# CP Violation of the CKM Matrix

+ 5 like - 0 dislike
1125 views

Considering the relation between the $SU(2)_{WEAK}$ partners of the $|u \rangle$, $|c \rangle$, and $|t \rangle$ mass eigenstates and the corrseponding downtype eigenstates

$\left( \begin{array}{c} J_{WEAK}^{-}|u \rangle\\ J_{WEAK}^{-}|c \rangle\\ J_{WEAK}^{-}|t \rangle\\ \end{array} \right) = \left( \begin{array}{ccc} V_{ud} & V_{us} & V_{ub} \\ V_{cd} & V_{cs} & V_{cb} \\ V_{td} & V_{ts} & V_{tb} \\ \end{array} \right) \left( \begin{array}{c} |d \rangle\\ |s \rangle\\ |b \rangle\\ \end{array} \right)$

I understand that it is allowed to multiply the six eigenstates by a complex phase factor without changing the probabilities.

But why does such a phase factor lead to CP violation (of the CKM Matrix) as Lumo mentions at the beginning of this article ? I`d like to see a mathematical argument (using equations) to better understand why this is.

+ 7 like - 0 dislike

The text by Lumo may have been a bit confusing but it's the other way around: the possibility to redefine the phases of the vectors leads to a reduction of independent angles and phases in the CKM matrix, but there's still one complex phase that can't be rotated away.

Imagine that you change the phases of the kets $u,c,t;d,s,b$ by six multiplicative coefficients, the exponentials of $i$ times the first six letters of the Greek alphabet, e.g. $$|u\rangle \to |u\rangle e^{-i\alpha}$$ and similarly for the other five states. That's equivalent to the following redefinition of $V=V_{CKM}$: $$V \to \pmatrix { e^{i\alpha}&0&0\\ 0&e^{i\beta}&0\\ 0&0&e^{i\gamma} } V \pmatrix { e^{-i\delta}&0&0\\ 0&e^{-i\epsilon}&0\\ 0&0&e^{-i\zeta} }$$ However, note that if you change all these six Greek letters by the same constant $$(\alpha,\beta,\gamma,\delta,\epsilon,\zeta) \to (\alpha+\omega,\beta+\omega,\gamma+\omega,\delta+\omega,\epsilon+\omega,\zeta+\omega)$$ the product of the 3 matrices in the right hand side above will be simply $V$ and $V$ will not change. So without a loss of generality, you may set $\zeta=0$ and there are only 5 independent phases of the ket vectors that may be used to redefine $V$.

Now, $V$ is a priori a general $U(3)$ matrix because it's a transition matrix between two orthonormal bases of the same 3-dimensional complex space. Such a matrix may be described by 9 real parameters. Why? It may be written as $V=\exp(iH)$ where $H$ is a general Hermitian matrix. And a general Hermitian matrix has 9 independent real parameters; literally 1/2 of the 18 parameters in the complex $3\times 3$ matrix. (It's the upper triangle above the main diagonal: the square that are fully in it are independent and complex; the entries on the diagonal have to be real, so only one real parameter, and the boxes below the diagonal are given by those above the diagonal because of the Hermiticity condition.)

So the space of a priori possible $U(3)$ matrices $V$ is 9-real-dimensional. However, the phase redefinition leads to identifications in this 9-dimensional space in such a way that each element is identified with a 5-dimensional space of physically equivalent values of the matrix $V$. Now, subtract $$9 - 5 = 4$$ and you see that the space of physically inequivalent matrices $V$ or the space of "equivalence classes" is 4-dimensional. If we got 3, it would probably mean that the matrix can be made real, an element of $SO(3)$, a 3-dimensional rotation, which depends on 3 angles. However, we obtained 4 parameters which means that we can't bring the general $U(3)$ matrix $V$ into a real form by the redefinition of phases of the six ket vectors.

It means that the most general matrix $V$ must still be allowed to be a complex matrix and there's no way to make the entries real without changing the physics. Now, there are various ways to parameterize the most general matrix $V$. One of the entries may be made $r\exp(i\delta_{CP})$ where the exponential is the CP-violating phase.

If you repeated the same exercise with 2 generations and not just 4, you would find out that the redefinition of the 4 (or 3) phases of the ket vectors for the quarks is enough to bring a general $U(2)$ matrix into the real form, i.e. into an element of $SO(2)$, and there would be no CP-violation. It's because a $U(2)$ matrix has $4$ real parameters and 3 of them may be redefined away by the phases, so the difference is just the single 1 angle in $SO(2)$. So three generations are the minimum number that allows CP-violation.

Just to be sure, a complex $V$ causes CP-violation because the CP symmetry kind of complex conjugates the fields in the Lagrangian or, equivalently, the parameters in the mass matrices. So if you calculate some typical "measure of CP-violation", it will depend on the angle $\delta_{CP}$ above.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Luboš Motl
answered Mar 16, 2012 by (10,278 points)
As an aside, you can factorize $U$ such that the CP violating component enters proportionally to the $\sin$ of only one of the mixing angles. As such, the small mixing angle of the CKM matrix mean that $\delta_{CP}$ can generate only a small CP violation, in contrast to recently established situation in neutrino mixing where $\theta_{1,3}$ is large enough to allow rather more CP violation.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user dmckee
nice explanation!

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Arnold Neumaier
Wow thanks a lot Lumo, did not want to bother You (directly) on TRF with this ..., LOL :-D. This nice and clear step by step explanation is exactly what I needed :-)

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
@dmckee Thanks for this interesting hint, I've heard about the large $\theta_{13}$ and now I better understand what for example is interesting about it.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.