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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

Jack
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    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? – Cosmas Zachos Oct 17 '20 at 01:45
  • "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?

    ....

     – Jack Oct 17 '20 at 02:11
  • ..."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

     – Jack Oct 17 '20 at 02:15
  • 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. – Cosmas Zachos Oct 17 '20 at 02:16
  • @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 https://www.physi.uni-heidelberg.de/~uwer/lectures/ExpProbes/script/chapter-9a.pdf – Jack Oct 17 '20 at 02:18
  • The CKM matrix ensures that a $B^+$ meson, composed of $u\bar b$ quarks, couples to a $W^+$ through them. The same $W^+$ also couples to $u\bar s$ and $u\bar d$, of course. It also couples to $c\bar b$, etc. Imagining that a quark is a mass eigenstate but not a flavor eigenstate is absurd. I think any and all considerations of neutrinos only serve to confuse you. you misread the answer to that question. WP on CKM would serve you better, or a mainstream text. – Cosmas Zachos Oct 17 '20 at 02:19
  • @CosmasZachos Again, from mainstream texts, I do not understand how you are saying it is absurd that mass and flavour eigenstates are different. Every text I can find on the matter seems to claim that a quark in a flavour eigenstate is in a superposition of mass eigenstates, and vice-versa, for another example slide 7 of this https://www.hep.phy.cam.ac.uk/~thomson/MPP/partIIIparticles/Handout_12_2011.ppt , what am I misunderstanding, and what about that answer have I misread? Is CKM and PMNS not analogous? – Jack Oct 17 '20 at 02:23
  • You appear confused. "Weak eigenstates" , the d', s', b', are not flavor (=mass) eigenstates. They are just non-propagating convenience linear combinations coupling to u,c,t via the W. (These, the ' combinations correspond to $\nu_e,\nu_\mu,\nu_\tau$, also non-propagating mirages, but this is clearly confusing you rather than helping...) – Cosmas Zachos Oct 17 '20 at 02:26
  • Here is my answer to basically the same type of confusion. It is a regular on this site, and has been answered half a dozen times. – Cosmas Zachos Oct 17 '20 at 02:32

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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.

benrg
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  • 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? – Jack Oct 17 '20 at 02:28
  • 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? – Jack Oct 17 '20 at 02:32
  • @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. – benrg Oct 17 '20 at 02:34
  • 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?

     – Jack Oct 17 '20 at 02:37
  • @Jack That's right. See this answer. – benrg Oct 17 '20 at 02:41
  • great I've accepted your answer. Though because of this being because of opposite conventions I am now slightly confused on something else, I'm unsure if it's enough to make another a question on it though. The Z boson couples to identical flavour eigenstates, so you can have for instance Z->uu~, or Z->nu_{e}nu_{e}~, but you can't have Z->ut~ or Z->nu_{e}nu_{mu}~... Since the flavour eigenstates are defined oppositely in these two cases, shouldn't it for the neutrino case be Z->nu_{1}nu_{1}~ ? – Jack Oct 17 '20 at 02:46
  • Basically, does the Z boson decay into pairs of weak eigenstates or pairs of mass eigenstates? – Jack Oct 17 '20 at 03:18
  • @Jack the $Z$ and $γ$ don't care about flavor, the interaction matrix is effectively diag(1,1,1) in any basis. $Z\to ν_e+\bar ν_e$ and $Z\to ν_1+\bar ν_1$ are both allowed. – benrg Oct 17 '20 at 03:26