From reading the answer in Difference between the CKM and the PMNS matrix , I gather that the transition $W\to ub$ where $u$ and $b$ mean flavour eigenstates is not possible, but it is possible where $u$ and $b$ means mass eigenstates. Is this understanding correct?

If this is so, then starting with a $B+$, which is in a flavour eigenstate of $u\bar b$, not a mass eigenstate (again is this correct?) then is the following Feynman diagram possible? I believed it was, but if the $b$ that comes from the $W$ is not a flavour eigenstate, then surely the annihilation vertex at the top of the diagram $b\bar b \to y$ is not possible, as electromagnetism does not change flavour and hence requires that both the $b$ and $\bar b$ are flavour eigenstates?

Thanks for any help.enter image description here

  • 1
    $\begingroup$ Quark flavors are defined by the masses of the respective quarks. What are you asking? A W connects u to b quarks. Your W seems to not conserve charge? $\endgroup$ Oct 17, 2020 at 1:45
  • $\begingroup$ "Quark flavors are defined by the masses of the respective quarks" I do not think I understand this. Isn't the whole point of the CKM matrix that the mass eigenstates and flavour eigenstates are distinct? "What are you asking?" I do not know what part of my question is not clear, could you let me know which part is confusing? .... $\endgroup$
    – Jack
    Oct 17, 2020 at 2:11
  • $\begingroup$ ..."A W connects u to b quarks." So a W can decay to a u and b that are in flavour eigenstates? If so is this not also the case for a W can decay to say a flavour eigenstate of an electron and muon neutrino? As far as I'm aware and the question I linked says, the CKM matrix and PMNS matrix are the same in mechanism. I have fixed the minor problem with the Feynman diagram $\endgroup$
    – Jack
    Oct 17, 2020 at 2:15
  • $\begingroup$ No, this is not the point of the CKM matrix. It mixes/straddles generations but a flavor of a quark is defined by its mass. $\endgroup$ Oct 17, 2020 at 2:16
  • $\begingroup$ @CosmasZachos I don't think I understand. Everything I've seen explains the CKM matrix as being due to the flavour eigenstates and mass eigenstates not being the same, for one example of many slide 1 of this physi.uni-heidelberg.de/~uwer/lectures/ExpProbes/script/… $\endgroup$
    – Jack
    Oct 17, 2020 at 2:18

1 Answer 1


The flavor eigenstates for the quarks are defined to be the same as the mass eigenstates. What you're talking about is a basis that diagonalizes the weak interaction matrix. There are many bases you could choose in principle, but the standard one is $u, c, t, d', s', b'$ where the primed particles are related to $d, s, b$ by the CKM matrix. $W\to\bar ub'$ (note the bar) can't happen because the amplitude is proportional to an off-diagonal entry of the diagonal interaction matrix.

Your Feynman diagram is allowed, but it would be disallowed if you replaced either or both of the $b$s by $b'$s.

  • $\begingroup$ Hi, thanks very much for the answer. I think I understand what you mean, do you mean the flavour eigenstates are not the same thing as the weak eigenstates, for quarks? $\endgroup$
    – Jack
    Oct 17, 2020 at 2:28
  • $\begingroup$ And if so is this not the same for the PMNS case? Are the flavour eigenstates and mass eigenstates the same for neutrinos as well? $\endgroup$
    – Jack
    Oct 17, 2020 at 2:32
  • $\begingroup$ @Jack Yes, the flavor eigenstates ($d,s,b$) are different from the weak eigenstates ($d',s',b'$). I didn't use the name "weak eigenstates" because I think it's misleading. The weak force doesn't single out that basis for the down-type quarks. It's only meaningful in combination with the standard basis for the up-type quarks. For neutrinos, the flavor basis = the "weak" basis ≠ the mass basis, but this is essentially just a convention and doesn't result from any fundamental difference in the physics of leptons vs quarks. $\endgroup$
    – benrg
    Oct 17, 2020 at 2:34
  • $\begingroup$ Thanks again, I'm pretty sure I understand now. So is the difference between this in the quark sector and the neutrino sector mainly just a case of how we choose to define the flavour eigenstates, and we make kind of the opposite choice in each case? i.e. in the quark sector we define the flavour eigenstates to be the mass eigenstates whereas in the neutrino sector we define the flavour eigenstates to be the weak eigenstates? Though we could for instance reverse the definition in the neutrino case to make it consistent with the quark case? $\endgroup$
    – Jack
    Oct 17, 2020 at 2:37
  • $\begingroup$ @Jack That's right. See this answer. $\endgroup$
    – benrg
    Oct 17, 2020 at 2:41

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