Some context:

Let us consider a spinless charged massive particle of mass $m$, charge $q$ in an electrostatic potential $V(x) = frac{m}{2q}omega^2x^2$.

The corresponding stationary Klein-Gordon equation is then:

$$
left(-dfrac{partial^2}{partial x^2} + m^2right)phi(x) = (E-qV(x))^2phi(x)
$$

One may right it in the following form:

$$
frac{(E^2-m^2)}{2m} phi(x) = left(frac{p_x^2}{2m} + W(x)right)phi(x) = Hphi(x)
$$

where $W(x) = frac{E}{2}omega^2x^2 – frac{1}{8}momega^4 x^4$

Then, we may want to evaluate the $E_n$ levels of energy. To do so, I had to consider the pertubative angle: write $H = H_0 + V_p$ where:

• $H_0 = frac{p_x^2}{2m} + frac{E}{2}omega^2x^2 = frac{p_x^2}{2m} + frac{1}{2}mtildeomega^2x^2$, $quad tilde omega^2 = frac{E}{m}omega^2$
• The perturbation $V_p = – frac{1}{8}momega^4 x^4$.

Finally, the energy levels $E_n$ satisfy, considering first order correction:

$$
dfrac{E^2_n – m^2}{2m} = tildeomega(n+1/2)+ langle n|V_p|n rangle = tildeomega(n+1/2) – frac{1}{8}m^2omega^4 langle n|X^4|n rangle
$$

Using the fact that $langle n|X^4|n rangle = frac{3}{4m^2tildeomega^2} (2n^2+2n+1)$ :

$$
dfrac{E^2_n – m^2}{2m} = sqrt{frac{E_n}{m}}omega(n+1/2) – frac{3 m}{32 E_n^2}omega^2 (2n^2+2n+1)quad (star)
$$

The question:

• Is $(star)$ correct?
• If so, solve $(star)$ for in the non-relativistic limit

I find it really difficult to solve it in the non-relativistic limit since we have $E_n$ at different powers… So it makes me wonder whether I made a mistake in $(star)$ or whether I really know how to take the non-relativistic limit properly (I tried without being alble to find any solvable equation in $E_n$ taking the first order for $mto infty$).