What Is the Ground State Energy of a Particle in a Modified Potential Well?

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Homework Statement

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.

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.

 

[vela]

Homework Statement

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.

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.

"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?

 

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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?

 

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so the ground state would be n=1?

 

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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

This seems wrong. Please help.

 

[vela]

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.

 

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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?

 

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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)

?

 

[vela]

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?

 

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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

 

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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?

 

[vela]

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

 

[vela]

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?

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?"

 

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

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 doesn't 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?

 

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

 

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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?"

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.

 

[vela]

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!

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.

 

[vela]

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.

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!

 

[vela]

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.

 

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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.

 

[vela]

That's right. You want to calculate
$$E_n^{(2)} = \sum_{k \ne n} \frac{ | H'_{nk} |^2}{E_n^{(0)} - E_k^{(0)}}$$where $H'_{nk}$ is the matrix element $\langle \phi_k^{(0)} | \hat{H'} | \phi_n^{(0)} \rangle$.

The sum is indeed an infinite series, but it converges. In fact, most of the matrix elements will turn out to be 0, so you can write down exactly what the sum is equal to.

Show us the integral corresponding to the the matrix element H'_k0 and your attempt to solve it.

 

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how does E(subscript k) converge if E(subscript k) is infinity?

 

[vela]

It's not infinite.

 

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what is it then? How can it not be infinite if k is all of the states and E(subscript n) is proportional to n^2. Taking n to be k, and sum from k=1 to infinity, E(subscript k) is infinite.

What have I not understood? Please help.

 

[vela]

Why don't you just do the calculations and see how it turns out?

I'm not trying to be coy, but it's just that your reasoning is so obviously wrong that I think it would help you more if you figured it out on your own.

 

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Ok, I have tried the denominator now:

the denomiator is:

[(n^2 -k^2)pi h-bar^2]/8ma

but as for the numerator I have no idea. If k represents all of the eigenstates then the matrix would go on forever. I have no intention of doing a matrix that goes on forever because it would take me forever to work out. So what have I misunderstood?

 

[vela]

Like I said, do the calculations and you will see your assumptions are wrong.

 

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If I do the calculations the answer turns out to be zero for the second order term.

k=odd
<psi(subscipt i)|H'|psi(subscript k>
=(1/a)(from -a to a) cos (pi x/2a) cos(k pi/2a) sin(3pi x/2a)
=0 because it is an odd function

k=even

<psi(subscript i)|H'|psi(subscript k)>
=(1/a)(from -a to a) cos(pi x/2a) sin (k pi x/2a) sin(3pi x/2a) dx
=(1/2a)∫(from -a to a) sin[((k+1)pi x)/2a]+sin[((1-k) pi x/2a)) sin (3 pi x/2a) dx

this is an odd function because

sin(A)sin(B)=(1/2)(cos(A-B)-cos(A+B))

Please help.

 

[vela]

k=even

<psi(subscript i)|H'|psi(subscript k)>
=(1/a)(from -a to a) cos(pi x/2a) sin (k pi x/2a) sin(3pi x/2a) dx

The product is an even function multiplied by two odd functions, so the integrand is even.