@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?
@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 $e^{-i\theta A } \lvert a, E \rangle$
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.
@SillyGoose also note that the energy eigenstate need not be simultaneously of both A and H for this to hold. take $He^{-iAa}|E\rangle=e^{-iAa}H|E\rangle=E e^{-iAa}|E\rangle$
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
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
@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?
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?
