Commutation Definition and 208 Threads

Does the usual commutation relations, e.g. between position and momentum, remains valid in relativistic quantum mechanics?

I am working on a problem for homework and am supposed to show that the angular momentum operator squared commutes with H and that angular momentum and H also commute. This must be done in spherical coordinates and everything I see says "it's straightforward" but I don't see it. At least not...

sometimes I see [\hat{q},\hat{p}] = i\hbar\widehat{\{q,p\}} + O(\hbar^2) what does the last term O(\hbar^2) mean? x=y

I am reading the first chapter of Sakurai's Modern QM and from pages 30 and 32 respectively, I understand that (i) If [A,B]=0, then they share the same set of eigenstates. (ii) Conversely, if two operators have the same eigenstates, then they commute. But we know that [L^2,L_z]=0...

Hi there, I need a help on one of the commutation proof, the question is, show that [FONT="verdana"][Lx,L^2]=0 cyclic where [FONT="verdana"]L=l1+l2 The expression simplifies to [FONT="verdana"][Lx,l1l2]+[Lx,l2l1] but I'm not sure if they are 0. Thanks for your help :D

I'm following a derivation (p85 of Symmetry Principles in Quantum Physics by Fonda & Ghirardi, for anyone who has it) in which the following assertion is made: "...we have \left[\mathcal{G}_p,\mathbf{r}_i\right] &=& \mathbf{v}_0t\mathcal{G}_p, \left[\mathcal{G}_r,\mathbf{p}_i\right] &=&...

Hi, I have a question, As it is said in QM, if two operators commute, they have so many common eigenstates that they form a basis. And the inverse is right. Now there is the question, if A,B,C are operators, [A,B]=0, [A,C]=0, then is "[B,C]=0" also right? If we simply say A and B, A and C...

Does anyone know of any tables that show the commutation relations of all QM opeartors? Any information would be appreciated.