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I'm studing quantum mechanics and I'm stuck on this problem. For 1D infinite potential well: $$V\left(x\right)=\left\{\begin{matrix} 0 \quad {\rm if} \quad 0<x<a \\+\infty \quad {\rm if} \quad x<0\:\lor\: x>a \end{matrix}\right.$$

The eigenvalue equation for energy is: $$H\psi=E\psi \quad \text{that is} \quad -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}=E \psi$$ with conditions: $$\psi \left( 0\right)=0 \quad \text{ and }\quad \psi\left( a \right)=0 $$ The solutions should be: $$\psi_{n}\left( x\right) = \sqrt{\frac{2}{a}}\sin\left( \frac{\pi n}{a}x\right)$$ From what I know the system can be associated with a Hilbert space $\mathcal{H}$ which must be complete, therefore every Cauchy sequence must have limit in $\mathcal{H}$. The sequence: $$ s_i=\sum_{k=1}^i \frac{4a \left( -1\right)^{k-1}}{\pi^2 \left( 2k-1\right)^2}\sin \left( \frac{\pi \left( 2k-1\right)}{a}x\right)$$ converge to a triangular function that is not differentiable in in $\frac{a}{2}$ and so we cannot apply the Hamiltonian operator to this "state". Were am I doing wrong?

I have recently started studying QM, I am using the following books:
"Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles" Robert Resnick,
"Modern Quantum Mechanics" J. J. Sakurai, Jim Napolitano,
but I'm thinking maybe they are not the best to use, do you have any advice for me? I would like a rigorous book but one that contains all the mathematical passages (or at least the main ones).

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Indeed you cannot apply $H$ to $\lim s_i$ naively. This is because $H$ is an unbounded operator, which is not defined on the whole Hilbert space. For the mathematical details look up "self-adjoint operators on Hilbert spaces". If you want to get a physical result without getting into the details, just decompose you state on the basis of eigenstates and use $H (\sum_n \alpha_n \psi_n ) = \sum_n \alpha_n E_n \psi_n$.

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    $\begingroup$ If I've done my math correctly for the given case, $\alpha_n E_n = 4 (-1)^{(n-1)/2} \hbar^2/m a$ (for odd $n$); so this sum does not converge. This, of course, reinforces your point about $H$ being an unbounded operator. $\endgroup$ May 10 at 18:40
  • $\begingroup$ Ok thanks, I'll look for unbounded operator then. $\endgroup$
    – Al1010
    May 10 at 21:21

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