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Adam Zalcman
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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 $(4)$.

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 $(4)$.

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 $(4)$.

Source Link
Adam Zalcman
  • 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 $(4)$.