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17

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

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} ...


13

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 ...


10

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 ...


9

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 ...


8

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. The dimension theorem for vector spaces tells us that any two bases for a vector space must have the same cardinality. This allows us to define the dimension ...


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. ...


7

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 ...


7

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 ...


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

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 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

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 $f$, ...


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

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

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 ...


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

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

The dimension of the Hilbert space of a free particle is countable. To see this, simply note that The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$. The dimension theorem guarantees that any two bases of of a vector space have the same cardinality, which allows us to define the dimension of a vector space as the cardinality ...


5

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 ...


4

The Dirac delta function is often defined as the following distribution: $$\int_a^b \delta(x - x_0) F(x)\mathrm{d}x = \begin{cases}F(x_0), & a < x_0 < b \\ 0, & \text{otherwise}\end{cases}$$ where $F$ is a suitable test function. Its derivative is then defined as $$\int_a^b \delta'(x - x_0) F(x)\mathrm{d}x = -\int_a^b \delta(x - x_0) ...


4

So, the properties of the derivative of the delta function can be shown relatively quickly though the following ansatz: Consider a function $\delta(x)$ such that $\delta(x) = \frac{1}{a^{2}}(x+a)$ if $-a<x<0$ and $\delta(x) = \frac{1}{a^{2}}(a-x)$ if $0<x<a$, and $\delta(x) = 0$ elsewhere. It is easy to see that $\delta(x)$ has area 1 ...


4

Vladimir's answer is off by factor of 2. The laplacian is $\nabla^2(\frac{1}{r^2}) = \frac{4}{r^4}$ A potential that falls of as $\frac{1}{r^2}$ is a dipole (In general, if it falls off as $r^{-n}$ its an ($2^{n-1}$)-pole, e.g $\frac{1}{r^3}$ bheaviour is quadrupole, etc). Is this a dirac delta? To find out, check : $$\int_{\mbox{All ...


4

That looks correct to me. Consider the basic property of the delta functions $$ \int dx f(x) \delta(x-a) = f(a). $$ Nothing forbids $f(x)$ to be a composite function, for example $f(x) \equiv g(x)\delta(x-b)$, so $f(a) = g(a) \delta(a-b)$. Hence we get, $$ \int dx f(x) \delta(x-a) \equiv \int dx \, g(x)\delta(x-b) \delta(x-a) = g(a)\delta(a-b). $$


3

@joshphysics gave an excellent illustration of why your first part, i.e. ⟨x|p^|x′⟩=−iℏ∂δ(x−x′)∂x? is consistent with quantum mechanics; Let's check your second part rather intuitively. Since in general: $$ \int xg(x)f'(x)dx=-\int f(x)\frac{d}{dx}(xg(x))dx=-\int f(x)(xg'(x)+g(x))dx $$ If $$ f(x)=\delta(x) $$ We conclude that: $$ \int ...


3

1) User joshphysics has already correctly answered OP's 1st question. 2a) Concerning OP's 2nd question, one derives $$i\hbar \delta(x-x^{\prime})~=~i\hbar\langle x | x^{\prime} \rangle ~=~\langle x | [\hat{x},\hat{p}] | x^{\prime} \rangle ~=~\langle x | \hat{x}\hat{p} | x' \rangle-\langle x | \hat{p} \hat{x} | x' \rangle$$ ...


3

In a $d$-dimensional Euclidean space (with positive definite norm), one has $$ \vec{\nabla} \cdot \frac{\vec{r}}{r^d} ~=~{\rm Vol}(S^{d-1})~\delta^d(\vec{r}), $$ cf. the divergence theorem and arguments involving either test functions and integration by part, or $\epsilon$-regularization, similar to methods applied in this Phys.SE answer. Here ${\rm ...


3

We start by mentioning a couple of standard formulas $$\tag{1} \psi(x)~=~\langle x | \psi \rangle, $$ and $$\tag{2} \langle x | y \rangle ~=~\delta(x-y).$$ The canonical commutation relation (CCR) is $$\tag{3} [\hat{x}, \hat{p}] ~=~i\hbar{\bf 1}. $$ The standard Schrödinger position representation reads $$\tag{4}\hat{x}~=~x, \qquad ...



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