First-order perturbation for a simple harmonic potential well

  • Thread starter rwooduk
  • Start date
  • #1
747
55

Homework Statement


The ground state of the wavefunction for an electron in a simple one-dimensional harmonic potential well is

[tex]\Psi _{0}(x)= \left ( \frac{m\omega }{\pi \hbar} \right )^{1/4} exp(-\frac{m\omega x^{2}}{2\hbar})[/tex]

By employing first-order perturbation theory calculate the energy shift [tex]E_{0}= \frac{\hbar \omega}{2} = 2eV[/tex] caused by the time-independant perturbation.

[tex]V(x)= V_{0} x^{3}[/tex]

with V0= 2eV and L= 5.10^-10 m

Homework Equations


E = E0 + V00

The Attempt at a Solution


The problem I'm having is that no integrals are given for this question (it's a past exam question). If I follow the process of finding V00

[tex]V_{00}= \left \langle \Psi _{0}| V_{0} x^{3} | \Psi _{0} \right \rangle [/tex]

[tex]
\int_{0}^{L} (\frac{m\omega }{\pi \hbar} \right ))^{1/2} x^{3} exp (-\frac{m\omega x^{2}}{\hbar})[/tex]

if I integrate by parts between 0 and L I get an integral of exp (-y^2) which ordinarily between minus infinity and infinity would be root pi, but between 0 and L gives ERF. Any Ideas of what to do in this situation?
 

Answers and Replies

  • #2
DrClaude
Mentor
7,551
3,893
You should be able to get that by integrating by parts, provided you choose the correct parts! What did you use?
 
  • Like
Likes rwooduk
  • #3
747
55
You should be able to get that by integrating by parts, provided you choose the correct parts! What did you use?
Using [tex]\int u dv = uv - \int vdu[/tex]

I set u = x^3 in order to reduce it to a lower power of x (and continue to so with the next by parts integrals) and dv = exp (-ax^2) where a is a constant.

The problem is when I integrate with 0 to L limits using substitution of y^2, it comes out as ERF.

http://www.integral-calculator.com/#expr=e^(-y^2)&intvar=y&ubound=l&lbound=0

Thanks for the reply.
 
  • #4
DrClaude
Mentor
7,551
3,893
Try ##u = x^2##.
 
  • Like
Likes rwooduk
  • #5
747
55
Try ##u = x^2##.
thanks, yes thats what I did but this gives an exponential of a square for dv, which when I integrate with 0 to L limits using substitution, namely exp (-y^2), it comes out as ERF.
 
  • #7
DrClaude
Mentor
7,551
3,893
thanks, yes thats what I did but this gives an exponential of a square for dv, which when I integrate with 0 to L limits using substitution, namely exp (-y^2), it comes out as ERF.
If you set ##u = x^2##, then ##dv = x e^{-a x^2} dx##, which can be easily integrated.
 
  • Like
Likes rwooduk
  • #8
747
55

Related Threads on First-order perturbation for a simple harmonic potential well

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
10
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
954
Top