Timeline for An optimization problem for one- dimensional Schrodinger operator
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 21, 2019 at 0:29 | comment | added | Michael Engelhardt | Have you solved this numerically? If not, I can do it pretty quickly, this problem is very simple. | |
| Nov 20, 2019 at 6:25 | comment | added | Michael Engelhardt | For the case $L\leq 1$, one can partially answer Q2 using the variational principle: Given the ground state for a given $a,b$, with eigenvalue $\lambda (a,b)$, the expectation value of any Hamiltonian with larger $a$ and smaller $b$ in that state is lower than $\lambda (a,b)$, because $x^4 \leq x^2 $. Hence, the ground state energy of the latter Hamiltonian is even lower, by the variational principle. This means that the minimal $\lambda (a,b)$ is achieved for $a=1$, $b=0$. For $L>1$, it's not so clear - maybe there's a clever way to rescale. | |
| Nov 20, 2019 at 5:57 | comment | added | BigM | Agreed. It's more common with negative sign. | |
| Nov 20, 2019 at 5:55 | history | edited | BigM | CC BY-SA 4.0 |
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| Nov 19, 2019 at 18:15 | comment | added | Christian Remling | I just realize that the second part of my comment applies to the operator $-d^2/dx^2+V$ (I'm so used to this minus sign that it registers subconsciously). However, similar remarks can also be made in your case. | |
| Nov 19, 2019 at 17:59 | comment | added | Christian Remling | The answer to Q1 has to be yes, I believe, and a standard source for these things would be Kato's book. Also, you can make $V$ arbitrarily negative on $[-1/2,1/2]$, say, by taking $a\to\infty$, so $\lambda(a)$ will be unbounded below. | |
| Nov 19, 2019 at 17:12 | history | asked | BigM | CC BY-SA 4.0 |