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?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Not the answer you're looking for? Browse other questions tagged or ask your own question.