Recent content by flyusx

In this particular case, since I wasn't introduced to the concept of dimensionless systems, would it be okay to use the ##\hbar=m=ω=1## method? In the future, when I see 'dimensionless', can I set ##\omega=\hbar=m=1## or would this be a really bad habit? Edit: There's an appendix in the back...

So ##\hbar=m=1## gets the right answer but the proper way to solve the problem is to define non-dimensional momentum and position operators. I'll try deriving the probability current. Probability density is given by $$\rho=\Psi^{*}\Psi$$ Using the product rule for derivatives gives...

I'm sort of confused. Are you saying ##\hbar=m=1## is coincidentally the right answer and you agree with vela in that: I must derive ##J## and ##\rho## from the new Schrodinger equation with respect to ##\tau## and ##u##, and use that in regards to my quantum state that is dependent on ##x##...

So ##\hbar=m=1## isn't valid? According to kurkman, ##\hbar=m=1## is what's done in Zettili's world.

I understand the point about getting dimensionless operators but I don't quite see how this would help. In this exercise, Zettili explicitly states the Hamiltonian is $$\hat{H}=\frac{1}{2}\hat{p}+\frac{1}{2}\hat{x}$$ and also explicitly states the momentum operator is $$\hat{p}=-i\frac{d}{dx}$$...

Thanks. I find that if I set ##\hbar=m=1##, I get the right answer.

If I set ##\hbar=m=1##, my answer obeys the continuity equation perfectly. Is this what 'dimensionless' implies (though t is not set to 1)? I uploaded a PDF of my work typed in Latex. Is my document not showing/do you prefer I type it up here?

I've tried to solve this problem (Zettili, Exercise 3.5) four times at this point. I believe my equation for the wave function at a later time ##t## is correct. The problem is my continuity equation is not satisfied; it does not equal zero. It's close but I'm off by a factor of ##m## and...

Yes, I am blind. Thank you.

Thanks, I checked out the section/page about solid angle. But <ψ|Α|ψ> would be zero as $$\int_{0}^{2\pi}\int_{0}^{2\pi}\sin(\theta)\cos^{2}(\theta)\;d\theta d\phi=0$$. <ψ|Β|ψ> would also remain zero. Is this reasonable, or are my integral bounds wrong?

A is anti-Hermitian, B is Hermitian. I'd figured that part by myself so I didn't bother to link it. Sorry. Does ##\sin(θ)## come from spherical integration? I've seen ##ρ^2\sin(θ)## for a volume integral but I can't conceptualise where the sine comes from.

I thought $$\langleψ(x)|\hat{A}|ψ(x)\rangle$$ meant $$\int\psi^{*}\hat{A}\psi\;dx$$, and ended up figuring out that one does integrate over all elements (for instance, x,y,z or in this case θ,φ). I know that in spherical coordinates, volume integration isn't simply dρdθdφ, but I thought that an...

Ah, so they're identical with the exception of a constant (the inner product of h2 and ψ).

So if a state ##|ψ\rangle## is measured and a value h2 is recorded, the state will be in ##|ψ'\rangle=|h2\rangle\langle h2|ψ\rangle##? But the probability calculation is identical when calculating with ##|h2\rangle##?

From Zettili's QM textbook, I have seen the |ψ'> representation used to solve Problem 3.8 whereas Problem 3.7 doesn't use this method. I've attached a PDF file of those two problems. In Problem 3.7(b), he projects each ##a## state onto ##\|φ_2\rangle## since that's the eigenvector associated...