Determine the ground state energy

1. Dec 8, 2011

blueyellow

1. The problem statement, all variables and given/known data

A particle of mass m is confined to move in a one-dimensional "infinite" potential well defined by V(x)=0, |x|< or=a, V(x)=infinity otherwise. The energy eigenvalues are E(subscript n)=((n^2)(pi^2)(h-bar^2))/(8m a^2), with n=1,2,3,... and the orthonormal eigenfunctions are the even and odd functions

psi(subscript n)=
[1/(sqrt a)]cos(n pi x/2a) for n=1, 3, 5...
[1/(sqrt a)]sin(n pi x/2a) for n=2, 4, 6 ...

The potential is modified between -a<x<a to V(x)=epsilon[(pi^2)(h-bar^2)/(8ma^2)]sin((3pi x)/2a) with epsilon<1

a) Determine the ground state energy to first-order in epsilon. (Note 2 sin Acos B=sin(A+B)+sin(A-B))
b) Determine the ground state energy to second order in epsilon.

3. The attempt at a solution

ground state, n=0

but I couldn't find the wavefunction for n=0 because it does not say what it is when n=0. Is the ground state energy just the potential? But the formula for the potential doesn't have an 'n' in it. Then what has the wavefunction got to do with anything? And I have no idea why 2sinA cos B needs to be used. I'm really confused, please help.

Last edited: Dec 8, 2011
2. Dec 8, 2011

vela

Staff Emeritus
"Ground state" doesn't mean n=0. The ground state is the state with the lowest energy. For which allowed values of n is the energy the lowest?

3. Dec 11, 2011

blueyellow

Thanks. I am still confused, though, because I do not know how to apply perturbation theory to it.

so H-hat=H(subscript 0)-hat +H'-hat

but what is H'-hat?

The question tells you what the modified potential is, but how is the potential related to H'-hat? I thought the Hamiltonian was concerned with the energy so how is it related to the potential?

4. Dec 11, 2011

blueyellow

so the ground state would be n=1?

5. Dec 11, 2011

blueyellow

E(subscript 1)^(1)=<psi (subscript 1)^(0)|H'-hat|psi(subscript 1)^0>

=integral from -a to a ((1/sqrt a) cos (pi x/2a))^2 (epsilon pi^2 h-bar^2)/(8ma^2) sin(3pi x/2a) dx

=integral from -a to a ((1/a) cos^2 (pi x/2a)) (epsilon pi^2 h-bar^2)/(8ma^2) sin(3pi x/2a) dx

=0

6. Dec 11, 2011

vela

Staff Emeritus
That's right. The first-order correction to the energy for those states is 0. You need to do the same thing for the sine states now.

7. Dec 11, 2011

blueyellow

Why do I need to bother with the sin states since n=1 is the ground state? The sin states are only for n=2, etc?

8. Dec 11, 2011

blueyellow

Why does 2sinA cos B=sin(A+B)+sin(A-B) need to be used anywhere? Am I supposed to say:

psi(subscript n)=(1/sqrt a) cos (n pi x/2a) +(1/sqrt a) sin (n pi x/2a)
<psi(subscript n)|psi(subscript n)>=(1/a) cos (n pi x/2a) sin(n pi x/2a)

using the above trig identity:

<psi(subscript n|psi(subscript n)>=(1/2a) sin(n pi x/a)

?

9. Dec 11, 2011

vela

Staff Emeritus
Sorry, I mistakenly thought you were supposed to calculate the energy correction for all states.

How did you do the integral without using the trig identity?

10. Dec 11, 2011

blueyellow

I typed it into Wolfram Alpha and it turned out to be zero. And then I found out it was because of it being an odd integral, supposedly

11. Dec 11, 2011

blueyellow

In some textbooks, the infinite well is from 0 to L rather than from -a to a. Does this mean that I can just integrate from 0 to a and then multiply that result by two?

12. Dec 11, 2011

vela

Staff Emeritus
The hint would make sense if you actually did the integral.

13. Dec 11, 2011

vela

Staff Emeritus
Nope. You can only do stuff that's mathematically justifiable. If you have valid mathematical reasons to do that, it's fine, but you can't simply say "these two problems are related so can I do this random thing?"

14. Dec 11, 2011

blueyellow

it still won't change the fact that the result is zero. So they want me to spend my time integrating, only to find out the result is zero. Four marks for that! And to find that perturbation doesn't perturb/change anything, if <phi|H'|phi>=0, it means that the ground state energy to first order in epsilon is just E(subscript n), since the perturbation doesnt affect it at all if it really equals zero. That would mean that the perturbation does not affect the wavefunction, so psi^(1)=psi^(0), therefore E(subscript n)^(2)=<psi(subscript 1)^(0)|H'-hat-E(subscript 1)^(1)|psi(subscript 1)^(1)>

question b)

E(subscript n)^(2)=<psi(subscript 1)^(0)|H'-hat - 0|psi(subscript 1)^(1)>
=<psi(subscript 1)^(1)|H'-hat-0|psi(subscript 1)^(1)>=0,
and if E(subscript n)^(2)=0, that means the second order perturbation is zero, so the second order perturbation doesn't change the ground state energy.

This must be wrong. Isn't the whole point of perturbation that it changes/perturbs something?

15. Dec 11, 2011

blueyellow

sorry if I sound angry - I don't mean to. Thanks again for your help.

16. Dec 11, 2011

blueyellow

But surely whether they call a point '0' or 'a' or 'L' or whatever doesn't matter in real life? It's just a way of naming some point, right? And -a to a is the same length as 0 to 2a? Maybe I should integrate it from 0 to 2a? Or is that integral going to equal zero also, haha.

17. Dec 11, 2011

vela

Staff Emeritus
Well, I wouldn't say that. They want you to find the answer using a correct method. If you can justify that the integral is indeed 0 using a valid argument, that's fine.

My point was that since you didn't actually do the integral yourself, you wouldn't have used the hint, so of course you wouldn't understand why the hint was given.

18. Dec 11, 2011

vela

Staff Emeritus
You're correct in thinking that any physical quantity you calculate should be independent of the choice of coordinate system; however, that's not the same as saying that the details of the calculation aren't affected by your choice of coordinate system.

There may be valid reasons why you can simply integrate from 0 to a and multiply that result by 2, but your reason for doing so wasn't one of them.

I don't have time to reply to the other parts of your posts right now. I'll be back later, but others may step in as well. Good luck in the meantime!

19. Dec 12, 2011

vela

Staff Emeritus
The fact that the first-order correction to the energy is 0 doesn't mean the correction to the wave function is 0. You should find a non-zero second-order correction to the energy.

20. Dec 12, 2011

blueyellow

for question b)

I found out that one has to do:

E(subscript n)^(2)
= Ʃ [k does not equal n underneath the sigma sign] [|<$\varphi$(subscript k)^(0)|H'-hat|$\varphi$(subscript n^(0)>|^2]/ (E(subscript n)^(0) - E(subscript k)^(0))

And k reperesents all of the states, or something. But if you consider all the n states and sum to infinity, then E(subscript k) ends up being infinity. Which has to be wrong. If anyone understands the formula or the 'k' thing, please help.