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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 someth …

If you have some text that deals with $$ g(x) = f(x)\delta(x-x_0), $$ with no integration, then $g(x)$ is also a distribution, to be used only in the form $$ \langle g, h\rangle = \int g(x) h(x) \math …

The correct way to do this is given here or here: $$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|},$$ where $a_{i}$ are the roots of the function $f(x …

The integrand is $f(p,q)=\infty$ for all points $(p,q)$ with $H(p,q)=E$. No, it's not. Thus, one can think of $f(p,q)$ as a distribution with infinitely many (uncountably infinite) Dirac functions …

It's not quite clear what the inner workings of your function are, because your $K$ does not match 't Hooft's, and you have not provided your working. However, there is a simpler test case which encap …

Yes, they can. It's not great form, but they can. For a simple example, consider the hamiltonian for a 1D particle in a potential $V(x)$, $$ \hat H=\frac1{2m}\hat{ p}^2+V(\hat{x}). $$ This holds equ …

Yes, it's perfectly possible. Start off with the obvious Ansatz, $$ \psi(x) = \begin{cases} A e^{\kappa x} & x<-a \\ B e^{\kappa x} + Ce^{-\kappa x} & -a<x<a \\ D e^{-\kappa x} & a<x \end{cases} $$ an …

First of all, I would encourage you to think of position of acting on momentum states to the left, that is, to commute them with the bra: $$ ⟨\mathbf p|\hat x=-i\hbar \frac{\partial}{\partial p_x}⟨\ma …

Your assertion that Usually, "white light" is described or defined as an uniform mixture of waves is pretty much completely incorrect: this is not how the term "white light" is treated in the li …

Your units don't add up. You're solving for the equation $$ \frac{\mathrm d^2 u}{\mathrm dx^2} = f $$ where $f = \rho/\epsilon$. If you take your solution at face value, then $u$ will have the dimensi …

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 …

The state $\psi(x) = \delta(x)$ is a perfectly valid state for the harmonic oscillator to occupy. (With caveats, though: it is not normalizable, so it's not a physically-accessible state. Still, it's …

But the potential is infinitely high at $x=0$ You've misunderstood the configuration. To the extent that this kind of language makes sense, at $x=0$ the potential is infinitely low. The delta-fu …

You don't need Dirac delta functions to describe a dipolar charge distribution. There's a very wide range of charge distributions whose electric fields are exactly dipolar (this Q&A describes one such …

The simplest way to solve this - and particularly, the way that minimizes the chances of messing it up - is to switch over to a single coordinate that inside the delta function. In your case it's easy …