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+---
+title: "Bloch sphere"
+firstLetter: "B"
+publishDate: 2021-03-09
+categories:
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-09T15:35:33+01:00
+draft: false
+markup: pandoc
+---
+
+# Bloch sphere
+
+In quantum mechanics, particularly quantum information,
+the **Bloch sphere** is an invaluable tool to visualize qubits.
+All pure qubit states are represented by a point on the sphere's surface:
+
+<a href="bloch.jpg">
+<img src="bloch-small.jpg" style="width:50%;display:block;margin:auto;">
+</a>
+
+The $x$, $y$ and $z$-axes represent the components of a spin-1/2-alike system,
+and their extremes are the eigenstates of the Pauli matrices:
+
+$$\begin{aligned}
+    \hat{\sigma}_z
+    \to \{\ket{0}, \ket{1}\}
+    \qquad
+    \hat{\sigma}_x
+    \to \{\ket{+}, \ket{-}\}
+    \qquad
+    \hat{\sigma}_y
+    \to \{\ket{+i}, \ket{-i}\}
+\end{aligned}$$
+
+Where the latter two states are expressed as follows in the conventional $z$-basis:
+
+$$\begin{aligned}
+    \ket{\pm}
+    = \frac{\ket{0} \pm \ket{1}}{\sqrt{2}}
+    \qquad \quad
+    \ket{\pm i}
+    = \frac{\ket{0} \pm i \ket{1}}{\sqrt{2}}
+\end{aligned}$$
+
+More generally, every point on the surface of the sphere
+describes a pure qubit state in terms of the angles $\theta$ and $\varphi$,
+respectively the elevation and azimuth:
+
+$$\begin{aligned}
+    \ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \ket{1}
+\end{aligned}$$
+
+We can generalize this further by describing points using the **Bloch vector** $\vec{r}$,
+with radius $r \le 1$:
+
+$$\begin{aligned}
+    \boxed{
+        \vec{r}
+        = \begin{bmatrix} r_x \\ r_y \\ r_z \end{bmatrix}
+        = \begin{bmatrix} r \sin\theta \cos\varphi \\ r \sin\theta \sin\varphi \\ r \cos\theta \end{bmatrix}
+    }
+\end{aligned}$$
+
+Note that $\vec{r}$ is not actually a qubit state,
+but rather an implicit description of one,
+meaning that it does not need to be normalized.
+The main point of the Bloch vector is that it allows us
+to describe the qubit using a [density operator](/know/concept/density-operator/):
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}
+        = \frac{1}{2} \Big( \hat{I} + \vec{r} \cdot \vec{\sigma} \Big)
+    }
+\end{aligned}$$
+
+Where $\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$ is the Pauli "vector".
+Now, we know that $\hat{\rho}$ represents a pure ensemble
+if and only if it is idempotent, i.e. $\hat{\rho}^2 = \hat{\rho}$:
+
+$$\begin{aligned}
+    \hat{\rho}^2
+    &= \frac{1}{4} \Big( \hat{I}^2 + 2 \hat{I} (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
+    = \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
+\end{aligned}$$
+
+You can easily convince yourself that if $(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$,
+then we get $\hat{\rho}$ again, and the state is pure:
+
+$$\begin{aligned}
+    (\vec{r} \cdot \vec{\sigma})^2
+    &= (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z)^2
+    \\
+    &= r_x^2 \hat{\sigma}_x^2 + r_x r_y \hat{\sigma}_x \hat{\sigma}_y + r_x r_z \hat{\sigma}_x \hat{\sigma}_z
+    + r_x r_y \hat{\sigma}_y \hat{\sigma}_x + r_y^2 \hat{\sigma}_y^2
+    \\
+    &\quad + r_y r_z \hat{\sigma}_y \hat{\sigma}_z + r_x r_z \hat{\sigma}_z \hat{\sigma}_x
+    + r_y r_z \hat{\sigma}_z \hat{\sigma}_y + r_z^2 \hat{\sigma}_z^2
+    \\
+    &= r_x^2 \hat{I} + r_y^2 \hat{I} + r_z^2 \hat{I}
+    + r_x r_y \{ \hat{\sigma}_x,  \hat{\sigma}_y \}
+    + r_y r_z \{ \hat{\sigma}_y,  \hat{\sigma}_z \}
+    + r_x r_z \{ \hat{\sigma}_x,  \hat{\sigma}_z \}
+    \\
+    &= (r_x^2 + r_y^2 + r_z^2) \hat{I}
+    = r^2 \hat{I}
+\end{aligned}$$
+
+Therefore, if the radius $r = 1$, the ensemble is pure,
+else if $r < 1$ it is mixed.
+
+Another useful property of the Bloch vector
+is that the expectation value of the Pauli matrices
+are given by the corresponding component of $\vec{r}$,
+for example for $\hat{\sigma}_z$:
+
+$$\begin{aligned}
+    \expval{\hat{\sigma}_z}
+    &= \Tr(\hat{\rho} \hat{\sigma}_z)
+    = \frac{1}{2} \Tr\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big)
+    = \frac{1}{2} \Tr\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big)
+    \\
+    &= \frac{1}{2} \Tr\big( r_x \hat{\sigma}_x \hat{\sigma}_z + r_y \hat{\sigma}_y \hat{\sigma}_z + r_z \hat{\sigma}_z^2 \big)
+    = \frac{1}{2} \Tr\big( r_z \hat{I} \big)
+    = r_z
+\end{aligned}$$