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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?

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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.

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  • $\begingroup$ My argument is: let f(x) or g(x) simply be a Dirac delta (or proportional to one). Would my problem then be that my vector space is ill equipped to handle deltas instead of functions? $\endgroup$ – J-J Dec 27 '18 at 0:23
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    $\begingroup$ @Jean-Jacq, You will again run into problems when you take the norm of a wavefunction which is a sum of delta functions since the square of a delta function isn't defined. This isn't an essential problem though, you just treat it in a finite dimensional vector space (i.e. a spatial lattice). You are running into problems because you are trying to mix the two approaches $\endgroup$ – octonion Dec 27 '18 at 0:29
  • $\begingroup$ Good point. I think you've convinced me that my definition has too many problems to work. $\endgroup$ – J-J Dec 27 '18 at 0:36

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