What is $\langle H \rangle$ in the infinite square well in the state $|\psi \rangle = | x_0 \rangle$, where $0 \le x_0 \le L$?
In the 1-dimensional infinite square well (with width $L$) the eigenvalues and the corresponding eigenfunctions are $$ E_n = \frac{\hbar^2 \pi^2 n^2}{2L^2 m},\qquad \psi_n (x) = \sqrt{\frac{2}{L}}\sin\left( \frac{n\pi}{L}x\right), \qquad x \in [0,L] $$ The general solution of the time-independent <PERSON> equation in the basis of the Hamitonian is $$ \psi (x) = \sum_{n=1}^\infty c_n\,\sqrt{\frac{2}{L}}\sin\left( \frac{n\pi}{L}x\right) $$
Question:
If the particle is in the state $\psi_(x) = \delta(x-x_0)$, where $0 \le x_0 \le L$, i.e. the particle is at the position $x =x_0$ at $t = 0$ with certainity, then what is the expectation value of the energy $\langle H \rangle$ in this state ? or even what is the probability of getting each of the eigenvalues $E_n$ upon measurement ?
I don't know if this is the right way but my attempt is as follows:
I. 669
The coefficients $c_n$ of the expansion using Fourier trick are $$ c_n = \int_0^L \sqrt{\frac{2}{L}}\sin\left( \frac{n\pi}{L}x\right)\delta(x-x_0)\,dx = \sqrt{\frac{2}{L}}\sin\left( \frac{n\pi}{L}x_0\right) $$ II. It folows that (I'm not sure if this makes sense) $$ \langle x_0|x_0\rangle = \sum_{n=1}^\infty |c_n|^2 = \delta(x_0 - x_0) = \infty $$ III. The expectation value of the energy is $$ \langle H \rangle = \sum_{n=1}^\infty E_n\,|c_n|^2 = \frac{2}{L}\sum_{n=1}^\infty E_n\,\sin^2\left( \frac{n\pi}{L}x_0\right) $$ or differently $$ \langle H \rangle = \int_{-\infty}^\infty \delta(x-x_0)(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2})\delta(x-x_0)\, dx = \cdots $$
I don't know how to check whether these two expressions of $\langle H \rangle$ are equivalent, how does the first converge and how to compute the second ?
ADDED:
The first expression should be more completely written as $$ \langle H \rangle = \frac{\langle \psi|H|\psi\rangle}{\langle\psi|\psi\rangle} = \frac{\sum_{n=1}^\infty E_n\,|c_n|^2}{\sum_{n=1}^\infty |c_n|^2} = \frac{\infty}{\infty} = \ ? $$ and I would suggest to get the limit as the limit of the sequence $$ \langle H \rangle_m = \left( \frac{\sum_{n=1}^m E_n\,|c_n|^2}{\sum_{n=1}^m |c_n|^2}\right)_{m\in\mathbb N} $$ Result: it will not converge 955