Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
I think there is a summation which can't be seen in this notation. You have tensors of rank 2 there. – Andyk Mar 13 '13 at 12:12
Can you expand on your comment a bit? I dont understand where do you suspect a summation. – TMOTTM Mar 13 '13 at 13:39
Something like this $\langle \mu \nu | \lambda \sigma \rangle =\langle \chi^{\lambda},\chi_{\mu}\rangle \langle \chi^{\sigma},\chi_{\nu}\rangle=\langle \chi^{\lambda},\chi_{\mu} \rangle \langle \delta_{\lambda}^{\sigma}\chi^{\lambda},\delta_{\nu}^{\mu}\chi_{\mu} \rangle=\delta_{\lambda}^{\sigma}\delta_{\nu}^{\mu}\langle \chi^{\lambda},\chi_{\mu} \rangle \langle \chi^{\lambda}, \chi_{\mu}\rangle$ – Andyk Mar 13 '13 at 17:47
So you have the deltas acting only on the second inner product (or the first, but only on one of them). Actually it would be better to change the indexes to $\lambda '$ and $\mu '$ in the second inner product in order to not confuse them. But there is no problem... – Andyk Mar 13 '13 at 17:53

$\mu$ 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.

share|cite|improve this answer
I get the mechanism of the Kronecker delta and now I also get how to read the first equation. So in the integral of the right hand side in the first equation, the second $\mu$ is implicitly the $\nu$, which I plug in for the $\nu$ in the first Kronecker delta. – TMOTTM Mar 13 '13 at 11:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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