meta user
network profile
| bio | website | joshphysics.com |
|---|---|---|
| location | Los Angeles | |
| age | 26 | |
| visits | member for | 4 months |
| seen | 3 hours ago | |
| stats | profile views | 1,539 |
I'm a physics graduate student at UCLA studying high energy theory. Check out my (new) website and blog at
I'm totally open to suggestions of interesting physics topics on which to write pedagogical blog entries. I'm passionate about making high quality physics content available to everyone. I'm also passionate about collaborative learning.
|
21h |
answered | Physical representation of volume to surface area |
|
21h |
comment |
Physical representation of volume to surface area By physical representation do you mean some sort of a physical description/interpretation of what sort of qualitative features this ratio indicates about the object at hand? |
|
23h |
comment |
Some Dirac notation unclarities @71GA Hmmm no I have not. By far my personal favorite quantum text is amazon.com/Quantum-mechanics-Volume-Claude-Cohen-Tannoudji/dp/… |
|
1d |
revised |
Why is the temperature zero in the ground state? deleted 1 characters in body |
|
1d |
comment |
Some Dirac notation unclarities @71GA Acting on a vector with a linear operator gives another vector. Taking the inner product of two vectors gives a number. Unfortunately I can't give you a comprehensive treatment of linear algebra and its physical interpretations in the context of quantum mechanics here; I think you would really benefit from reading a systematic treatment of bra-ket notation; all of these concepts are covered in a lot of books and notes etc. Yes I mistyped at the end. |
|
1d |
comment |
Some Dirac notation unclarities @71GA 1st The making a notational distinction between $\Psi$ and $\psi$ is a matter of taste. Some authors use the former to indicate extra time-dependence, I prefer not to. 2nd Yes. 3rd $\langle x\psi(t)\rangle$ denotes the inner product of a position basis vector and another vector, not the position operator acting on the vector. Yes, the function $\psi$ defined by $\psi(x,t) = \langle x|\psi(t)\rangle$ is the position space representation of $|\psi(t)\rangle$. |
|
1d |
comment |
Why is the temperature zero in the ground state? @annav I don't know; would it? Perhaps you could elaborate? |
|
1d |
answered | Some Dirac notation unclarities |
|
1d |
answered | Hamiltonion in 2-dimensions? |
|
1d |
comment |
Why is the temperature zero in the ground state? @user10001 I'm not sure. I don't even know if the argument can be extended to the case that the spectrum is discrete and the ground state is an accumulation point of the spectrum (although such a beast might just be pathological and physically ignorable). |
|
1d |
revised |
Why is the temperature zero in the ground state? added 27 characters in body |
|
1d |
comment |
Why is the temperature zero in the ground state? @yarnamc Well by making the wall diathermal, I imagine what you have in mind is that the two systems will come to equilibrium with one another and have the same temperature. But as soon as this happens, the small system will now be in the canonical ensemble, not the microcanonical ensemble, so it's not clear to me what your argument is. |
|
1d |
comment |
Do generators belong to the Lie group or the Lie algebra? @Christoph Hall is referring to what essentially amounts to topologically closed using a notion of convergence that he defines earlier. |
|
1d |
comment |
Why is the temperature zero in the ground state? @AlecS I went over the answer to make sure that "ground state" is replaced by "ground level" to make certain that their is no confusion even when the ground level is degenerate. If the ground level is degenerate, then the best one can say is that the density operator becomes the projector onto the lowest energy eigenspace. |
|
1d |
revised |
Why is the temperature zero in the ground state? added 51 characters in body |
|
1d |
revised |
Why is the temperature zero in the ground state? added 51 characters in body |
|
1d |
comment |
Do generators belong to the Lie group or the Lie algebra? @Christoph Yea I'm not sure how generally it holds. Hall defines a matrix Lie group as a closed subgroup of $\mathrm{GL}(n,\mathbb C)$, and then uses the convention I use in the response for $\mathrm{Ad}$... |
|
1d |
revised |
Why is the temperature zero in the ground state? edited body |
|
1d |
revised |
Why is the temperature zero in the ground state? added 110 characters in body |
|
1d |
answered | Why is the temperature zero in the ground state? |
