[SOLVED] Perturbation of the simple harmonic oscillatorby nowits Tags: harmonic, oscillator, perturbation, simple, solved 

#1
Dec307, 06:20 AM

P: 18

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 firstorder correction of the ground state. How does the correction change when a gets bigger? 2. Relevant equations [tex]E_0^1=<\psi_0^0H'\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. 



#2
Dec307, 06:22 AM

Sci Advisor
HW Helper
P: 2,886

Have you tried to compute the integral? It's a standard one. 



#3
Dec307, 07:10 AM

P: 18

I've never encountered an error function before in any homework problem, so I automatically assumed that I had done something wrong. 



#4
Dec307, 07:17 AM

Sci Advisor
HW Helper
P: 4,301

[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) [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). 



#5
Dec307, 07:40 AM

P: 18

Ok.
Thank you both! 


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