10:27 AM
Guys, in an exercise, how do you understand if there is inertia?
10:38 AM
Does it happen in our universe
If so, yes, there is inertia
10:52 AM
@Relativisticcucumber grrrr
@ClaudioMenchinelli +1 for my UG and it was all the profs except one :,(
ok i see in the chat history i have asked this 2x but i am still quite confused on the role of schurs lemma. i have revisted rep theory in qm and an more familiar with the spin case. i thought it was to label the spin irreps but now im questioning that since it seems this is done purely by experiments.
isnt the second point just true for anything that commutes w the ham? what does this have to do w rep theory at all?
11:16 AM
@Relativisticcucumber he is just stating that it is a consequence of the observation in the paragraph above the statemet
@Relativisticcucumber it is true that if $[A, H] = 0$, then (actually an iff) $A$ and $H$ are simultaneously diagonalizable and so share eigenstates. Which means that an $A,H$ eigenstate $\lvert a, E \rangle$ indeed will not change its energy upon an action such as $e^{-i\theta A } \lvert a, E \rangle$
i want to give up on schurs lemma but in every "physics and rep theory" resource it seems to be plopped in and left
and never revisited XD but i feel bad to just leave w o it
@SillyGoose yeah thats why iw as thinking
Hi guys, can anyone tell me if my approach for this exercise is correct?
A small sphere of mass m = 100g is attached to an ideal spring with elastic constant k = 19.6 N/m, rest length L = 40 cm, massless, whose second end is fixed at point A, as shown in the figure. The system is placed on a rough horizontal plane (dynamic friction coefficient μ= 0.5). If the spring is extended by a distance ∆l_0 = 20cm and the ball is then allowed to move under the action of the spring, the minimum distance from A reached by the ball in its motion is determined.
I would do it like this:
$E_p = \frac{1}{2}k\Delta l_0^2$
$F_f = \mu \cdot m \cdot g$
$W_f = F_f \cdot d$
$E_p = W_f$
$d_A= d + \Delta l_0$
@SillyGoose also note that the energy eigenstate need not be simultaneously of both A and H for this to hold. take $He^{-iAa}|E\rangle=e^{-iAa}H|E\rangle=E e^{-iAa}|E\rangle$
i suppose that is true, but i believe that it is always true that if $[A, H] = 0$, then any eigenstate of $H$ will also be an eigenstate of $A$
@SillyGoose no. e.g. degeneracy
assuming no degeneracy, that is correct
11:29 AM
I see
I think the statemetn I am lookign for is just that $[X, Y] = 0 \iff$ $X$ and $Y$ are simultaneously diagonalizable $\iff$ share a common eigenbasis
yeah. that is correct
which is consistent with not every eigenbasis for each operator being shared
also, how can i understand the difference between isomorphic and bijective for representations ?
specifically, i think there is a bijection between the projective reps of $SO(3)$ and reps of $SU(2)$, but are these necessarily isomorphic?
11:47 AM
Bijection just maps points uniquely, it doesn't have to imply that the group structure is preserved
There's a bijection between any Lie groups that isn't zero dimensional, since they all have the cardinality of the continuum
12:02 PM
bargmann's theorem (applied here) states that projective representations of $SO(3)$ "lift" to representations of $SU(2)$ where "lift" seems to mean the following. let $\rho : SO(3) \to GL(V)$. the lift of $\rho$ is a $\tilde{\rho}: SU(2) \to GL(V)$ such that $\pi \circ \tilde{\rho} = \rho$ where $\pi: SU(2) \to SO(3)$ is the projection
"Among these, the most universal and fundamental one is Ptolemy's law stating that the space of our world has three dimensions."
Who the hell calls it Ptolemy's law
this book states and proves bargmann's theorem it seems: link.springer.com/chapter/10.1007/978-3-540-70690-8_5
Oh god apparently there is a super obscure book of Ptolemy on the topics of dimensions
Peri diastaseos
12:19 PM
@SillyGoose oh no
i dont think i want to welcome that into my life
hm it seems like bargmann's theorem does not establish a bijection between the two sets of representations.
@SillyGoose damn it
is it: Let $G$ be a connected, simply-connected Lie group with trivial 2nd Lie algebra cohomology class. Then, for every projective representation $\rho: G \to GL(P(V))$, we can lift it to a representation of the universal cover of $G$, $\tilde{G}$, $\tilde{\rho}: \tilde{G} \to GL(V)$ such that $\pi \circ \tilde{\rho} = \rho$.
In particular, there may exist representations of the universal cover that are not in unique correspondence with representations of the original group.
So is the real correspondence is between projective representations $\rho$ and projected representations of the universal cover $\pi \circ \tilde{\rho}$ where $\pi: \tilde{G} \to G$ is the projection?
i mean certainly the above is a bijection, but i am wondering if bargmann's theorem makes the stronger claim that there is a bijection between representations of the universal cover and projective representations of the group
12:45 PM
> The gifted Ptolemy in his book On Dimension showed that there are not more than three dimensions;
Pretty short statement
Alas the book itself is lost
Alas, indeed.

2 hours later…
2:53 PM
@SillyGoose Bargmann's theorem is the statement that for $G$ simply-connected and $H^2(\mathfrak{g},R) = 0$, all projective representations of $G$ lift to unitary representations of $G$. So from "projective representations of group" = "projective representations of universal cover" and Bargmann's theorem you get "projective representations of group" = "unitary representations of universal cover". What exactly are you asking?
h o n k ~

7 hours later…
10:12 PM
Regarding fock states. Is there an intuitive explanation as to why a fock state doesn't have a definite phase? Can this be shown/derived mathematically?