Quantum Chemistry Kronecker Delta formality In semi-empirical quantum chemistry, one frequently encounters the so called zero differential overlap approximation
$$\langle \mu \nu | \lambda \sigma \rangle = \delta_{\mu\nu}\delta_{\lambda\sigma} \langle \mu \mu | \lambda \lambda \rangle .$$
Why is it rather not written as
$$\langle \mu \nu | \lambda \sigma \rangle = \delta_{\mu\nu}\delta_{\lambda\sigma} \langle \mu \nu | \lambda \sigma \rangle = \langle \mu \mu | \lambda \lambda \rangle $$
since on the right hand side of the first equation there are no $\nu$ nor $\sigma$ contained anymore.
So either all four variables plus the Kronecker Deltas (middle expression of second equation), or only the "remaining" variables after evaluation of the Kronecker deltas (last expression of second equation).
 A: $\mu$ and $\nu$ are dummy variables, they can take any values from $1$ to $N$. As such, you cannot evaluate the Kronecker delta, before specifying $\mu$ and $\nu$ are. For example, $\delta_{14} = 0$, and $\delta_{NN} = 1$, but that is only because you have been given what $\mu$ and $\nu$ are. 
So the first line still contains $\nu$ and $\sigma$! What the equation is saying is that, when $\mu = \nu$, and $\lambda = \sigma$, for any $\mu, \lambda = 1, \cdots, N$, then $\langle \mu \mu | \lambda \lambda \rangle = \langle \mu \mu | \lambda \lambda \rangle $, which is obviously true (though it doesn't tell you what the numerical value is). But if any one of those conditions is not true then $\langle \mu \nu | \lambda \sigma \rangle = 0$.
Now what you wrote doesn't make sense. If $\langle \mu \nu | \lambda \sigma \rangle = \delta_{\mu \nu} \delta_{\lambda \sigma} \langle \mu \nu | \lambda \sigma \rangle $ then $1 = \delta_{\mu \nu} \delta_{\lambda \sigma}$ (if $\langle \mu \nu | \lambda \sigma \rangle \neq 0$). But this is obviously a false statement, since if say $\mu = 1, \nu = 2$ then we have $ 1 = 0$.
The bottom line is that $\mu, \nu, \lambda, \sigma$ are dummy variables, and you cannot evaluate the Kronecker delta without being given what the two indices are.
