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.

  • $\begingroup$ 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? $\endgroup$
    – MonteNero
    Jan 29 at 19:32
  • $\begingroup$ 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. $\endgroup$ Jan 29 at 19:42

1 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 $U = e^{-i\theta/2}|0\rangle\langle 0|+e^{i\theta/2}|1\rangle\langle 1|$.

  • $\begingroup$ 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! $\endgroup$ Jan 30 at 8:13
  • $\begingroup$ 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. $\endgroup$
    – Danylo Y
    Jan 30 at 9:34
  • $\begingroup$ 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 $\alpha, \beta$. $\endgroup$
    – Danylo Y
    Jan 30 at 9:38

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