# quantized energies for a particle in a non-linear potential

Okay, so the question i'm trying to solve is to find the quantized energies for a particle in the potential:

$$V(x)=V_0 \left ( \frac{b}{x}-\frac{x}{b} \right )^2$$

for some constant b. I used the schrodinger equation with the substitution $$\psi=\frac{\tilde \psi}{\sqrt{x}}$$ as well as $$\tilde x=bx$$(to make it unitless) and substituted it into the schrodinger equation and obtained $$\tilde \psi''- \tilde \psi' x^{-1}+ \tilde \psi \left (\frac{3/4+V_0}{x^2}+V_0x^2-(E+2V_0 \right)=0$$

Then, I followed along with the derivation in pages 51-56 of Griffiths quantum textbook, by finding an integrating factor and trying to expand $$\psi$$ as a power series which led me to the following $$\psi(\tilde x)=f(x)e^-\frac{\tilde x^2}{2} \quad \text{Eq 2.77 in Griffiths 2nd ed}$$ $$f(x)=\sum_{n=0}^{\infty}C_nx^n$$ $$f'(x)=\sum_{n=0}^{\infty}nC_nx^{n-1}$$ $$f''(x)=\sum_{n=0}^{\infty}n(n-1)C_nx^{n-2}$$

This of course leads to $$\tilde \psi= f(\tilde x)e^{-\tilde x^2/2}$$ $$\tilde \psi '=e^{- x^2/2} (f'(x)-xf(x)) \quad \text{(dropped the tilde on the x) }$$ $$\tilde \psi ''=e^{-x^2/2}(f''(x)-2xf'(x)+f(x)(x^2-1))$$

I believe these are the correct $$\psi$$'s however my problem now is when I sub them back into the schrodinger equation I found above, I dont get the same cancellation that happens for the SHO, so I am struggling to find the recurrence relations in order to find the quantized energies. If anyone can help me find the energies I would be much appreciative.

• Offhand I'd wonder if this has a closed form of the sort you're trying to find. I suspect you're heading deep into the realm of hypergeometric equations, from whence none return sane. :-) – StephenG Sep 21 '18 at 0:08
• Thanks for the response, im pretty sure this problem has a closed form solution, I dont think I have to invoke hypergeometric equations for it, although it may be a tool of interest for down the road.... – user352879 Sep 21 '18 at 3:00
• Might be worth asking Mathematics SE for an opinion. – StephenG Sep 21 '18 at 3:32
Well, the standard practice is to pick coordinates suitable to one of the two minima of your potential; taking b, $$y\equiv x-b \qquad \Longrightarrow \quad V(y)=V_0 \left ( \frac{b}{y+b}-\frac{y+b}{b} \right )^2 \sim \frac {4V_0}{b^2} ~y^2 +...$$ for large b, so you have a bland oscillator.