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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. $$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x),$$ $$\tag{2} ... 25 You need nothing more than your understanding of$$ \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) $$So there you go. 16 This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-) The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, \delta-functions are not valid \mathcal{L}^2(\mathbb{R}) functions, so we need to define a kosher function which gives the \delta-function as a ... 14 It is a distribution. The easiest, cleanest way to think of it is as a linear functional \mathscr{H}\to\mathbb{R} on the Hilbert space \mathscr{H} of functions \mathbb{R}^N\to\mathbb{R} that are \mathbf{L}^2(\mathbb{R}^N). Input a function f\in\mathbf{L}^2(\mathbb{R}^N), and DiracDelta spits out f(0). It's a manifestly linear operator. ... 13 The Hilbert space {\cal H} of the one-dimensional harmonic oscillator in the position representation is the set L^{2}(\mathbb{R}) of square integrable functions \psi:\mathbb{R}\to\mathbb{C} on the real line. The Dirac delta distribution \delta(x-x_{0}) is not a function. In particular, it is not square integrable, cf. this Phys.SE post. One may ... 12 Take this \delta '(x) and apply in an arbitrary function f(x).$$ \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}. 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 ... 10 Although I think user Siva's argument is nice for intuition, I feel that the key mathetmatical point is being obscured; you just need to be careful about what you 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 ... 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 \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 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 V and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} ... 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, ... 9 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$$for any function g(a).$$LHS = \int \int g(a) \delta(x-a)da \ \delta(x-b) dx =\int g(x)\delta(x-b)dx =g(b) $$But RHS clearly = g(b) too. The result follows putting ... 9 The wavefunction has a discontinuity at x=-a, which gives a term -2aA i \hbar \delta(x+a) when you act with p. The contribution from this to the expectation value of momentum exactly cancels the imaginary value you have calculated. Two more-general points: The momentum operator is hermitian, which means its expectation value must be real (provided ... 8 I) It is worthwhile mentioning that there exists a basic approach well-suited to physics applications (where we usually assume locality) that avoids multiplying two distributions together. The idea is that the two inputs F and G in the Poisson bracket (PB)$$\tag{1}\{F,G\} ~=~ \int_M \!dx \left( \frac{\delta F}{\delta \phi(x)}\frac{\delta G}{\delta ...

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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 ...

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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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Mathematically spoken, since you want your wave functions to be square integrable, your wave functions must be in $L^2$ or some subspace thereof. However, you won't find a function in this space that has a support on a countable set of points, since the Lebesgue integral cannot see countable sets (measure 0), hence there cannot be a function (i.e. no wave ...

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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

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 $\int ... 6 Your question seems to confuse a number of things. Let me try to clear everything up: There is a difference between the Kronecker delta$\delta_{ij}$where$i,j$are discrete indices, and the Dirac delta$\delta(x-y)$, where$x,y$are continuous variables. Superficially, the latter can be thought of as the continuous version of the former. There is ... 6 I) Right, the differential form of Gauss's law $$\tag{1} {\bf\nabla} \cdot{\bf E}~=~ \frac{\rho}{\varepsilon_0}$$ uses the relatively advanced mathematical concept of Dirac delta distributions in case of point charges $$\tag{2} \rho({\bf r})~=~\sum_{i=1}^n q_i\delta^3({\bf r}-{\bf r}_i).$$ Note in particular, that it is technically wrong to claim (as ... 6 The physical meaning of a potential looking like the Delta function is a potential acting only near the origin$x = 0$, and along the rest of the axis the particle is free. I saw such potential-barriers used in nuclear physics. A potential like the derivative of the Delta function,$\delta '(x)$is an approximation of a potential that along all the$x$axis ... 6 There is a sense in which specifying an unknown probability distribution with known mean and variance by a gaussian with the corresponding mean and variance is the correct choice, and that is when those are the only things you are willing to say about the random variable. The gaussian distribution is the maximum entropy distribution with a given mean and ... 5 Ok, I have a solution for$\delta'(x)$based on a very crude limit. I'm going to neglect factors of$\hbar$,$m$, etc for the sake of eliminating clutter. Let$V_\epsilon(x)=\frac{\delta(x+\epsilon)-\delta(x)}{\epsilon}$. Then$\lim_{\epsilon\rightarrow 0}V_e(x)=\delta'(x)$. We'll solve the Schrodinger equation for finite$\epsilon$and then take the limit ... 5 For (most) physical purpuses, you can safely think of the Dirac delta function$\delta(x-x_0)$as some function not vanishing only around$x_0$, and with the property that its integral is normalized to one: $$\tag{1} \int dx \,\delta(x) = 1.$$ With this I mean that you can think of the delta as being a proper function, satisfying (1) and being not vanishing ... 5 Consider evolution of gaussian wave packet. Its wave function in position representation looks like: $$\Psi(\vec r,t)=\left(\frac a{a+i\hbar t/m}\right)^{3/2}\exp\left(-\frac{\vec r\cdot \vec r}{2(a+i\hbar t/m)}\right).\tag1$$ Corresponding relative probability density is $$P(r)=|\Psi|^2=\left(\frac a{\sqrt{a^2+(\hbar ... 5 Hint: Rather than using spherical coordinates, which are singular where the 3D Dirac delta function has support, work instead in Cartesian coordinates \vec{r}=(x,y,z) and use the defining property$$ \iiint_{\mathbb{R}^3}\! d^3r ~f(\vec{r})~ \delta^3(\vec{r})~=~f(\vec{0})$\$ of the 3D Dirac delta function.

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