## [SOLVED] Perturbation of the simple harmonic oscillator

1. The problem statement, all variables and given/known data
An additional term V0e-ax2 is added to the potential of the simple harmonic oscillator (V and a are constants, V is small, a>0). Calculate the first-order correction of the ground state. How does the correction change when a gets bigger?

2. Relevant equations
$$E_0^1=<\psi_0^0|H'|\psi_0^0>$$

3. The attempt at a solution
$$\alpha=\frac{m\omega}{\hbar}, E_0^1=\int ^{\infty}_{-\infty} (\frac{\alpha}{\pi})^{1/4}e^{-\alpha x^2/2}V_0e^{-ax^2}(\frac{\alpha}{\pi})^{1/4}e^{-\alpha x^2/2}dx=(\frac{\alpha}{\pi})^{1/2}V_0\int ^{\infty}_{-\infty} e^{(-\alpha -a)x^2}dx$$
So I suppose this is not what is wanted.
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 Quote by nowits 1. The problem statement, all variables and given/known data An additional term V0e-ax2 is added to the potential of the simple harmonic oscillator (V and a are constants, V is small, a>0). Calculate the first-order correction of the ground state. How does the correction change when a gets bigger? 2. Relevant equations $$E_0^1=<\psi_0^0|H'|\psi_0^0>$$ 3. The attempt at a solution $$\alpha=\frac{m\omega}{\hbar}, E_0^1=\int ^{\infty}_{-\infty} (\frac{\alpha}{\pi})^{1/4}e^{-\alpha x^2/2}V_0e^{-ax^2}(\frac{\alpha}{\pi})^{1/4}e^{-\alpha x^2/2}dx=(\frac{\alpha}{\pi})^{1/2}V_0\int ^{\infty}_{-\infty} e^{(-\alpha -a)x^2}dx$$ So I suppose this is not what is wanted.
What makes you think this is not right?
Have you tried to compute the integral? It's a standard one.

 Quote by nrqed Have you tried to compute the integral? It's a standard one.
$$\int e^{-\xi x^2}=\frac{\sqrt{\pi}\ erf(\sqrt{\xi}x)}{2\sqrt{\xi}}\ \ \ ?$$
I've never encountered an error function before in any homework problem, so I automatically assumed that I had done something wrong.

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## [SOLVED] Perturbation of the simple harmonic oscillator

That's right, the indefinite integral contains an erf.
But you have more information: you know the boundary conditions and (you should have) $\xi > 0$. Using
$$\lim_{x \to \pm \infty} \operatorname{erf}(x) = \pm 1$$
you can calculate it, and in fact it is just a Gaussian integral,
$$\int_{-\infty}^\infty e^{-\xi x^2} dx = \sqrt{\frac{\pi}{\xi}}$$
Remember this -- it's ubiquitous in physics (at least, every sort of physics that has to do with any sort of statistics, amongst which QM, thermal physics, QFT, SFT).
 Ok. Thank you both!