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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

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ arxiv.org/abs/hep-th/9812211 $\endgroup$ – Count Iblis Sep 27 '15 at 22:10

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