Normalisation in Harmonic Oscillators For a harmonic oscillator, I can write
$$
|\alpha \rangle = e^{-\frac{1}{2}|\alpha|^2} \Sigma_n \frac{\alpha^n}{\sqrt{n!}}|n\rangle = \sum_n\langle n|\alpha\rangle|n\rangle
$$
I can also write:
$$
|x \rangle = \sum_n\langle n|x\rangle|n\rangle
$$
Is the expression $\sum_n\langle n|x\rangle\langle x|n\rangle$ equal to one or delta function?
Here $\alpha$ is coherent state and $x$ is position.
 A: \begin{align}
\sum_n\langle n |x\rangle\langle x|n\rangle&=\sum_n\langle x|n\rangle\langle n |x\rangle\\
&=\langle x|\left(\sum_n\langle n|n\rangle\right)|x\rangle\\
&=\langle x||x\rangle\\
&=\langle x|x\rangle\\
&=\delta(x-x)=\delta(0)
\end{align}
To see why this normalisation is the case consider the identity operator in the x-basis.
$$=\int dx|x\rangle\langle x|$$
Squaring the identity operator must result in another identity operator so
\begin{align}
^2&=\left(\int dx|x\rangle\langle x|\right)\left(\int dy|y\rangle\langle y|\right)\\
&=\int dx\,dy |x\rangle\langle x|y\rangle\langle y|\overset !=
\end{align}
If we plug in $\langle x|y\rangle=\delta(x-y)$ we get the desired answer
\begin{align}
\int dx\,dy |x\rangle\langle x|y\rangle\langle y|&=\int dx\,dy |x\rangle\langle y|\delta(x-y)\\
&=\int dx|x\rangle\langle x|=
\end{align}
A: The expression
$$\sum_n\langle n|x'\rangle \langle x|n\rangle =\langle x|x'\rangle =\delta(x-x')=\delta_{x'}$$
$$\Rightarrow \sum_n\langle n|x'\rangle \langle x|n\rangle=\delta_x=\infty$$
