4.1.Qubits
In this chapter we want to understand the smallest dimensional (yet
extremely interesting) quantum system. Because of the equivalence
relation \(\ket{\psi} \sim c \ket{\psi}\), a one-dimensional Hilbert
space is trivial since it only describes a single state. Therefore the
smallest non-trivial system has dimension two, and in QI we refer to
such a system as a qubit. A standard orthonormal basis is
labelled \(\{ \ket{0}, \ket{1} \}\) (sometimes you will also find it
written as \(\{\ket{\uparrow},\ket{\downarrow}\}\) denoting the two
states of a spin 1/2 particle with spin “up” or “down”) so we see
the analogy with a classical bit which can take either value \(0\) or
\(1\) hence we usually call the basis \(\{ \ket{0}, \ket{1} \}\) the
computational basis. The difference for the qubit is that the
state can be a linear combination of \(\ket{0}\) and \(\ket{1}\). The
qubit is important since many general features can be understand in
this simple Hilbert space. Also, larger systems can often be described
a being built from several qubits – this is used in quantum
information and especially in quantum computation.
So let us try and characterise the most general pure state of the
qubit Hilbert space \(\mathcal{H}_q =\mbox{span} \{ \ket{0}, \ket{1}
\}\). Now if we have a qubit system the most general pure state can
be written as a linear combination of the two basis vectors
\[
\ket{\psi} = a \ket{0}+b\ket{1}\,,
\]
with \(a,b\in\mathbb{C}\) and \(|a|^2+|b|^2=1\) so that the state is already normalised \(\ip{\psi}{\psi}=1\).
The most general qubit state is labelled by two complex numbers, subject to a norm constraint and the fact that the overall phase is irrelevant. This leaves two real parameters, which parametrise a sphere.
Up to an irrelevant overall phase we can assume that \(a\in\mathbb{R}\),
hence any normalised pure state can be written in the form
\[ \ket{\psi} = \cos\left(\frac{\theta}{2}\right)\ket{0} + e^{i\phi}\sin\left(\frac{\theta}{2}\right)\ket{1} \]
where \(0 \le \theta \le \pi\) and \(0 \le \phi \lt{} 2\pi\).
Note that the normalisation condition \(|a|^2+|b|^2=1\) now becomes \(\cos\left(\frac{\theta}{2}\right)^2+ \vert e^{i\phi}\sin\left(\frac{\theta}{2}\right)\vert^2=1\).
This new coordinates \((\theta,\phi)\) are just a convenient
parametrisation of \(|a|^2+|b|^2=1\) with \(a\in\mathbb{R}\) in terms of
angles which correspond to the angles in spherical polar
coordinates. Indeed the equation \(|a|^2+|b|^2=1\) with
\(a\in\mathbb{R}\) can be rewritten as \(a^2+
(\mbox{Re}\,b)^2+(\mbox{Im}\, b)^2 =1\) which is indeed the equation
describing a 2-dimensional sphere, what we call \(S^2\). Therefore we
have a one-to-one mapping between pure qubits states and points on a
sphere (say of radius \(1\).) This is called the Bloch Sphere.
Now any point on the Bloch sphere can also be labelled by its position
vector, called the
Bloch vector\(\vec{r}=\left(\begin{matrix}x\\y\\z\end{matrix}\right)\) which has
\(|\vec{r}| =\sqrt{x^2+y^2+z^2}= 1\). The dictionary between
Cartesian coordinates and polar coordinates is very simple
\[
x = \sin(\theta) \cos(\phi)\,,\quad y=\sin(\theta) \sin(\phi)\,,\quad z= \cos(\theta)\,,
\]
\(\theta\) is the polar angle measured from the \(z\)-axis, while
\(\phi\) is the azimuthal angle measured around the \(z\)-axis as in
Figure
bloch.
There are six special states in the qubit system, which are eigenvectors of three special operators (to be discussed later).
There are six special states on the Bloch sphere whose cartesian
coordinates are simply
\(\left(\begin{matrix}\pm1\\0\\0\end{matrix}\right),\left(\begin{matrix}0\\\pm1\\0\end{matrix}\right),\left(\begin{matrix}0\\0\\ \pm1\end{matrix}\right)\)
given by
\begin{align*}
\vec{r}&= \left(\begin{matrix}1\\0\\0\end{matrix}\right)\,,\qquad (\theta,\phi) = (\pi/2,0)\,,\\
&\qquad\rightarrow\quad\ket{+}=\frac{1}{\sqrt{2}} (\ket{0}+ \ket{1})\to\frac{1}{\sqrt{2}} \left(\begin{matrix}1 \\ 1\end{matrix}\right)\,,\\[1ex]
\vec{r}&= \left(\begin{matrix}-1\\0\\0\end{matrix}\right)\,,\qquad (\theta,\phi) = (\pi/2,\pi)\,,\\
&\qquad\rightarrow\quad\ket{-}=\frac{1}{\sqrt{2}} (\ket{0}- \ket{1})\to\frac{1}{\sqrt{2}} \left(\begin{matrix}1 \\ -1\end{matrix}\right)\,,\\[1ex]
\vec{r}&= \left(\begin{matrix}0\\1\\0\end{matrix}\right)\,,\qquad (\theta,\phi) = (\pi/2,\pi/2)\,,\\
&\qquad\rightarrow\quad \ket{L}=\frac{1}{\sqrt{2}} (\ket{0}+i \ket{1})\to\frac{1}{\sqrt{2}} \left(\begin{matrix}1 \\ i\end{matrix}\right)\,,\\[1ex]
\vec{r}&= \left(\begin{matrix}0\\-1\\0\end{matrix}\right)\,,\qquad (\theta,\phi) = (\pi/2,3\pi/2)\,,\\
&\qquad\rightarrow\quad\ket{R}=\frac{1}{\sqrt{2}} (\ket{0}- i\ket{1})\to\frac{1}{\sqrt{2}} \left(\begin{matrix}1 \\ -i \end{matrix}\right)\,,\\[1ex]
\vec{r}& \left(\begin{matrix}0\\0\\1\end{matrix}\right)\,,\qquad (\theta,\phi) = (0, \cdot)\,,\\
&\qquad\rightarrow\quad\ket{0}\to\left(\begin{matrix}1 \\ 0\end{matrix}\right) \,,\\[1ex]
\vec{r}&= \left(\begin{matrix}0\\0\\-1\end{matrix}\right)\,,\qquad (\theta,\phi) = (\pi ,\cdot)\,,\\
&\qquad\rightarrow\quad\ket{1}\to\left(\begin{matrix}0 \\1\end{matrix}\right)\,.
\end{align*}
These states are special as they lie at the intersection of the Bloch
sphere with the cardinal axis and we will see later on they will each
have well defined measurements for three special observables for the
qubit system.
Show that besides the computational basis
\(\{\ket{0},\ket{1}\}\), the qubit Hilbert space \(\mathcal{H}_q\) can be
described in terms of the orthonormal basis \(\{\ket{+},\ket{-}\}\) or
by the orthonormal basis \(\{\ket{L},\ket{R}\}\).
4.2.Inside the Bloch sphere
We want to rewrite now the density matrix for a general pure state
written using polar angles to cartesian coordinates. To this end we
start with a generic pure state \( \ket{\psi} =
\cos\left(\frac{\theta}{2}\right)\ket{0} +
e^{i\phi}\sin\left(\frac{\theta}{2}\right)\ket{1} \) and we first
rewrite it using as above the standard basis representation for the
two basis ket vector \(\ket{0}\to \left(\begin{matrix}1
\\ 0 \end{matrix}\right)\,,\,\ket{1}\to \left(\begin{matrix}0
\\ 1 \end{matrix}\right)\) hence we have
\begin{align*}
&\ket{\psi}\to\left(\begin{matrix}\cos\left(\frac{\theta}{2}\right) \\ e^{i\phi}\sin\left(\frac{\theta}{2}\right) \end{matrix}\right) \,,\\
\,\hat{\rho} &=\ket{\psi}\bra{\psi} \to \rho = \left(\begin{matrix}\cos\left(\frac{\theta}{2}\right) \\ e^{i\phi}\sin\left(\frac{\theta}{2}\right) \end{matrix}\right) \left(\begin{matrix}\cos\left(\frac{\theta}{2}\right) & e^{-i\phi}\sin\left(\frac{\theta}{2}\right)\end{matrix}\right)\\
&\rho =
\frac{1}{2} \left(\begin{matrix}1+\cos(\theta) & e^{-i\phi}\sin(\theta)\\
e^{+i\phi}\sin(\theta) &1-\cos(\theta) \end{matrix}\right)=\frac{1}{2} \left(\begin{matrix}1+z& x-i y \\ x+ i y &1-z\end{matrix}\right)\,,
\end{align*}
where in the last line we used some simple trigonometric identities and finally the dictionary spherical \(\leftrightarrow\) cartesian coords.
It can be seen quite easily now that the density matrix can be written in terms of the Bloch vector \(\vec{r}=\left(\begin{matrix}x\\y\\z\end{matrix}\right)\) as
\(\rho = \frac{1}{2}\left( \mathbb{I}_2+ \vec{r}\cdot\vec{\sigma} \right)\), where we introduce the three matrices
\(\sigma_i\) called the Pauli \(\sigma\)-matrices
\[
\sigma_1 = \left(\begin{matrix}0&1\\ 1 & 0 \end{matrix}\right)\,,\quad \sigma_2 = \left(\begin{matrix}0&-i\\ i & 0 \end{matrix}\right)\,,\quad \sigma_3 = \left(\begin{matrix}1&0\\ 0 & -1 \end{matrix}\right)\,,
\]
A state with given Bloch vector can be mapped to a density matrix using the Pauli matrices.
so \(\vec{r}\cdot\vec{\sigma} = r_1 \sigma_1 +r_2 \sigma_2 +r_3 \sigma_3\).
As the density matrix is linear in the Bloch vector, mixed states also
have Bloch vector given in terms of a linear combination of the Bloch
vectors of the states in the ensemble. Since the coefficients are
probabilities, it is easy to see that the Bloch vector for a mixed
state must be the position vector of a point inside the Bloch sphere.
Let us check this in detail. Suppose we have the ensemble
\(\{(p_i,\rho_i)\}\) for some probabilities \(p_i\) and pure states
\(\rho_i\) defined by vectors \(\vec{r}_i\) on the Bloch sphere, i.e. \(|
\vec{r}_i|^2=1\). The density matrix for this mixed state is given by
\begin{equation}
\begin{aligned}
\rho &= \sum_i p_i \rho_i = \sum_i p_i \frac{\left( \mathbb{I}_2 + \vec{r}_i \cdot\vec{\sigma} \right)}{2} \\ &= \frac{1}{2}\left( \mathbb{I}_2 + \vec{r}\cdot\vec{\sigma} \right)\,,
\end{aligned}
\end{equation}
where we defined the vector \(\vec{r} = \sum_i p_i \vec{r}_i\).
As said above, since the density matrix is linear in the Bloch vector, mixed states as well can be represented in terms of a Bloch vector \(\vec{r}\).
Let us check now where this vector lies:
\begin{equation}
\begin{aligned}
| \vec{r} |^2 &= | \sum_i p_i \vec{r}_i |^2 = \sum_{ij} p_i p_j \,\vec{r}_i \cdot \vec{r}_j\\
&\leq \sum_{ij} p_i p_j |\vec{r}_i| | \vec{r}_j |\leq \sum_{ij} p_i p_j \leq 1\,,
\end{aligned}
\end{equation}
where we used the real Cauchy Schwartz inequality. Note that equality
holds if and only if all \(\vec{r}_i\) are collinear hence all the
density matrices \(\rho_i\) of the ensemble are given by the same
density matrix, hence we do not really have a mixed state but rather a
pure one.
For genuine mixed states the inequality is strict and so mixed states
are defined by point inside the Bloch sphere, i.e. \(| \vec{r} | \lt{}1\).
The Bloch sphere gives us a nice geometrical interpretation for the
full set of states of the qubit system: points on the sphere define
pure states, points inside the sphere define mixed states.
Remember the previous test to distinguish whether a density matrix
corresponds to a pure or a mixed state by considering whether
\(\Tr(\rho^2)=1\) or \(\lt{}1\) respectively. The condition \(\Tr(\rho^2)\lt{}1\)
for \(\rho\) to correspond to a mixed state has now for a qubit a
geometric interpretation:
\[ \Tr(\rho^2) = \frac{1}{2}\left( 1 + |\vec{r}|^2 \right) \le 1\,, \]
mixed states are
inside the Bloch sphere \(|\vec{r}|\lt{}1\) , pure
state are
on the Bloch sphere \(|\vec{r}|=1\).
4.3.Time-evolution of a qubit
Unitary transformations of a qubit are represented as rotations of the
Bloch sphere about the origin. This illustrates that unitary
transformations cannot transform pure states to mixed states, but also
we see that not all mixed states are related in this way. Indeed
\(\Tr(\rho^2) = (1 + |\vec{r}|^2)/2\) is invariant under unitary
transformations, and is a measure of how mixed a state is, with
\(\Tr(\rho^2) = 1\) for pure states to \(\Tr(\rho^2) = 1/2\) for the
“most mixed” state corresponding to the origin, i.e. \(\rho =
\mathbb{I}_2/2\). Of course, measurements are not unitary
transformations and they act as projectors hence they can transform
any state to a pure state.
Show that the quantity \(\Tr(\hat{\rho}^2) \) is invariant under time evolution.
Solution: We simply need to remember that a density matrix \(\hat{\rho}\) transforms under time evolution has \(
\hat{\rho} \to \hat{U} \hat{\rho}\hat{U}^\dagger\) hence
\[\hat{\rho}^2 \to \hat{U} \hat{\rho}\hat{U}^\dagger \hat{U} \hat{\rho}\hat{U}^\dagger =\hat{U} \hat{\rho}^2\hat{U}^\dagger\]
using the fact that time evolution is a unitary operation \(\hat{U}^\dagger \hat{U} =\hat{I}\).
We then conclude that \( \Tr(\hat{U} \hat{\rho}^2\hat{U}^\dagger)= \Tr(\hat{\rho}^2) \) using cyclicity of the trace. Since \(\Tr(\hat{\rho}^2) \) measures how mixed a state is (we will quantify better later on) it is then clear that time evolution cannot possibly change that.
Let \(\vec{r}_1\) and \(\vec{r}_2\) denote two distinct points on the
Bloch sphere, i.e. \(|\vec{r}_1|= |\vec{r}_1|=1\) and \(\vec{r}_1\neq
\vec{r}_2\), and consider the ensemble \(\{
(p,\vec{r}_1),(1-p,\vec{r}_2)\}\) with \(0 \leq p \leq 1\). The density
matrix corresponding to this ensemble is given by
\[
\rho = \rho_1 +\rho_2 = \frac{1}{2}(\mathbb{I}_2 + \vec{r}\cdot \sigma)\,,
\]
with \(\vec{r} = p \,\vec{r}_1+(1-p)\vec{r}_2\).
Note that geometrically the Bloch vector \(\vec{r}\) lies on the line between the points \(\vec{r}_1\) and \(\vec{r}_2\) and since \(0 \leq p \leq 1\) the point \(\vec{r}\) on this line will always fall within the Bloch sphere.
The state \(\rho\) will be pure if and only if \(p=1\), in which case
\(\rho=\rho_1\), or \(p=0\), in which case \(\rho=\rho_2\). Note in
particular that mixing can never produce a state farther from the
origin then the farthest initial state. Furthermore once we have
chosen a mixed state, i.e. a point inside the Bloch sphere, we can
find an infinite number of ways to write it as an ensemble of two pure
states! We just need to consider any line passing through this point
which will intersect the Bloch sphere in two points corresponding to
the two pure states of which this mixed state is an ensemble of.
In particular the “most” mixed of the qubit states is \(\rho =
\mathbb{I}_2 / 2\) which correspond of an ensemble of any two antipodal
points, for example the North and South poles \(\ket{0}, \ket{1}\)
states (but any other two antipodal points on the sphere will produce
the same), each one of them with probability \(50\%\).
For a qubit system it is simple to define a “distance” between
states given by the geometric distance of the relative positions using
the Bloch sphere, i.e. \(| \vec{r}_1 - \vec{r}_2|\) if the two states
are described by the Bloch vectors \(\vec{r}_1\,,\,\vec{r}_2\). This
turns out to be equal to what is called the
trace distance
between the two states
\[
D(\hat{\rho}_1,\hat{\rho}_2) = \frac{1}{2}\Tr \vert \hat{\rho}_1 -\hat{\rho}_2 \vert\,,
\]
where we need to give the definition of the operator \(| \hat{A} |\).
The operator \(| \hat{A} |\) is defined to be the positive operator
\[
\vert \hat{A} \vert = \sqrt{ \hat{A}^\dagger \hat{A} }\,.
\]
Note (check) that the operator \(\hat{A}^\dagger \hat{A}\) is both
hermitian and positive hence its square root is well
defined. For this, you just need to chose a basis that diagonalises
\(\hat{A}^\dagger \hat{A}\) with non-negative eigenvalues (because the
operator is hermitian and positive); in this basis the operator
\(\sqrt{ \hat{A}^\dagger \hat{A} }\) will then be diagonal with
eigenvalues given by the square root of the eigenvalues of
\(\hat{A}^\dagger \hat{A}\).
In practical terms if the operator \( \hat{A} \) is hermitian with
eigenvalues \(a_i\) then \( \hat{A}^\dagger \hat{A} =\hat{A}^2 \) and we
have that
\[
\Tr \vert \hat{A} \vert = \sum_i \vert a_i\vert\,.
\]
With this definition for \(\Tr \vert \hat{A} \vert\) we can then easily compute the trace distance between two qubit states
\begin{equation}
\begin{aligned}
D(\hat{\rho}_1,\hat{\rho}_2) &= \frac{1}{2}\Tr \vert \hat{\rho}_1 -\hat{\rho}_2 \vert =\frac{1}{4} \Tr \vert (\vec{r}_1 -\vec{r}_2)\cdot \sigma \vert \\
& =\frac{1}{2}\vert \vec{r}_1 -\vec{r}_2\vert\,,
\end{aligned}
\end{equation}
and indeed as anticipated the trace distance for qubit states
correspond precisely to the geometric distance in the Bloch sphere
representation.
The “trace distance” between two states
described by a density matrix corresponds to the geometric distance
between the two states in the Bloch sphere representation.
We can rewrite the trace distance between two states as
\begin{equation}
\begin{aligned}
D(\hat{\rho}_1,\hat{\rho}_2) &=\frac{1}{2}\Tr \vert \hat{\rho}_1 -\hat{\rho}_2 \vert \\
&= \frac{1}{2} \Tr \sqrt{ (\hat{\rho}_1 -\hat{\rho}_2)^\dagger (\hat{\rho}_1 -\hat{\rho}_2) } \\
&= \frac{1}{2} \Tr \sqrt{ (\hat{\rho}_1 -\hat{\rho}_2)^2 } = \frac{1}{2}\sum_i \vert \lambda_i\vert\,,
\end{aligned}
\end{equation}
where \(\lambda_i\) are the eigenvalues of the hermitian but not necessarily positive operator \(\hat{\rho}_1 -\hat{\rho}_2\).
This defines a
metric on the space of density matrices, i.e. something with the following properties
it is non-negative \(D(\hat{\rho}_1, \hat{\rho}_2) \geq 0\);
it is symmetric \(D(\hat{\rho}_1, \hat{\rho}_2) = D(\hat{\rho}_2, \hat{\rho}_1)\);
it satisfies triangle inequality \(D(\hat{\rho}_1, \hat{\rho}_3)\leq D(\hat{\rho}_1, \hat{\rho}_2)+D(\hat{\rho}_2, \hat{\rho}_3)\);
it separates points \(D(\hat{\rho}_1, \hat{\rho}_2) =0 \Leftrightarrow \hat{\rho}_1 = \hat{\rho}_2\).
Note that there are many notions of “distance” when discussing quantum mechanical states.
The trace distance is a possible meaningful notion of distance between states in that it tells us how distinguishable with measurements two states are, i.e. the closer they are the less “distinguishable” they are by simply measuring observables.
Another such notion is what is called fidelity although we will not cover this.
Compute the trace distance between the qubit states
\begin{align*}
\hat{\rho}_1 &= \frac{1}{4} \ket{0}\bra{0} + \frac{3}{4} \ket{1}\bra{1} \rightarrow \rho_1 =\left( \begin{matrix} \frac{1}{4} & 0\\ 0& \frac{3}{4}\end{matrix}\right)\,,\\
\hat{\rho}_2 &= \frac{2}{3} \ket{0}\bra{0} + \frac{1}{3} \ket{1}\bra{1} \rightarrow\rho_2=\left( \begin{matrix} \frac{2}{3} & 0\\ 0& \frac{1}{3}\end{matrix}\right)\,.
\end{align*}
If we compute the matrix \(\rho_1-\rho_2\) we obtain
\begin{equation}
\begin{aligned}
\rho_1-\rho_2 &=\left( \begin{matrix} \frac{-5}{12} & 0\\ 0& \frac{5}{12}\end{matrix}\right)\\
\Rightarrow D(\hat{\rho}_1,\hat{\rho}_2)&=\frac{1}{2}\Tr \vert \rho_1 -\rho_2 \vert \\
&= \frac{1}{2} \left( \left\vert \frac{-5}{12} \right\vert +\left\vert \frac{5}{12}\right\vert \right)= \frac{5}{12}\,.
\end{aligned}
\end{equation}
We could have also computed the two Bloch vectors associated to the two density matrices \(\vec{r}_1 = (0,0,-1/2)^T\) and \(\vec{r}_2 = (0,0,1/3)^T\) and in fact we have
\[
D(\hat{\rho}_1,\hat{\rho}_2) = \frac{1}{2} \vert \vec{r}_1 - \vec{r}_2 \vert = \frac{1}{2} \left\vert \left( \begin{matrix}0 \\ 0 \\ -\frac{5}{6}
\end{matrix}\right) \right\vert = \frac{5}{12} \,.
\]
4.4.Pauli Matrices
We will summarise here some key properties of the so-called
Pauli matrices, which were introduced above when we constructed
the density matrix corresponding to a generic qubit state. The
following can be checked easily.
\begin{align*}
\sigma_1^\dagger&= \sigma_1\,,\qquad\sigma_2^\dagger =\sigma_2\,,\qquad\sigma_3^\dagger=\sigma_3\,,\\
\mbox{Tr}(\sigma_1)&=\mbox{Tr}(\sigma_2)=\mbox{Tr}(\sigma_3)=0\,,\\
[\sigma_i,\sigma_j] & =\sigma_i\sigma_j-\sigma_j\sigma_i= 2 i\, \epsilon_{ijk}\, \sigma_k\,,\\
\{\sigma_i,\sigma_j\} & = \sigma_i\sigma_j+\sigma_j\sigma_i = 2\delta_{ij}\mathbb{I}_2\,,\\
\sigma_i \sigma_j &= \delta_{ij} \mathbb{I}_2 + i\,\epsilon_{ijk}\sigma_k\,,
\end{align*}
where \(\delta_{ij}\) is the Kronecker delta and \(\epsilon_{ijk}\) the Levi-Civita tensor.
So the Pauli matrices are Hermitian, traceless matrices, they actually form a basis for the vector space of \(2\times 2\) hermitian, traceless matrices (Ex. check that this is a vector space over the real numbers).
If we define the operators
\begin{align*}
X& = \frac{1}{2}(\mathbb{I}_2- \sigma_1) = \frac{1}{2}\left(\begin{matrix}1& -1\\ -1 & 1\end{matrix}\right)\,,\\
Y &= \frac{1}{2}(\mathbb{I}_2- \sigma_2)= \frac{1}{2}\left(\begin{matrix}1& i\\ -i & 1\end{matrix}\right)\,,\\
Z &= \frac{1}{2}(\mathbb{I}_2- \sigma_3)= \frac{1}{2}\left(\begin{matrix}0& 0\\ 0 & 2\end{matrix}\right)\,,\qquad
\end{align*}
we can easily see that the six “special” states defined above are precisely the eigenvectors of these three operators with eigenvalues either \(0\) or \(1\), i.e.
\begin{align*}
X \ket{+} = 0 \ket{+}\,,\qquad &X \ket{-} = 1 \ket{-}\,,\\
Y \ket{L} = 0 \ket{L}\,,\qquad&Y \ket{R} = 1 \ket{R}\,,\\
Z \ket{0} = 0 \ket{0}\,,\qquad&Z \ket{1} = 1 \ket{1}\,.
\end{align*}
Finally another important property of the Pauli matrices is that their exponential are unitary matrices
\begin{align*}
e^{ i \alpha \sigma_1} &= \left(\begin{matrix}\cos( \alpha) & i \sin(\alpha)\\ i \sin(\alpha) & \cos(\alpha) \end{matrix}\right)= \cos(\alpha) \mathbb{I}_2 +i \sin(\alpha) \sigma_1\,,\\
e^{ i \alpha \sigma_2} &= \left(\begin{matrix}\cos( \alpha) & \sin(\alpha)\\ - \sin(\alpha) & \cos(\alpha) \end{matrix}\right)= \cos(\alpha) \mathbb{I}_2 +i \sin(\alpha) \sigma_2\,,\\
e^{ i \alpha \sigma_3} &= \left(\begin{matrix}e^{i \alpha} & 0 \\0 &e^{- i \alpha} \end{matrix}\right)= \cos(\alpha) \mathbb{I}_2 +i \sin(\alpha) \sigma_3\,,
\end{align*}
where \(\alpha \in \mathbb{R}\).
Compute these exponentials and check that these are indeed unitary matrices.
More in general we can consider the unitary transformation
\[
U_\alpha(\vec{n}) = e^{i \alpha \vec{n}\cdot \sigma} = \cos(\alpha) \mathbb{I}_2 +i \sin(\alpha) \vec{n}\cdot \sigma\,,
\]
where again \(\alpha \in \mathbb{R}\) and \(\vec{n}\in\mathbb{R}^3\) is a \(3\)-dimensional unit vector, i.e. \(| \vec{n}|^2=1\).
This is a unitary transformation so it is a perfectly valid time evolution operator.
If we now act on the density matrix \(\rho = \frac{1}{2}\left( \mathbb{I}_2 + \vec{r}\cdot \sigma \right)\) of a state defined by the vector \(\vec{r}\) on the Bloch sphere we can see that
\[
U_\alpha(\vec{n}) \rho \,U_\alpha(\vec{n})^\dagger = \frac{1}{2}\left[ \mathbb{I}_2 + ( R_\alpha(\vec{n}) \vec{r}) \cdot \sigma\right]\,,
\]
where \(R_\alpha(\vec{n})\) is the \(3\times 3\) orthogonal matrix corresponding to a rotation of an angle \(2\alpha\) around the axis defined by the unit vector \(\vec{n}\) acting on the three dimensional vector \(\vec{r}\).
This means that if we prepare our qubit say in the state \(\ket{0}\) with Bloch vector \(\vec{r}=(0,0,1)^T\) by performing the unitary time evolution \(U_\alpha(\vec{n})\) we can transform this state in any other state on the Bloch sphere with Bloch vector \(\vec{r}'\) by just choosing some specific \(\alpha,\vec{n}\) such that the corresponding rotation of the Bloch vector \(R_\alpha(\vec{n})(0,0,1)^T =\vec{r}'\).
This will be extremely useful later on when discussing applications of entanglement.
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