Consider $V(x)$ a one dimensional polynomial, confining, symmetric double well potential i.e.

  1. $V(x)=V(-x)$ for all $x\in\mathbb{R}$
  2. $\displaystyle \lim_{x\to\pm\infty} V(x)=+\infty$
  3. $V(x)\in \mathbb{R}_{2n}[x]$
  4. $V$ has two local minima $x_1,x_2$ such that $x_1=-x_2$.

Then the operator $H=-\frac{d^2}{dx^2}+V(x)$ has pure discrete spectrum $(\lambda_i)_{i\geq0}$ and associated to each eigenvalue there is an eigenfunction $\phi_i$.

My questions are:

  1. I read that there are 2 eigenfunctions $\psi_1,\psi_2$ , associated to "each well of the potential" (localisation in one side or the other). What is the relation between those "more elementary eigenfunctions" $\psi_1,\psi_2$ and the ground state $\phi_0$. I understant that is a sort of linear combination, but, why we cannot say that it's $\psi_1$ the ground state instead of $\phi_0$?

  2. With the above assumptions on the potential, the question asked here has a positive answer? Or can we at least make explicit the growth of a bound $M_k$ depending on $k$? i.e. $\lvert\lvert \phi_k \rvert \rvert_\infty \leq M_k $ and $M_k\sim k^{\rm something}$


  • 1
    $\begingroup$ no, there is no eigenstate which is localised in just one side of the well, the eigenstates have equal weight in both wells; probably what you mean is that an approximation for the ground state $\phi_0$ consists of taking an eigenfunction of a single well, $\psi_1$ for the left well and $\psi_2$ for the right well, and then taking the even superposition of the two; the odd superposition will be the first excited state. $\endgroup$ Jun 2, 2017 at 14:06
  • $\begingroup$ For an even potential, the eigenfunctions are all even or odd. $\endgroup$ Jun 2, 2017 at 23:33

1 Answer 1


A power bound on $M_k$ is easy. Here's a crude argument, with no attempt made to obtain good bounds. If $\phi_k$ is a normalized eigenfunction with eigenvalue $E=E_k$, and $\phi_k(x)=M\gg 1$ at some local maximum, then $\phi_k$ must decrease to values $\le M/2$ within distance $\lesssim 1/M^2$, or otherwise we would pick up too much $L^2$ norm. Thus $-\phi''_k\gtrsim M^5$ somewhere, so from the equation satisfied by $\phi_k$ we see that we must have $E-V\gtrsim M^4$, or, since $V$ is bounded below, we can just say $E\gtrsim M^4$.

A large $E$ requires a large $k$, and we can get a quantitative estimate from oscillation theory. For a given large $E$, we will have $E-V\ge E/2$ on at least an interval of length $\gtrsim E^{+1/2n}$ (since $V(x)=cx^{2n}(1+o(1))$). The actual solution will have at least as many zeros as the one where we replace $E-V$ by $E/2$, and since this one is given by $\sin \sqrt{E/2}\, x$, we find $\gtrsim E^{(1+1/n)/2}$ zeros. By oscillation theory, there are thus at least this many eigenvalues below $E$.

In other words, $k\ge E^{(1+1/n)/2}$ or $k^{2n/(n+1)}\ge E$. It follows that $M_k\lesssim k^{n/(2(n+1))}$, as desired.

I think the stronger version $M_k=O(1)$ is a plausible conjecture, but I don't even know if this holds for the harmonic oscillator (this must be straightforward to check).

  • $\begingroup$ Dar Christian Remling, thank you for your answer. I don't understand clearly the "Thus $-\phi''(x) \geq M^5$. Could you please give me some (quite elementary) references on oscillation theory? Thank you again for your time! $\endgroup$
    – M. Veruete
    Jun 7, 2017 at 8:26
  • $\begingroup$ @M.Veruete: Since $\phi$ increases from values $\le M/2$ to $M$ on an interval of length $\lesssim 1/M^2$, we must have $\phi'\gtrsim M/(1/M^2)=M^3$ somewhere. Similarly, $\phi'$ itself varies from zero (at the maximum) to a value $\gtrsim M^3$ on that same interval, that gives the bound on $\phi''$. The wikipedia article on oscillation theory has some references, Teschl's book for example should work fine for an introduction: en.wikipedia.org/wiki/Oscillation_theory $\endgroup$ Jun 7, 2017 at 17:06

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