[SOLVED] Perturbation of the simple harmonic oscillator 1. The problem statement, all variables and given/known data An additional term V_{0}e^{-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 [tex]E_0^1=<\psi_0^0|H'|\psi_0^0>[/tex] 3. The attempt at a solution [tex]\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[/tex] So I suppose this is not what is wanted.
[tex]\int e^{-\xi x^2}=\frac{\sqrt{\pi}\ erf(\sqrt{\xi}x)}{2\sqrt{\xi}}\ \ \ ?[/tex] I've never encountered an error function before in any homework problem, so I automatically assumed that I had done something wrong.
That's right, the indefinite integral contains an erf. But you have more information: you know the boundary conditions and (you should have) [itex]\xi > 0[/itex]. Using [tex]\lim_{x \to \pm \infty} \operatorname{erf}(x) = \pm 1[/tex] you can calculate it, and in fact it is just a Gaussian integral, [tex]\int_{-\infty}^\infty e^{-\xi x^2} dx = \sqrt{\frac{\pi}{\xi}}[/tex] 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).