Reputation
2,655
Next privilege 3,000 Rep.
Cast close & reopen votes
Badges
1 15 39
Impact
~89k people reached

Nov
10
comment Derivation of the irreducible representations of SO(3)
The same principle applies to both $SO(3)$ and $SU(2)$. All this essentially comes down to the trivial fact that, if $x=y$, then we must have that $f(x)=f(y)$. In $SO(3)$, the identity element and the element that rotates by $\pi$ are the same element. Hence, they must do the same thing in a representation. Obviously, this principle is still true for $SU(2)$; however, in this case, we have distinct $g,h\in SU(2)$ such that $\rho (g)=\rho (h)=1\in SO(3)$ ($\rho :SU(2)\rightarrow SO(3)$ is the map I mentioned in my answer).
Nov
10
awarded  Favorite Question
Nov
9
comment Derivation of the irreducible representations of SO(3)
Also, it cannot be related to $L_i$ at all. $L_i$ is an element of the Lie algebra, and we know that $\mathfrak{su}(2)=\mathfrak{so}(3)$.
Nov
9
comment Derivation of the irreducible representations of SO(3)
This is exactly what I am suggesting. This has nothing to do with spin vs. orbital. If somebody has told you that spin in $SO(3)$ can change a sign after a rotation by $2\pi$, they were wrong. This can only happen for spin with $SU(2)$. The point is is that, in quantum mechanics, we don't really care about representations, but rather projective representations (basically representations up to phase). It turns out that projective representations of $SO(3)$ are in canonical one-to-one correspondence with actual representations of $SU(2)$.
Nov
9
comment Derivation of the irreducible representations of SO(3)
Suppose that the representation has spin $n/2$ with $n\in \mathbb{Z}^+$. Then, in particular, this representation will contain an element $v$ such that $S_zv=\frac{n}{2}v$. This is at the level of the Lie algebra. At the level of the Lie group, $R_z(\theta )=\exp (-i\theta S_z)$ is a rotation about the $z$-axis by an angle $\theta$. Hence, $R_z(\theta )v=\exp (-i\theta \frac{n}{2})$. For $\theta =2\pi$, we had better have that $R_z(2\pi )=R_z(0)$; however, using the formula above, $R_z(2\pi )=\exp (-i\pi)=-1$. Thus, this does not give us a representation of $SO(3)$ for half-integer spin.
Nov
9
revised Derivation of the irreducible representations of SO(3)
added 19 characters in body
Nov
9
comment Derivation of the irreducible representations of SO(3)
The spherical harmonics give you all the irreducible representations of $SO(3)$, but it does not give you all those of $SU(2)$. You'll note this does not make any direct reference to the Lie algebra of $SO(3)$
Nov
9
answered Integral over scalar product of eigenfunction of momentum operator and harmonic oscillator one
Nov
9
answered Calculation mistake some place in finding stress-energy tensor
Nov
9
revised Calculation mistake some place in finding stress-energy tensor
added 10 characters in body
Nov
9
answered Derivation of the irreducible representations of SO(3)
Nov
9
answered Why does the acceleration $g$ due to gravity not affect the period of a vertically mounted spring?
Nov
8
answered Neutrino-Neutron Interaction Feynman Diagram (W Boson Direction)
Oct
21
awarded  Notable Question
Oct
15
awarded  Nice Question
Sep
30
awarded  Explainer
Sep
28
revised What are great circles of 2-sphere?
deleted 2 characters in body
Sep
27
answered What are great circles of 2-sphere?
Jul
23
awarded  Good Question
Jul
22
awarded  Famous Question