I am trying to solve the following equation, this being a form of the Schrodinger equation for the harmonic oscillator, E=3 being the 1st energy level:
$u''(r) = (r^2 - 3) u(r)$
I am going trough a book on numerical analysis and this is solved with the Numerov method, the exact solution is only cited:
$A r e^{-r^2/2}$
How would I go about obtaining this solution? I tried Laplace transforms but this being a nonlinear equation, that method won't work, even Mathematica failed me
Thanks in advance
Define the operators $Q$ and $D$ by $(Qf)(r) = r f(r)$ and $(Df)(r) = f'(r).$ Note that $[D,Q] = DQ-QD = 1.$
Let $L = Q^2-D^2.$ The equation can then be written $Lu = 3u.$
Now set $A_- = Q-D$ and $A_+ = Q+D.$ Note that $L = A_- A_+ +1$ and that $[L, A_-] = 2 A_-.$
Let $u_0$ be a solution to $A_+ u_0 = 0,$ i.e. to $r u_0(r) + u_0'(r) = 0.$ This equation is trivial to solve; the solutions are $u_0(r) = C e^{-r^2/2},$ where $C$ is a constant. Note that $L u_0 = (A_- A_+ +1) u_0 = u_0.$
Then let $u_1 = A_- u_0$ and note that $$ L u_1 = L (A_- u_0) = [L, A_-] u_0 + A_- (L u_0) = (2 A_-) u_0 + A_- u_0 = 3 A_- u_0 = 3 u_1, $$ i.e. $u_1$ is a solution to our equation.
Since $u_0(r) = C e^{-r^2/2}$ we have $$ u_1(r) = r u_0(r) + u_0'(r) = 2C r e^{-r^2/2}. $$