# WKB method to calculate the ground eigenvalue of a Quartic Potential

I've come across a problem while studying the WKB method. I want to calculate the eigenvalues of a symmetric quartic double well potential. It could be any potential. I chose it to be $$V(x) = x^4 - 4x^2 +4$$ The hamiltonian with $$\hbar$$ = $$m$$ = $$1$$ gives $$H = -\frac{1}{2}\frac{d^2}{dx^2} + V(x)$$ and I plan to find the eigenvalues of the bound states given by the potential $$V(x)$$ represented bellow

$$\hskip1.7in$$

with returning points $$x_2 > x_1$$ and roots $$x=\pm \sqrt{2}$$

The quatization* problem for the double potential well with respect to even (odd) solutions is, for $$x>0$$ $$\theta \simeq (n + \frac{1}{2}) \pi \mp \frac{1}{2} e^{- \phi} \tag{1}$$

with $$\theta = \int_{x_1}^{x_2} p(x') dx'$$ $$\phi = \int_{0}^{x_1} |p(x')| dx'$$ $$p(x) = \sqrt{2m(E_n - V(x))} = \sqrt{2m(E_n - (x^4 - 4x^2 + 4))}$$

( *Introduction to Quantum Mechanics by David J. Griffiths, problem $$8.15$$ )

My problem lies exactly in solving eq. ($$1$$), since it involves integrals of the square-root of a quartic function $$\int_a^b \sqrt{2m(E_n - (x^4 - 4x^2 + 4))} dx$$

I used Mathematica but it couldn't compute a solution.

Is there any approximation or trick I could use to solve it analytically? If not, any software that could do the computation?

PS: After solving numerically the $$Schr\ddot{o}dinger$$ equation for the ground state eigenvalue I obtained $$E_0 \simeq 1.8$$ with $$\hbar$$ = $$m$$ = $$1$$ as stated above. With the WKB method I'm hoping to obtain a similar result.

• Why can’t you use numerical integration? Commented Apr 11, 2020 at 3:34
• Because E_n is an unknown. The main goal of eq. (1) is to get an expressions for E_n. At least for the ground eigenvalue E_0, I would consider a battle won. Commented Apr 11, 2020 at 4:21
• You integrate for various $E_n$ until you find one that works. There are systematic techniques for narrowing in on the right choice, similar to when you try to find the root of a polynomial numerically, Commented Apr 11, 2020 at 4:28
• I get $E_0=1.74646$. Commented Apr 11, 2020 at 4:44
• Look it up in Gradshtein&Ryzhik, though it is quite possible that such integrals are not solvable. Commented Apr 11, 2020 at 5:43

In this way I found the $$n=0$$ energies $$E_0^\text{even}\approx 1.74646$$ and $$E_0^\text{odd}\approx 2.07823$$ with a few lines of code. For the former, $$\theta\approx 1.43953$$ and $$\phi\approx 1.33739$$; for the latter, $$\theta\approx 1.73284$$ and $$\phi\approx 1.12677$$. Since this is only an approximation, it seemed pointless to go beyond standard precision.
The next energies with $$n=1$$ appear to be "above the hump", where I don't think your equations apply, because they give nonsense.
• I cut and pasted your code into Mathematica, except for the final semicolon. The output for me is $\{x\to 1.74646\}$. If you really have a semicolon at the end you're not going to see this. Commented Apr 11, 2020 at 21:33