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Suppose after some calculations I arrive at an equation of the form: \begin{equation} [\nabla^2 +V(\mathbf{r})]\psi(\mathbf{r})=k^2\psi(\mathbf{r}) \tag{1} \end{equation} As you can see here the Laplacian has the same sign as the eigenvalue. The eigenvalue $k^2$ will then be used to determine an energy defined as $E=-\dfrac{\hbar^2 k^2}{2m}$, which needs to be negative (in other words $k$ needs to be real).

(For anyone wondering, we're in the context of the Born Oppenheimer approximation, but I didn't want that to divert the attention for the main issue).

I cannot possibly regard equation (1) as a Schrodinger equation, even by changing the sign of the potential $V(r)$, because of the sign in the Laplacian.

My question is if there's a way to map this equation to a Schrodinger equation, in order to use known methods to solve it, and still obtain real values for $k$. (whereas an obvious choice would be to substitute $k$ with $ik$).

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    $\begingroup$ The operator $\nabla^2$ (together with some reasonable boundary conditions) is negative-definite. Therefore, an equation of the form $\nabla^2\psi=k^2\psi$ has no solution (for real $k$). You'd need $V(r)$ to be "very positive" for the equation to admit solutions at all. But if $V\gg \nabla^2$, you are not in the perturbative regime any more, so the whole picture becomes inconsistent. $\endgroup$ – AccidentalFourierTransform Mar 25 '18 at 14:35
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This is a long comment. How useful it will be depends on how the OP first encountered this issue.

As best I can tell, your question results from some sources featuring a sign error. Take Wikipedia, for example. The confusing sign convention you highlight appears below "written in the form[6]" (sadly the equations have no labels by which I can reference them), but the equation in question is split into "the two corresponding equations", in which the signs aren't surprising. (Having said that, those equations appear to have an error of their own, namely including a $2$ in front of each $\tau_{12}\nabla$ as if they've forgotten they were writing an anticommutator.)

To make this self-contained, I'll reproduce the equations in that source. The confusing sign appears in $$\frac{\hbar^2}{2m} (\nabla + \tau)^2 \Psi + (\mathbf{u} - E)\Psi = 0.$$The equations thus deduced are $$-\frac{\hbar^2}{2m} \nabla^2\psi_1 + (\tilde{u}_1 - E)\psi_1 - \frac{\hbar^2}{2m} [2\mathbf{\tau}_{12}\nabla + \nabla\mathbf{\tau}_{12}]\psi_2 = 0$$and similarly with $\psi_1\leftrightarrow\psi_2$.

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