One more thing

December 10, 2008 by Moshe

For the small minority of readers who are not entertained by ancient Jazz clips, here is a quick puzzle. Consider the quantum mechanics of a particle moving in a plane, paramatrized by coordinates x and y, and feeling the effect of the potential $V(x,y) = x^2 y^2$ , is the spectrum continuous or discrete?

(For a certain generation of string theorists this may bring back memories…)

The answer will come much later, as I am now hopping on a plane. Shame to leave really, looks like it is going to snow sometime soon.

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Posted in Physics, puzzle | 15 Comments

15 Responses

on December 10, 2008 at 11:10 pm matthiasr

I’d guess discrete, because all eigenstates will obviously be bound. On the other hand I’m not of a certain generation of string theorists…
on December 10, 2008 at 11:15 pm Robert

Do I have to say more than dWLN (for the bosonic as well as the susy version)?

Maybe you should tell your younger readers why this an important toy model: Assume you are at large |x|. Then |y| has to be small and is effectively 0 (maybe up to quantum fluctuations, see above). In this approximation however, the motion in the x directions does not notice the existence of the y degree of freedom very much, thus y can be pretty much ignored (“integrated out” in fancy language) for large |x|.

In field theory you would say that the field y has mass |x|. It is only close the the special value x=0 that y can move significantly and in that regime ignoring y would be quite wrong as “the field y becomes light”.

This model is the archetype of a singularity in moduli space: You should see x as a modulus and y as the massive field you are usually ignoring. It is only at special loci of the moduli space that you encouter a singularity which however can be cured by “integrating in” the degree of freedom that becomes light. A typical scenario would be a compactification where a cycle shrinks to zero volume and a brane wrapping that cycle becomes light and thus important. In that case, x would be the modulus of the compactification space that measures the volume of the cycle while y is the wrapped brane degree of freedom.

Let me end this comment with a short commercial: My little own contribution to the discussion of this model is the realisation that even the classical motion is trapped (due to a logarithmic effective potential) unless y is exactly 0 (that is y=y-dot=0), as discussed in hep-th/0009134 (which I did not get published in JHEP at that time which I still think is a pitty). This was later picked up by people discussion moduli trapping in cosmology.
on December 11, 2008 at 10:01 am Luboš Motl

It’s not hard for me to pretend that I’ve never read the particular paper by de Wit, Lüscher, and Nicolai, and I don’t know the particular conclusions of that paper.

On the other hand, the problem is much like the QM problems in the BFSS matrix model except that SUSY is missing – which could be lethal for the existence of threshold bound states if they would otherwise exist. Well, here it seems obvious that H can’t be zero, it is positive definite just like for a normal harmonic oscillator.

In Matrix theory, one gets both a threshold bound state as well as a continuum of multi-object states. The latter are gonna be absent because the SUSY cancellation is absent and the effective potential energy from the zero modes goes up.

The ground state wave function, if it exists, is positive real and concentrated around x=0 and y=0, two lines. For large y, we have a harmonic oscillator near x=0. Its frequency goes like abs(y). So the wavefunction for large fixed y and variable x should look like exp(-yx^2). Sorry if I am missing factors of two.

It should be possible to integrate the fast degree of freedom x out, in order to find a more detailed dependence on “y” around x=0, y large. Well, so the potential energy just gives us abs(y)/2 from the x-harmonic oscillator zero point energy (not cancelled by fermions, in this case). This is growing very positive, so everything’s gonna be bound. The y-dependence could carry some traces of the Airy function.

If Moshe added some fermions and cancelled the zero point energy abs(y)/2, that could be much tougher because the continuum portion of the spectrum could appear and quantum corrections (for a nonSUSY case) would likely make the model’s spectrum either discrete again or unstable (unbounded from below), but fortunately Moshe wasn’t that tough so the spectrum is pure discrete.
on December 11, 2008 at 12:36 pm Just Learning

Since stationary states can exist in a channel, I would suspect that at infinite distances there should be some discreteness in the spectrum.
on December 11, 2008 at 8:39 pm Jianyang

Without solving Schrodinger equation, I would guess the spectrum is discrete, and $E_n \sim n^{4/3}$ , or maybe $(n+const.)^{4/3}$ ?
on December 12, 2008 at 11:24 am Luboš Motl

If someone has solved Moshe’s problem, what about a slightly more complicated one? The potential is

H = p_x^2/2 + p_y^2 / 2 + V(x,y)
V(x,y) = x^2 y^2 / 2 – |y|/2 – |x|/2

The last terms are meant to remove the zero-point energies of the approximate harmonic oscillators for large x, y near 0, or vice versa.

Good luck, Lubos
on December 15, 2008 at 12:40 pm Just Learning

I’m still playing with this, but

$\frac{1}{2}\frac{\partial^2}{\partial x^2}\phi + \frac{1}{2}\frac{\partial^2}{\partial y^2}\phi = E\phi - x^2y^2\phi$

$\phi = e^{xy}$

and we assert that the permissible values for x and y are the set of unitary matrices, therefore

$x^2y^2 = I_n$

and n is equal to the smaller sized matrix

so

$\frac{1}{2}\frac{\partial^2}{\partial x^2}\phi + \frac{1}{2}\frac{\partial^2}{\partial y^2}\phi = E\phi - I_n\phi$

and

$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = 2(E - I_n)$

I’m sure there is something horribly wrong, but its a good first attempt I think
on December 15, 2008 at 4:17 pm Giotis

The Schrodinger equation is two dimensional and this complicates things. I have some doubts due to the fact that the particle could be anywhere at x if y=0 and anywhere at y if x=0. The potential is zero there. Thus if the wave-function is not square-integrable we should have a continuous spectrum.
on December 15, 2008 at 9:08 pm Luboš Motl

Giotis, it is really cute that after several people essentially reveal the complete solution with very detailed, pedagogical explanations, and remove pretty much all the mystery from this problem, you still write your own “solution” that completely neglects the uncertainty principle, quantum mechanics, and rational thinking in general.

It’s an operational proof that a correct solution, a pearl from the well-known proverb, is useless if it is served to consumers who are not ready to “get it”.

I suspect that the official outcome will be, much like in the case of the locus of the AdS/CFT, that all opinions about the equation are equally good. That may also include Just Learning who thinks that x,y in quantum mechanics are unitary matrices. At least Just Learning is sure (correctly) that his comments are horribly wrong. It’s good to be sure at least about one correct thing!
on December 18, 2008 at 2:18 am Just Learning

Let’s do this again

$\frac{\partial^2}{\partial x^2}\phi + \frac{\partial^2}{\partial y^2}\phi - 2x^2y^2\phi = -2E\phi$

$\phi = f(x)g(y)$

and

$x^2y^2 = h(x)h(y)$

after some rearranging

$\frac{1}{f(x)}\frac{\partial^2 f(x)}{\partial x^2} + \frac{1}{g(y)}\frac{\partial^2 g(y)}{\partial y^2} - 2h(x)h(y) = -2E$

also if we set the following conditions

$h(x)^2 = \frac{1}{f(x)}\frac{\partial^2 f(x)}{\partial x^2}$

$h(y)^2 = \frac{1}{g(y)}\frac{\partial^2 g(y)}{\partial y^2}$

then we see that

$h(x)^2 + h(y)^2 - 2h(x)h(y) = -2E$

some more rearranging reveals

$\sqrt{\frac{1}{f(x)}\frac{\partial^2 f(x)}{\partial x^2}} - \sqrt{\frac{1}{g(y)}\frac{\partial^2 g(y)}{\partial y^2}} = \sqrt{-2E}$

and we find that

$-\frac{1}{2}\frac{1}{f(x)}\frac{\partial^2 f(x)}{\partial x^2} = \left(\sqrt{E} \right)^2_x$

$-\frac{1}{2}\frac{1}{g(y)}\frac{\partial^2 g(y)}{\partial y^2} = \left(\sqrt{E} \right)^2_y$

which have well known general solutions, and

$-\frac{1}{2}\frac{1}{f(x)}\frac{\partial^2 f(x)}{\partial x^2} = \left(\sqrt{E} \right)^2_x = \frac{k^2_x}{2}$

$-\frac{1}{2}\frac{1}{g(y)}\frac{\partial^2 g(y)}{\partial y^2} = \left(\sqrt{E} \right)^2_y = \frac{k^2_y}{2}$

and

$h(x)^2 = -k^2_x$

$h(y)^2 = -k^2_y$

$h(x) = ik_x$

$h(y) = ik_y$

$h(x)^2 + h(y)^2 - 2h(x)h(y) = -2E$

$-k^2_x - k^2_y + 2k_xk_y = -2E$

Now the question is one of solving for the values of k in the x and y coordinates. I’ll work on that one.
on December 18, 2008 at 9:43 am Luboš Motl

Dear Just Learning, mathematics is cool but it is sometimes better to be a bit lazy and think what the steps mean and whether they are legitimate and helpful.

You can’t assume the wave function phi to be in the product form, f(x)g(y), because the problem doesn’t separate into the x-problem and y-problem.

Only if the Hamiltonian were a sum (not product!) of two Hamiltonians, H(x) + H(y), and if it were true even for the potential, i.e. V = V(x) + V(y), then you could split the wave function into a product.

Here, this is effectively true for large x and small y or vice versa, but only locally and all the functions (including the ground state) depend on both x,y, anyway, even for large x,y. You need to use something like the Born-Oppenheimer approximation which still separates the problem to two x,y problems but asymmetrically.

On the other hand, your equation x^2 y^2 = h(x) h(y) has a pretty simple solution: h(x)=x^2 and h(y)=y^2. But that won’t help you much to solve the problem.
on December 27, 2008 at 12:23 am Moshe

Oh yeah, the solution. Well, all the ingredients are already in this comment thread. Robert provided the canonical reference, Giotis the reason we may suspect the spectrum is continuous, and Lubos the reasons why it is in fact discrete.
on January 21, 2009 at 5:32 pm Lionel

Since I don’t know if anyone checks these old threads, I won’t write too much here. I read [1] and one argument is that H_1 >= H_2 where

$H_2 = \frac{1}{2}\left( -\nabla^2 + |x| + |y| \right)$

and since H_2 has discrete spectrum, so does H_1. I guess I must be dense, but if we were just comparing potentials, I would agree. Not so sure about comparing kinetic terms.

I suppose if the volume where E <= H_2 is finite for every E, then the volume for which E <= H_1 must be smaller or equal. But this is exactly the sort of criteria that the paper criticizes, so I was hoping for some more insight by talking about operators on Hilbert spaces, etc. Maybe I can make more sense out of the $latex Tr(e^{-t H}) condition.

[1] Simon, B. (1983). "Some quantum operators with discrete spectrum but classically continuous spectrum". Annals of Physics 146: 209-220. DOI:10.1016/0003-4916(83)90057-x.

Now I will cross my fingers and hope that all my LaTeX gets rendered properly. My suspicion is that it won’t.
on January 21, 2009 at 5:36 pm Lionel

Whowoulldathunk. Clearly PEBKAC.
on January 21, 2009 at 5:53 pm dberenstein

Dear Lionel:

If the volume of phase space that has energy E<E_0 for all choices of E_0 is finite, then the specrtum is discrete. This is a semiclassical estimate of the following type: each quantum state occupies a volume of area roughly given by the Planck constant to the appropriate power. If the volume is finite, then there can only be finitely many states with E<E_0 and we have an estimate of how many there are.

If there is a value E_0 for which the region E<E_0 has infinite volume, then the spectrum is almost certainly continuous and it should have at least one accumulation point. However, this depends very much on the details.