Eigenvalue of x^4 Potential: A Mystery Explored

[dingo_d]

So I'm working out on a potentials of the type x^2p, and I have a program that solves and gives the eigenenergies for a potential that I have (x^n in general).

I noticed that for a ground state the potential x^4 has the smallest eigenvalue : 0.667981 in units where \hbar=m=\omega=1.

I found out that there is an article that describes the physical background of this, but I haven't got the access to it :\

I've searched google and I haven't stumbled upon an explanation...

So any ideas why is the eigenvalue in the ground state of the x^4 potential lower than that of the potential x^2 (numerically)? When I set the power to something like x^20, or x^100 I see that the energies go up, as they should...

Is this just numerical quirk or some weird physics?

 

[dingo_d]

So no idea?

 

[jeppetrost]

This is quite funny...
Try using different polynomials, eg. 0.5*x^n or 8x^n and you will get different results.
I don't get it...

 

[Mike Pemulis]

I noticed that for a ground state the potential x^4 has the smallest eigenvalue : 0.667981 in units where LaTeX Code: \\hbar=m=\\omega=1 .

For a harmonic oscillator, it's 0.5, right? So 0.6... is actually higher. Am I missing something?

Edit: Yes, yes I am. Sorry for the dumbness. See SpectraCat's post.

 

[SpectraCat]

You have to be a little more specific about how you are setting up the problem. What are the coefficients for the potential, and in what units? What length scale are you using? What are the units on your kinetic energy operator (it looks like you are using atomic units, but I just want to be sure)? You are correct to suspicious about the answers you are getting ... consider the following points.

1) As the exponent increases, the potentials are actually getting flatter in the region around x=0. In fact, as the exponent approaches infinity, the potential approaches an infinite square between x=-1 and x=1, which has well-known energy levels that can be solved for analytically. The ground state of that potential will set the lower limit for the ground state energy for all of your potentials.

2) The x^2 potential also has analytical energy levels (harmonic oscillator). If hbar and omega are both 1, then you should get 0.5 (hartrees) for the ground state energy, right? However, but if that is the case, then your potential is not x^2 but rather 0.5*x^2, in the units you have chosen. If your potential is x^2 in atomic units then your force constant is 2 (hartrees per bohr radius squared) and omega should be sqrt(2).

3) If your problem is set up so that you have energy levels near the bottom of the well, the trend as the exponent is increased might not be what you expect.

4) If your numerical solver uses the Numerov-Cooley algorithm, or a similar one, then the eigenstate you find will depend on an initial guess of the energy. So, unless you are doing some sort of systematic search for eigenstates, you can easily miss one, particularly if they are closely spaced.

Hope this helps.

 

[dingo_d]

Wow that was thorough :)

the general problem was to use discretization to solve numerically the x^{2p} potential (p is positive integer). And as you've said, for p=1 you'll get SHO, and for p\to \infty you get infinite square well.

And using discretization method (where \hbar=m=\omega=1) I got the result for the ground state to be 0.667981 and for the SHO it's 0.707103. For infinite square well I get (which is almost exact to the analytical solution) 1.233700.

I have noticed, that for SHO the energy levels differs from the analytical solution as I take the higher states...

It's just that it said in the article that it's of some importance to the quantum field theorists, and I cannot seem to find anything about it :\

 

[jeppetrost]

I have found at least one problem in finding and comparing energies of x^(2n) potentials. It is, as Spectra was aiming at, I believe, that these different potentials all have different sizes (when converted back to units of joule). This makes it a little complicated to compare the energies directly, as you have to take some (n-dependent) factor along.

 

[SpectraCat]

Wow that was thorough :)

the general problem was to use discretization to solve numerically the x^{2p} potential (p is positive integer). And as you've said, for p=1 you'll get SHO, and for p\to \infty you get infinite square well.

And using discretization method (where \hbar=m=\omega=1) I got the result for the ground state to be 0.667981 and for the SHO it's 0.707103. For infinite square well I get (which is almost exact to the analytical solution) 1.233700.

I have noticed, that for SHO the energy levels differs from the analytical solution as I take the higher states...

It's just that it said in the article that it's of some importance to the quantum field theorists, and I cannot seem to find anything about it :\

Ok, it looks like you are probably doing everything correctly. I think the answer to your question is that the x^4 (quartic) potential represents some sort of optimum case between the potential having a wider classically allowed region, while also having shallow enough curvature to allow significant penetration into the classically forbidden region. For example, the reason that the SHO has a lower ground state energy than the 1D-PIB limit is that the SHO ground state has significant penetration into the classically forbidden region.
However, I can't say for sure whether there is some deeper physical meaning behind the fact that this happens for the quartic potential specifically .. there certainly might be, but nothing jumps out at me as to specifically why at the moment. Maybe a QFT expert could shed some light on this?

 

[dingo_d]

I think I got the fact why I get superb results for infinite square well solutions, and not that good approximation for 'softer' potentials.

Discretization works on a certain interval, and outside that interval the wave function is zero. Which is absolutely true for infinite square well (no tunneling into classical forbidden region), and for SHO I have some tunneling, so the interval should be very large for computer to neglect the tunneling effect.

But I'm still not quite sure as to why the x^4 potential gave smaller result...

You're right. Maybe some field theorist will come along and shed a light on it :D

 

[DrDu]

Maybe you should have a look at the WKB approximation for the solutions.

 

[dingo_d]

I used variational method for this problem, I guess I could try wkb to see what'll come out...