# Spin correlation function identity

The correlation function G between two spins is usually defined as

$$G=\langle \sigma_a \sigma_b\rangle - \langle \sigma_a\rangle \langle\sigma_b\rangle$$

The $\sigma$ are the value of the spins (+-1).That's supposed to be equal to

$$= \left\langle\ (\sigma_a - \langle \sigma_a\rangle) (\sigma_b - \langle \sigma_b\rangle) \ \right\rangle$$

And I'm not quote sure why that is. I tried multiplying the second version out and thought that the negative terms would disappear when averaged. That leaves the almost correct formula, but without the minus sign. So where is the trick?

Sources for these equations can be found in for example here on page 4

• is stems from the fact that spins are independent. Take a look on correlation in general, more precise, on how expectation $E[(X-\mu_X)(Y-\mu_Y)]$ computed here: en.wikipedia.org/wiki/… – aaaaa says reinstate Monica Mar 22 '15 at 21:35

Following is computation of $E[(\sigma_a - E[\sigma_a])(\sigma_b-E[\sigma_b])]$:
$E[(\sigma_a - E[\sigma_a])(\sigma_b-E[\sigma_b])]=\\ E[\sigma_a\sigma_b-\sigma_aE[\sigma_b]-E[\sigma_a]\sigma_b+E[\sigma_a]E[\sigma_b]]=\\ E[\sigma_a\sigma_b]-E[\sigma_a]E[\sigma_b]-E[\sigma_a]E[\sigma_b]+E[\sigma_a]E[\sigma_b]=\\ E[\sigma_a\sigma_b]-E[\sigma_a]E[\sigma_b]$
Note that $E[const]=const$, and $E[a+b]=E[a]+E[b]$
$E$ here is expectation or mean
• Ah, thanks. I thought that $E[\sigma]$ would be 0, because spins etc. But this makes sense. – Basti Mar 22 '15 at 22:01
• $E[\sigma]$ might be 0, but in general case it doesn't matter. Here QM just uses regular statistics/math – aaaaa says reinstate Monica Mar 22 '15 at 22:28