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.


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

  • 2
    $\begingroup$ 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. $\endgroup$ – Steven Mathey Sep 23 '15 at 10:54
  • 1
    $\begingroup$ @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. $\endgroup$ – Anant Saxena Sep 23 '15 at 11:06
  • 2
    $\begingroup$ 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. $\endgroup$ – Steven Mathey Sep 23 '15 at 11:08
  • 1
    $\begingroup$ 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. $\endgroup$ – Steven Mathey Sep 23 '15 at 11:20
  • 2
    $\begingroup$ arxiv.org/abs/hep-th/9812211 $\endgroup$ – Count Iblis Sep 27 '15 at 22:10

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.