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!
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It is equal to $1$ or $0$, $1$ when $r=r'$ and $0$ when $r \neq r'$ – Courage Mar 24 '16 at 08:20
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@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" (https://en.wikipedia.org/wiki/Rigged_Hilbert_space). – Wolpertinger Mar 24 '16 at 08:28
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1Possible duplicate: http://physics.stackexchange.com/q/89958/2451 and links therein. – Qmechanic Mar 24 '16 at 09:19