We have thus $$ -\frac{d^2}{d\rho^2} u(\rho) + \frac{mk}{\hbar^2} \alpha^4\rho^2u(\rho) = \frac{2m\alpha^2}{\hbar^2}E u(\rho) . $$ The constant \( \alpha \) can now be fixed so that $$ \frac{mk}{\hbar^2} \alpha^4 = 1, $$ or $$ \alpha = \left(\frac{\hbar^2}{mk}\right)^{1/4}. $$ Defining $$ \lambda = \frac{2m\alpha^2}{\hbar^2}E, $$ we can rewrite Schroedinger's equation as $$ -\frac{d^2}{d\rho^2} u(\rho) + \rho^2u(\rho) = \lambda u(\rho) . $$ This is the first equation to solve numerically. In three dimensions the eigenvalues for \( l=0 \) are \( \lambda_0=3,\lambda_1=7,\lambda_2=11,\dots . \)