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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 $$(4)$$.

• 11.7k
• 2
• 12
• 49

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 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 $$(4)$$.