Timeline for Schrödinger eigenfunctions are bounded
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2017 at 9:27 | vote | accept | M. Veruete | ||
| May 16, 2017 at 16:17 | comment | added | Christian Remling | @RichardMontgomery: I don't really have a precise answer, but I can't imagine that having any consequences on this issue. In fact, maybe even a totally explicit (and real analytic) version of my example such as $V(x)=(2+\sin x^2) e^{x^2}$ would work. | |
| May 16, 2017 at 13:14 | comment | added | Richard Montgomery | If one imposes analyticity to V, then what? | |
| May 16, 2017 at 3:05 | comment | added | Richard Montgomery | Yes. I did not read your last sentence Christian. Using your trick I see you can do as you say, having a potential which is non-negative, piecewise constant, and zero on a countable collection of intervals $I_j$ whose size tends to zero. There will be eigenfunctions supported on each interval and their max will go like $\sqrt{2/|I_j|}$. | |
| May 15, 2017 at 16:31 | comment | added | Christian Remling | @RichardMontgomery: You completely misunderstood my example. I construct a whole line potential $V$ that contains parts that look like $V=c_n$ on $I_n=(a_n,b_n)$, $V\gg c_n$ on two intervals of length $L_n\gg 1$ to the left and right of $I_n$. This will have approximately the eigenfunctions I stated, but for arbitrarily small $b_n-a_n$. | |
| May 15, 2017 at 2:30 | comment | added | Richard Montgomery | The bound M of the question is a bound $M = M(V)$. For your example,there is such an M: $M = \sqrt{2/(b-a)}$, also valid for the higher eigenvalues which are this M multiplied by $sin(k\pi (x-a)/(b-a))$, $k=2,3,...$ so your `counterexample' is in fact an example of this bounded phenomenon. | |
| May 14, 2017 at 22:02 | history | answered | Christian Remling | CC BY-SA 3.0 |