# Why is the inner product of position eigenstates not normalised? [duplicate]

I have read that $$<{\bf r}|{\bf r}'> = δ({\bf r}-{\bf r}').$$

I don't understand how this is correct, I want to say it is equal to 1 or 0, rather than an unnormalised delta function.

Clearly I am missing something!

• It is equal to $1$ or $0$, $1$ when $r=r'$ and $0$ when $r \neq r'$ – Oswald Mar 24 '16 at 8:20
• @TheGhostOfPerdition it is most certainly not 1 when $r=r'$. The reason is that the position basis is not trace class and not square-normalisable. In that sense they form a so called "rigged Hilbert space" (en.wikipedia.org/wiki/Rigged_Hilbert_space). – Wolpertinger Mar 24 '16 at 8:28
• Possible duplicate: physics.stackexchange.com/q/89958/2451 and links therein. – Qmechanic Mar 24 '16 at 9:19