Bloch Sphere#

The Bloch sphere is an extremely useful and accurate way of visualizing the information in a qubit. Mathematically it is not so transparent, but it is very intuitive. Formally, we represent a quantum state \(\ket{\psi}\) in spherical coordinates as a point with angles \(\theta\) and \(\phi\) on the unit sphere \(r=1\):

\[\begin{gather*} \ket{\psi} = e^{\I\gamma} \Bigl( \cos\frac{\theta}{2}\ket{0} + e^{\I \phi}\sin\frac{\theta}{2}\ket{1} \Bigr) = e^{\I \gamma} \begin{pmatrix} \cos\frac{\theta}{2}\\ e^{\I \phi}\sin\frac{\theta}{2} \end{pmatrix}, \\ \begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} r\sin\phi\;\cos\phi\\ r\sin\theta\;\sin\phi\\ r\cos\theta \end{pmatrix}. \end{gather*}\]

https://upload.wikimedia.org/wikipedia/commons/6/6b/Bloch_sphere.svg — Fig. 2 Bloch sphere representation of#

\[\begin{gather*} \ket{\psi} = \cos\frac{\theta}{2}\ket{0} + e^{\I \phi}\sin\frac{\theta}{2}\ket{1} \end{gather*}\]

# :tags: [hide-input]

from qiskit.visualization import plot_bloch_vector


def get_vec(psi):
    """Return the Bloch vector from psi."""
    gamma = np.angle(psi[0])
    psi = np.exp(-1j*gamma) * np.asarray(psi)
    phi = np.angle(psi[1]) - np.angle(psi[0])
    theta = 2*np.arctan2(psi[0].real, 
                         (np.exp(-1j*phi)*psi[1]).real)
    x = np.sin(theta)*np.cos(phi)
    y = np.sin(theta)*np.sin(phi)
    z = np.cos(theta)
    return (x, y, z)
    
fig = plt.figure(figsize=(3*10, 10))
ax = fig.add_subplot(1, 3, 1, projection='3d')
plot_bloch_vector(get_vec([1, 1]), ax=ax, title=r"$|Ψ⟩ = \frac{|0⟩ + |1⟩}{\sqrt{2}}$")
ax = fig.add_subplot(1, 3, 2, projection='3d')
plot_bloch_vector(get_vec([1, -1]), ax=ax, title=r"$|Ψ⟩ = \frac{|0⟩ - |1⟩}{\sqrt{2}}$")
ax = fig.add_subplot(1, 3, 3, projection='3d')
plot_bloch_vector(get_vec([1, 1j]), ax=ax, title=r"$|Ψ⟩ = \frac{|0⟩ + \mathrm{i}|1⟩}{\sqrt{2}}$")

../_images/8c729b9a95716ae0feeaa0a233ce7f3f3af2ffe097e42acc8f40e3df8fd1214b.png

Note

Some important questions:

Why does the phase \(\gamma\) not affect the position on the Bloch sphere?
Why is the angle \(\theta/2\), half of what one would normally use?
Related: Why are the “orthogonal” vectors \(\ket{0}\) and \(\ket{1}\) not at right angles?

[Williams, 2011] has a nice discussion of these confusions in §1.3.3.

Intuitive Picture#

The Bloch sphere is nice because it not only provides a complete and faithful representation of a single qubit state, but arbitrary single-qubit operations (gates) have a simple picture. Specifically, any single-qubit gate can be expressed in terms of a clockwise rotation \(\vec{\theta} = \theta \uvec{n}\) of \(\theta\) radians about an axis \(\uvec{n}\) as:

\[\begin{gather*} \mat{U} = \mat{R_{\vec{\theta}}} = \exp\left(\vec{\theta}\cdot \frac{\vec{\mat{\sigma}}}{2\I}\right) = \cos\Bigl(\frac{\theta}{2}\Bigr)\;\mat{1} - \I \sin\Bigl(\frac{\theta}{2}\Bigr) (n_x \mat{X} + n_y \mat{Y} + n_z \mat{Z}). \end{gather*}\]

In terms of spin-½ particles, this is easily implemented using an external magnetic field pointing in the \(\uvec{n}\) direction.

In component form:

\[\begin{gather*} \mat{U} = \begin{pmatrix} u_0 & -u_1^*\\ u_1 & u_0^* \end{pmatrix},\\ =\begin{pmatrix} \cos\Bigl(\frac{\theta}{2}\Bigr) - \I\sin\Bigl(\frac{\theta}{2}\Bigr)n_z & - \I\sin\Bigl(\frac{\theta}{2}\Bigr)(n_x - \I n_y)\\ - \I\sin\Bigl(\frac{\theta}{2}\Bigr)(n_x + \I n_y) & \cos\Bigl(\frac{\theta}{2}\Bigr) + \I\sin\Bigl(\frac{\theta}{2}\Bigr)n_z \end{pmatrix} \end{gather*}\]

Alternatively, redefine \(\theta = \omega\), express \(\hat{n}\) in polar coordinates, and let \(\eta\) be the phase of \(u_0\):

\[\begin{gather*} \hat{n} = \begin{pmatrix} \sin\theta \cos\phi\\ \sin\theta \sin\phi\\ \cos\theta \end{pmatrix}, \qquad e^{\I\eta} = \frac{\cos(\frac{\omega}{2}) - \I\sin(\frac{\omega}{2})n_z} {\sqrt{\cos^2\tfrac{\omega}{2} + \sin^2\tfrac{\omega}{2}\;\cos^2\theta}},\\ \mat{U} = \begin{pmatrix} \cos\tfrac{\omega}{2} - \I\sin\tfrac{\omega}{2}\cos\theta & - \I\sin\tfrac{\omega}{2}\sin\theta e^{-\I\phi}\\ - \I\sin\tfrac{\omega}{2}\sin\theta e^{\I\phi} & \cos\tfrac{\omega}{2}+ \I\sin\tfrac{\omega}{2}\cos\theta \end{pmatrix}. \end{gather*}\]

The magnitude of \(u_0\) (the denominator of \(e^{\I\eta}\) above) can be simplified by introducing a new angle \(\chi\), which describes how the vector \(\ket{0}\) is rotated:

\[\begin{gather*} \sin\chi = \sin\tfrac{\omega}{2}\sin\theta,\\ \begin{pmatrix} u_0\\ u_1 \end{pmatrix} = e^{\I\eta} \begin{pmatrix} \cos\chi\\ -\I\sin\chi\; e^{\I\phi} \end{pmatrix} = \mat{U}\ket{0}. \end{gather*}\]

QuTiP#

import qutip
b = qutip.Bloch()
b.make_sphere()
display(b)

---------------------------------------------------------------------------
ModuleNotFoundError                       Traceback (most recent call last)
Cell In[3], line 1
----> 1 import qutip
      2 b = qutip.Bloch()
      3 b.make_sphere()

ModuleNotFoundError: No module named 'qutip'

Bloch Sphere

Contents

Bloch Sphere#

Intuitive Picture#

QuTiP#