Resonance of Schrödinger operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:29:40Z http://mathoverflow.net/feeds/question/110405 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110405/resonance-of-schrodinger-operator Resonance of Schrödinger operator Shanlin Huang 2012-10-23T09:15:23Z 2012-10-24T13:30:13Z <p>Consider the dispersive estimates for the Schrödinger flow <code>$$e^{itH}P_{c},\quad H=-\Delta+V \quad \text{on}\quad \mathbb{R}^n,n\ge 1$$</code> where <code>$P_{c}$</code> is the projection onto the continuous spectrum of $H$, and we will be most concerned with whether it has the form <code>$$\|e^{itH}P_{c}\|_{L^1\to L^{\infty}}\leq C |t|^{-\frac{n}{2}}$$</code> In order to get this estimate, some decay and regularity condition must be put on the potential $V$, an important assumption is that zero is neither an eigenvalue nor a resonance.</p> <p>If $0$ is a eigenvalue, then it's easy to see that the above estimates may fail. My question is then if zero is a resonance but not an eigenvalue, why will the estimates above go wrong?</p> <p>Zero is said to be a resonance in the sense that if the operator <code>$(I-V\Delta^{-1})^{-1}$</code> is bounded on <code>$L^1$</code>(why not on <code>$L^2$</code> ?),see the paper of <a href="http://arxiv.org/pdf/0704.1200.pdf" rel="nofollow">Vodev</a>,I found this is less illuminating for me, so I want to know if there are some better understanding of this definition to make it more intuitive.</p> <p><strong>Edit</strong> As Terry and Delio have commented,the key point is the asymptotic expansions of the resolvents around the zeero energy.for odd dimension,with <code>$\Im z&gt;0$</code>,one can write <code>$$(-\Delta+V-z)^{-1}=\frac{A_{-1}}{z}+\frac{A_{-\frac12}}{z^{\frac12}}+A_{0}+O(z)$$</code> (for even dimension,the $log z$ terms are included)where $A_{-1}$ is the projection onto the eigenspace of $H$,and $A_{-\frac12}$ is related to both eigenspace and resonance functions. So in order to get the optimal decay ($t^{-\frac{n}{2}}$)for large t,one need <code>$A_{-1}=A_{-\frac12}=0$</code>,that is zero is neither an eigenvalue nor a resonance .</p>