# How is $j=1/2$ representation, $U(R(\theta,\hat{\bf n}))=e^{i{\sigma}\cdot{\hat {\bf n}}\theta/2}$, is a projective representation of ${\rm SO}(3)$?

A projective unitary representation of $${\rm SO(3)}$$ satisfies $$U(R_1)U(R_2)=e^{i\phi(R_1,R_2)}U(R_1R_2)\tag{1}$$ where $$R_1,R_2\in {\rm SO(3)}$$. How to show that the $$j=1/2$$ representation, $$U(R(\theta,\hat{\bf n}))=e^{i{\sigma}\cdot{\hat {\bf n}}\theta/2}$$, is a projective representation of $${\rm SO}(3)$$ i.e., satisfies the condition $$(1)$$. To do this, one has to show that for $$R_1R_2=R_3\Rightarrow U(R_1)U(R_2)=e^{i\phi}U(R_3).\tag{3}$$ Any suggestions how to show this or at least check this?

1. OP describes projective representations in terms of a 2-cocycle, see section 3 below. An alternative description is in terms of a quotient $$PSU(2)~:=~ SU(2)/\mathbb{Z}_2~\cong~SO(3),\tag{A}$$ where $$SU(2)$$ denotes the 2-dimension $$j=1/2$$ non-projective defining/fundamental/spinor representation and $$\mathbb{Z}_{2}~\cong~\{\pm {\bf 1}_{2 \times 2}\}.\tag{B}$$ In other words, in this latter description the 2-dimensional representation of $$SO(3)$$ is double-valued, i.e. there are 2 branches $$\pm U$$ represents the same $$SO(3)$$ rotation.

2. Let $$\vec{\alpha}=\theta\hat{\bf n}$$ be a rotation-vector in the axis-angle representation $$(\hat{\bf n},\theta)$$. The opposite branch is given by the axis-angle representation $$(-\hat{\bf n},2\pi\!-\!\theta)$$. To describe a general $$SO(3)$$-element ($$SU(2)$$-element) it is enough to consider a rotation-vector $$\vec{\alpha}\in \mathbb{R}^3$$ with length $$|\vec{\alpha}|\leq \pi$$ ($$|\vec{\alpha}|\leq 2\pi$$), respectively. Note that the $$4\pi$$-periodicity of $$SU(2)$$ becomes the familiar $$2\pi$$-periodicity of $$SO(3)$$. See also e.g. this & this related Phys.SE posts.

3. From the non-projective defining representation of $$SU(2)$$, we have $$U(\vec{\gamma})~=~U(\vec{\alpha})U(\vec{\beta}) ,\tag{C}$$ cf. e.g. this Phys.SE post. As mentioned before, we may assume that $$|\vec{\alpha}|,|\vec{\beta}|,|\vec{\gamma}| \leq 2\pi$$. However, if we only want to use rotation-vectors with lengths $$\leq \pi$$ (corresponding to $$SO(3)$$-rotations), we might have to use the opposite branch. Such a transition costs a non-trivial 2-cofactor in eq. (C).

References:

1. G 't Hooft, Introduction to Lie Groups in Physics, lecture notes; chapters 3 + 6. The pdf file is available here.
• The definition I used is the one given in Weinberg's QFT. But with this definition, I have a difficulty seeing which $U$ matrix satisfies a relation of the form $U(R_1)U(R_2)=e^{i\phi}U(R_1 R_2)$. Maybe I am confused by the notation, here. @Qmechanic
– SRS
Commented May 31, 2021 at 14:04
• Yes, Weinberg uses a 2-cocycle. Commented May 31, 2021 at 14:37
• Can you please tell how the notation $U(R_1)U(R_2)=e^{i\phi}U(R_1R_2)$ makes sense? For which $U$ matrices this relation is true? @Qmechanic
– SRS
Commented May 31, 2021 at 14:40
• @SRS You cannot find SU(2) matrices for which this holds, else it would not be representation of SU(2). Commented May 31, 2021 at 17:45
• You are probably right. But I am confused by the notation of Weinberg : $U(R_1)U(R_2)=e^{i\phi(R_1,R_2)}U(R_1R_2)$. What is $U$ here? @ZeroTheHero
– SRS
Commented May 31, 2021 at 18:07

Take $$\hat{\bf{n}}_1=\hat{\bf{n}}_2$$, and $$\theta_1+\theta_2=2\pi$$. As an $$SO(3)$$ element, you should have $$U(R_1)U(R_2)=U(2\pi)=1$$ but here you get $$-1$$.

Thus, you get $$U(R_1)U(R_2)=e^{i\phi}1$$, where $$e^{i\phi}=-1$$ (or $$\phi=\pi$$)

• Where is the phase $e^{i\phi}$? We should get $U(R_1)U(R_2)=e^{i\phi}U(R_1R_2)$. Right?
– SRS
Commented May 31, 2021 at 12:20
• You took $R_1=R(\theta_1,\hat{\bf n}_1)$ and $R_2=R(\theta_2,\hat{\bf n}_2)$. With $\hat{\bf n}_1=\hat{\bf n}_2=\hat{\bf n}$, and $\theta_1=2\pi-\theta_2=\theta$, $$R_1R_2=R(\hat{\bf n},\theta)R(\hat{\bf n},2\pi-\theta)=R(\hat{\bf n},2\pi)=1.$$ Now, $$U(R_1)U(R_2)=e^{i{\sigma}\cdot{\hat {\bf n}}\theta/2}e^{i{\sigma}\cdot{\hat {\bf n}}(2\pi-\theta)/2}=e^{i{\sigma}\cdot{\hat {\bf n}}\pi}=-1$$ and $$U(R_3)=U(R(\hat{\bf n},2\pi))=e^{i{\sigma}\cdot{\hat {\bf n}}\pi}=-1$$ which says $U(R_1)U(R_2)=U(R_3)$. So I dot yet get this. Any error? @ZeroTheHero
• Of course in SU(2) the product is just $U(R_1)U(R_2)=U(R_1R_2)$ else it would not be a representation of SU(2). It is when you make an interpretation of this product as product of SO(3) elements that the phase appears (in the sense that $\theta$ labels an element in SO(3)). That's why you have a projective representation of SO(3) (not of SU(2)). There are no irreducible representations of SO(3) of dimension 2 so it cannot be that $e^{-i\theta\sigma_z/2}$ is in SO(3), yet the SU(2) elements combine "almost" as SO(3) elements. Commented May 31, 2021 at 13:31