# Tag Info

Accepted

Accepted

### Dirac Delta in definition of Green function

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 ...
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### How is Gauss' Law (integral form) arrived at from Coulomb's Law, and how is the differential form arrived at from that?

@Qmechanic's already provided a nice answer. I would like to provide another one. Consider a charge $q$ be enclosed by any surface (not necessarily a sphere). Something like this - Now, you write ...
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### Physical meaning of the Jacobian in relation to Dirac delta function

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\...
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### Divergence of $\frac{ \hat {\bf r}}{r^2} \equiv \frac{{\bf r}}{r^3}$, what is the 'paradox'?

Now what is the paradox, exactly? The paradox is that the vector field $\vec{v}$ considered obviously points away from the origin and hence seems to have a non-zero divergence, however, when you ...
• 3,438
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### Modeling a pure dipole as a function similar to a Dirac delta function

The (distributional) derivative of the delta function does what you want. By definition, $$\int \mathrm dx \ \delta'(x)f(x) = -f'(0)$$ which is motivated by the standard integration by parts formula. ...
• 70.3k

### Is continuity of the wavefunction "put in by hand" for the Dirac delta potentials?

Were $\psi(x)$ not continuous at $x=0$ then $\psi''(x)$ would contain the derivative of a $\delta$-function, and there is nothing else in the equation $H\psi=E\psi$ that could cancel it, so a ...
• 54.7k
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### Where does it become apparent in real scalar QFT that the field has to be an operator-valued distribution, as opposed to an operator-valued function?

Since the commutation relations are $$[\phi(x,t),\pi(y,t)] = \mathrm{i}\delta(x-y)$$ at least one of $\phi$ and $\pi$ must be a distribution, too, since functions are closed under multiplication and ...
• 127k

### Dirac delta function and correlation functions

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 ...
• 127k
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• 54.7k
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### Wave Function Collapse and the Dirac Delta Function

That is a good question; this exposes an idealization in the textbook formalism of Quantum Mechanics which is known not to apply exactly in real life. Notably, you realized that if the probability ...
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### Why does a delta-function potential well have only 1 bound state?

Here is an argument. On one hand, in 1D the $n$th bound state has $n\!-\!1$ nodes. But a bound state in the delta function well is in a classically forbidden region (and hence exponentially decaying)...
• 208k
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### Dirac delta function and correlation functions

In addition to the correct mathematical interpretation appearing in the other answer by ACuriousMind, perhaps a good physically minded viewpoint is to observe that objects like $\delta(t)$ have ...
• 74.8k
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### This vector potential gives a magnetic monopole field, what's wrong with it?

Yes, you have problems with this potential due to the singularity. Notice that you do want the singularity in $r=0$, as you are talking about a point charge (and the electric potential is singular in ...
• 1,824
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### Trouble with position operator in quantum mechanics

By saying $X|x\rangle = \lambda |x\rangle$ and then integrating over $x$ without allowing for the fact that $\lambda$ depends on $x$, you're essentially saying that the action of $X$ on all $|x\rangle$...
• 134k

### Derivative of delta function

It’s not a typo. The distribution $\delta’$ is odd meaning $\delta’(y-x)=-\delta’(x-y)$.
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### Is continuity of the wavefunction "put in by hand" for the Dirac delta potentials?

Well, let's carefully analyze the system: Let us assume the integral form $$\psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent ...
• 208k
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### Is the Dirac $\delta$-function necessarily symmetric?

"Delta function" is not a function, but a distribution. Distribution is a prescription for how to assign number to a test function. This distribution may but does not have to have function ...
• 39.2k
You have the definition of the vector potential. $$\mathbf{B}=\nabla \times \mathbf{A}$$ According to Stokes' theorem this is equivalent to \iint_S \mathbf{B}\ d\mathbf{S} = \oint_{\partial S} \...