In order to represent the single qubit $|\psi\rangle$ we use an unitary vector in a $\mathbb{C}^2$ Hilbert space whose (one of the) orthonormal base is $(|0\rangle, |1\rangle)$.

We can draw $|\psi\rangle$ using a Bloch ball. However, I found this notation quite confusing, because orthogonal vectors are spatially antiparallel (brief explanation in this Physics Stackexchange question).

Block sphere

Do you know any different graphical representation for a single qubit?


4 Answers 4


In the link included in your question, about another question written by user098876, "Understanding the Bloch sphere", Daniel makes a helpful comment:

"Drawing points on the sphere to represent the state of a quantum two-level system does not mean that you should think of those points as real vectors in 3D space. – DanielSank Sep 3 '15 at 20:17".

Oversimplified explanation: It's a two-side plane (or two planes) projected on a sphere.

"I found this notation quite confusing, because orthogonal vectors are spatially antiparallel (brief explanation in this Physics Stackexchange question). Do you know any different graphical representation for a single qubit?"

There are a number of efforts underway to provide a more general representation that extends from qubits to qudits. This explanation and representation using a Majorana sphere isn't so different, it's still a sphere, but perhaps it's less confusing:

For qubits on a Majorana sphere see: "N-qubit states as points on the Bloch sphere".

"Abstract. We show how the Majorana representation can be used to express the pure states of an N-qubit system ... In conclusion, the Majorana representation is useful when spin-$S$ particles are studied, while the alternative representation is preferable when the states of an $N$-qubit system are discussed. Besides helping to visualize $N$-qubit states and the way they transform in rotations and other operations, the latter representation may also help to identify some special $N$-qubit states, like the Majorana representation did in the context of spinor Bose-Einstein condensates.".

See: "Majorana representation, qutrit Hilbert space and NMR implementation of qutrit gates":

Page 1:

"The Bloch sphere provides a representation of the quantum states of a single qubit onto $\mathcal S^2$ (a unit sphere in three real dimensions), with pure states mapped onto the surface and the mixed states lying in the interior. This geometrical representation is useful in providing a visualization of quantum states and their transformations, particularly in the case of NMR-based quantum computation, where the spin-$\frac{1}{2}$ magnetization and its transformation through NMR rf pulses is visualized on the Bloch sphere. There have been several proposals for the geometrical representation for higher-level quantum systems however, extensions of a Bloch sphere-like picture to higher spins is not straightforward. A geometrical representation was proposed by Majorana in which, a pure state of a spin ‘$s$’ is represented by ‘2$_s$’ points on the surface of a unit sphere, called the Majorana sphere.

The Majorana representation for spin−$s$ systems has found widespread applications such as determining geometric phase of spins, representing $N$ spinors by $N$ points, geometrical representation of multi-qubit entangled states, statistics of chaotic quantum dynamical systems and characterizing polarized light. A single qutrit (three-level quantum system) is of particular importance in qudit-based ($d$-level quantum system) quantum computing schemes. A qutrit is the smallest system that exhibits inherent quantum features such as contextuality, which has been conjectured to be a resource for quantum computing. NMR qudit quantum computing can be performed by using nuclei with spin s > $\frac{1}{2}$ or can be modeled by two or more coupled spin-$\frac{1}{2}$ nuclei. In this work we use the Majorana sphere description of a single qutrit, where states of a qutrit are represented by a pair of points on a unit sphere, to provide insights into the qutrit state space.

Page 5:

The magnitude of the magnetization vector $\vert-$$\vec{M}$$\vert$ in a pure ensemble of a single qutrit can assume values in the range $[0,1]$. On the contrary, the pure ensemble of a qubit always possesses unit magnitude of the magnetization vector associated with it. The geometrical picture of the single qutrit magnetization vector is provided by the Majorana representation. The value $\vert-$$\vec{M}$$\vert$ depends upon the length of the bisector $OO'$ and lies along the $z$-axis and is rotationally invariant. Thus corresponding to a given value of the length of the bisector, one can assume concentric spheres with continuously varying radii, whose surfaces are the surfaces of constant magnetization. The radii of these spheres are equal to $\vert-$$\vec{M}$$\vert$, that vary in the range $[0,1]$.

Page 10:


A geometrical representation of a qutrit is described in this work, wherein qutrit states are represented by two points on a unit sphere as per the Majorana representation. A parameterization of single-qutrit states was obtained to generate arbitrary states from a one-parameter family of canonical states via the action of $SO(3)$ transformations. The spin-$1$ magnetization vector was represented on the Majorana sphere and states were identified as ‘pointing’ or ‘non-pointing’ depending on the zero or non-zero value of the spin magnetization. The transformations generated by the action of $SU(3)$ generators were also integrated into the Majorana geometrical picture. Unlike qubits, the decomposition of single-qutrit quantum gates in terms of radio-frequency pulses is not straightforward and the Majorana sphere representation provides a way to geometrically describe these gates. Close observations of the dynamics of points representing a qutrit on the Majorana sphere under the action of various quantum gates were used to obtain the rf pulse decompositions and basic single-qutrit gates were experimentally implemented using NMR.

Majorana Sphere - Dogra, Dorai, and Arvind

FIG. 1. A qutrit on the Majorana sphere is represented by two points $P_1$ and $P_2$, connected with the center of the sphere by lines shown in red and blue respectively. $\theta_1$, $\phi_1$ are the polar and azimuthal angles corresponding to point $P_1$ ($\theta_2$, $\phi_2$ are the angles for point $P_2$). (a) Roots of the Majorana polynomial are shown in the plane $z = 0$ by points $P'_1$ and $P'_2$, whose stereographic projection give rise to the Majorana representation. Three examples are shown corresponding to the Majorana representation of single-qutrit basis vectors $(b)\;\vert+1\rangle$, $(c)\;\vert0\rangle$ and $(d)\;\vert-1\rangle$. One of the points is shown as a solid (red) circle, while the other point is represented by an empty (blue) circle.

See: "Majorana Representation of Higher Spin States" (.PDF) by Wheeler (Website) or "Wigner tomography of multispin quantum states":

What does it look like using Tomography - "In this paper, we theoretically develop a tomography scheme for spherical functions of arbitrary multispin quantum states. We study experimental schemes to reconstruct the generalized Wigner representation of a given density operator (representing mixed or pure quantum states)."

Compare that to the complexity of the Bloch sphere depicted in: "Bloch-sphere representation of three-vertex geometric phases". The shape is the same it's all how you visualize the projection used.

Here's a less busy image:

Bloch sphere

Think of the Bloch sphere cut in half by a very large sheet of paper. At the edge of the paper (infinity) any point on the top of the sheet draws a line to (infinity) the top of the ball (the bottom of the ball for the underside of the sheet). Points nearest the center of the paper (mixed states) draw lines to the center of the sphere. That represents the distance up to infinity on a tiny ball, a qubit/qudit is finite so the paper is not so big.

Now draw points on the 2D paper, draw lines from the paper to the ball, remove the paper, and look at or through the clear ball to see the other endpoint of the line.

A much more accurate and difficult explanation is offered in the links above.

  • $\begingroup$ Thank you for your answer. Please, can you add a very brief description of how to represent a qubit (not qutrit) on a Majorana sphere? Then I'll mark this answer as accepted because it answers perfectly my question. $\endgroup$
    – incud
    Commented Apr 9, 2018 at 6:51
  • $\begingroup$ @incud - Added another paper at the top that's a bit easier going and directly qubit oriented. $\endgroup$
    – Rob
    Commented Apr 9, 2018 at 14:14

Adding to what @pyramids conveyed in their answer:

A qubit's state is generally written as $\alpha|0\rangle + \beta|1\rangle$, where $\alpha, \beta \in \Bbb{C}$, and $|\alpha|^2+|\beta|^2=1$.

$\Bbb{C}^2(\Bbb{R})$ is a four-dimensional vector space, over the field of real numbers. Since any $n$-dimensional real vector space is isomorphic to $\Bbb{R}^n(\Bbb{R})$, you can represent any qubit's state as a point in a $4$-dimensional real space, too, whose basis vectors you can consider to be $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$. In such a case a qubit's state would be represented as $a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1)$.

Say, $\alpha = a + i b$ (where $a,b\in \Bbb{R}$) and $\beta = c + id$ (where $c,d\in \Bbb{R}$). You need the condition $|a+ib|^2+|c+id|^2=1\implies a^2+b^2+c^2+d^2=1$ to be satisfied, which implies the state of the qubit would be a point on a 3-sphere.

As you know, it is difficult to efficiently represent a $4$-dimensional space on a $2$-dimensional surface like a paper, or your screen. Hence, you don't see that representation used often. Bloch sphere is pretty much the most efficient representation out there (for a single qubit), since it reduces one degree of freedom (of the complex numbers $\alpha,\beta$ each of which have two degrees of freedom) due to the fact that a qubit's state is usually normalized to a magnitude of $1$ i.e. $|\alpha|^2+|\beta|^2=1$.

Now, using the Hopf coordinates let's say:

$$\alpha = e^{i\psi}\cos(\theta/2)$$

$$\beta = e^{i(\psi+\phi)}\sin(\theta/2)$$

Here, $\theta$ can run from $0$ to $\pi$ whereas, $\psi$ and $\phi+\psi$ can take values between $0$ to $\pi$.

In case you're wondering why $\theta/2$ is being used instead of $\theta$ have a look at the answers on this excellent thread on Physics Stack Exchange.

Okay, even now you notice three degrees of freedom $\psi,\phi,\theta$, whereas in a unit radii sphere, you only have two angles which you can change to get the different states of a qubit.

Notice that $\phi$ is basically the "relative phase" between $\alpha$ and $\beta$. On the other hand $\psi$ does not contribute to the "relative phase" of $\alpha,\beta$. Also, neither $\phi$ nor $\psi$ contribute to the magnitude of $\alpha,\beta$ (since $|e^{i\varphi}|=1$ for any angle $\varphi$). Since $\psi$ contributes neither to "relative phase" nor to the "magnitudes" of $\alpha,\beta$ it is said to have no physically observable consequences and we can arbitrarily choose $\alpha$ to be real by eliminating the factor of $e^{i\psi}$.

Thus we end up with:

$$\alpha = \cos(\theta/2)$$ and $$\beta=e^{i\phi}\sin(\theta/2)$$ Where $\theta$ can run from $0$ to $\pi$, and $\phi$ can run from $0$ to $2\pi$.

This practical simplification allows you to represent a qubit's state using just $2$ degrees of freedom on $3$-dimensional spherical surface having unit radius, which again can again efficiently be "drawn" on a $2$-dimensional surface, as shown in the following image.

enter image description here

Mathematically, it is not possible to reduce the degrees of freedom any further, and so, I'd say there is no other "more efficient" geometrical representation of a single qubit than the Bloch sphere.

Source: Wikipedia:Bloch_Sphere

  • $\begingroup$ Typo. You mean 3-sphere. The $n$ in n-sphere indicates the dimension of the sphere itself not the Euclidean space where it is embedded. $\endgroup$
    – AHusain
    Commented Apr 7, 2018 at 17:23
  • $\begingroup$ Wonderful answer, much more than what I was looking for. However, I don't understand completely when you say that with 4 parameters/d.o.f. I need a 3-sphere in $\mathbb{R}^4$ (I agree); than with 3 parameters I need a 2-sphere in $\mathbb{R}^3$ (I agree); than with 2 parameters I still need a 2-sphere in $\mathbb{R}^3$ (couldn't I use a 1-sphere/circle?) $\endgroup$
    – incud
    Commented Apr 8, 2018 at 12:03
  • $\begingroup$ @incud You would have only one degree of freedom in a circle with unit radius i.e. the angle w.r.t. to a certain reference line. $\endgroup$ Commented Apr 8, 2018 at 12:06
  • $\begingroup$ @Blue My fault, I got confused. I wasn't thinking about the unit radius of the circle. Thank you for your answer $\endgroup$
    – incud
    Commented Apr 8, 2018 at 12:39
  • $\begingroup$ Why are you imposing the need for the states to correspond to points on a sphere (in some dimension) of unit radius? As conveyed in the answer by @groupsgroupsgroups, if you only think about pure states, there's no reason to do this. but you make no mention of mixed states... $\endgroup$
    – DaftWullie
    Commented May 22, 2018 at 11:01

The Bloch sphere historically came about to describe spins where up and down can actually be viewed as being (anti)parallel rather than (mathematically) orthogonal.

You can naturally (and perhaps more naturally!) depict a qubit's state in a way that orthogonal states are indeed orthogonal. Then a pure 1-qubit state occupies a point on the surface of a 4-dimensional sphere.


(Firstly, the "reputation points" requirement is stupid - this remark should be a comment on the previous post.)

A single qubit in a pure state has 2 real degrees of freedom, not 3, when you quotient out both magnitude and phase (i.e., complex normalization). So, most reasonable two-dimensional surfaces could be used (e.g., the 2-sphere or anything topologically equivalent).

Finding a useful representation is another story. The Bloch sphere has a natural extension to mixed states (which have 3 degrees of freedom), whereas this does not appear to be the case otherwise..

  • 2
    $\begingroup$ Welcome to quantum computing SE! While the 'required rep' thing can be a nuisance at times, it does (seem to) help more than it hinders, so that's probably going to stay. Instead of commenting on the other post, you can suggest an edit to fix the issue. In any case, I'll leave a comment pointing to this answer and it'll get sorted out hopefully soon $\endgroup$
    – Mithrandir24601
    Commented Apr 7, 2018 at 22:45

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