# Dirac delta function and correlation functions

I was reading this article, which introduces the Delta function as a general sequence of integrable functions, i.e. if $$\displaystyle\int_{-\infty}^{+\infty} g(x)~\mathrm dx = G,$$ where $g(x)$ could be any function that gives a finite value for $G$, then $$\displaystyle \lim_{\gamma \to \infty} \gamma g(\gamma x)=G\delta(x).$$ Using this definition, the authors claim that the value of delta function is not necessarily infinite at $x=0$, because we can choose $g(x)$ in a way that $g(0) =0$.

Now for a white (or Gaussian) noise, let's call it $\zeta(t)$, the correlation function is defined as: $$\langle \zeta(t) \zeta(t^\prime )\rangle= K \delta(t-t^\prime),$$ which can be understood in an intuitive way as stating that the noise is uncorrelated for different times, and has a strength of $K$.

What happens to this correlation function if we choose the "Delta-generating sequence" of functions so that $\delta(0)=0$? Should we just give a meaning to this type of correlation functions (involving a Delta) only when they are inside an integral? (and if it's the case, then assume we want to integrate this correlation from $t'=0$ to $t'=t$, then upon using different sequences of functions (to produce Delta), the final result of the integration may be different)

• One gets the feeling that the work presented in that article is a bit questionable. The Dirac delta function is supposed to be unique, having certain properties that imply that the value of a Dirac delta function at the origin is undefined. – flippiefanus Sep 30 '16 at 10:26
• @flippiefanus The article's derivations seem quite rigorous (in a physical sense of course!). But I have seen this type of Delta-manipulations in other papers too. (For e.g., one was integrating a function multiplied by derivative of delta function over a finite region, then after using integration by parts, which gave a term with $\delta(0)$, claimed that we can regularize Delta function in a way that its value at the origin becomes zero.) – dedekindCuttage Sep 30 '16 at 10:42

In addition to the correct mathematical interpretation appearing in the other answer by ACuriousMind, perhaps a good physically minded viewpoint is to observe that objects like $\delta(t)$ have always to be interpreted in the sense of the average value, using some smearing or averaging function (in QFT we use the same interpretation regarding field operators).

If you would like to know how $\delta(t)$ is made in a neighborhood of a point $t_0$, you should use a function $f$ concentrated in that neighborhood of $t_0$. The only information you may obtain is the integral $\int \delta(t)f(t) dt$. This way you see that the integral is non zero only if the neighborhood where $f$ does not vanish includes $t=0$, in this sense $\delta(t)$ is concentrated around $t=0$. A more precise shape of $\delta(t)$ is unaccessible.

As a matter of fact this is the way several practical instruments work, computing some sort of average values around a precise (though directly unaccessible) value $t_0$.

The white noise correlation has to be interpreted within the same framework. To check the correlation you should use an averaging function $f(t,t')$ describing the sensibility of your instrument. What you can see/measure is $$\int \langle \zeta(t) \zeta(t^\prime )\rangle f(t,t') dtdt'$$ If the correlation is the white noise one, the integral vanishes as soon as $f(t,t) = 0$.

Adopting this approach, every question regarding the locally precise shape of $\delta$ is meaningless.

• Nice, but I'm a little confused yet. Following this approach, we should treat the white noise somewhat differently from other types of noises (with non-$\delta$ correlations, for eg., with a power-law correlation), and it seems a little spurious. I think that we are trying to avoid a mathematical subtlety in the cost of physical intuition (perhaps I'm wrong). I have read many times, for eg. in Landau texts (I can give an exact address if you want), that the delta function is derived from the discrete (Kronecker) delta, which can be interpreted more easily, but is nonsensical mathematicaly – dedekindCuttage Sep 30 '16 at 12:06
• Well I am both a mathematician and a physicist :), I just tried to give some physical interpretation to the mathematical theory. An interpretation more precise than the intuitive, but as we know misleading, idea that a delta is a very peaked Gaussian function. – Valter Moretti Sep 30 '16 at 14:02
• Indeed, this physical interpretation made a lot more sense. As a last check, these $\delta$ functions must always be interpreted in an integral (as an average). Maybe sometimes we can manipulate them outside integrals, but the averaging process must be at the back of our minds. True? (and it's tempting to deduce that it's not always necessary to think about them as the more abstract objects like distributions :) ) – dedekindCuttage Sep 30 '16 at 15:12
• Yes, it is as you wrote. The mathematical notion of distribution is nothing but the abstract translation of the averaging procedure you mention... – Valter Moretti Sep 30 '16 at 15:44
• @Sebastiano It sounds a bit far from my competence, I suggest you to write here (in PSE) a corresponding question, so that many persons can answer in principle.. – Valter Moretti Aug 2 '19 at 12:06

Saying that $\delta(0) = 0$ is completely non-sensical since the Dirac delta function is not a function to begin with. When we physicists write $$\int \delta(x)f(x) \mathrm{d}x = f(0) \tag{1}$$ when that's all the "definition" of the delta "function" you actually need. Formally, the $\delta$ function is a tempered distribution, something that assigns numbers to test functions. The "integral notation" eq. (1) is just a mnemonic because in many respects this assignment "behaves like an integral", e.g. it obeys a variant of integration by parts. The formal definition of the delta "function" is just $f\mapsto \delta[f] = f(0)$ where you are already notationally prohibited from trying to feed a position like $x=0$ as $\delta(0)$ to it.

While it is true that one can represent the $\delta$ function as the limit of certain other functions (these are called nascent delta functions), this limit is not taken in the space of functions, but in the space of distributions, so the result is not a function. I do not have access to the specific article you are reading, but in general, manipulating the value of the delta function at specific points does not make any rigorous sense. As so often in physics, this does not necessarily mean the result obtained is wrong, but it should be proven by other means in order to trust it.

• Ok, I understand your point. It is frequently said that the value of the Delta function at the origin is infinite. I think the authors were somehow trying to show that it's not necessarily the case (or as you mentioned, it's even nonsensical). But my problem (which I explained in my comment above) remains. Consider the problem of finding the answer to $\int_{0}^{T} f(t) \delta'(t) dt$. Upon integration by parts (and treating the $\delta$ as an ordinary function), there appears a term with $\delta(0)$. What should be done with terms like this? – dedekindCuttage Sep 30 '16 at 11:07
• And another concern is, how to give a meaning to a correlation function which involves a Delta function? – dedekindCuttage Sep 30 '16 at 11:10
• @dedekindCuttage: The definition of $\delta'$ is that it is the distribution that gives $-f'(0)$. It does not make sense to integrate $\int_0^T$ - the integrals involving a $\delta$ are not actual integrals, and do not possess an integration range in the proper sense. A more rigorous notation would be $\delta'[f] = -f'(0)$ as the definition of the derivative, which again makes the problem posed completely non-sensical – ACuriousMind Sep 30 '16 at 11:11
• As I'm not a mathematician, the exact definition of the $\delta$ function is not really that important for me (call me arrogant!). I'm actually trying to figure out what is meant by these types of correlation in the context of physics (or stochastic processes, if you like). And I've seen many times that physicists (including myself!) manipulate these $\delta$-functions separately, without acting on any particular functions (as integration or the $\delta'$ action you mentioned). I just got curious that is there any standard procedure for these treatments, and I ended up reading that article. – dedekindCuttage Sep 30 '16 at 12:18
• @dedekindCuttage you can only manipulate them as freely if only you consider them a limit of nascent delta functions. You can manipulate the integral under the limit, and then finally take this limit. You'll then get what physicists get. But you should still be aware that certain operations aren't quite valid, like integration with at least one limit at $0$. This will become apparent with some asymmetric nascent delta functions. – Ruslan Sep 30 '16 at 18:36