Tag Info

Hot answers tagged

21

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} ...


20

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.


14

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 ...


11

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 ...


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 \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} ...


9

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

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 ...


8

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 ...


8

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 presumably sits. ...


8

The joshpysics's answer is good. I'm only want to tell about some details. Let's have field $$ \mathbf A = g\frac{\mathbf r}{|\mathbf r|^{3}}. \qquad (.1) $$ It has singularity at zero. We can eliminate it by modification $(.1)$ by $$ \mathbf A = g\frac{\mathbf r}{|\mathbf r|^{3}} \to g\frac{\mathbf r}{(r^{2} + a^{2})^{\frac{3}{2}}}. $$ Then we can take ...


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 ...


7

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 ...


6

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 ...


6

We put the permittivity $\varepsilon=1$ to one from now on. Let us first rephrase the question a bit. Instead of starting from the potential $$\Phi=\frac{1}{r} \qquad \mathrm{and} \qquad \Phi=\frac{1}{2r^2}, \qquad r\neq 0, $$ respectively, let us assume that the electric field has be given as $$\vec{E}=\frac{\vec{r}}{r^3} \qquad \mathrm{and} ...


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

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

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 ...


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.


5

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.


5

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 ...


5

The answer is no, the generalized function (=distribution) $$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}=0 \qquad\mathrm{a.e.} $$ is almost everywhere (a.e.) equal to the ordinary zero function $0:\mathbb{R}\to\mathbb{R}$ that sends $t\mapsto 0$. Proof. Consider a test function $f\in C^{\infty}_c(\mathbb{R})$, i.e., an infinitely ...


5

There is a heuristic way to look at this. The Dirac delta function corresponds to a spike when its argument is zero. You can view it as the limit of a sequence of Gaussian functions whose areas are all one but whose width goes to zero. The derivative of a Gaussian function looks like this: So in the limit, the derivative of the Dirac function is something ...


5

OP wrote (v1): Does this imply that $$\tag{1}\langle x | \hat{p} | x^{\prime} \rangle = 0$$ whenever $x \neq x^{\prime}$? For two fixed values of $x \neq x^{\prime}$, the answer is Yes $^1$. But don't try to integrate (1) over $x$ or $x^{\prime}$, i.e. treat $x$ and $x^{\prime}$ as running parameters. Is the momentum operator diagonal in position ...


5

As DJBunk mentions the delta function is symmetric $$\delta(x)=\delta(-x)$$ so you certainly have $$\delta(x-x')=\delta(x'-x).$$ But you should also know that in general we have $$\langle a | b \rangle = \langle b | a \rangle^* $$ and since in this case the inner product is real, you will also have $$\langle x | x' \rangle = \langle x' | x \rangle .$$ So it ...


5

If e.g. $b=1$, it is obvious that the delta-functional – you should really call it $\Delta$ and not $\delta$ – "cancels" against the integral, giving $$\int L(\phi,-a\phi) {\mathcal D}\phi $$ where I just substituted the right value for $\phi'$. Similarly for $a=1$. Now, for general $a,b$, the Jacobian is just the simple power $b^{-N}$ where $N$ is the ...


5

Continuous Fourier analysis, which contains both the Fourier transform and the Fourier series, and which is used in e.g. signal processing, naturally picks the average value of the left and right limits, cf. the Dini-Dirichlet criterion. For the Heaviside step function, this means that $$\tag{1} H(0)~=~\frac{1}{2} \left(\lim_{x\to 0^-} H(x)+ \lim_{x\to 0^+} ...


5

The Hilbert dimension of the Hilbert space of a free particle is countable. To see this, note that The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$. An orthonormal basis of a Hilbert space $\mathcal H$ is any subset $B\subseteq \mathcal H$ whose span is dense in $\mathcal H$. All orthornormal bases of a given non-empty ...



Only top voted, non community-wiki answers of a minimum length are eligible