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51

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. $$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x),$$ $$\tag{2} \delta_{\epsilon}(x)~:=~\frac{1}{\epsilon}... 36 You need nothing more than your understanding of$$ \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. 35 Yes. The Dirac delta always has the inverse dimension of its argument. You can read this from its definition, your first equation. So in one dimension \delta(x) has dimensions of inverse length, in three spatial dimensions \delta^{(3)}(\vec x) (sometimes simply written \delta(\vec x)) has dimension of inverse volume, and in n dimensions of momentum \... 27 This question was 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 ... 25 The Hilbert space {\cal H} of the one-dimensional harmonic oscillator in the position representation is the set L^2(\mathbb{R})={\cal L}^2(\mathbb{R})/{\cal N} (of equivalence classes) of square integrable functions \psi:\mathbb{R}\to\mathbb{C} on the real line. The equivalence relation is modulo measurable functions that vanish a.e. The Dirac delta ... 22 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. ... 18 Since the energy spectrum does not depend on the absolute position \vec{r}=\vec{a} of the delta potential, we may assume that \vec{a}=\vec{0}. Therefore, in its current formulation (v1), OP is effectively saying that The attractive 1D delta potential V(x) = -A\delta(x), A>0, has exactly one bound state. The same is true for the 3D delta ... 17 \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. 17 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 ... 16 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 ... 15 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 ... 15 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 \pi(...

15

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 ...

14

The previous answers are all correct, but I thought I'd give a more conceptual explanation for why the delta-function basis is the "wrong" basis in which to expand when counting degrees of freedom. Since the situation is much, much more complicated in QFT, for simplicity I'll only consider first-quantized wavefunctions for a system with a fixed, finite ...

14

This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law \begin{equation} \nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0} \end{equation} If we integrate this over some volume $V$ and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} \int_V\mathrm{d}^3\vec{x}\;\nabla\cdot\...

14

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 $\delta '(x) \rightarrow -\delta (x) \frac{\mathrm{d}}{\mathrm{d}x}$.

14

Your question has been answered again and again, and again, albeit indirectly and elliptically--I'll just be more direct and specific. The point is you skipped variables: in this case, t, and so the expression you wrote ("according to books, $\hat{L}K(x;x') = \delta(x-x')~$"), is nonsense, as you already properly found out; unless you included t in the ...

13

Let us for simplicity consider $n$ point charges $q_1$, $\ldots$, $q_n$, at positions $\vec{r}_1$, $\ldots$, $\vec{r}_n$, in the electrostatic limit, with vacuum permittivity $\epsilon_0$. Now let us sketch one possible strategy to prove Gauss' law from Coulomb's law: Deduce from Coulomb's law that the electric field at position $\vec{r}$ is $$\tag{1} \... 13 Suppose I want to show$$\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)\; dafor 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 \... 13 \newcommand{\bx}{\mathbf{x}} \newcommand{\psih}{\hat{\psi}} This is worked out on p. 20 of Fetter and Walecka. I'll add just a little extra detail here. Your particle density operator n(\bx)=\sum_\alpha\delta(\bx-\bx_\alpha) is in first-quantized form and \hat{n}(\bx)=\psih^\dagger(\bx)\psih(\bx) is second-quantized. When a general one-body ... 13 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}$$when that's all the "definition" of the delta "function" you actually need. Formally, the \delta function is a tempered distribution, something that assigns ... 13 The notation$$\delta_{x_0}(\phi) = \phi(x_0)$$makes sense to me. It's clear that \delta_{x_0} is a linear^{[a]} functional, i.e. it takes a function \phi as an input and produces a number \phi(x_0) as output. Suppose we'd like an integral representation of \delta_{x_0}. One way to do it is with a limit, e.g.$$\delta_{x_0}(\phi) \equiv \phi(x_0)...

13

We can handle easily integrals where the vector variable of integration, let $\:\mathbf{u}\:$, is the argument of the $\:\delta-$function, for example \begin{align} \iiint\delta^{3}\left(\mathbf{u}\right)\rm{L}\left(\mathbf{u}\right) \mathrm d^{3}\mathbf{u}&=\rm{L}\left(\mathbf{0}\right) \tag{01}\\ \iiint\delta^{3}\left(\mathbf{...

12

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, ...

12

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 $$\mathrm{pv}(1/x)(\varphi) = \lim_{\epsilon\to 0+}\int_{... 12 The question doesn't make sense, since the terms "transverse" and "longitudinal" don't apply to scalar fields. They refer to the relationship between the polarization of a wave and its propagation direction, but a scalar wave has no polarization; its value at every point is just a number. On the other hand, if you're using the scalar wave equation to ... 11 It depends what you want to calculate. As you rightly note, delta functions are not dimensionless, so that including one in your integral will change its dimensionality: you will be calculating something rather different! Most of the time this won't matter if you do it right, but you do need to think about what you want to calculate. The integral \int f(x,... 11 1) As OP basically notes, an n-dimensional delta function transforms under change of variables f:\mathbb{R}^n \to \mathbb{R}^n with (the absolute value of) an inverse Jacobian$$ \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, ... 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 The first equation,\frac{1}{x-x_0+i\epsilon}=P\frac{1}{x-x_0}-i\pi\delta(x-x_0)$$is actually a shorthand notation for its correct full form, which is$$\underset{\epsilon\rightarrow0^+}{lim}\int_{-\infty}^\infty\frac{f(x)}{x-x_0+i\epsilon}\,dx=P\int_{-\infty}^\infty\frac{f(x)}{x-x_0}\,dx-i\pi f(x_0) and is valid for functions which are analytic in the ...

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