Simple perturbation theory

1. Sep 14, 2004

JohanL

we have a particle in an infinite one-dimensional square well potential

[V(x)=0 for 0<x<L and V(x) is infinite otherwise]

and introduce a small potential (perturbation) in the middle of the
square well potential. Then the first order energy correction
for the ground state is 100 times lower than the first order correction
to the first excited state. I dont understand why...

The wave function without the small potential step in the middle
is zero in the middle for the first excited state and maximum in the middle
for the ground state. Has the diffrence in energy corrections something
to do with this?

2. Sep 14, 2004

humanino

Welcome JohanL !

I don't know exactly the parameters, so it is difficult to answer. Maybe the reason is that the lower energy state is especially stable, whereas excited states already have dynamics included.

I hope better answers will come.

3. Sep 14, 2004

JohanL

Thank you!

Im just a beginner in perturbation theory...
The wavefunctions are just the usual standing waves in
[V(x)=0 for 0<x<L and V(x) is infinite otherwise]

and then a small step of width a=L/10 and height h is introduced in the
middle of the square well potential, i.e.
V(x)=h for 0.5(L-a)<x<0.5(L+a)

Then with perturbation theory I calculate the first orde energy corrections
to the unperturbed eigenvalues and I get that the corrections
is 100 times larger for the first excited state than for the ground state.
and i dont understand why...

4. Sep 14, 2004

humanino

OK, now all the parameters are here.

I don't know if my interpretation is correct. If I figure out something better, I'll let you know. The calculations barely tell you any deep reason. Do they ?

5. Sep 14, 2004

vanesch

Staff Emeritus
Do you mean <psi|V|psi> where V is your potential and psi the unperturbed state (sin(pi nx/L) ?

So I guess you calculate something like Integral( h sin^2(pi n x/L) dx) between x = 0.5(L-a) and 0.5(L+a) with appropriate normalisations I don't know by heart ?

cheers,
Patrick.

6. Sep 14, 2004

JohanL

Yes, thats exactly what i have done.
and when i have calculated the integral i set n=1 for
the energy correction for the ground state and then n=2
for the correction for the first excited states.
and the corrections i get is
E(n=1)=0.0016 h
E(n=2)=0.19 h

and then i need a qualitative interpretation for why they differ
so much...

7. Sep 14, 2004

vanesch

Staff Emeritus
I get something else.

psi_n(x) = sqrt(2/L) Sin[ n Pi x/L ]

I then find (using Mathematica) that the integral of psi_n(x)^2 from (L-a)/2 to (L+a)/2 equals:

a/L - Cos[n Pi] Sin[ a n Pi/L] / (n Pi)

this gives for the first energy correction:

a/L + Sin[a Pi/L]

for the second:

a/L - Sin[2 a Pi/L] / (2 Pi)

....

(all this times h of course).
Filling in L -> 1, a -> 1/10, I find numerically:
{0.198363, 0.00645107, 0.185839, 0.0243173, 0.163662, 0.0495449}

for n = 1, 2, 3, 4, 5 and 6.

Now maybe I'm wrong ! I can send you the (simple) mathematica sheet if you want.

cheers,
Patrick.

8. Sep 14, 2004

nrqed

Hi there.

I can tell without doing the calculation that your results can't be right.

Your perturbation is in the center of the well. Around that point, the ground state probability density has a maximum whereas the second excited state has a node. Therefore, the ground state will be much more affected by the perturbation than the first excited state.

Pat

9. Sep 14, 2004

JohanL

Thank you Patrick and Pat!
Now I understand!