3:27 AM
@Sanjana yay :D
h o n k ~
@Asklepian usual language is already too imprecise for every day life. i don't think it can be even close to precise for describing physical phenomena. i think what you are describing makes sense as a starting point for discovering or researching novel ideas. however, i don't think it makes sense as a way of actually learning existing knowledge. but that is my own opinion
3:46 AM
if one would like to discovery learn all of quantum mechanics then that might be a life long task :P
or longer
4:18 AM
agreed ðŸ’¯

2 hours later…
6:06 AM
6:30 AM
@ACuriousMind i dont see a key difference between these hamiltonians tho?
@Asklepian unless you dont speak english well and want to use it to clean up your grammar, im not sure why one would even "use chatgpt to clean up a question" -- chat gpt doesnt know your question better than you do.
is $U^\dagger S_+(0) U = S_+(0)\exp(itf(S_z))$ a consequence of BCH formula or something?

4 hours later…
10:15 AM
@Relativisticcucumber Sorry, I phrased that badly. All I wanted to say was that the transverse Ising model is a toy model, and in reality you have all three $V_i$ not zero.
@SillyGoose Much simpler: For any function $g(S_z)$, you have that $g(S_z)S_+ = S_+ g(S_z+1)$. Proof: Just evaluate both sides on eigenstates of $S_z$.
@ACuriousMind Hi. I have a difficulty. If I have a thin rigid bar ACB of mass m bent into an "L" shape at point C with the arms AC and CB long L_AC and L_CB < L_AC, respectively, and placed vertically with the two ends A and B resting on the ground, how do I work out where the COM is in order to take moments? I found that the coordinates of COM are $x_COM= \frac{(L_AC)^2}{2(L_AC + L_BC)}$ and $y_COM= \frac{(L_BC)^2}{2(L_AC + L_BC)}$, but how do I find where COM is?
Please don't ping me with random mechanics questions; you can just post the question to the room, if someone is interested in it, they'll answer
right now I'm not in the mood for another round of Newtonian mechanics, sorry
@ACuriousMind OK, sorry...
11:03 AM
hi
Guys, sorry but if the value of $\theta$ is missing in an exercise, can I assign it?
can i assume a typical angle to illustrate the calculation? Like $\theta = 30^{\circ}$?
11:58 AM
@Pizza i think you can leave it arbitrary and then discuss the dynamics of the system qualitatively. idk what ur specific q is asking tho
this can be considered a lindblad master equation, right? as in i can solve it by assuming there exists a liouvillian, finding it, and insisting that $\dot{\rho} = \mathcal{L}\rho$, right?
@ACuriousMind ok ok thanks
ACM: thank you

3 hours later…
2:45 PM
@Sanjana I think you need the $1/(r-m^{1/d})$ term instead. The cheap way to do this is to note that $r=[(r-m^{1/d})+m^{1/d}]$ and then use binomial expansion on the numerator; you just need that one term, so the specific term can be easily picked out. The denominator ought to factorise somewhat trivially about that point, and so you can read out that one single Laurent series term that you are most interested about.
2:55 PM
@Obliv One tries not to exclude the mathematically illterate from physics discourse because Faraday existed, but even Faraday, in his great wisdom, is not able to see that his own theory is flawed, whereas Maxwell, upon mathematicising that same theory, immediately saw that Ampère's law required a correction. So, no, it is not even at the level of QM or relativity; you needed maths before them.
@Asklepian Would you advise a person who is otherwise equally as educated as you are, who had years of informal reading on law, to represent themselves in a court of law?
@Asklepian When you asked this on the main site, I gave you a comment. Did you read that? On a broader note, did you read and understand that deleting questions on an SE site does not hide that those questions exist to moderators, and that the downvotes that have been accrued to those questions continue to act on your account?
Sorry, I meant when you sent me the question inside a chatroom. I had replied to that.