Bloch sphere: how to represent qubits

In quantum mechanics and specifically quantum computing, the Bloch sphere is a geometrical representation of two-levels quantum systems as qubits.

In this article we will define the Bloch sphere and show how it can be used to easily represent quantum kets and quantum density operators. We will show how there is a correspondence between the points on the Bloch sphere and pure states, and a correspondence between the points inside the Bloch sphere (Bloch ball) and mixed states (check out this article for more details about the difference between pure and mixed quantum states). Finally we see how to implement rotations in the Bloch sphere picture and how these correspond to unitary operators short of a global phase.

Article contents

Bloch sphere representation of kets

    


As said, the Bloch sphere can be used to represent two-levels quantum systems as qubits. Let's start by seeing how this representation naturally arises when trying to represent a generic qubit.

We can write a generic qubit in the form

\begin{equation}
|\psi\rangle = a|0\rangle + b|1\rangle
\end{equation}

where $a$ and $b$ are complex numbers, therefore we can write more expplicitly as

\begin{equation}
|\psi\rangle = r_a e^{i \phi_a}|0\rangle + r_b e^{i \phi_b}|1\rangle
\end{equation}

Let's now use the fact that the state has to be normalized, we have

\begin{equation}
|r_a e^{i \phi_a}|0\rangle + r_b e^{i \phi_b}|1\rangle|^2 = r^2_a + r^2_b = 1
\end{equation}

this means that we can express these two coefficients, without loss of generality, as

\begin{equation}
\begin{cases}
 r_a = cos(\frac{\theta}{2})  \\
 r_b = sin(\frac{\theta}{2})
\end{cases}
\end{equation}

where we have chosen $\frac{\theta}{2}$ instead of $\theta$ for the sake of simplicity in the following. We also know that a quantum state is defined short of a global phase, which means that we can rewrite the qubit simply as 

\begin{equation}
|\psi\rangle = cos(\frac{\theta}{2})|0\rangle + sin(\frac{\theta}{2}) e^{i (\phi_b-\phi_a)}|1\rangle \equiv cos(\frac{\theta}{2})|0\rangle + sin(\frac{\theta}{2}) e^{i \phi}|1\rangle
\end{equation}

therefore we have found a way of representing a generic qubit using two real angles rather than two complex numbers. From the definition of $\theta$ and $\phi$ it is clear that in order to obtain a full representation we only need $\theta \in [0, \pi]$ and $\phi \in [0, 2\pi]$. Now these two angles in these ranges identify the surface of a sphere, this is the Bloch sphere!

    
the 3-d vector identified by these two angles is called Bloch bector and has the following components

\begin{equation}
\vec{b} = (sin\theta cos\phi, sin\theta sin\phi, cos\theta)
\end{equation}

Bloch sphere representation of density operators

Let’s now look at how the Bloch sphere can be likewise used to represent density operators. Let’s take again the ket

    
\begin{equation}
|\psi\rangle = cos(\frac{\theta}{2})|0\rangle + sin(\frac{\theta}{2}) e^{i \phi}|1\rangle
\end{equation}

and let's use it to write the corresponding density matrix

\begin{align}
\hat{\rho} &= |\psi\rangle \langle \psi| = \big(cos(\frac{\theta}{2})|0\rangle + sin(\frac{\theta}{2}) e^{i \phi}|1\rangle\big)\big(cos(\frac{\theta}{2})\langle 0| + sin(\frac{\theta}{2}) e^{i \phi}\langle 1|\big)\\ &= cos^2(\frac{\theta}{2}) |0\rangle \langle 0| + cos(\frac{\theta}{2})sin(\frac{\theta}{2}) e^{i \phi}|0\rangle \langle 1| + cos(\frac{\theta}{2})sin(\frac{\theta}{2}) e^{-i \phi}|1\rangle \langle 0| + sin^2(\frac{\theta}{2})|1\rangle \langle 1|
\end{align}

let’s now write this formula in matrix notation

    
\begin{equation}
|\psi\rangle = cos(\frac{\theta}{2})|0\rangle + sin(\frac{\theta}{2}) e^{i \phi}|1\rangle
\end{equation}

and let's use it to write the corresponding density matrix

\begin{equation}
\hat{\rho} = \begin{pmatrix} cos^2(\frac{\theta}{2}) & cos(\frac{\theta}{2})sin(\frac{\theta}{2})e^{i\phi} \\ cos(\frac{\theta}{2})sin(\frac{\theta}{2})e^{-i\phi} & sin^2(\frac{\theta}{2}) \end{pmatrix}
\end{equation}

we can simplify this matrix by using the half angle formulas. We get

    
\begin{equation}
\hat{\rho} = \frac{1}{2}\begin{pmatrix} 1+cos\theta & sin\theta e^{i\phi} \\ sin\theta e^{-i\phi} & 1-cos\theta \end{pmatrix}
\end{equation}

Pauli matrices

Before continuing, let’s briefly introduce Pauli matrices.

    
\begin{equation}
\hat{\sigma}_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \hspace{1cm} \hat{\sigma}_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \hspace{1cm} \hat{\sigma}_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\end{equation}

these matrices have various roles in quantum mechanics and, along with the identity $\hat{I}$ form a basis for the $2x2$ complex matrices, which means that we can express every density matrix (for one qubit) as a linear combination of these four matrices. In particular

\begin{equation}
\hat{\rho} = a_0 \hat{I} + a_1 \hat{\sigma}_x + a_2 \hat{\sigma}_y + a_3 \hat{\sigma}_z
\end{equation}

where the property $Tr\hat{\rho} = 1$ has to hold, therefore $a_0 = \frac{1}{2}$ necessarily.

Now let's see how we can use these four matrices to represent the density operator obtained above. It is easy to see that we can express this operator as

\begin{equation}
\hat{\rho} = \frac{1}{2} \big[\hat{I} + cos\theta cos\phi \cdot \hat{\sigma}_x + cos\theta sin\phi \cdot \hat{\sigma}_y + cos\theta \cdot \hat{\sigma}_z \big]
\end{equation}

we see that the coefficients $a_1$, $a_2$ and $a_3$ are exactly the components of the Bloch vector! This means that we can use the Bloch sphere (through the Bloch vector) to represent density matrices and kets at the same time. In particular we will have 

\begin{equation}
\hat{\rho} = \frac{1}{2} \big[\hat{I} + \textbf{b} \cdot \hat{\boldsymbol{\sigma}} \big]
\end{equation}

where $\hat{\boldsymbol{\sigma}}$ indicates the vector of Pauli matrices (each component is a matrix).

Pure and mixed states on the Bloch sphere

We have just seen how the Bloch sphere can be used to represent easily quantum pure states for two-levels systems as qubits, both through the ket representation and through the density operator representation. Let’s now see how this picture can be extended to mixed states as well.

    
We know that varying $\theta \in [0, \pi]$ and $\phi \in [0, 2\pi]$ we have a complete representation of pure qubits, which means that there doesn’t exist a qubit which can’t be represented this way (in fact we have derived this representation from the most general qubit). However we know that, including mixed states, there is no state which can't be written as 

\begin{equation}
\hat{\rho} = \frac{1}{2} \big[\hat{I} + \textbf{b} \cdot \hat{\boldsymbol{\sigma}} \big]
\end{equation}

because $\{\hat{I}, \hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z\}$ is a basis for the $2x2$ complex matrices space. Therefore all the qubits (pure and mixed) must lie somewhere in the 3D space where the Bloch sphere lies. In particular, given every qubit, there must be a vector $\hat{a}$ such that the qubit can be written as $\hat{\rho} = \frac{1}{2} \big[\hat{I} + \textbf{a} \cdot \hat{\boldsymbol{\sigma}} \big]$.

We already know that the pure states lie on the Bloch sphere, namely $|\textbf{a}|^2 = 1$, where are the mixed states?

Proposition: there are no states outside the Bloch sphere

    


Let's get back to the density operator

\begin{equation}
\hat{\rho} = a_0 \hat{I} + a_1 \hat{\sigma}_x + a_2 \hat{\sigma}_y + a_3 \hat{\sigma}_z
\end{equation}

we know that to comply with the laws of quantum mechanics $\hat{\rho}$ must be positive semidefinite, which means its eigenvalues must be non-negative. Let's compute the eigenvalues of $\hat{\rho}$. To do so, we need to rewrite as a matrix, it can be easily verified that one gets

\begin{equation}
\hat{\sigma}_x = \frac{1}{2}\begin{pmatrix} 1 + a_3 & a_1 - i a_2 \\ a_1 + i a_2 & 1 - a_3 \end{pmatrix}
\end{equation}

and the eigenvalues are

\begin{equation}
\lambda_{1, 2} = \frac{1 \pm (a^2_1 + a^2_3 + a^2_3)}{2}
\end{equation}

in order for both eigenvalues to be non-negative the following condition must hold 

\begin{equation}
a^2_1 + a^2_3 + a^2_3 ⋜ 1
\end{equation}

which means that there are no physical states outside of the Bloch sphere.

Now we know that there is nothing outside of the sphere and the points on the surface are in a 1-1 correspondence with pure states, therefore all the mixed states must lie inside of the Bloch sphere. It is also easy to see that there are no points inside the Bloch sphere which have no corresponding mixed qubit, in fact for each vector inside the sphere we can construct a legitimate density operator.

Unitary operators

We have seen how the Bloch sphere can be used to efficiently represent pure and mixed qubits, let’s now briefly look at how unitary operators act in the Bloch sphere picture.

    


It's easy to show that the Pauli matrices generate rotations of the Bloch vector around the respective axis, namely

\begin{equation}
\hat{R}_x(\theta) = e^{-i \frac{\theta}{2}\hat{\sigma}_x}\\
\hat{R}_y(\theta) = e^{-i \frac{\theta}{2}\hat{\sigma}_y}\\
\hat{R}_z(\theta) = e^{-i \frac{\theta}{2}\hat{\sigma}_z}
\end{equation}

while a general rotation by an angle $\chi$ around a generic axis $\vec{n}$ is given by

\begin{equation}
\hat{R}_{\textbf{n}}(\chi) = e^{-i\frac{\chi}{2}\textbf{n}\cdot\hat{\boldsymbol{\sigma}}}
\end{equation}

where $\textbf{n}$ is a normalized vector.

Look here for a proof of these equations.

Since this operator can rotate the Bloch vector to any point on the Bloch sphere, which include all the possible states of a qubit short of a global phase, any unitary transformation can be written as a generic Bloch rotation with a global phase shift, namely

    
\begin{equation}
\hat{U} = e^{i\xi}\hat{R}_{\textbf{n}}(\chi)
\end{equation}

Summarizing

    
To summarize, consider the following equation 

$\hat{\rho} = \frac{1}{2} \big[\hat{I} + \textbf{a} \cdot \hat{\boldsymbol{\sigma}} \big]$

where the vector $\hat{a}$ spans a 3D space. All the vectors $\hat{a}$ such that $|\hat{a}|^2 = 1$ identify a sphere, the Bloch sphere, and the points on the sphere have a $1-1$ correspondence with pure qubits through the equation written above.

All the vectors $\hat{a}$ such that $|\hat{a}|^2 ⋜ 1$ identify a ball (Bloch ball) and the points in the ball are in a 1-1 correspondence with mixed qubits through the same equation, while there are no qubits outside the Bloch sphere.

Any rotation of the Bloch vector can be implemented by the operator $\hat{R}_{\textbf{n}}(\chi)$, which corresponds to a generic unitary transformation short of a global phase.

Bibliogaphy

[1] Nielsen, M.A. and Chuang, I., 2002. Quantum computation and quantum information.