# Bit From Trit, and Lubos’ Booboo

Black holes and the standard model is a bit out of whack, as far as symmetry goes. I’ll discuss the issues, and then discuss how I think the problem is solved.

Quasinormal modes of vibration of black holes.
One of the curiousities of attempts to unify general relativity and the standard model is the quasinormal modes of vibration of black holes. This requires a little explanation.

It’s long been known that black holes exponentially approach a condition where the only numbers that characterize them is their mass, their spin, and their electric charge. This is sometimes called the “black holes have no hair” theorem, also known as Price’s Theorem, but named the “no hair theorem” by John Wheeler.

Suppose we begin with a black hole that has just a little hair, that is, we begin with a slightly perturbed black hole, one that is not quite symmetric. Over time, this hair will disappear. In the process of disappearing, the perturbation will change. As the perturbation dies out, it is possible that the perturbation will act as a sine wave multiplied by an exponential decay: $\exp(-at) \;\sin(bt)$, where a and b are constants with units of 1/time. If so, the constant b defines a characteristic frequency of this particular perturbation. And this is a fundamental characteristic of the black hole.

Of course there are a lot of ways that a black hole could be non symmetric. In analyzing the exponential decay towards perfect symmetry, we need to organize the perturbations according to symmetry principles. In so doing, we choose a orthonormal set of perturbations. This provides the “normal” in the word “quasinormal”. For example, for a very simple object, a complete set of quasinormal modes of vibration could be:
$\exp(-at) \;\sin(bt)$
$\exp(-a't) \;\cos(bt)$
These are orthonormal because the cosine and sine are orthonormal. That is, the orthonormality is to be considered while ignoring the negative exponential part.

For the black hole, the quasinormal mode of vibration problem is purely a problem in classical general relativity, no quantum mechanics is involved. But since the quasinormal mode decays exponentially to the no hair situation, one eventually finds that quantum mechanics must apply, and the perturbation of the black hole must be radiated away as a particle. Therefore, quasinormal modes are important to quantum gravity. The emitted particle is part of the Hawking radiation of the black hole, the part which is non symmetric in a way. But we would expect that the rest of the Hawking radiation would follow the same laws as the quasinormal mode particles. And these efforts align with Bohr’s Correspondence Principle: classical mechanics should be obtained for very large quantum numbers. In this case, quantum transition frequencies should be equal to classical oscillation frequencies, so the frequencies of the quasinormal modes can be thought of as transition frequencies.

Quantum mechanically, one can think of the quasinormal modes as “unbound states” in that they are not quite bound to the black hole and eventually head off to infinity, they are radiated. There is a similar but oppositely oriented calculation for the bound states of a black hole. These eventually head off to the singularity at the center, that is, they head off to zero. I’ve linked to the elegant article by Anthony Lasenby, Chris Doran, Jonathan Pritchard, Alejandro Caceres, and Sam Dolan of the Cambridge Geometry group, which uses Clifford (Geometric) Algebra to calculate the bound states. I’d love to redo those calculations for the quasinormal modes; I’d have to suspect it is easier. In either case, bound or unbound solutions, one solves something like Schroedinger’s equation. The energy turns out to be complex, with the imaginary part giving the exponential decay.

In solving a Schoredinger’s equation for a particle in quasinormal modes around a black hole we seem to be ignoring the fact that there is no known way to unite general relativity and quantum mechanics. But actually, we are not doing anything so radical. We’re assuming the perturbations are small. So, we’re ignoring the gravitational effect that the small particle has on the big black hole; we are not actually unifying GR and QM in this calculation. And our black hole is not an infinitesimal one, we can suppose that it is large enough that quantum effects do not dominate it (other than the radiation we’re computing here).

Black Hole Entropy and Quantum Statistics
One of the consequences of Hawking radiation is that black holes have a temperature, and this temperature turns out to be related to their area. Temperature is a concept from classical thermodynamics. One can use quantum mechanics to provide a theoretical foundation for temperature in terms of statistical mechanics. This is the Bose-Einstein and Fermi-Dirac statistics of quantum mechanics, and the classical Maxwell-Boltzmann statistics.

In physics, “statistics” tell how states are to be counted up, and whether or not different states can exist with identical quantum numbers. If the quantum numbers can be shared, these are Bose-Einstein statistics, if they cannot be shared, that is, if the Pauli exclusion principle applies, the statistics are Fermi-Dirac. A triumph of modern quantum mechanics is the Spin Statistics Theorem, which says that particles with spin that is integer must follow Bose-Einstein statistics, and particles whose spin is half an odd integer must follow Fermi-Dirac statistics.

Lubos’ Booboo

In 2002, Lubos Motl solved the Schroedinger’s problem for all spin cases, and computed the frequencies $\nu$. Then he did a very natural thing, for which he may have some regrets.

A black hole is characterized by a temperature. In quantum mechanics, a frequency corresponds to an energy. And if you have an energy and a temperature, you can use the theory of quantum statistics to compute an “occupation number”, that is, the probability that a particle is found in that energy state. For this calculation, Lubos interpreted this as an interaction between the particle represented by the quasinormal mode frequency, and a background of Hawking radiation.

And Lubos’ result for spin 1/2 and spin 1 matched the expected quantum statistics. That is, spin-1/2 and spin-1 perturbations of the black hole defined frequencies that oscillated at a rate that was natural for quantum statistics and the temperature of the blackhole. But spin 0 and spin 2 were unexpectedly wrong. First of all, instead of having the expected Bose-Einstein statistics they instead had statistics implied by the Pauli exclusion principle, that is, Fermi-Dirac statistics, and second of all, an extra ln(3) suggested that the states needed to be counted by 3s. Instead of these spin-0 and spin-2 particles filling one state, it filled three. In Lubos’ words:

Why do we fail to obtain the same Bose-Einstein factor as we did for odd j? Instead, we calculated a result more similar to the half-integer case, i.e. Fermi-Dirac statistics with the number 3 replacing the usual number 1; let us call it Tripled Pauli statistics. Such an occupation number (51) can be derived for objects that satisfy the Pauli’s principle, but if such an object does appear (only one of them can be present in a given state), it can appear in three different forms. Does it mean that scalar quanta and gravitons near the black hole become (or interact with) J = 1 links (triplets) in a spin
network that happen to follow the Pauli’s principle?

In Lubos’ calculations, it really isn’t a spin-0, spin-1/2, spin-1, or spin-2 particle that is being modeled in classical GR. So it’s easy to ignore Lubos’ “tripled Pauli statistics” and this has universally been done, except perhaps by me and a few compatriots. For example, in their excellent paper approved of by Lubos himself, Jose Natario and Ricardo Schiappa write:

The perturbations come in three types: tensor type perturbations, vector type perturbations and scalar type perturbations. It should be noted that this nomenclature refers to the tensorial behavior on the sphere, Sd−2, of each Einstein–Maxwell gauge invariant type of perturbation, and is not related to perturbations associated to external particles. For instance, one should not confuse vector type perturbations with perturbations associated to the propagation of a spin–1 vector particle in the background spacetime, or scalar type perturbations with perturbations associated to the propagation of a spin–0 scalar particle.

And they do not bother with the spin-1/2 case at all.

Complete Sets of Mutually Unbiased Bases of the Pauli Algebra

And yet Lubos’ analysis works perfectly for spin-1/2 and spin-1 calculations. So instead let’s assume that Lubos’ booboo is a clue on how gravity should be combined with quantum mechanics. We need to have particles that have Fermi-Dirac statistics, but where each state is filled three different ways.

The fundamental Fermi-Dirac particles we know about are spin-1/2 and their representation is with the Pauli algebra, that is, Pauli’s spin matrices. If we wish to use this same algebra to represent “tripled Pauli statistics,” we need to take each state and split it into three states.

The clue on how to do this is the concept of “mutually unbiased bases” of a Hilbert space. A basis for a Hilbert space is a complete set of orthonormal states. For the Pauli algebra, the dimensionality of the Hilbert space is 2, so a complete set of orthonormal states has two elements. These would be the eigenvectors for spin in the $u = (u_x,u_y,u_z)$ direction, that is, spin +u and spin -u. The usual choice of basis for the Pauli algebra is spin +z and spin -z.

“Mutually unbiased” says given any two basis sets for the Hilbert space, the transition probabilities are all equal. Another way of saying this is that there is no shared knowledge of the state between two mutually unbiased bases. “Complete” means that you can’t add any more bases to your set of mutually unbiased bases because the new additions would inevitably have shared knowledge (i.e. not equal probabilities) when compared with at least one of the bases already included.

For the Pauli algebra, the largest number of mutually unbiased bases one can name is 3. The three bases must be spin in three geometrically orthogonal directions. So the usual example of a complete set of MUB for the Pauli algebra are the three bases: {spin +x, spin -x}, {spin +y, spin -y}, and {spin +z, spin -z}. That these bases are mutually unbiased can be verified by noting that the transition probabilities between any state taken from any one of them, and a state taken from either of the other two bases, will always be one half. That is, the amplitude between two states will be

(times an arbitrary complex phase of course) whenever u is an eigenvector of spin taken from one of the bases, (example spin +x), and v is an eigenvector of spin taken from one of the other baes (example spin -y). Since the transition probability is the squared magnitude of the amplitude, the transition probability will be 1/2. Note that the transition amplitudes of a state with itself is 1, and the transition amplitude between two different basis states is 0.

Looked at from the point of view of a complete set of MUB for the Pauli algebra, Lubos’ inane comment about “tripled Pauli statistics” suggests we should use the Pauli algebra MUBs to model elementary particles. Since this is not suitable for any known particles, we must conclude that these MUB spin-1/2 particles are preons.

Since one of these “MUB particles” must fill three different states, it is natural to associate it with the 3-d version of Feynman’s checkerboard model of Dirac particles. Feynman’s original notes were for the case of 1+1 dimension. The generalization to 3+1 dimensions is obvious, but in violation of Lorentz symmetry and so is annoying to old school physicists.

In the 3-d Feynman checkerboard model, the electron is splt into a left handed and a right handed portions which interact with each other through a coupling that defines the mass. Each of these two half particles obeys tripled Pauli statistics. Suppose spin is oriented in the (1,1,1) direction and that the electron is stationary. Then the right half of the electron fills the +x, +y, and +z basis states of the usual Pauli MUB, while the left half of the electron fills the -x, -y, and -z basis states.

These two sets of states, +x, +y, +z, and -x, -y, -z, are orthogonal in that one is for left handed electrons and the other is for right handed electrons. But the left handed electron fills all three states, though it is only found in one at any given time. Thus we have tripled Pauli statistics.

Transition Probabilities and Coupling Constants

If we are to make calculations with a quantum field theory, we need to define coupling constants for the interactions between the particles. The usual method is to assume a symmetry, and when this doesn’t work so good, assume that the symmetry is broken.

A natural way of choosing coupling constants is to use the transition amplitudes. This is not possible when all particles are taken from states that are orthonormal because then the transition amplitudes are all either zero or 1. But when the particles are taken from an MUB, a non trivial set of transition amplitudes are automatically defined.

Given that the transition probabilities are all equal, how can I say that the transition amplitudes are “non trivial?” Well the transition amplitudes include complex phases that do not appear in the transition probabilities and these complex phases are complicated enough to give a rich structure to the transition amplitudes among the basis elements of a complete set of MUBs.

The problem with discussing these complex phases is that they depend on the choice of arbitrary complex phase in the states. The transition amplitude is not well defined until we choose a particular phase convention for the spin states u and v.

Density matrices, the Vacuum and Phases

As a proponent of density matrix / density operator theory, my favorite solution to this problem is to use (pure) density matrices to define a phase convention. The primary advantage of density matrices over spinors is that density matrices have no arbitrary complex phase. In fact, anything you write down using density matrices is a quantum observable, and this includes complex quantities such as quantum phase or Berry-Pancharatnam phase.

We convert all our spinor states into pure density matrices by mulitiplication by the Hermitian conjugate, i.e.

Next, we select a state “0” which can be anything at all, provided that it is not orthogonal (i.e. does not annihilate) to any state in the complete set of MUBs. This state “0” is called the vacuum due to a paper on the “measurement algebra” by Schwinger, which uses similar notation.

Now we do all calculations with respect to complex multiples of the vacuum. First, since none of the states annihilates the vacuum, all matrix elements of basis elements with the vacuum are non zero. To get the density matrix equivalent of spinor transition amplitudes, we will need to use the phases of these things, but avoid their magnitudes. To do this, we will divide by the magnitudes, so abbreviate:

Now all of our complex matrix elements will be transformed into complex multiples of the vacuum density matrix. Since the vacuum is normalized, these complex multiples of the vacuum will add, subtract and multiply just like complex numbers. The function that takes us from complex multiples of the vacuum density matrix to the complex numbers is the trace. And the reverse function is just multiplication of the vacuum state by the complex number. So we could do all this using the trace function, but I think it’s more elegant written purely in density matrices.

The matrix element will be replaced by the following, which, in fact, is a multiple of the vacuum state:

Since we’re dividing by ( |u| |v| ), the above product will have the right magnitude for the transition amplitude. The factor

and similar for v, are complex phases and so are just a convenient phase choice for the bra or ket. Note that the phase choice is compatible with the complex conjugacy relation between bras and kets, and so it is consistent. Thus, we have written the spinor amplitude entirely with density matrices, and the choice of vacuum amounts to a particular phase choice for the states in the MUBs.

Conclusion

That’s probably enough for this blog post. The story of “bit from trit” continues in the details of the Koide mass formulas discussed elsewhere. To make a long story short, the phases in the transition amplitudes for the Pauli MUBs is the source of the angle $\pi/12$ that appears in the neutrino mass formula.

Finally, I should point out that among the quasinormal modes of vibration, the most bizarre are those called “asymptotic”. Taking dimension d=5 is gives the only dimension where these asymptotic quasinormal modes of vibration are avoided. Since these are rather bizarre solutions, this can be interpreted as evidence for d=5, which happens to agree with the evidence from density matrix theory and Geometric (Clifford) algebra.

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### 5 responses to “Bit From Trit, and Lubos’ Booboo”

1. Grammarian

“Black holes and the standard model is a bit out of whack, as far as symmetry goes.”

It should read “Black holes and the standard model are a bit out of whack…” since you are talking about two things…like “Jack and Jill are crazy, they’re getting water from a hill!”

2. Dear Carl, thanks for your comments. It is nice to be intrigued by the trit speculations, as you call them, and I could tell you some stories that are somewhat more detailed. It’s still fair to say that this idea hasn’t led to any satisfactory semi-complete picture, so it’s status is just somewhat better than the status of various other discrete theories of gravity.

Also, you overestimate the reliability of my calculation for general spins. While the formulae naturally depend on “j”, I haven’t really solved the Dirac equation on the Schwarzschild background or other things. There are other papers that solve it explicitly, and it’s disappointing if Ricardo et al. don’t do it. But from some viewpoints, their paper is still the most complete one.

3. Pingback: Quantum Cloning « Mass

4. carlbrannen

I guess I should add that the TV character said to be modeled after Luboš Motl, Sheldon Cooper, now has a movement suggesting he get an emmy. Congratulations, I guess.