quantum chemistry kronecker delta formality?

μandνare dummy variables, they can take any values from1toN. As such, you cannot evaluate the Kronecker delta, before specifyingμandνare. For example,δ14=0, andδNN=1, but that is only because you have been given whatμandνare.

So the first line still containsνandσ! What the equation is saying is that, whenμ=ν,andλ=σ, for anyμ,λ=1,⋯,N, then⟨μμ|λλ⟩=⟨μμ|λλ⟩, which is obviously true (though it doesn't tell you what the numerical value is).But if any one of those conditions is not truethen⟨μν|λσ⟩=0.

Now what you wrote doesn't make sense. If⟨μν|λσ⟩=δμνδλσ⟨μν|λσ⟩then1=δμνδλσ(if⟨μν|λσ⟩≠0). But this is obviously a false statement, since if sayμ=1,ν=2then we have1=0.

The bottom line is thatμ,ν,λ,σare dummy variables, and you cannot evaluate the Kronecker delta without being given what the two indices are.