119. Fun with energy gap for QCD Born-Oppenheimer Hamiltonian
Save This Post as PDFLet us try to solve the Exercise 2 in this post about the Wilson loop. Since we, scientists, are all lazy (as you are!), that’s what I would like to do: I will give three different solutions of the exercise with three different answers 
Then, if you are interested to learn the subject (and you probably are, becase the subject is quark confinement, right?), you can choose a solution you like most (I will even introduce a poll 
) and kindly explain in comments why do you think that two other solutions are wrong The goal of the experiment is to show how applicable is the democracy approach (as advocated by Lee Smolin ) to science.
Let me remind you the formulation of the problem. Calculating the expectation value of the Wilson loop, we have found it to be given by
,
where 
is the matrix element
of the Born-Oppenheimer Hamiltonian for the bound system of two heavy quarks. I would like to estimate the gap in the energy spectrum of this Hamiltonian.
Solution 1. Non-relativistic BO Hamiltonian. Let us write the Born-Oppenheimer Hamiltonian for the quarks as
, (1)
where 
is the mass of the quark on one of the ends of the Wilson line.
As we agreed, the notion of interquark potential is only applicable when quarks are non-relativistic (i.e., extremely heavy). We then expand the kinetic term in powers of 
and get in the leading approximation
,
where the 
term is omitted. To find the gap in the spectrum, we can solve the corresponding Schroedinger equation with the boundary condition 
. The expression for the gap in the spectrum of the Born-Oppenheimer Hamiltonian is then given by
. (2)
As you see, the gap is much smaller than 
and weakly depends on the quark mass.
Solution 2. Relativistic BO Hamiltonian. Although using the notion of interquark potential is not a very bright idea, if quarks are relativistic, let me still write down the relativistic Hamiltonian (1) and try to solve it explicitly.
This gives
(3)
where 
is radial quantum number, 
is the angular momentum (I took the WKB conditions to simplify the final answer (3)). The result (3) is known as the Regge trajectory for hadrons. (Study of Regge trajectories was quite a business back in 1970s, but it is no longer…)
As we see from (3), the gap is of the order
, (4)
and it does not depend on . I should actually tell you that the Regge trajectory is what is seen in Nature, in real QCD
Solution 3. Rotating non-relativistic string.
As we have found in this post, the maximal length of QCD string is
in both heavy and light quark limits. Considering a flux tube between two heavy non-relativistic quarks, we find that excited states of the corresponding Born-Oppenheimer Hamiltonian correspond to excited (oscillated) or/and rotated flux tube. The gap in the spectrum of the Hamiltonian can be therefore estimated as
. (5)
It does not depend on the mass of the heavy quark.
So, what do you think is the correct answer for the value of the mass gap in the Born-Oppenheimer Hamiltonian – (2), (4) or (5)? Let me again remind you what is the setup: two heavy, non-relativistic, quarks are connected by the flux tube, the Wilson loop expectation value is given by the area law and the interquark potential grows linearly with distance between the quarks.
What is the correct estimation for the mass gap in spectrum of the BO Hamiltonian for two heavy quarks?
- Given by the Solution 2 (Regge trajectory) (52%, 14 Votes)
- Neither (19%, 5 Votes)
- Given by the Solution 3 (rotating/oscillating flux tube of the maximal length) (15%, 4 Votes)
- All of them, in different regimes (11%, 3 Votes)
- Given by the Solution 1 (non-relativistic quarks + quantum mechanics) (3%, 1 Votes)
Total Voters: 27
Loading ...P.S. The poll will be closed on Dec 6.
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hi Dima
1) How did you get (2)?
2) I think that all solutions are correct, they are just applicable in different regimes.
Dear Instanton
1) Eigenfunctions of the Hamiltonian are the Airy functions. Knowing zeros of the Airy function, you can easily estimate the gap between energy levels in the Hamiltonian.
2) Care to explain why?
Cheers,
Dmitry.
Dear Dmitry,
a very nice homework. Truth to be told, I am not sure about the answer. But it is true that Instanton’s answer, currently shared by 14% of the readers, is the only acceptable answer in democracy where every solution can always be correct. Saying that one solution is more correct than others is a case of discrimination.
Also, I didn’t quite understand whether the homework was physical or mathematical. Are you asking which of them is more correct in the real world ie. the full QCD? Otherwise, mathematically you seem to be solving three different problems, pretty well?
But when you say that the quarks are heavy and nonrelativistic, doesn’t it eliminate the relativistic solution 2? 
Also, if you define the quarks to be the heavy degrees of freedom in BO and you define the flux tube to be the light degrees of freedom, doesn’t it mean that you say that the lightest energy gaps must come from the string’s vibration? That would make solution 3 sound pretty good except that isn’t the flux tube always a relativistic string?
For instanton: the scaling in (2) with the cubed root can be seen by dimensional analysis, without knowing Airy functions. You solve a quantum Hamiltonian, H = p^2 / M + sigma X. Now, to find the scalings, choose a new variable y=Qx and its dual p_y=x/Q (scaled in the opposite way) in such a way that H = K (py^2 + y). By mapping these two things with two coefficients, you have K/Q^2 = 1/M, KQ = sigma. Dividing the latter equation by the former, you have Q^3 = sigma M, Q=(sigma M)^(1/3). And K = (sigma^2 / M)^{1/3}. This K is Dmitry’s gap of H because the remaining factor, py^2 + y, has spacing of order one by dimensional analysis: there are no parameters here.
Best wishes
Lubos
Dear Lubos
You are right to the point
although I would prefer if readers figure it out for themselves… The homework is mathematical rather than physical, because what we consider is the flux tube between two almost infinitely heavy quarks. This indeed rules out (2) from my point of view (otherwise, we would actually be unable to speak about interquark potential).
I don’t know, is it so for any length? The tension is \Lambda^2, that’s more or less the mass of the unit length of the tube. The excitation energy is of the order
1/\sigma/L^3 << \Lambda.
Cheers
Dear Dmitry,
you don’t need to be afraid of the fate of democracy in physics. Right now, the solution 2 has 14 votes i.e. 61 percent so it is very close to be the clear winner. My preferred stringy answer 3 only has my vote.
So in this discussion, you shouldn’t be afraid to eliminate answer 2 because it will probably win, anyway. So much for democracy in physics!
The reason why the “relativistic” answers to non-relativistic problems are often wrong was probably known to Schr?dinger when he abandoned the Klein-Gordon equation he discovered, after it gave him a wrong Hydrogen spectrum. To be sure, the correct full relativistic theory must give the right non-relativistic limit but it seems that this BO truncated theory wasn’t quite the right one, so it probably disagrees with the non-relativistic answer that is closer to the truth even in the non-relativistic limit. Still, it’s hard to see what’s exactly wrong with it at the same level of accuracy. It might be just your looking at different types of modes, see below.
Concerning the non-relativistic nature of the fluxtube, it would be interesting if you wrote what you exactly mean by its being non-relativistic. As you say, the tension is close to the mass per unit length (times squared speed of light), both at the QCD scale: isn’t it really exact? Note that the piano string, a good non-relativistic string, has the mass per unit length times c squared vastly exceeding the tension, doesn’t it? Most of the latent mass is “useless” in the pianos, I thought.
By a relativistic string in the narrowest sense, I clearly mean a string described by a Nambu-Goto-like action. Does it imply that all of its classical motion has relativistic typical speeds? Well, it probably does. But shouldn’t it be true for the QCD flux tube, too? Or does it break down for strings stretched between distant branes? It probably does. Isn’t it the fundamental string of a dual holographic string theory which is relativistic? Isn’t there still an exact 2D Lorentz symmetry inside the 2D flux tube world sheet?
For open strings stretched between two parallel distant branes (and D7-branes are usually the stringy pictures for heavy quarks), the latent mass would be proportional to the distance. But the excitations would carry stringy alpha-prime energies independent of the separation, similar to our unloved answer 2.
So if you’re going to argue that 2 is wrong, and I feel it is, it would be interesting to see where the relativistic counting for the string goes awry. Of course, there are many other details – for your pedagogical – that confuse meson-outsiders like me. By the gap, you probably mean the minimum spacing among the different types.
In the non-relativistic Airy problem, you have automatically adopted the vibrational (radial) modes, using the molecular jargon, not the vibrational modes. From the molecular perspective, that’s confusing because the rotational modes have lower energies (microwave spectrum) than the vibrational (IR), don’t they?
For the non-relativistic flux tube 3, you seem to abandon the radial fluctuations, that would probably give a huge gap, and you choose the vastly smaller transverse excitations of the non-relativistic string. That’s logical but the confusion remains why you threw away the low-E rotational modes in the Airy problem.
Best wishes
Lubos
Dear Lubos
As you understand, the goal was to show that democracy fails, so I will keep the solution 2 in the poll, since the poll is quite an illustration as it is.
See for example this Merlin-Paton paper. (or Zwiebach’s book, Chapter 4
). Let us discretize the tube and consider elementary pieces of the tube as non-relativistic particles.
Indeed, so it seems that the string is always relativistic, but…
…but still I have an impression that long strings with heavy quarks on their ends should be non-relativistic. Unfortunately, I do not see at this point how to prove it. If I stick to quantum mechanics, the gap is small for large L, and only very high energy states correspond to heavy quarks moving with relativistic velocities.
As usual.
Yes, and that’s, I think, the reason why Instanton asked his question (1). The problem is 3-dimensional, and the solution is for one-dimensional problem, so it’s wrong.
Rotational modes have both smaller energies and much smaller gaps between them. Let us consider a molecule consisting of two similar atoms of the mass M. Then, the energy of vibrational mode is proportional to 1/\sqrt{M} (Bohr-Sommerfeld), while the energy of rotational mode – to 1/M.
Yes, and that’s the solution I would vote for (I am not going to vote till the Dec 5, although I have an impression that it is good time to close the poll after this discussion).
Cheers
hi Dima and Lubos
> Yes, and that?s, I think, the reason why Instanton asked his question (1)
Yes, I did realize that you propose a 1d solution for a 3d problem.