Register to reply

Perturbation of the simple harmonic oscillator

Share this thread:
nowits
#1
Dec3-07, 06:20 AM
P: 18
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
[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.
Phys.Org News Partner Science news on Phys.org
'Smart material' chin strap harvests energy from chewing
King Richard III died painfully on battlefield
Capturing ancient Maya sites from both a rat's and a 'bat's eye view'
nrqed
#2
Dec3-07, 06:22 AM
Sci Advisor
HW Helper
P: 3,017
Quote Quote by nowits View Post
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
[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.
What makes you think this is not right?
Have you tried to compute the integral? It's a standard one.
nowits
#3
Dec3-07, 07:10 AM
P: 18
Quote Quote by nrqed View Post
Have you tried to compute the integral? It's a standard one.
[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.

CompuChip
#4
Dec3-07, 07:17 AM
Sci Advisor
HW Helper
P: 4,300
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).
nowits
#5
Dec3-07, 07:40 AM
P: 18
Ok.

Thank you both!


Register to reply

Related Discussions
Simple Harmonic Oscillator Introductory Physics Homework 1
Simple Harmonic Oscillator Introductory Physics Homework 0
1-D simple harmonic oscillator Advanced Physics Homework 3
QM simple harmonic oscillator Advanced Physics Homework 3
Simple Harmonic Oscillator Introductory Physics Homework 8