Orthonormality and completeness in infinite dimensions: 2 different definitions [duplicate]

Question

In finite dimensional vector spaces, orthonormality is defined as $\langle x_i|x_j \rangle=\delta_{ij}$ and the completeness relation is given simply by $$I = \sum_i |x_i\rangle\langle x_i|.$$

To me, the most obvious extension of this finite dimensional case to infinite dimensions is to keep orthonormality as $\langle x|y \rangle=\delta_{xy}$ (where $x$, $y$ are real), while the completeness relation becomes $$I = \sum_{i=1}^{\infty} |x_i\rangle\langle x_i| = \int |x\rangle\langle x|.\qquad(\ast)$$

Now in quantum mechanics, orthonormality of basis vectors $|x\rangle$ and $|y\rangle$ in an infinite dimensional vector space is usually defined as $\langle x|y \rangle=\delta(x-y)$, while the completeness relation is given by $$I = \int |x\rangle\langle x| dx.$$

It seems to me, however, that my extension of orthonormality and completeness to infinite dimensions is equivalent to the usual extension.

As a demonstration that my version of completeness is consistent with my definition of orthonormality, let $| f\rangle$ be some arbitrary linear combination (integral) of basis vectors $|x\rangle$: $$|f\rangle = \int f(x) |x\rangle dx.$$ Applying my version of the completeness relation to it: \begin{align} \int |x\rangle\langle x|f\rangle &= \int |x\rangle\langle x|\left(\int f(y) |y\rangle dy\right)\\ &= \int\int f(y)|x\rangle\langle x|y\rangle dy\\ &= \int\int f(y)|x\rangle\delta_{xy} dy\\ &= \int f(y)|y\rangle dy\\ &= |f\rangle \end{align} so equation $(\ast)$ does work as the completeness relation.

Is there anything wrong with defining things this way? If not, then why is the usual definition of orthonormality and completeness preferred?

Answer 1

Here's the main thing that goes wrong: What happens when you take an inner product? $$\langle f|g\rangle = \int\int f^*(x)g(y)\delta_{xy} dx dy$$ You are doing a two dimensional integral over an integrand that vanishes everywhere except on the 1d subspace $x=y$, which is a set of measure zero, so the integral vanishes. This is why we need delta functions to define scalar products when our states are written as an integral over a basis.