# Solutions to Schrödinger equation parameter dependence

This is somewhat unrelated to what I normally do in mathematics, which is why it may be obvious to some of you, but I was puzzled by this:

If we look for classical solutions on $[0,1]$ to

$$-y''(x) = \lambda y(x)$$ with initial conditions $y(0)=1, y'(0)=0$ then the solution is $y_{\lambda}(x)=\cos(\sqrt{\lambda}x).$ Now consider $f(\lambda)=y_{\lambda}(1)$ Now, if we allow $\lambda$ to be complex (which is why I call it $z$ in the following) and always pick the correct branch of the square root, then a plot showed me that we have in fact $\frac{1}{\left\lvert f(z) \right\rvert} =O\left({\left\lvert \textbf{Im}(z) \right\rvert^{-1}}\right)$ by which I mean that there is a constant $C_R>0$ such that

$\frac{{\left\lvert \textbf{Im}(z) \right\rvert}}{\left\lvert f(z) \right\rvert} \le C_R$ for $z \in \mathbb{C}\backslash \mathbb{R}\cap B(0,R)$ and $B(0,R)$ is an arbitrary ball in the complex numbers.

Numerically, I found some evidence that if $V$ is an even and let's say smooth function with respect to $\frac{1}{2}$ then the same holds for solutions to $-y''(x)+V(x)y(x)=\lambda y(x)$ with the same initial conditions. I am not really sure how to show this theoretically, if it is even true?

Best

Kinzlin

• Are you sure that you wrote your initial condition correctly? – Alexandre Eremenko Feb 9 '17 at 7:55
• @AlexandreEremenko sorry, yes you are right – Kinzlin Feb 9 '17 at 7:59
• The solutions are entire functions of $z$ (for arbitrary $V$), there's no need to "pick a branch of the square root." – Christian Remling Feb 9 '17 at 19:00

For the equation $-y''+V(x)y=zy$, and the solution defined by $y(0,z)=1,\; y'(0,z)=1$, if we set $f(z)=y(1,z)$, then $$f(z)=\cos\sqrt{z}+O(|z|^{-1/2}\exp(|\Im z|),\quad z\to\infty.$$ The only assumption is that $V$ is continuous on $[0,1]$. See, for example, Levitan, Sargsjan, Introduction to spectral theory, AMS 1975, Chapter I, Lemma 2.2.
• @Kinzin: The constant in $O$ is explicit. Just look at the proof in the book. Also z can approach the real axis and even to be on the real axis; asymptotics works. – Alexandre Eremenko Feb 9 '17 at 16:04
• Continuity of $V$ is not needed for this, it works the same way for locally integrable $V$. – Christian Remling Feb 9 '17 at 18:57