I remember reading this somewhere... Is there an elegant proof for this?

  • 4
    $\begingroup$ see quantumcomputing.stackexchange.com/a/16552/55 $\endgroup$
    – glS
    Jul 6, 2021 at 23:02
  • 2
    $\begingroup$ It helps if you add text to the question too... $\endgroup$ Jul 7, 2021 at 4:21
  • $\begingroup$ Yes, all of them can be written in the form, $\exp \left[ -i \theta \hat{n} \cdot \vec{\sigma} \right]$, where $\hat{n}$ is a unit vector on the Bloch sphere. $\endgroup$ Jul 7, 2021 at 6:17

3 Answers 3


TL;DR: Yes, ignoring the unobservable global phase, every single-qubit unitary corresponds to a unique rotation of $\mathbb{R}^3$ and vice versa.

Single-qubit unitaries and rotations

Let us first pin down the two objects in question. The first one - the set of single-qubit unitaries - is sometimes imprecisely described as the group $U(2)$ of $2 \times 2$ unitary matrices. However, since the global phase is unobservable, we must identify all unitaries that differ by a scalar factor. Now, scalar factors are precisely the elements of the center $Z(U(2))$ of $U(2)$. Thus, the identification corresponds to taking the quotient of $U(2)$ by its center and the resulting object is called the projective unitary group

$$PU(2) := U(2)/Z(U(2)).$$

The second object is the group $SO(3)$ of rotations in $\mathbb{R}^3$ or equivalently, the group of $3\times 3$ real orthogonal matrices with unit determinant.

Proof by group theory

Recall that the second isomorphism theorem states that for any group $G$, a subgroup $S\subset G$, and a normal subgroup $N\triangleleft G$, the intersection $S\cap N$ is a normal subgroup of $S$ and that

$$ (SN)/N \cong S/(S\cap N).\tag1 $$

Now, set $S := SU(2)$ and $N := \{e^{i\theta} I|\theta\in[0,2\pi)\}=Z(U(2))$ and note that $SN = U(2)$ and $S\cap N = \{I, -I\}\cong \mathbb{Z}_2$. Substituting into $(1)$, we get

$$ U(2)/Z(U(2)) \cong SU(2)/\mathbb{Z}_2, $$

but $U(2)/Z(U(2))=PU(2)$ by definition, so

$$ PU(2) \cong SU(2)/\mathbb{Z}_2.\tag2 $$

Finally, it is well known that $SU(2)$ is a double cover of $SO(3)$

$$ SU(2)/\mathbb{Z}_2 \cong SO(3).\tag3 $$

Combining $(2)$ and $(3)$, we get

$$ PU(2) \cong SO(3)\tag4 $$

which says that the group $PU(2)$ of single-qubit unitaries up to global phase is isomorphic to the group $SO(3)$ of rotations of $\mathbb{R}^3$.

Explicit construction

The connection between the $2$-dimensional complex vector space and the $3$-dimensional real vector space is established by the linear bijection $\vec{ }: \mathfrak{su}(2)\to\mathbb{R}^3$ that assigns to any $2\times 2$ traceless skew-Hermitian matrix its expansion in the basis $iX$, $iY$, $iZ$ where $X$, $Y$ and $Z$ are Pauli matrices. It is easy to check that

$$ \vec{a} \cdot \vec{b} = \frac{1}{2}\mathrm{tr}(a^\dagger b) \tag5 $$ $$ \vec{a} \times \vec{b} = \frac{1}{2i}[a, b].\tag6 $$

Now, for a $2\times 2$ unitary $U$, define the real $3\times 3$ matrix $\Phi(U)$ by

$$ \Phi(U)\vec{a} = \vec{b} $$

where $b=UaU^\dagger$. We will show that $\Phi$ accomplishes the isomorphism in $(4)$.

If $V=e^{i\theta}U$ is another representative of $U$'s equivalence class in $PU(2)$ then $\Phi(U)=\Phi(V)$, so $\Phi$ is well-defined on $PU(2)$. Further, by substituting into $(5)$, we see that

$$ (\Phi(U)\vec{a})\cdot(\Phi(U)\vec{b}) = \frac12\mathrm{tr}(Ua^\dagger U^\dagger UbU^\dagger) = \frac12\mathrm{tr}(a^\dagger b) = \vec{a}\cdot\vec{b} $$

so $\Phi(U)$ preserves the Euclidean inner product in $\mathbb{R}^3$ and thus $\Phi(U)\in O(3)$. Similarly, by computing the triple product using $(5)$ and $(6)$ we see that

$$ \begin{align} \Phi(U)\vec{a}\cdot\left((\Phi(U)\vec{b})\times(\Phi(U)\vec{c})\right) =& \frac{1}{4i}\mathrm{tr}(Ua^\dagger U^\dagger [UbU^\dagger, UcU^\dagger]) \\ =& \frac{1}{4i}\mathrm{tr}(Ua^\dagger U^\dagger U[b, c]U^\dagger) \\ =& \frac{1}{4i}\mathrm{tr}(a^\dagger [b, c]) \\ =& \vec{a}\cdot(\vec{b}\times\vec{c}) \end{align} $$

so $\Phi(U)$ is orientation-preserving and thus $\Phi(U)\in SO(3)$. Therefore, $\Phi$ is a map from $PU(2)$ to $SO(3)$.

It is clear that $\Phi$ is a homomorphism. Moreover, if $\Phi(U)\vec{x}=\vec{x}$ for all $\vec{x}\in\mathbb{R}^3$ then $Ux = xU$ for all $x\in\mathfrak{su(2)}$ which implies that $U$ commutes with all $2\times 2$ complex matrices and thus in particular $U\in Z(U(2))$. Therefore, $\Phi$ is injective. Finally, it is easy to check that for any unit vector $\vec{n}\in\mathbb{R}^3$ and angle $\phi$, the $3\times 3$ real orthogonal matrix

$$ \Phi\left(I\cos\frac{\phi}{2} -i(n_xX+n_yY+n_zZ)\sin\frac{\phi}{2}\right) $$

effects a rotation by angle $\phi$ around $\vec{n}=(n_x, n_y, n_z)$. Therefore, $\Phi$ is surjective. Consequently, $\Phi: PU(2) \to SO(3)$ accomplishes the isomorphism in $(4)$.

  • $\begingroup$ thanks for the elaborate answer. not sure what you mean by 'vice versa' though in your TLDR section. $\endgroup$ Jul 7, 2021 at 17:26
  • 2
    $\begingroup$ I meant that the correspondence is one-to-one, i.e. each single-qubit unitary corresponds to exactly one rotation and each rotation corresponds to exactly one single-qubit unitary. $\endgroup$ Jul 7, 2021 at 18:48

A couple of additional remarks:

  1. The "Bloch sphere" is a representation in $S^2$ (the unit sphere) of qubit (pure) states. Such a representation is possible because there is a bijection $\mathbb{CP}^1\simeq S^2$, via the standard mapping between states and Bloch vectors, given e.g. in this other answer.

  2. "Single-qubit unitaries" are unitary $2\times 2$ matrices. This group is denoted with $U(2)$. More precisely, one should consider these operators as defined modulo their determinant, so that their action on elements in the projective space $\mathbb{CP}^1$ is well-defined. This results in the projective unitary group $PU(2)$. The relation between these has already been discussed in this other answer.

  3. We thus have on the one hand states in $\mathbb{CP}^1$ which are acted upon by projective unitaries in $PU(2)$ (or less formally, states in $\mathbb C^2$ acted upon by unitaries in $U(2)$). The question is: how does the action of a (projective) unitary $U$ on some $\psi\in\mathbb{CP}^1$ translate via the bijection $\mathbb{CP}^1\simeq S^2$? More precisely, if $\varphi:\mathbb{CP}^1\to \mathbb R^3$ transforms between ket states and Bloch vectors, we are looking for the map $\Phi:{\rm PU}(2)\to{\rm Lin}(\mathbb R^3)$ such that, for any $U\in\mathrm{PU}(2)$ and $\psi\in\mathbb{CP}^1$, we have $$\varphi(U.\psi) = \Phi(U).\varphi(\psi).$$ The map $\Phi$ must also be a homomorphism for this equation to be consistent, and as shown in the other answer, we can see that the correct choice is $\Phi(U)\in{\rm SO}(3)$. Another way to state this is to say that ${\rm PU}(2)$ acts on $S^2$ via ${\rm SO}(3)$ matrices.


Consider a single-qubit unitary. This has two eigenvalues $e^{i\theta_1}$ and $e^{i\theta_2}$. So $V=e^{-i(\theta_1+\theta_2)/2}U$ has eigenvalues $e^{\pm i\phi}$ where $\phi=(\theta_1-\theta_2)/2$. Note that the two eigenvectors of $V$ are (i) orthogonal, meaning that they are anti-parallel on the Bloch sphere, defining a single axis, and (ii) these are the states on the Bloch sphere that are unchanged by $V$. So these are going to tell us about the axis of rotation.

If you want to see that it's a rotation that happens around this axis, recall that for any pair of states $|\psi\rangle$ and $|\phi\rangle$, these are evolved to $V|\psi\rangle$ and $V|\phi\rangle$, but the inner product is preserved. So, if one of those states is the axis of the Bloch sphere, you see that the only states you can move to form a circle around the sphere at constant angle with the rotation axis. Moreover, the whole sphere has to move uniformly in order to preserve inner products between all pairs of states on the surface (the only other option is something like a reflection which is not completely positive).

More formally, we can write $V$ in the form $$ V=\cos\phi I+i\sin\phi N $$ where $N$ is a $2\times 2$ matrix with eigenvalues $\pm 1$. Since $N$ must be Hermitian and traceless, we can decompose it in terms of the Pauli basis, recognising that it must be of the form $\vec{n}\cdot\vec{\sigma}$ where $\vec{n}\in\mathbb{R}^3$ and $\vec{n}\cdot\vec{n}=1$ (to get the eigenvalues right). The Bloch vector $\vec{n}$ is the axis of rotation and $\phi$ is the angle of rotation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.