for different things. Some people call anything that describes a quantum state a "wavefunction", I would use it for everything in $L^2$, and apparently you found someone who only uses it for Schwartz functions! $\endgroup$
    – ACuriousMind
    Sep 18, 2021 at 12:27
  • $\begingroup$ Isn't the wave packet the most general expression for a quantum state? The Gaussian wave packet for instance describes coherent states and it is a Schwartz function $\endgroup$
    – ric.san
    Sep 18, 2021 at 13:16

2 Answers 2

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$L^2$ is the completion of $\mathscr S$ in $||\cdot||_2$:

Whenever you study wave-functions in Quantum Mechanics, you start with some function space $\mathcal F(\mathbb R^n, \mathbb C) \times \mathbb R$ and the $L^2$-norm $$||\psi(t)||^2_2 = \int d\mathbf x~ |\psi(\mathbf x, t)|^2 $$ There are different function spaces for different situations, for example $\mathcal C_c^2(\mathbb R^n, \mathbb C)$ (twice differential fcts. with compact support), $\mathcal C_c^\infty(\mathbb R^n, \mathbb C)$ (smooth fcts. with compact support) or $\mathscr S(\mathbb R^n, \mathbb C)$ (as you defined it). All of these spaces are dense subsets of $L^2$, meaning they contain enough functions to approximate any $L^2$-function to arbitrary precision.

However, Hilbert spaces are complete space. This means:
The Hilbert space $\mathscr H(\mathbb R^n, \mathbb C)$ of QM is the completion of the function space $\mathcal F(\mathbb R^n, \mathbb C)$!

But since $\mathcal F(\mathbb R^n, \mathbb C) \subset L^2(\mathbb R^n, \mathbb C)$ is dense, the only completion in the $L^2$-norm is $L^2$ itself, i.e.: $$\mathscr H(\mathbb R^n, \mathbb C) := \overline{\mathcal F(\mathbb R^n, \mathbb C)} = L^2(\mathbb R^n, \mathbb C). $$

Why don't we start with $L^2(\mathbb R^n, \mathbb C)$ ?

$L^2$ may be the largest space of functions with finite $L^2$-norm, but it consequently has the weakest properties. Simple quantum mechanical operators are not necessarily defined on it. For example, consider $\psi \in L^2(\mathbb R)$. Then, its $\hat x$ and $\hat p$ expectation values might not be defined: $$ \langle \hat x \rangle_\psi = \langle \psi(x), \hat x \psi(x) \rangle = \int dx~ x |\psi(x)|^2 = \infty,~ \langle \hat p\rangle_\psi = \int dx~ \overline{\psi(x)} \psi'(x) = \infty. $$ It could be that $\psi'(x)$ is not defined in the first place. This is why one starts with a safer space (e.g. $\mathscr{S}$) and then considers the completion of the operator ($\hat x$ or $\hat p$ or some other) on this smaller space.

I hope that clarifies things!

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The $L^2$ Norms actually do define norms on the Schwarz space. After all Schwarz space is naturally identified as a subspace of $L^2$ as you have already pointed out! The reason why Schwarz space is just a Frechet space and not a Banach/Hilbert space is that Schwarz space is not complete w.r.t. the $L^2$ norms (its completion is the space of all $L^2$ functions).

Another problem is that if you want to talk about probability densities, you of course need to restrict yourself to normed states i.e. not all Schwarz space functions are valid wave functions.

If you are interested in the Functional Analytic details of QM I recommend you to look for the notion of rigged Hilbert spaces. The article The role of the rigged Hilbert space in Quantum Mechanics gives a really nice introduction to the topic. Furthermore, the book "Quantum theory for Mathematicians" by Brian C. Hall is also a wonderful gift for every person who is interested in the mathematical nuances of QM.

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