Today when I opened physics stack exchange chemistry I found that my reputation spiked up to 495 whereas yesterday it was around 335, for which I don't know how it happened? I just want to make sure if everything is OK as I don't know any reasons for this.
I suppose we can't change the way reputation works on just our site so eventually if this became a serious proposal change it would have to happen on the mother Meta. But before I go and get flamed there, I figured I'd start a discussion here.
It seems that on our site in particular, we don't h...
@EmilioPisanty I somehow felt it must get trouble. Despite numerous attempts to come up with two simple qm problems that are both analytically solvable and related by a perturbation that is accessible to p. t. theory I couldn't find one. I think its for a reason, since in that case you would be able to equate the infinite p. t. series with an (simple) term, which seems to be against all we know from number theory...
@EmilioPisanty I just know that you do not have recursion free algorithms for obtaining a general $n$th order perturbation correction term of say the energy. For that you have e.g. also Feynman diagrams. And this stuff is connected with modular forms.
@EmilioPisanty I make a claim: There is NO QM problem with a closed form solution (not like the Dirac-spike one with $\tan x=x$) that is a modification of another closed form solution problem were the two of them can be related by perturbation theory with a convergent perturbation series.
@EmilioPisanty What is the infinite order PT correction and the analytic solution to this problem?
afiak both of them are inaccessible.
@EmilioPisanty mind I am not talking about a finite order PT solution at complete basis limit. I am talking about the infinte order P.T. series limit each order at complete basis limit.
A pet peeve of mine is when I type an answer to a question and then someone else posts an answer that does not provide any new information to what I already posted.
@yuvrajsingh At the UG level usually vibrations and waves are covered sufficiently in intro physics I would say. And then if other topics need more in-depth coverage then they are covered as needed
I other words, those topics come up in different contexts
I am not familiar with any intro books that are solely focused on vibrations and waves
@yuvrajsingh everyone is bad at this, take a clean browser in incognito mode and try hours of pain... That's how you get it for free (illegal way) or pay to watch (easy, legal way)
@yuvrajsingh I'd suggest you not to go through online random websites, because they are not safe, a lot of malwares, so, wait for it on TV (UTV mostly gets them )
@Qmechanic are the rumours of a swag pack when you hit 100k true?
given that it's been a year since you hit the milestone
I don't really agree with the change in scoring, and it's not like I've been in a rush to get to 100k (or I wouldn't have slowed down my progress handing bounties out left and right =P), but it seems I've been put close to the line, and if there's swag to be got then I do want it =P
From $\mathbf{L}=\mathbf{Q}\times \mathbf{P}$ we have $L_z=Q_xP_y-Q_yP_x$. Then, introduce the following new operators (assuming units of $\hbar=1$):
\begin{align}
q_1=\frac{Q_x+P_y}{\sqrt{2}},\\
q_2=\frac{Q_x-P_y}{\sqrt{2}},\\
p_1=\frac{P_x-Q_y}{\sqrt{2}},\\
p_2=\frac{P_x+Q_y}{\sqrt{2}}.
\end{al...
@EmilioPisanty See *Integer versus half-integer angular momentum*, Ian R. Gatland, American Journal of Physics 74, 191 (2006); doi: 10.1119/1.2166372 for a more direct argument.
@EmilioPisanty I solved it (particle on the ring + mag. field) and in deed its the first example I have found, that works like "analytical problem + perturbation = analytcal problem". The point is that only degenerate PT theory for one $\pm l$ state is required all other contributions are $=0$. And moreover the magnetic field breaks the $\pm l$ degeneracy. That was -- to me -- unexpected. The levels are $E_l = \frac{B^2 R_0^2}{4} - l B + \frac{l^2}{R_0^2} + V_0$.
Most surprising is that the perturbation does not change the states but only the levels, we then get linear terms in $l$.
$R_0$ is the radius of the ring, and atomic constants throughout. The state functions are $\psi = \frac{1}{\sqrt{2\pi}} \exp{i l \theta}$.
for $l \in \Bbb Z$.
Fun ting is that all integrals are like $\langle l | i B \partial_\theta | k\rangle = \pm \delta_{|k|,|l|}$ so thats why all order (deg.) PT series that contribute stop at the first term.
$E_l$ in the field free case ($B=0$) is exactly the levels of ther particle on the ring and first order PT yields the linear correction $-lB$ in its lower (of the two) eigenvalues form degenerate PT.