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Apr 26, 2021 at 8:15 comment added Carlo Beenakker @ChristianRemling --- true, thanks for pointing this out; perhaps from the physics point of view a scaling with the separation of the wells is somewhat more natural than a scaling with Planck's constant...
Apr 26, 2021 at 6:31 history edited Carlo Beenakker CC BY-SA 4.0
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Apr 25, 2021 at 22:12 comment added Christian Remling I think from a formal point of view, the problem here is that $h\to 0$ in the OP doesn't in an obvious way correspond to pushing the wells apart by sending $f\to\infty$. Rather, it seems equivalent to a rescaling $V(x)\to V(hx)$ of a potential (which does push them out to infinity, but also changes their shape).
Apr 25, 2021 at 19:24 history edited Carlo Beenakker CC BY-SA 4.0
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Apr 25, 2021 at 19:09 history edited Carlo Beenakker CC BY-SA 4.0
deleted 133 characters in body
Apr 25, 2021 at 19:03 history edited Carlo Beenakker CC BY-SA 4.0
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Apr 25, 2021 at 18:06 comment added asv In my question the potential is symmetric (even). By non-degenerate I meant that the second derivative at the minima is positive.
Apr 25, 2021 at 17:49 history answered Carlo Beenakker CC BY-SA 4.0