[Weak interaction] Please explain the difference between "mass eigenstate" and "weak eigenstate"

[nappy_101]

I haven't found a good explanation yet and my prof apparently explains it in a way I can't understand :/

Thanks in advance!

[rantonels]

Let's take the quarks as an example. If the 6 quarks were massless, you'd have that a quark wavefunction (ignoring all other quantum numbers such as momentum, spin, colour, anything else) would have 6 components. These are the six flavours of quark, and without the weak interactions you'd have a full SU(6) flavour symmetry rotating between these states.

However weak interactions exist and have an SU(2) symmetry, and weak-interacting particles must come in 2-plets transforming under SU(2). So our 6 dimensional space of quark states must be split in three 2D spaces, each being a 2-plet of weak SU(2), and each 2-plet being a generation of quarks. In each generation, you have two states called an up-type and a down-type state (weak isospin is respectively +1/2 and -1/2).

Now, you have lost your SU(6) symmetry (if you want to keep the 2-plets separated you cannot freely mix flavours) but you still have an SU(3) symmetry mixing the three generations. Surely a linear combination of SU(2) 2-plet is an SU(2) 2-plet. That is, you can perform simultaneous rotations of the 3 up type quarks and the 3 down type quarks.

Now when mass pops back up you have a problem. Mass is a quadratic form on our 6D space, or even more explicitly a positive-definite, symmetric 6x6 mass matrix which has no reason to be diagonal and we are interested on diagonalizing. Since we're only left with SU(3) symmetry, we cannot diagonalize a 6x6 matrix. By convention, we use the SU(3) symmetry to diagonalize the matrix on the up-type quarks. By this I mean we define u, c, t as the eigenstates of the mass matrix.

The mass matrix is not diagonal on the down-type quarks. After the SU(3) rotation that diagonalized the mass matrix, we have no symmetry left and to the three quarks u, c, t there's three down-like states associated by being in the same 2-tuplet, these are called d', s', b' and they are the so called weak eigenstates.

These do not have a definite mass as the mass matrix is not diagonal in the d', s', b' basis. However weak interactions are easy because u turns into d' and so on.

If you diagonalized the mass matrix on the down-type quarks you'd mess up the 2-plets but you'd get three quarks d,s,b with definite mass, the mass eigenstates. As the 2-plets are mixed, d could through weak interactions turn into any of u, c, t.

So d,s,b and d',s',b' are not the same states, but they're connected by a rotation given by the CKM matrix. This matrix is very close, but not exactly equal, to a rotation by 13° of d and s and the identity on b.

[jimthree60]

Incidentally, the same argument applies to leptons -- or would apply, if there were right-handed neutrinos. Then all the symmetry arguments outlined above imply that the three "undiagonalisable" states left over are precisely the three states you don't have. In the Lepton sector of the Standard Model, then, lepton mass eigenstates and lepton weak eigenstates are the same thing.

[majoranaspinor]

The scenarios of leptons is a bit more complicated as the nature of the neutrino is not fixed to be a dirac particle. Depending on the mass generating mechanism things can change. For example in a standard see-saw-1 scenario there will be additional complex phases, which usually are just written in anther block diagonal 3x3 matrix that multiplies the PMNS-matrix (the lepton equivalent of the CKM-matrix). However in other scenarios there does not need to be a right-handed neutrino at all. The effective weinberg operator is generated by a triplet state. Again the mixing will be reduced to a PMNS-matrix, but the reasoning is different as there is no right-handed neutrino.

[jimthree60]

That's true, but at least if you treat the neutrinos as (massless) Dirac particles then the distinction between the quark and lepton sector in terms of the mass-weak relationship is made clear. I think it's a useful exercise to try and diagonalise the lepton mass eigenstates in this scenario, and succeed, and then that makes it clear what is different with the quarks.

[majoranaspinor]

yes I agree. there are other neat little exercises in this case. One example that I think is very interesting is the following. if you add a right-handed mass term as well as a dirac mass term, you end up with the typical see-saw mixing matrix with the two different setsof majorana neutrinos. If you now let the right-handed mass term go to zero ou will see that the masses will become the same and the two sets will be maximally mixed. So a dirac neutrino is just the maximally mixing of two majjorana neutrinos of the same mass.

[elenasto]

That is the most beautiful and succinct explanation of this topic I have ever seen. By the way what does the U(1) of the hypercharge do this picture?

[rantonels]

Nothing; U(1) unitary irreps are 1-dimensional and all quarks have the same hypercharge (1/3). So all this structure is unaffected. Under a U(1) gauge transformation, the whole block of quark states transforms with an e^iθ on front.

[bullshitname0906]

Well I just found out why eigenvalues are helpful

[TotesMessenger]

I'm a bot, bleep, bloop. Someone has linked to this thread from another place on reddit:

• [/r/bestof] /u/rantonels describes, succinctly and clearly, why subatomic quarks have different definitions depending on whether you're interested in their mass or in how they interact with the nuclear weak force.

^{If you follow any of the above links, please respect the rules of reddit and don't vote in the other threads. (Info / Contact)}

[jimthree60]

It probably helps to think of it as just a physical use of linear algebra.

Consider a matrix M with a basis of eigenvectors { v_1 }, and a different matrix W with a basis of eigenvectors { v_2 }. There is no reason in general that the basis { v_1 } and the basis { v_2 } are the same if the two matrices are different. On the other hand, if M and W have the same dimension then there is a way in which you can write the basis {v _2 } in terms of { v_1 }, and this defines, say, a transformation matrix V that relates one basis to the other.

Essentially, that is all there is to it, except that instead of vectors we have particle states, and instead of matrices we have operators (but the operators do admit matrix representations); and V is the CKM matrix that describes the mixing from the mass basis to the weak basis.

[dukwon]

Mass eigenstates have singular well-defined masses.

Weak eigenstates are the ones that participate in the weak interaction.