Is the fact that weak eigenstates are not mass eigenstates completely arbitrary? Or is there a deeper reason for the existence of the PMNS and CKM matrices?
Answer
Yes, there is a very good reason why leptons and quarks mix. It would be shocking if they didn't mix. Just to avoid confusions, we're not saying that leptons and quarks mix with each other: they don't because leptons are eigenstates of color or baryon charge with different eigenvalues than quarks which are also eigenstates: this difference prevents mixing between leptons and quarks because the mass terms would violate the charge conservation (for color and/or baryon charge - the former would be an internal inconsistency; the latter would be consistent internally but it would imply that the proton would decay rapidly).
However, leptons mix with leptons and quarks mix with quarks.
The theory - whether you mean quantum field theory, string theory, or anything else that may be found in the future - produces several generations of quarks. Let's talk about upper quarks, $u_1, u_2, u_3$ - normally called $u,c,t$. For each of them, the quantum field theory produces a 4-component Dirac spinor. It is actually more correct to decompose it into 2-component spinors, but let us use the 4-component formalism here.
If those three Dirac spinors have some kinetic terms, they may always be diagonalized so that the kinetic terms have the form i$$\sum_i \overline{u_i}\partial^\mu \gamma_\mu u_i$$ There could be some general matrix $I_{ij}$ multiplying $\overline{u_i}$ and $u_j$ but by a proper transformation of the $u$ fields, it can be set to the identity.
However, once you bring the kinetic terms into this form - by choosing the right combinations - there is not much freedom left. Generally, the underlying quantum field theory or string theory or whatever will also produce arbitrarily bilinear interactions of the form $$-\sum_{i,j} M_{ij}\overline{u_i} u_j$$ The matrix $M_{ij}$ may be chosen Hermitean because the antihermitean part would cancel, anyway, as the product of two $u$'s is Hermitian - it gets complex-conjugated under the exchange of $i,j$. But aside from the Hermiticity, $M_{ij}$ is a completely unknown and arbitrary matrix. There is no reason for it to be diagonal; even if it is a generic Hermitian matrix with off-diagonal terms, it satisfies all symmetries and consistency conditions we normally require. In particular, it conserves the electroweak SU(2) isospin, U(1) hypercharge, SU(3) color, the angular momentum (only spin in this case), and the rest of the Lorentz symmetry (and also the baryon number). It's not surprising because from the viewpoint of the values of all these charges, the three generations of the quarks are indistinguishable, so coupling $\overline{u_1}$ with $u_2$ is as good as coupling it with $u_1$.
Now, the kinetic term is still invariant under SU(3) rotations of the three quark flavors. You may use this surviving symmetry to diagonalize the matrix $M_{ij}$. In the right basis, it is diagonal. However, you won't be able to diagonalize the lower-type quark mass matrix at the same moment.
A similar mass matrix exists for the down-type quarks. However, the Standard Model has an SU(2) electroweak isospin symmetry which is very important, at any energy scale. For a chosen triplet of fields $u_1,u_2,u_3$ of the up-type quarks for which the mass matrix was diagonal, we have their SU(2) partners - obtained by applying an SU(2) generator - $d_1,d_2,d_3$. However, the mass matrix for these three down-type quarks, which is $M^d_{ij}$, another Hermitian matrix, can be and probably will be non-diagonal once again. Right now, we don't have any symmetry left.
So it means that the mass of the three down-type quarks is expressed by a $3\times 3$ Hermitian matrix that is not necessarily diagonal, and there is no useful residual symmetry left that would allow us to diagonalize the matrix. So we must accept it as a fact: the three mass eigenstates among the down-type quarks will be different from the SU(2) partners of the three mass eigenstates among the three up-type quarks. You can't do anything about it. The CKM matrix is what relates these two bases.
The case of the leptons is analogous if one includes the masses for the neutrinos.
It would be shocking to find the matrix to be diagonal in the same basis because there is no real "need" - such as symmetry or renormalizability or anomaly cancellation - that would dictate that the off-diagonal elements of the mass matrix have to vanish. Because it doesn't have to happen, it probably won't happen - this principle is known as Gell-Mann's totalitarian (or anarchic) principle. All terms that are allowed by the symmetries and consistency will occur with nonzero coefficients - most likely with coefficients that are "natural" i.e. comparable to one (or the typical mass scale). In particular, this is also true for the off-diagonal elements of the mass matrices relating several generations of particles that are otherwise equivalent when it comes to their conserved charges. So the mixing generally occurs.
Nature seems to satisfy the anarchic principle, at least with some nonzero coefficients - not necessarily comparable to one. So it follows that it produces two pretty much random Hermitian matrices for the up-type quarks and down-type quarks, and the bases in which these two matrices can be diagonalized won't have the property that one basis may be obtained by SU(2)-rotating the other one. That's why the CKM matrix is not equal to the identity and one needs to discuss the mixing. Similarly for the PMNS matrix for the leptons. "Guessing" that the matrices should be diagonal is as unnatural as guessing that the wealth of Bill Gates is exactly 56,789,012,345 dollars. There's no reason why a "nice" guess like that should be true, so it will probably not be true. The same thing holds for the guess that the off-diagonal entries of the mass matrices vanish.
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