Is the function PU(2) and SO(3) induced by the Bloch sphere bijective?

Question

I have difficulty understanding the fact that, as written in this reference,

every single-qubit unitary corresponds to a unique rotation of R3 and vice versa.

If I understand well, this means there is a bijection between

• the space $PU(2)$ of unitary transforms of $\mathbb{C}^2$ up to a multiplication of an element of norm 1
• $SO(3)$, the space of rotations of $\mathbb{R}^3$.

I also think I have seen that, If I call $\mathsf{T} : PU(2) \rightarrow SO(3)$ this transformation, it has the following property : for all $x \in \mathbb{C}^2$ and all $f \in PU(2)$, $\phi(f(x)) = \mathsf{T}(f)(\phi(x))$ where $\phi : \mathbb{CP}^1 \rightarrow S(\mathbb{R}^3)$ is the bijection of the Bloch sphere.

My problem is that, I have the impression $\mathsf{T}$ cannot be a bijection because of the following "demonstration"

1. It is sufficient to know the image of $|1>$ and $|0>$ in order to completely determine an unitary transformation. Thus, for $f,g \in PU(2)$ : $f(|1>)=g(|1>) \land f(|0>)=g(|0>) \Rightarrow f=g$.
2. $|1\!>$ and $|0\!>$ are sent to 2 colinear vectors $u_1=(0,0,1),u_0 = (0,0,-1)$ of the sphere by the morphism $\phi : \mathbb{CP}^1 \rightarrow S(\mathbb{R}^3)$ that define the bloch sphere. Thus, it is not sufficient to know the image of $u_1$ and $u_0$ through a rotation (element of $SO(3)$) to completely define the rotation. More precisely, $\exists A, B \in SO(3) : \, A(u_1)=B(u_1) \land A(u_0)=B(u_0) \land A\neq B$
3. Taking these $A,B$. I use the bijectivity of $\mathsf{T}$ to note $A=\mathsf{T}(f_A), A=\mathsf{T}(f_B)$, for some $f_A, f_B \in PU(2)$ and I see that: $\phi(f_A(|0\!>)) = \mathsf{T}(f_A)(\phi(|0\!>))) = A(u_0) = B(u_0) = \ldots = \phi(f_B(|0\!>))$ that implies, by bijectivity of $\phi$, that $f_A(|0\!>)=f_B(|0\!>)$. The same reasonning led to $f_A(|1\!>) = f_B(|1\!>)$, thus $f_A=f_B$, which contradict the fact that $A \neq B$ because $\mathsf{T}(f_A)= A, \mathsf{T}(f_B)=B$.

My question is basically : where am I wrong?

Thanks for your help and to anyone reading it.

Answer 1

The element $U \in PU(2)$ is not uniquely determined by the pair $|u_0\rangle = U|0\rangle$, $|u_1\rangle = U|1\rangle \in \mathbb{CP}^1$. There is a freedom, $U = e^{i\alpha}|u_0\rangle\langle 0| + e^{i\beta}|u_1\rangle\langle 1|$.

The correspondence between $\mathsf{T}$ and $\phi$ can be visualized as the following. Any $U \in PU(2)$ can be represented as $U = e^{-i\theta/2}|u_0\rangle\langle u_0|+e^{i\theta/2}|u_1\rangle\langle u_1|$ where $|u_0\rangle$, $|u_1\rangle$ form an orthonormal basis. Then $\mathsf{T}(U)$ is the rotation around $\phi(|u_0\rangle)$, $\phi(|u_1\rangle)$ axis by angle $\theta$, clockwise if we look from $u_1$ to $u_0$.

So, the freedom of picking a rotation around z axis corresponds to the freedom of $\theta$ in $U = e^{-i\theta/2}|0\rangle\langle 0|+e^{i\theta/2}|1\rangle\langle 1|$.