2
$\begingroup$

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 $M_k\sim k^{\rm something}$

Thanks!

$\endgroup$
  • 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. $\endgroup$ – Carlo Beenakker Jun 2 '17 at 14:06
  • $\begingroup$ For an even potential, the eigenfunctions are all even or odd. $\endgroup$ – Christian Remling Jun 2 '17 at 23:33
4
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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 '17 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 $\endgroup$ – Christian Remling Jun 7 '17 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.