# Non-unique zero function in the function space (Hilbert space)

This is precisely why the $L^2(\mathbb R)$ is **not** simply the space of square-integrable functions from $\mathbb R$ to $\mathbb C$ (which we might call $SI(\mathbb R)$).

$SI(\mathbb R)$ consists of all functions $f:\mathbb R\rightarrow\mathbb C $ such that $\int_\mathbb{R} |f(x)|^2 dx$ exists and is finite. But as you note, if you try to make it into a Hilbert space, you run into problems. The solution is to define an equivalence relation $\sim$ on $SI(\mathbb R)$, whereby $f \sim g$ if $f(x)\neq g(x)$ only on a set of Lebesgue measure zero - that is, $f\sim g$ if they agree almost everywhere.

From there, define $L^2(\mathbb R)$ as the quotient set $SI(\mathbb R)/\sim$, whose elements are *equivalence classes* of square-integrable functions under the equivalence relation $\sim$. This resolves the ambiguity - the functions $f(x)=0$ and $g(x)=\begin{cases}1 & x=0\\ 0& x\neq 0\end{cases}$ are different elements of $SI(\mathbb R)$, but they are two equivalent representatives of the *same* element of $L^2(\mathbb R)$.

This is an important point that is usualy swept under the rug in introductory classes.

The elements of the functional Hilbert space used in Quantum mechancs (called $L^2[{\mathbb R}]$ in the mathematics literature) are not really functions, but rather *equivalence classes* of functions such that $f_1\sim f_2$ if $f_1$ and $f_2$ differ by a function $\zeta(x)$ of zero length i.e if $f_1(x)=f_2(x)+ \zeta(x)$ where $\int |\zeta(x)|^2 dx=0$ . Since all the "zero functions" differ by functions of zero length they are regarded as being the "same," so the "zero vector" becomes unique.

As a consequence, wavefunctions $\psi(x)$ do not have actual values at any point $x$. It is only integrals over regions that have numerical values. This leads to further problems such as what do we mean by boundary conditions $\psi(x)=0$ in the Schroedinger equation? These are issues that are answered in books on functional analysis, but are regarded too difficult for undergraduate QM courses.

The mistake is in assuming that the Hilbert space's base set $V$ "physically" *consists of functions* directly. It does *not.* Forming the space of Lebesgue square-integrable functions is just the first step to build the Hilbert space.

The second step is to *identify* functions which differ only on sets of Lebesgue measure zero as being the same function: that is, to define an equivalence relation

$$f \sim g := \left[\mu_L\left(\{ u \in \mathbb{R} : f(u) \neq g(u) \}\right) = 0\right]$$

where $\mu_L$ is the Lebesgue measure and we're measuring the size of the set of points on which the two functions are equal, and forming a Boolean expression inquiring if the measure is zero. You then take the quotient of the set of all such functions by this relation.

Thus the members of the Hilbert space - the ket vectors - are **not** functions, but *equivalence classes* $[f]_\sim$ **of** functions $f$ under this relation. The zero element is not $u \mapsto 0$, but rather $[u \mapsto 0]_\sim$ (using anonymous function notation). Thus an function like $\mathbf{1}_{S_C}$, the indicator function of the Cantor set $S_C$, is also in $[u \mapsto 0]_\sim$ and hence is an alternative representative for that same equivalence class and thus an alternative *representation* of the zero vector (ket), not the *formal definition* of it. Or to put to formal language,

$$|\rangle := [u \rightarrow 0]_\sim$$

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