When we try to see if the weak sector is $CP$ invariant, we $CP$ transform all the fields in the charged interactions terms and we get a condition involving the elements of the CKM matrix and the arbitrary phases of the $CP$ transformed fields:

$$V_{ij} = V^*_{ij} \, e^{i(\xi_W + \phi_{d_j} - \phi_{u_i})}$$

Then, the argument goes: there are $9$ parameters in $V_\text{CKM}$ because it is a general $3 \times 3$ unitary matrix. These $9$ parameters are split in $3$ "angles" and $6$ phases, the $3$ angles being the ones you get if you restrict to an element of $\text{SO}(3)$.

To make the above condition hold, we need to fix the $CP$ phases to cancel the $6$ CKM phases. We have $7$ $CP$ phases (1 from $W$, $3$ from the downs and $3$ from the ups), so it seems like we can do it.

But then we say "actually we have only $5$ independent $CP$ phases, because there are $2$ residual global symmetries corresponding to baryon number and electric charge". Therefore, $1$ phase remains in the CKM and the condition can never hold.

I don't understand the last point: why does the presence of $\text{U}(1)_\text{B}$ and $\text{U}(1)_\text{Q}$ global symmetries reduces the number of $CP$ phases I can fix to cancel the CKM phases?


1 Answer 1


It is straightforward to see that only 5 of the 7 phases are independent, i.e. the unitary $3\times 3$ CKM matrix effectively only has a residual$^1$ $\frac{U(1)^3\times U(1)^3}{U(1)}$ flavor symmetry, cf. e.g. this Phys.SE post.


$^1$ The kinetic term for $Q=\begin{pmatrix}u_L \cr u_d \end{pmatrix}$ has a $U(3)\times U(3)$ flavor symmetry, which is broken by the Yukawa terms.

  • $\begingroup$ How can a U(1) be a "flavour symmetry"? Aren't flavour symmetries transformations that change flavour but keep something invariant? U(1)'s don't change flavour, they shift the phase of each quark flavour independently $\endgroup$
    – Siupa
    Commented Mar 17, 2023 at 11:59
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Mar 17, 2023 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.