Perturbation of the simple harmonic oscillator

[nowits]

[SOLVED] Perturbation of the simple harmonic oscillator

Homework Statement

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?

Homework Equations

E_0^1=<\psi_0^0|H'|\psi_0^0>

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.

 

[nrqed]

Homework Statement

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?

Homework Equations

E_0^1=<\psi_0^0|H'|\psi_0^0>

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.

 

[nowits]

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.

 

[CompuChip]

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).

 

[nowits]

Ok.

Thank you both!