Let $V:R^d\to R_+$ be with a compact support. The Schrödinger operator $H_a=-\Delta - a V$ acting in $L^2(R^d)$ has then (at most) finitely many negative eigenvalues. Denote the number of negative eigenvalues by $N(a)$, then one has clearly $N(a)\to+\infty$ for $a\to+\infty$, and there exists a sequence of critical coupling constants $(a_n)$ at which $a\mapsto N(a)$ has jumps.

My question is as follows:

Assume that $a=b$ is *not* critical, i.e. that $a\mapsto N(a)$ is constant for $a\in(b-\epsilon,b+\epsilon)$. Does it follow that the equation $(-\Delta-b V)u=0$ has no non-trivial bounded solutions? Maybe the "boundedness" should be understood is a suitable sense ("not growing very fast" or "suitably decaying decaying at infinity") depending on the dimension $d$.

Informally, the question is: how to characterize the critical values of $a$ in terms of the solutions of $(-\Delta-a V)u=0$. I am mostly interested in the case $d=2$.

I suspect that the answer should be well known but I do not manage to find a text saying it explicitly.