# Solving the quantum an-harmonic oscillator pertubatively?

## Background

Generally while solving the quantum an-harmonic oscillator:

$$-\frac{d^2 y}{dx^2} + k_1 x^4 y + k_2 x^2 y= E y$$

Most people (I've googled) on the internet always solve this using:

$$\underbrace{-\frac{d^2 y}{dx^2} + k_2 x^2 y}_{H_o y} + \epsilon \underbrace{k_1 x^4 y}_{H' y} = E y$$

Where $H_o$ is the unperturbed Hamiltonian and $H'$ is the perturbation.

However, I thought of first using a smart-ansatz,

$y= \exp(- \sqrt k_1 |x|^3/3) \alpha(x)$

where $|x|=$ to the modulus of x. Hence, $$y''= \begin{cases} (k_1 x^4 \alpha - 2x \sqrt k_1 \alpha - 2 \sqrt k_1 x^2 \alpha' + \alpha'') e^{(- \sqrt k_1 x^3/3)} & x \geq 0 \\ (k_1 x^4 \alpha + 2x \sqrt k_1 \alpha + 2 \sqrt k_1 x^2 \alpha' + \alpha'') e^{( \sqrt k_1 x^3/3)} & x \leq 0 \end{cases}$$

Substituting in the an-harmonic oscillator equation:

$$- \alpha'' \pm 2 \sqrt k_1 x^2 \alpha' \pm 2x \sqrt k_1 \alpha + k_2 x^2 \alpha = E \alpha$$

My idea is to now apply perturbation theory:

$$\underbrace{- \alpha'' \pm 2x \sqrt k_1 \alpha + k_2 x^2 \alpha}_{H_o \alpha} \pm \epsilon \underbrace{2 \sqrt k_1 x^2 \alpha'}_{H' \alpha} = E \alpha$$

We can recognize that the series is $H_o$ is the shifted harmonic oscillator.

## Questions

How do I calculate the solution to the unperterbed piecewise differential equation?

$$(E + k_1/k_2) \alpha = \begin{cases} - \alpha'' + k_2(x - \sqrt{k_1}/k_2)^2 \alpha & x \geq 0 \\ - \alpha'' + k_2(x+ \sqrt{k_1}/k_2)^2 \alpha & x \leq 0 \end{cases}$$

How does one calculate the second order pertubation?

$$E^1 = 0$$

(See my attempt for why)

How do I calculate the radius of convergence of the resulting perturbation series?

$$(E + k_1/k_2) \alpha = \begin{cases} - \alpha'' + k_2(x - \sqrt{k_1}/k_2)^2 \alpha - \epsilon 2 \sqrt k_1 x^2 \alpha' & x \geq 0 \\ - \alpha'' + k_2(x+ \sqrt{k_1}/k_2)^2 \alpha + \epsilon 2 \sqrt k_1 x^2 \alpha' & x \leq 0 \end{cases}$$

If it is not, how does it compare asymptotically to the general approach? Also is there any deeper meaning to this (getting another harmonic oscillator $+$ perturbation term upon substitution with $e^{- \sqrt k_1 |x|^3/3)} \alpha(x)$ )?

## My Attempt

For the unperterbed case:

$$\alpha^{0}_n = \begin{cases} H_n(x - \sqrt{k_1}/k_2) \exp(- \sqrt k_2(x - \sqrt{k_1}/k_2)^2/2) & x \geq 0 \\ H_n(x + \sqrt{k_1}/k_2) \exp(- \sqrt k_2 (x + \sqrt{k_1}/k_2)^2/2) & x \leq 0 \end{cases}$$

Where $H_n$ is the n'th hermite polynomial.

We note $\alpha$ must be an even function as $y$, $\exp(-|x|^3)$ are even functions. The pertubation however is odd! Integrating the even functions over an odd operator we get:

$$\alpha = \alpha_o^n + \epsilon \alpha_1^n + \epsilon^2 \alpha_2^n + ...$$

$$E^1 = \int_{- \infty}^\infty \alpha^o_n H' \alpha^o_n dx = 0$$

Now we apply 2nd order pertubation theory:

$$E^2 = \sum_{n \neq m} \frac{|<\alpha_o^n | H' | \alpha_o^m >|^2}{E^0_n - E^0_m}$$

... Still working on it

• I'm not sure that your ansatz is that smart, sorry. First, the absolute value makes the second derivative of $\alpha(x)$ discontinuous at $x=0$ (since we know that $y(x)$ is a smooth function). This will make everything messy and complicated. Second, if you set $\epsilon=0$, you are back to the unperturbed problem in both cases. This means, for example, that the unperturbed solution for $\alpha(x)$ will be $\alpha_0(x) = \text{e}^{\sqrt{k_1}\left|x\right|^3/3} y_0(x)$. I did not do the calculation, but I expect that the solution for $\alpha(x)$ will simply undo your ansatz order by order. – Steven Mathey Sep 23 '15 at 10:54
• @StevenMathey I did some of my own calculations and I get: $\alpha_o = A_o \exp(-\kappa (x-\sqrt k_1/k_2)^2$ for $x>0$ and $\alpha_o = A_o \exp(-\kappa (x+\sqrt k_1/k_2)^2$ for $x<0$ ... where $\kappa$ is a constant. – drewdles Sep 23 '15 at 11:06
• If you are interested in different ways to do perturbation theory, I recommend that you look into this. The idea is simple and nice. There is a good lecture online here and this book will contain even more details. – Steven Mathey Sep 23 '15 at 11:08
• You are right, sorry. I assumed that $\epsilon$ in your second equation is the same as in the sixth. Then the perturbation series are different. – Steven Mathey Sep 23 '15 at 11:20
• arxiv.org/abs/hep-th/9812211 – Count Iblis Sep 27 '15 at 22:10