# Non-dimensionalize Schroedinger's equation for this potential

I am having trouble non-dimensionalize this S.E. in order to solve numerically.. the potential is $$V(x)=-V_{0}/(1+x^2/L^2)$$

we know that $A = V_{0}/\hbar \omega$ is dimensionless, and $B = E/\hbar \omega$ is also dimensionless, $\omega$ is the frequency. I used the $\omega$ to the one in the harmonic oscillator and got a dimensionless equation, but I only got 5 negative eigenvalues which show something is wrong, because there are infinite bound states.

• What is A here? Is it an assumed value or something you need to evaluate? If you want to make an equation dimensionless, basically you want to replace all dimensionful quantities with a scale factor times a base value - for instance, $x=\eta L$. Now you replace that in the expression and integrate over the dimensionless $\eta$. – levitopher Dec 4 '15 at 20:00
• @tjkt - because there are infinite bound states. Are you sure? That's not an infinite well... – Gert Dec 4 '15 at 20:16
• The reason I think this is not right is because of this post I made yesterday...physics.stackexchange.com/q/222125 – tjkt Dec 4 '15 at 20:17
• @tjkt: I missed that thread. Interesting. – Gert Dec 4 '15 at 21:30
• @tjkt The reason that you only find five negative eigenvalues is your numerical accuracy. For more details, see my comments in the answer of your previous question. – Praan Dec 5 '15 at 4:14

\begin{eqnarray*} H &=&H_{0}+V(x)=\frac{p^{2}}{2m}+V(x)=-\frac{\hbar ^{2}}{2m}\partial _{x}^{2}+V(x) \\ V(x) &=&-V_{0}\frac{1}{1+x^{2}/L^{2}} \end{eqnarray*} Setting $x=Ly$ we have \begin{equation*} H=-\frac{\hbar ^{2}}{2m}\frac{1}{L^{2}}\partial _{y}^{2}-V_{0}\frac{1}{ 1+y^{2}}=\frac{\hbar ^{2}}{2mL^{2}}\{-\partial _{y}^{2}-2mL^{2}V_{0}\frac{1}{ 1+y^{2}}\} \end{equation*} so the important part is \begin{equation*} h=h_{0}+h_{1}=-\partial _{y}^{2}-V_{1}\frac{1}{1+y^{2}},\;V_{1}=2mL^{2}V_{0} \end{equation*}
$V(x)$ is relatively compact wrt. to $p^{2}$ so the essential spectra of $H$ and $H_{0}$ coincide. This implies that their continuous spectra coincide and equal $[0,\infty )$. In general $H$ can have continuum-embedded, i.e. non-negative eigenvalues. In addition, if there are an infinite number of eigenvalues they must accumulate in $0$ and there are no other accumulation points. There exist advanced techniques to show that in this case there are no non-negative eigenvalues. This is what in general can be said. But it suggests that you should look for eigenvalues close to $0$. This is to be expected since the potential is shallow.