Time-independent Schrodinger equation

Question

the equation : (-h/2m)y'' + U(x)y = (E)y

How do you put the time-independent Schrodinger equation in the form of:

y'' + G(x)y' + P(x)y = 0

Answer 1

I'm assuming you want the equation in one spatial variable:

Start with the wave equation $v^{2}\frac{\partial^2 u}{\partial x^2}=\frac{\partial u^2 }{\partial t^2}$, write $u(t,x)=\psi (x)f(t)$. Substitute into the differential equation and you will get

1. $f\frac{\mathrm{d^{2}} \psi }{\mathrm{d} x^{2}}=\frac{1}{v^{2}}\psi \frac{\mathrm{d^{2}}f }{\mathrm{d} x^{2}}$.

Now, substitute the (unnormalized) solution $f(t)=e^{i\omega t}$, and you obtain $\frac{\mathrm{d^{2}}\psi }{\mathrm{d} x^{2}}=\frac{-\omega ^{2}}{v^{2}}\psi$.

From here it's just a question of writing $E=\frac{p^{2}}{2m}+V(x)$. Then, using the DeBroglie relation $p\lambda=h$ with $\lambda \nu=v$ and $\omega =2\pi \nu$, succesive substitutions will give

$\frac{-h^{2}}{8\pi^{2}m}\frac{\mathrm{d^{2}}\psi }{\mathrm{d} x^{2}}+V\psi =E\psi$.

*This procedure works just as well using the vector $r=(x,y,z)$, if you replace $m$ by the reduced mass $\mu$.

** if you do not know the solution $f$ to the wave equation, then you can discover it by noting that if you divide 1. by $f$ and $\psi $ then the LHS is a function of $x$ only and the RHS of $t$ only. This allows you to separate 1. into two ordinary differential equations which may be solved separately. For this, you need some additional conditions.

Answer 2

Lets try $$ y(x) = \mathrm{e}^{g(x)}v(x) $$ we find $$ y' = g'y + \mathrm{e}^{g}v'\\ y'' = g''y + g'y' + g'\mathrm{e}^{g}v' + \mathrm{e}^{g}v'' = \mathrm{e}^{g(x)}\left[g''v + g'^2v + 2g'v' + v''\right] $$ thus you get $$ \mathrm{e}^{g(x)}\left[\frac{-h}{2m}\left(g''v + g'^2v + 2g'v'+ v''\right) + U(x)v - Ev\right] = 0 $$ this leads to $$ -\frac{h}{2m}v''- \frac{2h}{2m}g'v'+\left(U(x) - E -\frac{h}{2m}\left(g'' + g'^2\right)\right)v = 0 $$ or $$ v'' +2g'v' + \left(g''+g'^2-\frac{2m}{h}\left(U(x)-E\right)\right)v =0 $$