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## Hot answers tagged distributions

32

Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function. One can e.g. represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e. $$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x),$$ $$\tag{2} \delta_{\epsilon}(x)~:=~\frac{1}{\epsilon}... 28 You need nothing more than your understanding of$$ \int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a) $$Just treat one of the delta functions as f(x)\equiv\delta(x-\lambda) in your problem. So it would be something like this:$$ \int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda) $$So there you go. 20 As it seems that you are not completely satisfied with Ron Maimon's answer, I'll put it in a bit different way. The problem is that in your derivation you have a hidden ambiguity.$$\langle{\psi}_p\vert \hat{x}\vert\psi_p\rangle = \infty \;\;\;\;\; \Rightarrow \;\;\;\;\; \langle[\hat{x},\hat{p}]\rangle=...=(p-p)\langleψ_p|\hat{x}|ψ_p\rangle = 0\cdot\infty = ...

18

This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-) The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, $\delta$-functions are not valid $\mathcal{L}^2(\mathbb{R})$ functions, so we need to define a kosher function which gives the $\delta$-function as a ...

14

The Hilbert space ${\cal H}$ of the one-dimensional harmonic oscillator in the position representation is the set $L^{2}(\mathbb{R})$ of square integrable functions $\psi:\mathbb{R}\to\mathbb{C}$ on the real line. The Dirac delta distribution $\delta(x-x_{0})$ is not a function. In particular, it is not square integrable, cf. this Phys.SE post. One may ...

14

It is a distribution. The easiest, cleanest way to think of it is as a linear functional $\mathscr{H}\to\mathbb{R}$ on the Hilbert space $\mathscr{H}$ of functions $\mathbb{R}^N\to\mathbb{R}$ that are $\mathbf{L}^2(\mathbb{R}^N)$. Input a function $f\in\mathbf{L}^2(\mathbb{R}^N)$, and DiracDelta spits out $f(0)$. It's a manifestly linear operator. ...

14

This is notation from Distribution Theory in Functional Analysis. The theory of distributions is meant to make things like the Dirac Delta rigorous. In this context, just to give you one overview, a distribution is a functional on the space of test functions. We define the space of test functions over $\mathbb{R}$ as $\mathcal{D}(\mathbb{R})$ being the ...

12

Take this $\delta '(x)$ and apply in an arbitrary function $f(x)$. $$\int_{a}^{b} \delta'(x) f(x)\ \mathrm{d}x = f(x) \delta(x) |_{a}^{b} - \int_{a}^{b} \delta(x) f'(x)\ \mathrm{d}x = -f'(0)$$ Then $\delta '(x) \rightarrow -\delta (x) \frac{\mathrm{d}}{\mathrm{d}x}$.

12

The wavefunction has a discontinuity at $x=-a$, which gives a term $-2aA i \hbar \delta(x+a)$ when you act with $p$. The contribution from this to the expectation value of momentum exactly cancels the imaginary value you have calculated. Two more-general points: The momentum operator is hermitian, which means its expectation value must be real (provided ...

11

One needs to be careful about what one mean by the "size" of a vector space. A theorem of functional analysis tells us that any two Hilbert bases for a Hilbert space must have the same cardinality. This allows us to define the Hilbert dimension of a Hilbert space as the cardinality of any Hilbert basis. The Hilbert space for the one-dimensional harmonic ...

11

For each $r>0$, the divergence of the magnetic field of the monopole is zero as you have already checked; \begin{align} \nabla\cdot\mathbf B(\mathbf x) = 0, \qquad \text{for all $\mathbf x\neq \mathbf 0$}. \end{align} But what if we also want to find the divergence of this field at the origin? After all, that is where the point source sits. We might ...

11

$\delta'$ is the charge density that generates a dipole. That is, the charge density of two nearby point charges of equal and opposite magnitude in the limit as they get closer and closer to each other. Imagine approximating the delta function with a smooth bump function, and it becomes clear what is going on.

10

Suppose I want to show $$\int \delta(x-a)\delta(x-b)\; dx = \delta(a-b)$$ To do that , I need to show $$\int g(a)\int \delta(x-a)\delta(x-b) \;dx \;da = \int g(a)\delta(a-b)\; da$$ for any function $g(a)$. \begin{align}\textrm{LHS}& = \int \int g(a) \delta(x-a)\;da \ \delta(x-b) \;dx\\ &=\int g(x)\delta(x-b)\;dx \\&=g(b) \end{align} But $\... 10 The$\delta$function is not continuous, so it's a priori not differentiable. In fact, it's not even well-defined as an ordinary real-valued function, but can be made so in terms of distributions - linear maps on a space of test functions given by$f\mapsto\int\delta f=f(a)$. It's possible to sensibly define derivatives of distributions by looking at ... 10 This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law $$\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}$$ If we integrate this over some volume$Vand apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} \int_V\mathrm{d}^3\vec{x}\;\nabla\cdot\... 10 I) It is worthwhile mentioning that there exists a basic approach well-suited to physics applications (where we usually assume locality) that avoids multiplying two distributions together. The idea is that the two inputsF$and$G$in the Poisson bracket (PB) $$\tag{1}\{F,G\} ~=~ \int_M \!dx \left( \frac{\delta F}{\delta \phi(x)}\frac{\delta G}{\delta \pi(... 10 You can think of the Dirac distribution as being an element in some operator valued Banach algebra. One can then use the Riesz calculus (http://en.wikipedia.org/wiki/Holomorphic_functional_calculus) to define an arbitrary holomorphic function of this Dirac delta (or more general distributions). In particular (much like the Cauchy-Integral Formula) we have, ... 10 This is not a peculiar physicist notation oddly enough. The notation allows one to interpret 1/x as a distribution (which makes sense since it's being added to the delta distribution on the right hand side of the equation). For a suitable test function \varphi, one defines this distribution as$$ \mathrm{pv}(1/x)(\varphi) = \lim_{\epsilon\to 0+}\int_{... 8 The delta function$\delta(x)$has unit area, but the function$\delta(2x)$is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is $$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$ where the$x_i$are the roots of$f$. In ... 7 I) Problem. We put (four pi times) the permittivity$4\pi\varepsilon=1$equal to one from now on (in the SI unit system) for simplicity. Let us first rephrase the question a bit. Instead of starting from the potential $$\tag{1} \Phi~=~\frac{1}{r} \qquad \mathrm{and} \qquad \Phi~=~\frac{1}{2r^2}, \qquad r\neq 0,$$ respectively, let us assume that the ... 7 It depends what you want to calculate. As you rightly note, delta functions are not dimensionless, so that including one in your integral will change its dimensionality: you will be calculating something rather different! Most of the time this won't matter if you do it right, but you do need to think about what you want to calculate. The integral$\int f(x,...

7

As is customary in such question, I will point to this paper, which excellently discusses the problems the Dirac formalism has. Now, in your concrete example, the problem lies in the energy/momentum states $| p_0 \rangle$ themselves, which are non-normalizable, since the wave function associated is the Fourier transform of $\delta(p-p_0)$, which means that $... 6 The nature (and glory) of the dirac delta function is that the volume integral $$\int_{\Delta V} dV' \delta ( \boldsymbol{r-r'} ) = \left\{ \begin{array}{cc} 1 & \text{if } \Delta V \text{ contains } \boldsymbol{r}\\ 0 & \text{if } \Delta V \text{ does not contain } \boldsymbol{r} \end{array} \right.$$ Using this function, you can write the ... 6 First equation is wrong, it should say$\rho(\vec{r}) = \delta(\vec{r} - \vec{r}')q$. (Note that you had two errors). You treat it like a normal charge density$\rho(\vec{r})$, if you integrate the density over any volume you get the total charge within that volume. 6 I'll give you the derivation from my book which includes a nice way to see how the delta functions arise: .............................................................................................................................................................. We can derive the potential field$\vec{A}$and the electromagnetic fields$\vec{E}$and$\vec{...

6

It looks like a delta-function. However, because $\epsilon / (\epsilon^2+t^2)$ - you should omit one $\epsilon$ in the numerator, to get the right integral equal to one, by the way - decreases too slowly as $|t|\to\infty$, as $1/t^2$, it will only work as the Dirac delta distribution for test functions that decrease at infinity or at least increase slower ...

6

1) As OP basically notes, an $n$-dimensional delta function transforms under change of variables $f:\mathbb{R}^n \to \mathbb{R}^n$ with (the absolute value of) an inverse Jacobian $$\tag{1} \delta^n(f(x))~=~ \sum_{x_{(0)},f(x_{(0)})=0 }\frac{1}{|\det(\partial f(x_{(0)}))|} \delta^n(x-x_{(0)}),$$ where the sum $\sum$ is over all zeroes $x_{(0)}$ of $f$, ...

6

The distribution approach nicely described by joshphysics and PhysiXxx wholly answer your question and show why your proof doesn't work, but there another way to reason with the correct part of your proof. It is, of course, ultimately mathematically equivalent. Simply work out the flux through a spherical shell $\mathcal{S}$ centred on the origin; from the ...

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