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I'm studying quantum mechanics and I'm considering the usual time-independent Schrödinger equation

\begin{equation*} -\left(\frac{\hbar^{2}}{2m}\right)\left(\nabla'\right)^{2}u_{E}(\mathbf{x}') + V(\mathbf{x}')u_{E}(\mathbf{x}') = Eu_{E}(\mathbf{x}') \end{equation*}

where

\begin{equation*} \left(\nabla '\right)^{2} \end{equation*}

is the Laplacian. The textbook I'm reading goes on to mention that, upon imposing the boundary condition

\begin{equation*} u_{E}(\mathbf{x}')\to 0\hspace{1pc}\mbox{ as }\hspace{1pc} \left|\mathbf{x}'\right|\to 0, \end{equation*}

it is the case that the Schrödinger equation admits nontrivial solutions only for a discrete set of values of $E$. Although the text doesn't go on to mention it, I know this is due to the spectral theory of the Schrödinger PDE. I'd be grateful for references that discuss this.

Edit:I would mention that the textbook I'm using is the 3rd edition of Modern Quantum Mechanics by Sakurai and Napolitano, just in case anyone wanted to know.

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  • $\begingroup$ You'll need to specify the full domain of the problem to get a concrete answer. Also, what page of Modern Quantum Mechanics is this on? $\endgroup$
    – Matthew Cassell
    Commented Feb 13, 2023 at 9:45
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    $\begingroup$ Never mind, it's on page 93, equation (2.188). You wrote the condition wrong, the condition is $$u_{E}(\mathbf{x}') \to 0 \ \text{as } \lvert \mathbf{x}' \rvert \to \color{red}{\infty}$$ This is required for the function to be square integrable and hence normalisable. $\endgroup$
    – Matthew Cassell
    Commented Feb 13, 2023 at 9:52

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