Recent content by lackrange

Bump. Any ideas at all would be great.

It's on page 4 (this isn't homework, just something I stumbled upon, it's also in Griffith's 8.17):http://www.physics.udel.edu/~msafrono/425/Lecture%2018.pdf . Can someone help me understand this solution? What exactly is happening...are there particles at h/2 that are smashing against the...

Right. But that is my question. If the system happens to be in an eigenstate of xp+px (I am not claiming they are complete, there just needs to be one), then could you make a measurement of xp+px?

You wouldn't be measuring x and p separately and then putting them together to get xp+px...you would just be making one measurement, xp+px.

I believe there are subtleties. I don't think the operators are necessarily hermitian on the whole Hilbert space. xp+px is hermitian, but I don't believe it has a complete set of eigenstates. So maybe the operator A isn't defined on the whole Hilbert space..

Say you have some operator A with an incomplete set of eigenstates, but the state of the system is such that it happens to be expressible as a sum (possibly infinite, or integral, whatever) of the eigenstates of A, and let's say the eigenvalues are real and whatever is necessary...we may assume...

Ok. So basically if you could measure a particle to be at position b, then the position operator has to have an associated eigenket |b>, and whatever states these eigenkets span, H' also spans? I think this would have to be provable..otherwise when you use the variational principle, how would...

That's basically what I'm asking. So is the fact that the energy eigenstates are complete part of the postulate, or is it just a result of the position and momentum eigenstates being complete? If it's the latter, is it obvious?

So one of the postulate of quantum mechanics is that observables have complete eigenfunctions. Can someone let me know if I am understanding this properly: Basically you postulate for example, position kets |x> such that any state can be represented by a linear combination of these states...

I never thought about this stuff much before, but I am getting confused by a couple of things. For example, would the state ∣E,l,m⟩ be the tensor product of ∣E⟩, ∣l⟩, ∣m⟩, ie. ∣E⟩∣l⟩∣m⟩? I always just looked at this as a way to keep track of operators that had simultaneous eigenfunctions in a...

I never thought about this stuff much before, but I am getting confused by a couple of things. For example, would the state ∣E,l,m⟩ be the tensor product of ∣E⟩, ∣l⟩, ∣m⟩, ie. ∣E⟩∣l⟩∣m⟩? I always just looked at this as a way to keep track of operators that had simultaneous eigenfunctions in a...

A 30 is at least good enough, 33 is more than good enough. This isn't med school, undergraduate math and physics isn't competitive.

A harmonic function in a region is zero on an open portion of the boundary, and its normal derivative is also zero on the same part, and it is continuously differentiable on the boundary. I have to show that the function is zero everywhere, but I have no idea how. I have tried this for hours...

anyone?

Problem: Find the characteristics of xyu_x+(2y^2-x^6)u_y=0 So I rewrote this as u_x+\frac{2y^2-x^6}{xy}u_y=0 and then set this as \frac{du}{dx}=0\implies \frac{dy}{dx}=\frac{2y^2-x^6}{xy} I solved this, and found that the characteristics were \frac{y^2+x^6}{x^4}=C where C is a constant, and...