The known elementary particles range in mass from about 0.0004eV for the lightest neutrino to around 170 billion eV = 1.7 x 10^11 for the heaviest quark, the top. This is a ratio of about 400,000,000,000,000 to 1.
On the other hand, the energy scale available from Einstein’s theory of gravitation (which relates mass to energy) suggests that the natural mass for a typical particle should be the Planck mass, about 2.43 × 10^27 eV, 11 orders of magnitude larger than even the top quark and 25 orders of magnitude larger than the lightest neutrino mass.
From the point of view of the Planck energy, all particles known to man have mass very close to zero. Let’s write the Hamiltonian for the system as a first order Hamiltonian , in which the energies (and therefore the masses) of all our usual particles are zero, plus a perturbation , which will provide a correction to the zero energies. For the full Hamiltonian we have:
where is a small number.
My favorite text for this sort of thing is my first quantum textbook, Quantum States of Atoms, Molecules and Solids by Morrison, Estle, and Lane. This is a somewhat too advanced book to use as a reasonable introduction to quantum mechanics. I don’t know what Dr. Petschek was thinking about when he chose it and, with the passage of time, I guess I’ll never know. What’s worse, I took the class my first year in college. Nevertheless, it has an excellent collection of calculation techniques.
Let be the exact solutions to the energy eigenvalue problem of the unperturbed Hamiltonian, that is, when lambda is zero. Let be the corresponding energies. By our assumption, this is zero. According to (4.21), the 1st order correction to the energies is:
By assumption, is zero. The next term amounts to an energy correction that acts to change these energies. But this isn’t what we want. If we wanted to just dial in an energy, we’d have put this in at the start with the term. So we look at the 2nd order correction to the energies.
Leaving off the 0th and 1st order terms which we are supposing are zero, and leaving off the lambd factor, the 2nd order perturbed energy is (MEL eqn 4.33 ) :
[Oh great! WordPress images are broken!] In the above, the sum over k is taken over states that are different from n. I’ve left in the in the denominator even though it’s zero because almost everybody is used to seeing it there. And there is a problem with this definition because we’re assuming the energies are degenerate (i.e. all the known elementary particles have zero energy). We will discuss the resolution to the degeneracy problem in a moment. For now, let’s just look at the form of the above. The in the denominator is of the order of the Planck energy. The numerator has units of energy squared, so overall, the term has units of energy. And the perturbation lambda appears as a square.
Koide Mass Formulas
The matrix element appearing in the numerator is squared. Suppose we assume that is a simple object that satisfies some symmetry law for example. Since the matrix elements are squared, such a symmetry may not be obviously apparent in the energies (masses) of the elementary particles but instead in the square roots of their masses. If the masses of the elementary particles are each determined by just one of the matrix elements, then we should look for symmetry in the square roots of the masses of the elementary particles. If we think of the matrix elements as being the components of a vector, then the mass appears as the squared length of the vector, as noted by Michael Rios.
As it turns out, there is a mass formula for the charged leptons that uses square roots. It’s the Koide mass formula. In units of MeV, the charged lepton masses are given by . As Kea recently noted, I’ve found several similar formulas for the masses of the mesons and baryons. I’m busily writing up a paper on this subject at a deeper level than this note. Now to return to the problem of the degenerate masses.
For the case at hand, the energies are degenerate so we have to go to degenerate perturbation theory. When energies are degenerate, we have a choice in how to split the space of wave functions. Different splits amount to different linear superpositions. MEL discusses this in section 4.6 by using the example of two degenerate energies.
I’m tired of typing LaTeX, especially since the tools I’m using can’t show “|” or even “]”. And the wordpress preprocessor mangles angle brackets… Call our two states 1 and 2. Let H’ be the correction to the Hamiltonian. Since H doesn’t distinguish between the eigenvalues, we need to find the eigenvectors of H’. Accordingly, find the matrix elements, (1|H’|1), (1|H’|2), (2|H’|1), and (2|H’|2) and write these as a 2×2 matrix. The eigenvectors we want are the eigenvectors of this 2×2 matrix.
The final result is given as equation ( 4.88 ) in MEL. Unlike the non degenerate case, the calculations have no division by an energy and are not squares of matrix elements. The result is that the energies are given by:
For a system that has some energies degenerate and other energies non degenerate (such as would arise in a system consisting of the known particles with degenerate (zero) energies plus Planck particles with non degenerate energies), one adds, in addition to the above contribution from the degenerate particles, another contribution from the non degenerate particles in the same form as the squared matrix elements described earlier.
The approach of the standard model has been that mass should arise from interactions between the known particles (including the Higgs). For this sort of theory, formulas that are linear in energy, such as the above, are natural to use, rather than the Koide formula, which is linear in the square root of energy.
As far as explaining the elementary particle masses, the immediate advantage in replacing the energies with their square roots is that the hierarchy problem is reduced to only half as large. The tau has a mass 3500 times the mass of the electron but Koide’s formula brings them into the same linear form as the square roots differ by the more manageable ratio of about 60.
The difference in masses between the neutrinos and the charged leptons is more of a problem at around 500,000,000 to 1. I think that the difference in scale is due to the power of a coupling constant.