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

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?

• I'm not exactly sure what your points are. But how do you send two orthogonal kets $|0\rangle$ and $|1\rangle$ into two colinear vectors given that we deal with unitary/orthogonal transformations? Jan 29 at 19:32
• This is by the bijection $\phi$ given by the Bloch sphere, that is actually not linear. And then, the transformation $\mathsf{T}$ send any unitary operator of $\mathbb{C}^2$ into a rotation of $\mathbb{R}^3$. I added some precisions about the links between $\phi$ and $\mathsf{T}$ in my question. Jan 29 at 19:42

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 $$U = e^{-i\theta/2}|0\rangle\langle 0|+e^{i\theta/2}|1\rangle\langle 1|$$.
• Thanks for this enlightening answer! I have 2 questions about it. First : do you have a reference about it? Second, If I understand well, in the expression of $U$ with $\theta$, the change of $\theta$ does not change $U$ as an element of $PU(2)$, and it is the only change that make it to be the same element in $PU(2)$? For example, $U' = e^{i\alpha }e^{-i\theta/2} |u_0><u_0| + e^{i\beta}e^{i\theta/2} |u_1><u_1|$ would not be the same element (unless for some choices of $\alpha, \beta$)? I fell I am still confused so a reference would be of great help! Jan 30 at 8:13
• Nielsen&Chuang, Quantum Computation and Quantum Information, 4.2 Single qubit operations. The change of $\theta$ does change $U$ as an element of $PU(2)$. In $PU(2)$ only $U$ and $e^{i\alpha}U$ are the same. Jan 30 at 9:34
• That's the point, $U' = e^{i\alpha }e^{-i\theta/2} |u_0\rangle\langle u_0| + e^{i\beta}e^{i\theta/2} |u_1\rangle\langle u_1|$ is not the same element of $PU(2)$ in general, but $U'|u_0\rangle$, $U'|u_1\rangle$ are the same elements in $\mathbb{CP}^1$ for different $\alpha, \beta$. Jan 30 at 9:38