The Bloch Sphere

Definition¶

The Bloch sphere is a geometrical way to represent both $d=2$ dimensional quantum systems (qubits) and the unitary dynamics of those $d=2$ dimensional systems.

All possible qubit states, both pure and mixed, can be represented as a point on or inside of a unit sphere. Qubit dynamics can then be represented as rotations on the unit sphere.

States¶

Any qubit density operator $\rho \in \mathcal{H}^2$ has a Bloch sphere representation as

where $\mathbb{I}$ is the $d=2$ dimensional identity operator, $\bm{n} \in \mathbb{R}^{3}$ is the Bloch vector, and $\bm{\sigma}$ is a vector of Pauli operators, such that

where

The elements of the Bloch vector can be found as

The $\frac{1}{2} \mathbb{I}$ terms fixes the normalisation, such that $\textrm{tr}\big[ \rho \big] = 1$.

Alternatively, in terms of spherical polar coordinates, the Bloch vector elements are given by

where the norm of the Bloch vector is the taken as the radius. Each set of three coordinates $( \theta, \phi, \vert n \vert)$ represent a uniquely qubit state.

Properties¶

1. $\vert \bm{n} \vert = 1$ if and only if $\rho$ is a pure state. If $\vert \bm{n} \vert < 1$ then $\rho$ is a mixed state. Hence, points on the surface of the sphere are pure states, whilst points in the interior of the sphere are mixed states. The smaller the norm of the Bloch vector, the more mixed a qubit state is.
2. The qubit maximally mixed state, $\rho = \frac{1}{2} \mathbb{I}$, has Bloch vector $\bm{n}_{mm} = (0,0,0)^{t}$. With $\vert \bm{n}_{mm} \vert = 0$ it can be seen that the qubit maximally mixed state sits at the centre of the unit circle.

Orthogonality¶

Given a pure quantum state $\ket{\psi}$ with Bloch vector $\bm{n}$, the orthogonal state $\ket{\psi'}$, such that $\braket{\psi|\psi'}=0$, has Bloch vector $-\bm{n}$.

Hence, orthogonal pure state correspond to the opposite point on the bloch sphere. Perpendicular Bloch vectors do not mean orthogonal quantum states.

Higher Dimensions¶

There exists generalisations of the Bloch vector form to higher dimensional states, but the nice geometrical interpretation of points in a unit sphere is lost.

In order to get a higher dimensional Bloch vector form, an operator basis is needed. Typically, in a $d$ dimensional space, an operator basis is a set of operators $d^2-1$ operators $\{ A_i \}$ such that

1. $\mathbb{I} \in \{ A_i \}$.
   • The set contains the identity operator
2. $\textrm{tr} \big[ A_i^{\dagger}A_j \big] = N \delta_{ij}$, where $N \in \mathbb{R}$ and $\delta_{ij}$ is the delta function.
   • The set of operators is orthogonal according to the Hilbert-Schmit inner product.

In the qubit case, these set of operators were the identity and Pauli operators, $\{ \mathbb{I}, X, Y, Z \}$.

Given $\{ A_i \}$ forms a basis, any matrix can be decomposed into $\{ A_i \}$. For a density operators $\rho \in \mathcal{H}^{d}$, it’s density matrix with respect to $\{ A_i \}$ can be written as

where $\bm{b} \in \mathbb{R}^{d^2-1}$ and $\bm{\Sigma}$ is a vector whose elements are the operators in the basis $\{ A_i \}$. Hence, $\rho$ is written as linear combination of operators in the basis.

As before, the $\frac{1}{d} \mathbb{I}$ term fixes the normalisation, such that $\textrm{tr}\big[ \rho \big] = 1$, and the components of the Bloch vector can be found as $b_i = \textrm{tr} \big[ \rho A_i \big]$.

However, not all Bloch vectors, $\bm{b}$, correspond to valid density operators, as they did in the qubit case.

There are multiple different operator basis one could choose and hence no unique Bloch vector for a given state. See the following for more details on different operator basis one could choose: Bertlmann & Krammer (2008), Asadian et al. (2016).

Dynamics¶

Unitaries¶

Unitary operators cause rotations of points on the Bloch sphere.

The Pauli operators, $\{ X, Y, Z \}$, give rise to rotations around the $x, y$, and $z$ axis respectively when exponentiated,

General rotations around an axis defined by the Bloch vector $\bm{n} = (n_x, n_y, n_z )^{t}$ are given by

as $\big(\bm{n} \cdot \bm{\sigma} \big)^2=\mathbb{I}$.

Using this general rotation, any single qubit unitary can be created by adding an additional phase factor,

where $a, \Theta \in \mathbb{R}^{1}$.

Channels¶

Quantum channels that are not unitary will act to take points on the surface of the Bloch sphere into the interior - pure states are taken to mixed states.

More generally, the set of output states from a channel can be visualised on the Bloch sphere. Consider the quantum depolarising channel,

which can be interpreted as returning the state unchanged with probability $p$, and returning noise with probability $1-p$. The action of this channel is a uniform contraction of the Bloch sphere, with the radius of the new sphere depending on $p$. If $p=0$, the set of possible output states becomes just a point as the channel will output a single state - the maximally mixed state - for all input states.

Consider instead the dephasing channel,

which acts to dampen the off-diagonal terms of the density matrix with respect to some basis. If the off-diagonal terms with respect to the standard basis are dampened, the set of possible output states is an ellipsoid, a sphere contracted along the $x$ and $y$ axis wih the complete $z$ axis contained. This highlights the fact that any state that is diagonal in the standard basis is left unchanged by the dephasing channel. If the dephasing is performed with respect to a different basis, then the ellipsoid with be rotated compared to the above.