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Dirac's delta function.

  1. Nov 2, 2011 #1
    I'm having some difficulties understanding WHY sign function's derivative actually is dirac's delta function? Or more specifically why the derivative equals one at zero and NOT infinite, as the sign function's "actual" derivative at zero equals infinite? Atleast it would make sense.
    Thanks for your answers =)
  2. jcsd
  3. Nov 2, 2011 #2
    First, the Dirac pulse is the derivative of the Heaviside step function, which is not exactly the same as the standard definition of the sign function.

    The Dirac delta function actually IS infinity at x = 0 and not 1. However, the integral of the Dirac delta function on an interval containing zero is 1, which is why the integral from -inf to x is the Heaviside step funtion. Do you see the correspondence now?

    You might confuse the Dirac delta function with its discrete equivalent, the Kronecker delta function.

    DISCLAIMER: I'm aware that the above is far from rigorous, but I don't have the knowledge and I didn't think it was necessary in this context to make it so.
  4. Nov 2, 2011 #3
    Normally the integral over a point is zero, but in this case, is it just defined to be one? How the f- can you do such a definition? And if I derivate sign function (or Heaviside step function) and I get that infinite at zero, can I just go "oh well it looks like dirac's delta function thus I can just say it's not infinite but one at zero"!? I'm confused.
  5. Nov 2, 2011 #4
    It's not mathematically well-defined. To make it precise, you can use distributions. The delta function isn't a normal function. It's a distribution, which is a functional on the space of test functions (compactly supported, C infinity functions).

    Alternatively, you can think of the delta function as just a limit. You take a triangle that has area one and then compress while keeping the area constant, so that it becomes an infinitely big spike. And you can smooth things out if you want to approximate it with smooth functions. So, while there isn't a function with that property, there are functions that behave arbitrarily close to it. So, the delta function is what we call a "weak limit" of those guys.

    Actually, what you want to capture in the formal definition is the sifting property of the delta function: if you integrate a function times the delta function, you get the function value at zero.

    So the idea of distributions is that integrating against a function is a functional, and you can generalize that by just considering all functionals, rather than ones that just come from integrating. So, you can't get a function that gives you what you want by integrating, but maybe you can get a functional that does what you want. That turns out to work.

    I'll let wikipedia take it from here:

  6. Nov 2, 2011 #5


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    DemoniWaari, you need to tell us how much rigor you want in our answers. There is a way to justify every property of the Dirac. In fact there are many ways. However, they all require knowledge of topological vector spaces, Lebesgue integration, dual of vector spaces and/or weak limits of functionals. If you don't have the necessary background, well...
  7. Nov 3, 2011 #6
    Yeah I don't have enough expertice in order to fully understand this. But I think homeomorphic's answer was enough for me, though I still can't understand it fully but that's just fine, I'll get to it later on with my studies (hopefully).

    Anyway thanks guys, now I'm a little bit smarter!
  8. Nov 3, 2011 #7
    But the question that was asked can be answered at an elementary level, just not rigorously.

    The DISCRETE delta function is equal to one at zero. The usual delta function IS infinite at zero. And that's the answer. Plus, the sign function can be written in terms of the step function, so the step function is good enough to find the derivative of that.

    Also, it's not that hard to take derivatives of distributions without being an expert on topological vector spaces and weak limits and all that, but it's probably still more trouble than it's worth. It will make more sense if you have more background knowledge. Until then, it's easier to proceed non-rigorously, I think.
  9. Nov 3, 2011 #8


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    That is not enough.
    Let's suppose we define a function, I'm going to call it [itex]\phi[/itex] for now, from the real numbers to the extended reals where
    [itex]\phi(x) = 0[/itex] for all non-zero x, and
    [itex]\phi(0) = \infty[/itex].
    This is completely valid. Now let's calculate [itex]\int_{-\infty}^{\infty} \phi(x) \, dx[/itex] with Riemann sums. [itex]\phi(x^*_i)[/itex] is going to be 0 or infinity. So the Riemann sum (if it existed) can never be equal to 1 and we break the sifting property, which is the most important property of Dirac delta! The sifting property is also needed in the definition itself.

    The thing is, you can get away with non-rigor if you want. But it'll break the intuition you spent years establishing, and at some point you will start asking questions. And OP is at that point. There are self-contained books on this topic, some much easier to read than others, but it looks like OP has already left. Oh well.

    Edit: At the end of the day, the confusion arises because we write
    [itex]\int_{\mathbb{R}} f(x) \delta (x) \, dx[/itex]
    If we had chosen a different notation that didn't look like the integral, and reduces to the integral for ordinary functions, nobody would have a problem. Sometimes I think bra-ket needs more love.
    Last edited: Nov 3, 2011
  10. Nov 4, 2011 #9
    Of course, it's not enough, rigorously. I said above that it was not mathematically well-defined. You are over-analyzing it. I'm thinking like a physicist, here. I am a mathematician, but I can enter physicist mode and work with things that aren't mathematically well-defined. If we insisted on everything making perfect mathematical sense, then we wouldn't have quantum field theory! In fact, the situation with the delta function was precisely the same until Schwarz came along and was able to formulate it rigorously.

    I was well aware of that and that is why I fired away at the poor guy with my distributions.

    I disagree that it will break the intuition that you spent years establishing. Why? Will it fly away? Nope. It will still be there. Is the intuition wrong? Not really, provided you are careful enough. It's just not quite precise. It suffices just to know that you are taking some liberties with rigor. Part of what I tried to convey earlier is that you can get functions that come arbitrarily close to doing what the delta function does. That's part of the intuition. So, there can be more rigor in a non-rigorous approach than you seem to presume.

    Well, since I know something about Hilbert spaces, I sort of have an inner product in the back of my mind. But if you think of it as a weak limit, the integral doesn't seem like such a stretch. At the end of the day, the confusion only arises because people just don't think deeply enough. And that's different from rigorously enough.

    Back when I was an electrical engineering student, there were delta functions all over the place. I didn't know anything about distributions. I knew that it didn't quite make mathematical sense, but I always had that intuition that the delta function was some sort of limit. Which it is. A weak limit. Turns out, there's nothing wrong with the way I was thinking of it, and it could be made perfectly precise and rigorous.

    You can't ask engineers and probably most physicists to learn distributions. Forget it. They won't be interested. And they can get by just fine without it.
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