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Consider the Schrodinger operator in $n$ dimensions with a potential $V$, which grows rather quickly as $\mid x\mid$ tends to infinity, but with negative potential in a bounded region, for example, a single potential well. The potential I have in mind is rather simple. For instance, in dimension 2, $V(x)=(x^2+y^2)^2-4x$. Suppose $\psi$ is a normalized positive ground state corresponding to the smallest eigenvalue $E_0$ : $$-\Delta\psi+V\psi=E_0\psi \, ,$$ where $\Delta$ is the Laplacian on square integrable functions. Also, I assume the Schrodinger operator is positive so that $E_0$ is non-negative. The ground state should decay as $\mid x\mid$ tends to infinity. Also, it should localize and concentrate around the negative potential well. I assume that $$\parallel\psi\parallel=1 \, .$$ Let $U$ be the region where $V<E_0$. I expect the norm of $\psi$ on $U$, i.e. $\int_U\mid\psi(x)\mid^2d^nx$, to be large and close to 1. My question is: is it true? If it is true, how to estimate the norm of $\psi$ on $U$?

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    $\begingroup$ Maybe you mean $V<E_0$ instead of $V<0$ for the region $U$? This would correspond to the classically allowed region $\endgroup$
    – Marcel
    Feb 2, 2017 at 21:06
  • $\begingroup$ If the statement of the problem is changed as suggested by Marcel and Willie Wong, then the question becomes how to estimate the probability of being in the classically allowed region. Have you tried straightforward approaches such as picking a basis (maybe the harmonic oscillator basis), finding the matrix elements of the hamiltonian in that basis, determining the ground state numerically, and then numerically integrating to find the desired probability? $\endgroup$
    – user21349
    Feb 3, 2017 at 5:46
  • $\begingroup$ Yes, Marcel and Willie. I should refine my question: let $U$ to be the region where $V$ is less than the smallest eigenvalue $E_0$ and ask how to estimate the norm of the eigenfunction in $U$. $\endgroup$
    – Wai
    Feb 3, 2017 at 11:19

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It is certainly not true in general that $\int_{V\le E_0}|\psi|^2$ will be close to $\|\psi\|^2$ for the ground state $\psi$. (What is true along these lines has been explained by Willie in his answer.)

In fact, this is completely hopeless. We can take $V=-\delta$ in one dimension, which will lead to negative eigenvalues (like any negative potential in one or two dimensions). More specifically, the ground state energy is $E_0=-1/4$, with eigenfunction $\psi=e^{-|x|/2}$. However, $\{ V<E_0 \}$ is just a single point, so supports no norm.

If the delta distribution bothers you, then you can of course also approximate by smooth functions and get arbitrarily close to this situation with a $V\in C_0^{\infty}$.

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  • $\begingroup$ Christian, here I am not interested in the Dirac $\delta$ distribution, since I consider $V$ to be a potential function that tends to infinity as $\mid x\mid$ tends to infinity. It is like $\mid x\mid^4$. For this type of potential, it is very possible that the ground state localizes around the region $V<E_0$. $\endgroup$
    – Wai
    Feb 3, 2017 at 11:36
  • $\begingroup$ @wai: No, in fact this type of assumption ($V\simeq x^4$ for large $x$) isn't going to help, you can still build the same type of counterexample: $V=-\delta$ near the origin, $V=x^4$ for large $x$ $\endgroup$ Feb 3, 2017 at 15:41
  • $\begingroup$ Christian, I am not interested in the $\delta$ function here becasue I consider only rather smooth potential. For smooth potentials, the eigenfunctions are smooth. Your example of ground state is not even differentiable at 0, because of the $\delta$ function potential. The $\delta$ potential is like a infinite potential well at the origin, but has no width. It seems that it may not localize in $U$ simply because the well has not width, as you just explained. It is an exceptional limiting case that for me is not very useful. Can you think of counter-examples with smooth potentials? $\endgroup$
    – Wai
    Feb 3, 2017 at 19:26
  • $\begingroup$ Christian, also if you use smooth functions to approximate the $\delta$ function potential, then the smooth functions represent some potential wells that have widths. So, in this case, there is no reason why the corresponding ground state may not localize significantly in the potential well, since now the wells have non-zero widths. $\endgroup$
    – Wai
    Feb 3, 2017 at 19:33
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    $\begingroup$ @wai: Yes, one can do this with an explicit calculation if you approximate by piecewise constant functions, but it's also a general ODE fact that the solutions will converge locally uniformly to those of $V=-\delta$ if $V_n\, dx \to -\delta$ as measures, in weak $*$ sense. $\endgroup$ Feb 3, 2017 at 22:39
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Why should it be true? If $\psi$ solves $$-\triangle \psi + V\psi = E_0\psi,$$ then it also solves $$- \triangle \psi + (V + \lambda)\psi = (E_0 + \lambda)\psi$$ for any $\lambda \in\mathbb{R}$. You can easily arrange $V + \lambda$ and $V$ to have very different corresponding set $U$.

(For instance, just think about $V$ being the quantum harmonic oscillator, but with a tiny little $\lambda$ subtracted. Certainly the norm of $\psi$ is not large on $U$.)


What you should be looking at is the set where $E_0 > V$. The classical results in this direction is due to Agmon, and discussed in his Lectures on Exponential Decay of Solutions of Second-Order Elliptic Equations. An example of a theorem is something like this (the details are not entirely correct but it gets you the flavor):

There exists a weight function $\rho$ that behaves like the distance function to the set on which $V < E_0$ such that for every $\epsilon > 0$, $$\int e^{(1-\epsilon)\rho(x)} |\psi(x)|^2 dx < \infty $$

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  • $\begingroup$ Willie, the theorem you mentioned seems not to answer my question directly. It only implies the exponential decay of the eigenfunction. However, it does not say that the probability to find the eigenstate in $U$ is large or close to 1. I'll look at the book you mentioned. $\endgroup$
    – Wai
    Feb 3, 2017 at 11:31

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