Recent content by EquationOfMotion

All edge lengths are 2r - I think it should be half an octahedron. Issue is that it's not the volume of the area rotated, unless I'm very much mistaken. It's a square pyramid with a cap.

Huh maybe it's not called that? Cone with a square base, in this case all sides are the same length so I guess it'd be half an octahedron. Ok wait it's a square pyramid.

I'm assuming the way to go about it is to integrate in spherical coordinates, but I have no idea what the bounds would be since the bottom edges of the square pyramid are some function of r, theta, and phi.

I was looking up stuff for relativistic two-body decay, and I came across this pdf: https://indico.cern.ch/event/391122/contributions/928962/attachments/782786/1073126/twoBodyDecay.pdf which says that because ##p_2^2=P^2−2P\cdot p_1+p_1^2## we apparently have ##m_2^2=M^2−2ME_1+m_1^2##. But this...

Ooh, so it does always work as long as you know what the Hermitian conjugate is, but you of course *need* the correct conjugate. Thanks!

Doesn't moving around the operators in an inner product work regardless of whether or not it's Hermitian?

So say our inner product is defined as ##\int_a^b f^*(x)g(x) dx##, which is pretty standard. For some operator ##\hat A##, do we then have ## \langle \hat A ψ | \hat A ψ \rangle = \langle ψ | \hat A ^* \hat A | ψ \rangle = \int_a^b ψ^*(x) \hat A ^* \hat A ψ(x) dx##? This seems counter-intuitive...

So more correctly, we'd have ##(\frac{\partial p}{\partial t})Ψ##. But of course, you take the time derivative of ##p## first, which means this becomes ##(0)Ψ = 0##? Or is there something else I'm missing?

I think ##\frac{d \langle p \rangle}{dt} = -\langle \frac{\partial V(x)}{\partial x} \rangle## is the Ehrenfest theorem. The Wikipedia page however notes that were quantum expectation values to be consistent with Newtonian mechanics, we'd have ##F = -\frac{\partial V}{\partial x}##. Unless I'm...

Are you allowed to directly take partial derivatives of operators? I was under the impression you needed a test function.

Is there any good physical or graphical intuition for why ##\frac{d \langle p \rangle}{dt} = -\frac{\partial V(\langle x \rangle)}{\partial x}##? Classically this is apparently true. Thanks.

Doesn't this mean ##\frac{\partial}{\partial t} \hat{p} f(x,t)## for any function ##f(x,t)##? Any intuition on this?

I am indeed not being careful with partial and total derivatives, apologies for the confusion. More correctly, we have <[p, H]>/(ih) + <∂p/∂t> = <[p, V]>/(ih) + <∂p/∂t> = <[p, V]>/(ih), which implies <∂p/∂t> = 0.

Do we not have that <[p, H]>/(ih) + <dp/dt> = <[p, V]>/(ih) + <dp/dt>? Why not?

"Suppose we wanted to know the instantaneous change in momentum p. Using Ehrenfest's theorem, we have <[p, H]>/(ih) + <dp/dt> = <[p, V]>/(ih) since the operator p commutes with itself and has no time dependence." Given H = p^2/(2m) + V and p = -ih(d/dx), [p, H] = [p,V], which implies that...