Two-level systems, quantum mechanical systems whose state space is \(\CC^2\), are relatively simple yet still rich enough to exhibit most of the peculiarities of the quantum world. Moreover, they are physically important – we will consider nuclear magnetic resonance and the ammonia maser as examples.

Throughout we will assume, unless explicitly stated otherwise, that the column vectors and matrices representing state vectors and observables respectively are with respect to the standard basis of \(\CC^2\).

Properties of Pauli matrices

It’s straightforward to verify that the three Pauli matrices,
\begin{equation}
\boldsymbol{\sigma}_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}\qquad\boldsymbol{\sigma}_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\qquad\boldsymbol{\sigma}_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix},
\end{equation}
each square to the identity,
\begin{equation}
\boldsymbol{\sigma}_i^2=\mathbf{I},\qquad i=1,2,3
\end{equation}
and that they are all traceless,
\begin{equation}
\tr\boldsymbol{\sigma}_i=0,\qquad i=1,2,3.
\end{equation}
From these two facts it follows that each Pauli matrix has two eigenvalues \(\pm1\). We can compute the commutators and find
\begin{equation}
[\boldsymbol{\sigma}_i,\boldsymbol{\sigma}_j]=2i\epsilon_{ijk}\boldsymbol{\sigma}_k.
\end{equation}
Likewise the anti-commutators are
\begin{equation}
\{\boldsymbol{\sigma}_i,\boldsymbol{\sigma}_j\}=2\delta_{ij}\mathbf{I},
\end{equation}
and since the product of any pair of operators is one-half the sum of the anti-commutator and the commutator we have
\begin{equation}
\boldsymbol{\sigma}_i\boldsymbol{\sigma}_j=\delta_{ij}\mathbf{I}+i\epsilon_{ijk}\boldsymbol{\sigma}_k.
\end{equation}
A simple consequence of this is the rather useful relation,
\begin{equation}
(\mathbf{u}\cdot\boldsymbol{\sigma})(\mathbf{v}\cdot\boldsymbol{\sigma})=(\mathbf{u}\cdot\mathbf{v})\mathbf{I}+i(\mathbf{u}\times\mathbf{v})\cdot\boldsymbol{\sigma}.
\end{equation}

Hermitian and Unitary operators on \(\CC^2\)

Any \(2\times2\) matrix \(\mathbf{M}\) representing a linear operator on \(\CC^2\) can be expressed as a linear combination of the identity matrix and the three Pauli matrices,
\begin{equation}
\mathbf{M}=m_0\mathbf{I}+\mathbf{m}\cdot\boldsymbol{\sigma}
\end{equation}
where \(m_0\) and the components \(m_1,m_2,m_3\) of the vector \(\mathbf{m}\) are complex numbers and \(\boldsymbol{\sigma}\) is the vector with components the Pauli matrices, \(\boldsymbol{\sigma}=(\boldsymbol{\sigma}_1,\boldsymbol{\sigma}_2,\boldsymbol{\sigma}_3)\). It follows that
\begin{equation}
m_0=\frac{1}{2}\tr\mathbf{M},\quad m_i=\frac{1}{2}\tr(\mathbf{M}\boldsymbol{\sigma}_i).
\end{equation}

The condition for a matrix \(\mathbf{Q}\) to be Hermitian is that \(\mathbf{Q}^\dagger=\mathbf{Q}\). Thus we must have
\begin{equation}
\mathbf{Q}^\dagger=q_0^*\mathbf{I}+\mathbf{q}^*\cdot\boldsymbol{\sigma}=q_0\mathbf{I}+\mathbf{q}\cdot\boldsymbol{\sigma}=\mathbf{Q},
\end{equation}
where \(\mathbf{q}^*\) indicates the vector whose components are the complex conjugate of the vector \(\mathbf{q}\). It follows immediately that any Hermitian operator, that is, any qubit observable \(Q\), can be represented by a matrix
\begin{equation}
\mathbf{Q}=q_0\mathbf{I}+\mathbf{q}\cdot\boldsymbol{\sigma}
\end{equation}
where \(q_0\) and the components of the vector \(\mathbf{q}\) are all real.

The condition for a matrix \(\mathbf{U}\) to be unitary is \(\mathbf{U}^\dagger\mathbf{U}=\mathbf{U}\mathbf{U}^\dagger=\mathbf{I}\).

Theorem Any qubit unitary transformation \(U\) can, up to a choice of phase, be represented by a matrix \(\mathbf{U}\) given by
\begin{equation}
\mathbf{U}=\exp{(-i\theta\mathbf{n}\cdot\boldsymbol{\sigma})}
\end{equation}
where \(\theta\) and \(\mathbf{n}\) can be interpreted respectively as an angle and a unit vector in 3-dimensional space.

Proof We begin with the general form
\begin{equation}
\mathbf{U}=u_0\mathbf{I}+\mathbf{u}\cdot\boldsymbol{\sigma}
\end{equation}
in which \(u_0\) and \(\mathbf{u}\) are an arbitrary complex number and complex valued vector respectively. We will impose the condition \(\mathbf{U}^\dagger\mathbf{U}=\mathbf{I}\) but observe that this condition leaves an overall choice of phase unconstrained. Using this flexibility we can take \(u_0\) to be real then
\begin{align*}
\mathbf{U}^\dagger\mathbf{U}&=(u_0\mathbf{I}+\mathbf{u}^*\cdot\boldsymbol{\sigma})(u_0\mathbf{I}+\mathbf{u}\cdot\boldsymbol{\sigma})\\
&=u_0^2\mathbf{I}+2u_0\Real\mathbf{u}\cdot\boldsymbol{\sigma}+(\mathbf{u}^*\cdot\boldsymbol{\sigma})(\mathbf{u}\cdot\boldsymbol{\sigma})\\
&=u_0^2\mathbf{I}+2u_0\Real\mathbf{u}\cdot\boldsymbol{\sigma}+\mathbf{u}\cdot\mathbf{u}^*\mathbf{I}-i(\mathbf{u}\times\mathbf{u}^*)\cdot\boldsymbol{\sigma}=\mathbf{I}.
\end{align*}
Similarly, we have,
\begin{equation*}
\mathbf{U}\mathbf{U}^\dagger=u_0^2\mathbf{I}+2u_0\Real\mathbf{u}\cdot\boldsymbol{\sigma}+\mathbf{u}\cdot\mathbf{u}^*\mathbf{I}+i(\mathbf{u}\times\mathbf{u}^*)\cdot\boldsymbol{\sigma}=\mathbf{I}.
\end{equation*}
This means that we must have
\begin{equation}
\mathbf{u}\times\mathbf{u}^*=0,\label{eq:condition one}
\end{equation}
\begin{equation}
u_0\Real\mathbf{u}=0\label{eq:condition two}
\end{equation}
and
\begin{equation}
u_0^2+\mathbf{u}^*\cdot\mathbf{u}=1.\label{eq:condition three}
\end{equation}
Equation \eqref{eq:condition one} implies that
\begin{equation}
(\Real\mathbf{u}+i\Imag\mathbf{u})\times(\Real\mathbf{u}-i\Imag\mathbf{u})=-2i\Real\mathbf{u}\times\Imag\mathbf{u}=0
\end{equation}
that is, \(\Real\mathbf{u}\parallel\Imag\mathbf{u}\) so that we must be able to write \(\mathbf{u}=\alpha\mathbf{v}\) for some complex number \(\alpha\) and real vector \(\mathbf{v}\). The second condition, \eqref{eq:condition two}, tells us that either \(u_0=0\) or \(\mathbf{u}\) is pure imaginary. So that together, \eqref{eq:condition one} and \eqref{eq:condition two} imply that either \(\mathbf{u}=i\mathbf{v}\) or \(u_0=0\) and \(\mathbf{u}=\alpha\mathbf{v}\). In the latter case \eqref{eq:condition three} then implies that \(|\alpha|^2|\mathbf{v}|^2=1\) so we can write
\begin{align*}
\mathbf{U}&=\alpha\mathbf{v}\cdot\boldsymbol{\sigma}\\
&=e^{i\phi}|\alpha||\mathbf{v}|\frac{\mathbf{v}}{|\mathbf{v}|}\cdot\boldsymbol{\sigma}
\end{align*}
which up to a choice of phase has the form
\begin{equation}
\mathbf{U}=i\mathbf{n}\cdot\boldsymbol{\sigma}
\end{equation}
for some real unit vector \(\mathbf{n}\).
In the case that \(u_0\neq0\), since
\begin{equation}
u_0^2+\mathbf{v}\cdot\mathbf{v}=1
\end{equation}
we can write \(u_0=\cos\theta\) and \(\mathbf{v}=-\sin\theta\mathbf{n}\) for some angle \(\theta\) and a (real) unit vector \(\mathbf{n}\), that is, in either case, we have that up to an overall phase,
\begin{equation}
\mathbf{U}=\cos\theta\mathbf{I}-i\sin\theta\mathbf{n}\cdot\boldsymbol{\sigma}.
\end{equation}
Finally, observing that \((\mathbf{n}\cdot\boldsymbol{\sigma})^2=\mathbf{I}\) the desired matrix exponential can be written as
\begin{align*}
\exp{(-i\theta\mathbf{n}\cdot\boldsymbol{\sigma})}&=\mathbf{I}-i\theta\mathbf{n}\cdot\boldsymbol{\sigma}+\frac{i^2\theta^2}{2!}(\mathbf{n}\cdot\boldsymbol{\sigma})^2-\frac{i^3\theta^3}{3!}(\mathbf{n}\cdot\boldsymbol{\sigma})^3+\dots\\
&=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}+\dots\right)\mathbf{I}-i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}+\dots\right)\mathbf{n}\cdot\boldsymbol{\sigma}\\
&=\cos\theta\mathbf{I}-i\sin\theta\mathbf{n}\cdot\boldsymbol{\sigma}
\end{align*}\(\blacksquare\)

Unitary Operators on \(\CC^2\) and Rotations in \(\RR^3\)

The \(2\times2\) unitary matrices form a group under the usual matrix multiplication. This is the Lie group \(U(2)\). The \(2\times2\) unitary matrices with determinant 1 form the special unitary group \(SU(2)\). The previous theorem tells us that a general element of \(SU(2)\), which we’ll denote \(\mathbf{U}(\mathbf{n},\theta)\), can be written as
\begin{equation}
\mathbf{U}(\mathbf{n},\theta)=\exp{\left(-i\frac{\theta}{2}\mathbf{n}\cdot\boldsymbol{\sigma}\right)}.
\end{equation}
That is, the phase has been chosen such that \(\det\mathbf{U}(\mathbf{n},\theta)=1\).

The group of rotations in 3-dimensional space, denoted \(SO(3)\), consists of the all \(3\times3\)-orthogonal matrices with determinant 1. The reason for choosing the half-angle \(\theta\) is made clear in the following result.

Theorem There is a 2-to-1 group homomorphism from \(SU(2)\) to \(SO(3)\), \(\mathbf{U}(\mathbf{n},\theta)\mapsto\mathbf{R}(\mathbf{n},\theta)\) where \(\mathbf{R}(\mathbf{n},\theta)\) is the rotation through an angle \(\theta\) about the axis \(\mathbf{n}\).

Proof If we denote by \(H\) the set of traceless, Hermitian, \(2\times2\) matrices then it is not difficult to see that this is a real 3-dimensional vector space for which the Pauli matrices are a basis. The map \(f:\RR^3\mapto H\) given by \(f(\mathbf{v})=\mathbf{v}\cdot\boldsymbol{\sigma}\) is then an isomorphism of vector spaces. Defining an inner product on \(H\) according to
\begin{equation}
(\mathbf{M},\mathbf{N})_H=\frac{1}{2}\tr(\mathbf{M}\mathbf{N})
\end{equation}
this isomorphism becomes an isometry of vector spaces since,
\begin{equation*}
(f(\mathbf{v}),f(\mathbf{w}))_H=\frac{1}{2}\tr((\mathbf{v}\cdot\boldsymbol{\sigma})(\mathbf{w}\cdot\boldsymbol{\sigma}))=\mathbf{v}\cdot\mathbf{w}=(\mathbf{v},\mathbf{w})_{\RR^3},
\end{equation*}
that is, \(H\) and \(\RR^3\) are isometric. Now to any \(2\times2\) unitary matrix \(\mathbf{U}\) we can associate a linear operator \(T_\mathbf{U}\) on \(H\) such that \(T_\mathbf{U}\mathbf{M}=\mathbf{U}\mathbf{M}\mathbf{U}^\dagger\). This is clearly an isometry on \(H\) and so the corresponding linear operator on \(\RR^3\), \(f^{-1}\circ T_\mathbf{U}\circ f\) which we’ll denote \(\mathbf{R}_\mathbf{U}\) and is such that
\begin{equation}
\mathbf{R}_\mathbf{U}\mathbf{v}\cdot\boldsymbol{\sigma}=\mathbf{U}\mathbf{v}\cdot\boldsymbol{\sigma}\mathbf{U}^\dagger\label{eq:rotation from unitary}
\end{equation}
for any \(\mathbf{v}\in\RR^3\) and must be an isometrty, that is, an orthogonal operator. In fact, since
\begin{align*}
\tr\left((\mathbf{R}_\mathbf{U}\mathbf{e}_1\cdot\boldsymbol{\sigma})(\mathbf{R}_\mathbf{U}\mathbf{e}_2\cdot\boldsymbol{\sigma})
(\mathbf{R}_\mathbf{U}\mathbf{e}_3\cdot\boldsymbol{\sigma})\right)&=\tr\left((R_\mathbf{U})_1^i(R_\mathbf{U})_2^j(R_\mathbf{U})_3^k\boldsymbol{\sigma}_i\boldsymbol{\sigma}_j\boldsymbol{\sigma}_k\right)\\
&=2i\epsilon_{ijk}(R_\mathbf{U})_1^i(R_\mathbf{U})_2^j(R_\mathbf{U})_3^k\\
&=2i\det{\mathbf{R}_\mathbf{U}}
\end{align*}
and also
\begin{align*}
\tr\left((\mathbf{R}_\mathbf{U}\mathbf{e}_1\cdot\boldsymbol{\sigma})(\mathbf{R}_\mathbf{U}\mathbf{e}_2\cdot\boldsymbol{\sigma})
(\mathbf{R}_\mathbf{U}\mathbf{e}_3\cdot\boldsymbol{\sigma})\right)&=\tr\left(\mathbf{U}\boldsymbol{\sigma}_1\mathbf{U}^\dagger\mathbf{U}\boldsymbol{\sigma}_2\mathbf{U}^\dagger\mathbf{U}\boldsymbol{\sigma}_3\mathbf{U}^\dagger\right)\\
&=\tr(\boldsymbol{\sigma}_1\boldsymbol{\sigma}_2\boldsymbol{\sigma}_3)\\
&=2i
\end{align*}
we see that \(\mathbf{R}_\mathbf{U}\in SO(3)\). Also we observe that given two unitary matrices \(\mathbf{U}_1\) and \(\mathbf{U}_2\),
\begin{align*}
\mathbf{R}_{\mathbf{U}_1\mathbf{U}_2}\mathbf{v}\cdot\boldsymbol{\sigma}&=(\mathbf{U}_1\mathbf{U}_2)\mathbf{v}\cdot\boldsymbol{\sigma}(\mathbf{U}_1\mathbf{U}_2)^\dagger\\
&=\mathbf{U}_1\left(\mathbf{R}_{\mathbf{U}_2}\mathbf{v}\cdot\boldsymbol{\sigma}\right)\mathbf{U}_1^\dagger\\
&=(\mathbf{R}_{\mathbf{U}_1}\mathbf{R}_{\mathbf{U}_2}\mathbf{v})\cdot\boldsymbol{\sigma}
\end{align*}
So defining the map \(\Phi:SU(2)\mapto SO(3)\) such that
\begin{equation}
\Phi(\mathbf{U}(\mathbf{n},\theta))=\mathbf{R}_\mathbf{U},
\end{equation}
we have a group homomorphism. The kernel of this map consists of unitary matrices \(\mathbf{U}(\mathbf{n},\theta)\) such that
\begin{equation*}
\mathbf{U}(\mathbf{n},\theta)(\mathbf{v}\cdot\boldsymbol{\sigma})=(\mathbf{v}\cdot\boldsymbol{\sigma})\mathbf{U}(\mathbf{n},\theta)
\end{equation*}
for any vector \(\mathbf{v}\). It follows that \(\mathbf{U}(\mathbf{n},\theta)\) must be a multiple of the identity matrix and since \(\det\mathbf{U}(\mathbf{n},\theta)=1\) it can only be \(\pm\mathbf{I}\). Thus, \(\ker\Phi=\{\mathbf{I},-\mathbf{I}\}\) and so the homomorphism is 2-to-1. To confirm the nature of the spatial rotation corresponding to \(\mathbf{U}(\mathbf{n},\theta)\), chose \(\mathbf{v}=\mathbf{n}\) in \eqref{eq:rotation from unitary} to see that \(\mathbf{R}_\mathbf{U}\mathbf{n}=\mathbf{n}\) so that \(\mathbf{R}_\mathbf{U}\) is a rotation about the axis \(\mathbf{n}\). To determine the angle \(\gamma\) of rotation we note that if \(\mathbf{m}\) is a vector perpendicular to \(\mathbf{n}\) then \(\cos\gamma=(\mathbf{R}_\mathbf{U}\mathbf{m})\cdot\mathbf{m}\) and we have
\begin{align*}
\cos\gamma&=(\mathbf{R}_\mathbf{U}\mathbf{m})\cdot\mathbf{m}\\
&=\frac{1}{2}\tr\left(\left(\cos\frac{\theta}{2}\mathbf{I}-i\sin\frac{\theta}{2}\mathbf{n}\cdot\boldsymbol{\sigma}\right)(\mathbf{m}\cdot\boldsymbol{\sigma})\right.\\
&\qquad\times\left.\left(\cos\frac{\theta}{2}\mathbf{I}+i\sin\frac{\theta}{2}\mathbf{n}\cdot\boldsymbol{\sigma}\right)(\mathbf{m}\cdot\boldsymbol{\sigma})\right)\\
&=\frac{1}{2}\tr\left(\left(\cos\frac{\theta}{2}\mathbf{m}\cdot\boldsymbol{\sigma}-\sin\frac{\theta}{2}(\mathbf{m}\times\mathbf{n})\cdot\boldsymbol{\sigma}\right)\right.\\
&\qquad\times\left.\left(\cos\frac{\theta}{2}\mathbf{m}\cdot\boldsymbol{\sigma}+\sin\frac{\theta}{2}(\mathbf{m}\times\mathbf{n})\cdot\boldsymbol{\sigma}\right)\right)\\
&=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}\\
&=\cos\theta
\end{align*}
so that the unitary operator \(\mathbf{U}(\mathbf{n},\theta)\) corresponds to a spatial rotation about the axis \(\mathbf{n}\) through an angle \(\theta\). We therefore denote the rotation \(\mathbf{R}(\mathbf{n},\theta)\).\(\blacksquare\)

The Bloch sphere revisited

We have seen that any qubit observable, \(Q\), can be represented as a matrix

\begin{equation*}
\mathbf{Q}=q_0\mathbf{I}+\mathbf{q}\cdot\boldsymbol{\sigma}
\end{equation*}

where \(q_0\in\RR\) and \(\mathbf{q}\in\RR^3\). Recall the Bloch sphere,

in which a general qubit state, \(\ket{\psi}\), is given by,

\begin{equation*}
\ket{\psi}=\cos(\theta/2)\ket{0}+e^{i\varphi}\sin(\theta/2)\ket{1}.
\end{equation*}

It can be useful to denote this state vector by \(\ket{\mathbf{n};+}\), where \(\mathbf{n}\) is the unit vector with polar coordinates \((1,\theta,\phi)\), that is,
\begin{equation*}
\ket{\mathbf{n};+}=\cos(\theta/2)\ket{0}+e^{i\varphi}\sin(\theta/2)\ket{1}.
\end{equation*}
where
\begin{equation*}
\mathbf{n}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)
\end{equation*}
and
\begin{equation*}
\ket{\mathbf{n};-}=\sin(\theta/2)\ket{0}-e^{i\varphi}\cos(\theta/2)\ket{1}.
\end{equation*}
corresponding to the antipodal point on the Bloch sphere (\(\theta\mapto\pi-\theta\) and \(\phi\mapto2\pi+\phi\)). Indeed, \(\ket{\mathbf{n};\pm}\) are precisely the eigenvectors of the observable \(\mathbf{n}\cdot\boldsymbol{\sigma}\),
\begin{equation}
(\mathbf{n}\cdot\boldsymbol{\sigma})\ket{\mathbf{n};\pm}=\pm\ket{\mathbf{n};\pm}.
\end{equation}
For example,
\begin{align*}
(\mathbf{n}\cdot\boldsymbol{\sigma})\ket{\mathbf{n};+}&=\begin{pmatrix}\cos\theta&&e^{-i\phi}\sin\theta\\ e^{i\phi}\sin\theta&&-\cos\theta\end{pmatrix}\begin{pmatrix}\cos\frac{\theta}{2}\\e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}\\
&=\begin{pmatrix}\cos\theta\cos\frac{\theta}{2}+\sin\theta\sin\frac{\theta}{2}\\
e^{i\phi}(\sin\theta\cos\frac{\theta}{2}-\cos\theta\sin\frac{\theta}{2})\end{pmatrix}\\
&=\begin{pmatrix}\cos\frac{\theta}{2}\\e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}=\ket{\mathbf{n};+}
\end{align*}
Note of course that \(\braket{\mathbf{n};+|\mathbf{n};-}=0\), that is \(\ket{\mathbf{n};+}\) and \(\ket{\mathbf{n};-}\) are orthogonal as state vectors in the Hilbert space \(\CC^2\) though of course \(\mathbf{n}\) and \(-\mathbf{n}\) are certainly not orhthogonal vectors in \(\RR^3\)!

We’ve seen that there is a 2-to-1 homomorphism from \(SU(2)\) to \(SO(3)\) such that \(\mathbf{U}(\mathbf{n},\theta)\mapsto\mathbf{R}(\mathbf{n},\theta)\) where
\begin{equation*}
\mathbf{U}(\mathbf{n},\theta)=\exp\left(-i\frac{\theta}{2}\mathbf{n}\cdot\boldsymbol{\sigma}\right)=\cos\frac{\theta}{2}\mathbf{I}-i\sin\frac{\theta}{2}\mathbf{n}\cdot\boldsymbol{\sigma}
\end{equation*}
and the rotation \(\mathbf{R}(\mathbf{n},\theta)\) is such that
\begin{equation*}
(\mathbf{R}(\mathbf{n},\theta)\mathbf{v})\cdot\boldsymbol{\sigma}=\mathbf{U}(\mathbf{n},\theta)\mathbf{v}\cdot\boldsymbol{\sigma}\mathbf{U}(\mathbf{n},\theta)^\dagger,
\end{equation*}
which we confirmed was a rotation of \(\theta\) about the axis \(\mathbf{n}\). This means that for an arbitrary vector \(\mathbf{v}\in\RR^3\),
\begin{equation}
\mathbf{R}(\mathbf{n},\theta)\mathbf{v}=\cos\theta\mathbf{v}+(1-\cos\theta)(\mathbf{v}\cdot\mathbf{n})\mathbf{n}+\sin\theta\mathbf{n}\times\mathbf{v}.
\end{equation}

In terms of the state vector notation \(\ket{\mathbf{n};\pm}\) relating unit vectors in \(\RR^3\) to states on the Bloch sphere we have that
\begin{equation}
\mathbf{U}(\mathbf{m},\alpha)\ket{\mathbf{n};+}=\ket{\mathbf{R}(\mathbf{m},\alpha)\mathbf{n};+}
\end{equation}
since
\begin{align*}
\left((\mathbf{R}(\mathbf{m},\alpha)\mathbf{n})\cdot\boldsymbol{\sigma}\right)\mathbf{U}(\mathbf{m},\alpha)\ket{\mathbf{n};+}&=\mathbf{U}(\mathbf{m},\alpha)\mathbf{n}\cdot\boldsymbol{\sigma}\ket{\mathbf{n};+}\\
&=\mathbf{U}(\mathbf{m},\alpha)\ket{\mathbf{n};+}
\end{align*}

Thus, as would have been anticipated, the unitary operator \(\mathbf{U}(\mathbf{m},\alpha)\) rotates the Bloch sphere state \(\ket{\mathbf{n};+}\) by an angle \(\alpha\) around the axis \(\mathbf{m}\).