particle physics - What is the correct definition of the Jarlskog invariant?

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

$\bullet$ 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?

$\bullet$ 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

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 $\sum_k \epsilon^{ijk}$.