Why aren't u, c, and t quarks a mixture of mass eigenstates?

[MarekS]

Weak eigenstates d', s', b' are a mixture of mass eigenstates. For example,
s'=V_cd*d+V_cs*s+V_cb*b

This doesn't seem to be the case for u, c, t quarks. For example, there is no
c'=V_su*u+V_sc*c+V_st*t

Why is that?

 

[Steely Dan]

From a purely practical or phenomenological point of view, the CKM matrix elements are useful, for example, for making predictions on decay rates (if someone who knows more about QFT than me -- which is probably most people on this forum -- wants to chime in from that side of things, please do). If you have some decay that proceeds via the weak interaction, you take your QED calculation for the decay rate and multiply by the square of the CKM matrix element for that interaction to get the actual rate. For the decays that always involve a coupling between one down-type quark and one up-type quark, you can use the CKM matrix element regardless of which way the decay is going (from up-type to down-type, or vice versa). So you only need to create this mixing for one of the two sets (up-type or down-type) to explain measured decay rates. Why it was chosen to be the down-type and not the up-type quarks is quite arbitrary; it could have been done in reverse, but that's not the convention.

 

[MarekS]

Thanks for the reply.

Yes, V_ud (say) can be used in both directions, so a CKM matrix that transforms u, c, t to u', c', t' would contain the same elements as a d,s,b one and hence be superfluous.

However, the weak force sees a down-type quark as a linear superposition of down-type quarks. Does it then see an up-type quark as a linear superposition of up-type quarks?

 

[Steely Dan]

Thanks for the reply.

Yes, V_ud (say) can be used in both directions, so a CKM matrix that transforms u, c, t to u', c', t' would contain the same elements as a d,s,b one and hence be superfluous.

However, the weak force sees a down-type quark as a linear superposition of down-type quarks. Does it then see an up-type quark as a linear superposition of up-type quarks?

In the language of CKM, no. What would be correct to say is that what we previously called any given down-type quark is itself a superposition of the d, s, b quarks, and this affects decays in both directions. You can either have up-type quarks be a linear superposition of u, c, t or down-type quarks be a linear superposition of d, s, b, but not both (at least, not without changing the CKM framework).

 

[MarekS]

You can either have up-type quarks be a linear superposition of u, c, t or down-type quarks be a linear superposition of d, s, b, but not both (at least, not without changing the CKM framework).

I can't see why I can't have both. Also, I don't actually understand why a down-type quark is a linear superposition of down-type quarks when the interaction takes from a down-type quark to an up-type quark (and vice versa). I think it would make sense for an up-type quark to be a linear superposition of down-type quarks. For instance,

u'=V_ud*d+V_us*s+V_ub*b

Then, in a weak interaction, the up quark, u', changes to a d, s, or b because it can as it is part d, s, and b.

 

[MarekS]

I can't see why I can't have both. Also, I don't actually understand why a down-type quark is a linear superposition of down-type quarks when the interaction takes from a down-type quark to an up-type quark (and vice versa). I think it would make sense for an up-type quark to be a linear superposition of down-type quarks. For instance,

u'=V_ud*d+V_us*s+V_ub*b

Then, in a weak interaction, the up quark, u', changes to a d, s, or b because it can as it is part d, s, and b.

Never mind. I thought this through and I understand now. Thanks for your help.