How to interpret the definition of optical autocorrelation function for electric field? For EM waves the "temporal coherence" can be described quantitively using the autocorrelation function
$$\Gamma(\vec{r_1},\tau )=\lim_{T \to \infty }\frac{1}{2T}\int _{{-T }}^{{+T }}E(\vec{r_1},t+\tau)E^{*}(\vec{r_1},t )dt$$
Where $E^*$ denotes the complex conjugate of the electric field.
My question is: how does taking the mean over time of the product between the electric field in a point at $t+\tau$ and its complex conjugate in the same point at $t$ measure the autocorrelation of the field? 
Each single product being summed in the integral is, in general, a complex quantity and, if there is no autocorrelation at all the sum of all components in the integral should give $0$ as a result. This is the only "intuitive" thing because in that case all these terms should "compensate" and give zero. Nevertheless in the other cases I do not see the link between the concept of autocorrelation and the definition above. 
In other words how does the quantity $E(\vec{r_1},t+\tau)E^{*}(\vec{r_1},t )$ characterize the autocorrelation, from a physical point of view?
 A: The word correlation refers to the similarity that exists when comparing to object. So autocorrelation then refers to the similarity of an object with itself. However, that sounds silly, why wouldn't an object be similar to itself? So what it means in the context of functions is the extent to which this function is similar to shifted versions of itself. Mathematically it is determined (for real-valued deterministic finite energy functions $g(t)$) by computing 
$$ R_f(\tau) = \int_{-\infty}^{\infty} g(t) g(t+\tau)\ dt . $$
The autocorrelation function is always the largest at the origin $R_f(0)\geq R_f(\tau)$ and it is symmetric $R_f(-\tau)=R_f(\tau)$. So as one can expect, the highest similarity is found with no shift.
[As an aside, note that this differs from the convolution, which is defined by 
$$ C_f(\tau) = \int_{-\infty}^{\infty} g(t) g(\tau-t)\ dt . ]$$
What happens when the functions are not finite energy functions, as in the case of signals of finite power? Then the first expression would blow up. So then we use a limit process
$$ R_f(\tau) = \lim_{T\rightarrow\infty} \frac{1}{T}\int_{-T/2}^{T/2} g(t) g(t+\tau)\ dt . $$
If the function is not real-valued then we need to incorporate a complex conjugate to take care of the phase, which would otherwise leading to destructive interference
$$ R_f(\tau) = \lim_{T\rightarrow\infty} \frac{1}{T}\int_{-T/2}^{T/2} g(t) g^*(t+\tau)\ dt . $$
What if it is not a deterministic function, but rather a random function? In that case we need to define a random process and compute the ensemble average 
$$ R_f(\tau) = \langle g(t) g^*(t+\tau) \rangle . $$
One can think of the ensemble as a large set of random functions that share the same salient properties. The ensemble average is then a summation
$$ R_f(\tau) = \frac{1}{N}\sum_n^N g_n(t) g_n^*(t+\tau) . $$
If the ensemble is ergodic, then one can replace the ensemble average with the integral expression above applied to just one of the elements of the ensemble.
This is the basic theory in a nutshell and one can apply it to the electric field of an optical wave. Physical electric fields are real-valued function, but they are often assumed to be complex for computational convenience, with the understanding that one can relate the result thus obtain to what woudl be obtained with with real-valued functions.
A: There are several ways to see the motivation for complex-conjugating one of the $E$'s in the definition of the autocorrelation function. One way is by considering the $\tau \to 0$ limit. In this case we get
$$ \Gamma({\bf r}, \tau = 0) = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^T |{\bf E}(r, t)|^2 dt = \left \langle |{\bf E}(r, t)|^2 \right \rangle,$$
where the angle brackets denote a time average. If we didn't complex-conjugate either $E$ in the definition, then we would instead get $\left \langle {\bf E}(r, t)^2 \right \rangle$. For a simple plane wave, this quantity would time-average out to zero because we'd get destructive interference from the oscillating phase factor. But clearly any function should intuitively have a nonzero correlation with itself at the same point in spacetime, so this definition wouldn't be very natural.
Another, more abstract motivation is that complex conjugation is very closely related to time-reversal (e.g. for plane waves oscillating periodically in time as $E_0({\bf r})\, e^{-i \omega t}$, they are equivalent), so roughly speaking, $E^*(t)$ "takes you back" by a time $t$, and $E(t + \tau)$ "takes you forward" by a time $t + \tau$, so the net result of their product is to "go forward in time" by an amount $\tau$, as we would intuitively expect for an autocorrelation function. (This explanation probably only makes sense to those familiar with the Heisenberg picture of quantum mechanics.)
A: The conjugation is a merely operational way to cancel out the imaginary parts. Whenever you multiply two complex numbers, you have to be careful and conjugate one of them if your goal is to work only with the real part.
As you know, $\vec{E}$ is a real vector function. We add the imaginary part just for simplicity  $E\propto \cos \phi \ \rightarrow E\propto (\cos \phi + i \sin \phi)\equiv e^{i\phi} $.
In fact, we should actually write the integral as
$ \int \Re e({\vec{E}(t+\tau))} \cdot \Re e({\vec{E}})\ dt $
But that's tedious because there are many cosines multiplied and it forces us to use trigonometrical identities. It's much easier to use complex numbers, for which we just add a meaningless sine of the same function summing. Since integrals and derivatives are linear, there's no problem with that, but products requiere either taking real parts or conjugating one. Both produce the same result.
Now, if the question is why we take the integral of 
$\Re e({\vec{E}(t+\tau))} \cdot \Re e({\vec{E}})\ dt $
It refers to a "statistics and probability theory" theorem, for which if two variables are really random, then
$ \overline{(A\cdot B)}= \bar{A}\bar{B} $
where the bar denotes mean. That happens if the covariance is 0. Consequently, the product inside the integral tells us about the covariance between those two values, and, consequently, about the correlation. 
