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Apparently expressions such as $$ \int \delta (x) f(x)dx = f(0)\tag{1}$$ are widely used in Physics.

After a little discussion in the Math SE, I realized that these expression are absolutely wrong from the mathematical point of view.

My question is: can this lack of rigorousness in Physics affect results (i.e. leading to mistakes or wrong results) or is it OK to use these expressions as long as we keep in mind that the Dirac-delta function is a distribution?

PS: I'm adding an other example of fallacious expressions: the Fourier transform of the $\delta$ function being $1$ ($\hat \delta = 1$). It turns out the Fourier transform of the Dirac is not the function $1$, but the regular distribution associated with the function $1$.

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    $\begingroup$ As long as $\int$ is short for $\int_{\Bbb R^n}$ where $x$ is $n$-dimensional, I see no issue, at least not if such definite integrals are defined distributionally. $\endgroup$
    – J.G.
    Commented Sep 7, 2022 at 8:57
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    $\begingroup$ When I studied university physics, that first expression was practically given as the definition of the delta function. But in physics teaching it's generally implicit that any mathematical tools actually work in the context they're used in, without having to precisely define the cases where they don't. $\endgroup$
    – Dronir
    Commented Sep 8, 2022 at 6:15

5 Answers 5

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The expression $$ \int \delta(x)f(x)\mathrm{d}x = f(0)$$ is not wrong, you simply need to read the left-hand side of the equation as what a mathematician would write something like $\langle \delta, f\rangle$, i.e. the application of the $\delta$-distribution to a function. That is, given a function or distribution $g$, we write its application/inner product with a test function $f$ as $\int g(x)f(x)\mathrm{d}x$ - because for an actual function that literally is how the associated distribution is defined. It's just notation - what is wrong is to believe that $g(x)$ on its own has any meaning in general, since distributions do not have values at points.

This means that this notation does not really distinguish between a function $f(x)$ and the distribution defined by $\int f(x) \cdots \mathrm{d}x$, and so saying that the Fourier transform of $\delta$ is the constant function 1 is perhaps sloppy, but not wrong.

There is a difference between being a bit sloppy (as in these cases) and being meaningfully wrong, as when saying things like $\delta(x) = 0$ for all $x\neq 0$. And even then you might sometimes find instances in physics where things like $\delta(0)$ are written and you might want to say it's all wrong because that doesn't mean anything but again, there are interpretations of this notation that make sense (in that example that $\delta(0)$ is essentially code for an infinite volume limit of a finite theory that we don't really want to spell out in detail).

Being rigorous in the mathematical sense is not a binary state - we're not either completely rigorous or completely wrong, but almost always something in between, and walking on the boundary between physics and mathematics requires us to be careful with our judgements: Many things that seem "wrong" can be reinterpreted in terms of shorthand notation, others are secretly right but simply have unstated hypotheses, some might really be unsalvageable.

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  • $\begingroup$ ... and part of being a physicist is to either understand those symbolic notations for the math that is behind them, or, alternatively, accept some stuff as given and just perform calculations. $\endgroup$
    – kricheli
    Commented Sep 8, 2022 at 7:04
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    $\begingroup$ Saying $\delta(x) = 0$ for $x \neq 0$ is no more wrong than the integral notation at the top. It's just a symbolic way of talking about the support of the delta distribution (perhaps more intuitive for the less mathematically minded). $\endgroup$
    – kricheli
    Commented Sep 8, 2022 at 7:07
  • $\begingroup$ "It's just a symbolic way of talking about the support of the delta distribution" - There is no way to talk about the support, because the support is not defined, is it? $\endgroup$
    – Filippo
    Commented Sep 8, 2022 at 7:59
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    $\begingroup$ @Filippo There is a definition of support of a distribution that makes kricheli's comment correct (one more data point for the "can be reinterpreted in terms of shorthand notation" part). $\endgroup$
    – ACuriousMind
    Commented Sep 8, 2022 at 8:28
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    $\begingroup$ As @ACuriousMind comments, there is a notion of "support" of a distribution, and, then, by mild abuse of language, we can rigorously say that a distribution "is zero" off its support. To say or pretend that it has pointwise values $0$ off its support is not entirely misleading. Similarly, there is the notion of "smooth/regular" regions, where locally the distribution is given by integration against a smooth function, etc. Some care is necessary to avoid seeming paradoxes... :) Also, "wavefront sets"... $\endgroup$ Commented Sep 8, 2022 at 16:57
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It depends on the context: there are rigorous mathematical derivations, where one should be careful about not being sloppy. However, there are also physical derivations, where the delta function is just a well-understood substitute for a continuous function (e.g., a Lorentz curve) or a substitute for something of a finite but very small width/duration.

Example 1
A retarded Green's function for a free particle is $$ g^r(\omega)=\frac{1}{\omega - \epsilon_k +i0^+}=\mathcal{P}\frac{1}{\omega-\epsilon_k}-i\pi\delta(\omega-\epsilon_k) $$ However, once a realistic situation is considered, the Green's function typically acquires finite broadening, $$ G^r(\omega)=\frac{1}{\omega - \epsilon_k +i\gamma}=\frac{\omega-\epsilon_k}{(\omega-\epsilon_k)^2+\gamma^2}-\frac{i\gamma}{(\omega-\epsilon_k)^2+\gamma^2}. $$ In this case the delta-function (and more generally the infinitesimal $0^+$) is merely the symbolic shorthand for the anticipated result. (See this answer for an example of such a derivation.)

Example 2
Delta function potential barrier considered in basic quantum mechanics can be viewed as a rectangular barrier of height $V_0$ and width $a$. Delta function formulation arises from taking $a$ to zero, $a\rightarrow 0^+$ while keeping product $V_0 a$ finite. However, one may avoid using the delta function by writing the boundary conditions for a finite barrier, obtaining the characteristic equation, and only then taking the limit $a\rightarrow 0^+$. This is more tedious, but foolproof.

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One existing answer asserts that the notation $\int \delta(x) f(x) dx$ can be taken as an assertion about a distribution not a function. That's fine and I don't disagree. But here I will agree with Roger Vadim's answer and present it in a more simple way.

We can, if we wish, regard $\int \delta(x) f(x) dx$ as a limit: $$ \int \delta(x) f(x) dx \equiv {\rm Lim}_{w\rightarrow 0} \int g(x,w) f(x) dx $$ where $g(x,w)$ is a suitable well-behaved function (e.g. Gaussian) with a peak at $x=0$ and width $w$ (say full width at half maximum, or something like that) and area $1$. In this way of looking at it, the presence of $\delta(x)$ in the expression on the left hand side is telling the informed reader that this kind of limit-process is taken as understood.

Indeed it is often the case in physics that this very idea is exactly how the delta-function is being used. Then a mathematician came along, misread the physicist's notation, and brought in some apparatus to make it ok according to some other definition of $\delta(x)$.

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  • $\begingroup$ Warning: One may think that all nascent dirac deltas are "suitable" representations, but it turns out that such limits have different values depending on the chosen sequence. $\endgroup$
    – Filippo
    Commented Sep 12, 2022 at 13:59
  • $\begingroup$ @Filippo Can you give a reference for a case in which this issue caused disagreement between calculation and an experimental result? $\endgroup$
    – John Doty
    Commented Sep 12, 2022 at 14:19
  • $\begingroup$ @Filippo Your comment is a useful reminder, but I note that the integral considered in the question you refer to goes from 0, i.e. the very place where the function has its peak. Clearly that is a special case. $\endgroup$ Commented Sep 12, 2022 at 14:41
  • $\begingroup$ @AndrewSteane Indeed the comment resulted from a discussion whether expressions of the form$$\int_D\delta(f(x))dx$$should be regarded as being defined or not. As I have show in my first comment, they are not defined in the context of distrubution theory, because the limit depends on the chosen nascent dirac delta. However, we can rigorously evaluate such integrals if we consider a list of "suitable" representations. $\endgroup$
    – Filippo
    Commented Sep 12, 2022 at 15:17
  • $\begingroup$ I was thaught this stuff in my first semester, but unfortunately I didn't ask if this has any applications to theoretical physics... $\endgroup$
    – Filippo
    Commented Sep 12, 2022 at 15:20
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Part of the problem may be the teaching in physics. You'll probably be using the "delta function" in your first years at university, when there is simply no way that your prof can bother you with full-blown distribution theory, which you might learn in the third year or later when studying mathematics. (At least this was the case where I studied.)

So there need to be some simpler notations that are more intuitively accessible for things that happen in distribution theory. And then, as a physicist, you either learn to accept those rules for calculation and learn to work with them, or you learn more mathematics etc. and understand what's behind those notations. The fact that they don't tell the full story does not necessarily mean that what you do with them is wrong (although this may happen).

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    $\begingroup$ "Full-blown distribution theory" - Of course there are some theorems with elaborate proofs (as always), but I would say that one can easily learn the definitions and use those theorems. $\endgroup$
    – Filippo
    Commented Sep 8, 2022 at 7:56
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    $\begingroup$ You can't expect everyone using the delta function in physics to learn definitions that mathematicians use when studying distribution theory. One should not generalize too much as in "one can easily learn...". $\endgroup$
    – kricheli
    Commented Sep 8, 2022 at 10:11
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    $\begingroup$ @kricheli I would add that it's also missing the point a bit. A baker doesn't need to know how to grow wheat. $\endgroup$ Commented Sep 9, 2022 at 21:22
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A circle is locus of points having fixed distance from a fixed point, or we can say a circle is, $r = f(r, \theta) = constant$. The purpose is to show that any function as a continuous is sum of dimensionless points having value of a function at that point. This is the idea of delta function, they act as test point for response of a system which is generally expressed as differential equation (non-homogenous) and whose solution is we are seeking. But sometimes source is not expressed as well behave function, they abruptly change the value, like step function or other function like lump of charges accumulated.

Though expression in question is not expressing its whole sense, but we employ it. Suppose source or cause $f(x)$ (delta function is generally source) are distributed in region R, located at a distance x from origin and we are interested in its response or effect at origin, which is expressed as a relation,

L $u(0) = f(0)$, where L is operator

$\int{\delta(x)}f(x)\text{d}x = f(0)$, from question,

and $\delta(-x)=\delta(x)$

But, L $G(0,x) = \delta(x)$, where $G(0,x)$ is response of unit test source at $x$ to origin, known as green's function.

Thus, L $u(0) = \int$L $ G(0,x)f(x)\text{d}x$

or, L $u(0)=$ L $\int G(0,x)f(x)\text{d}x$

$\implies u(0) = \int G(0,x)f(x)\text{d}x$

Thus, with the help of delta function as it is based on principle of superposition, we have solution of non-homogenous differential equation for any arbitrary source, if its delta response can be find. It has numerous applications in physics from potential of quantum tunneling to find potential of gravitation or elecrostatic or response of forced harmonic oscillator. With the help of Fourier transformation, its delta response can be find.

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