What is the correct definition of the Jarlskog invariant?

Question

In this lecture on neutrino physics, Prof. Feruglio defines the Jarlskog invariant as $$J=\text{Im}(U_{\alpha i}^{*} U_{\beta i}^{\,} U_{\alpha j}^{\,} U_{\beta j}^{*})\tag{1}$$ where $U$ is the neutrino mixing matrix with elements $U_{\alpha i}$. Here, $\alpha$ labels neutrino flavours ($e,\mu$ or $\tau$) and $i$ labels neutrino mass eigenstates such that

$$|\nu_\alpha\rangle=\sum_{i=1}^{3}U^*_{\alpha i}|\nu_i\rangle.$$ On the other hand, this well-cited paper defines $$J=\text{Im}(U_{e2}^{\,}U_{e3}^{*} U_{\mu 2}^{*}U_{\mu 3}^{\,}).\tag{2}$$

• Clearly, these two definitions are different because, in general, none of the entries of $U$ is zero. Which one of these definitions is correct and why?
• Moreover, what does the expression (1) mean? But it implies a sum over $\alpha,\beta, i$ and $j$? The expansion of this term would be different upon whether there are these sums in the definition or not.

Answer 1

Nonono! Absolutely no sums in (1).

(1) is the same as (2), or rather, the 9 equivalent ways of writing (1) include (2) as well. I'll only anchor this to M Schwartz's text (29.91-2) for the combinatorically identical quark sector, which I know you have based essentially this question on, before.

Greek indices denote flavor and Latin ones mass eigenstates, so, then e~1, μ~2, τ~3. I'll also tweak your (1) a bit to comport with Schwartz's cycle. Again, do not sum over repeated indices!

Define the 4-tensor $$(\alpha,\beta;i,j)\equiv \text{Im}(U_{\alpha i} U_{\beta j} U^*_{\alpha j} U_{\beta i}^{*})~,$$ so it is evident by inspection that $$ (\beta,\alpha;i,j)=-(\alpha,\beta;i,j)=(\alpha,\beta;j,i). $$ You then see that, up to antisymmetry, there are only 3×3 non-vanishing components, which, remarkably, from the unitarity of U, can be shown to be all identical in magnitude, to wit, $$ (\alpha,\beta;i,j)= J ~ \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}_{\alpha \beta} \otimes \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}_{ij}, $$ such that, $$ J=(e,\mu;2,3)=(e,\mu;1,2)=(e,\mu;3,1)=(\mu,\tau;2,3)=(\mu,\tau;1,2)=(\mu,\tau;3,1)\\ =(\tau,e;2,3)=(\tau,e;3,1)=(\tau,e;1,2). $$

• Unitarity, $\sum_i U_{\alpha i}U^*_{\beta i}=\delta ^{\alpha \beta}$, enters and controls by imposing all rows and columns of the above written matrix to sum to zero, so instead of 3 independent parameters there is only one, and ditto for the left matrix in the tensor product: they must necessarily both be of the type $\sum_k \epsilon^{ijk}$.