# Why are quantum states represented by Bloch or Q-Spheres?

I'm new to quantum computing.

Why are quantum states/qubits represented on spheres: either as Bloch spheres or as Q-Spheres.

Is it just a convenient graphical representation, or is there something deeper going on?

Thanks!

This is a special case of my answer here with $$d = 2$$, but I think it is worth elaborating. For the qubit Hilbert space $$\mathbb{C}^2$$, the corresponding space of qubit states is the complex projective line $$\mathbb{C}\textbf{P}^1$$. The complex projective line is equivalently the Riemann sphere, and this is where the Bloch sphere picture comes from.
• A qudit has Hilbert space $\mathbb{C}^d$, where $d$ represents dimensions. So a qubit is a special case of a qudit where $d = 2$ because a qubit has two basis vectors $|0\rangle$ and $|1\rangle$. Geometrically, a qudit has state space $\mathbb{C}\textbf{P}^{d - 1}$. So a qubit has state space $\mathbb{C}\textbf{P}^{2 - 1} = \mathbb{C}\textbf{P}^{1}$, which just so happens to be a sphere. Dec 28, 2023 at 2:19
Quantum bits are mathematically represented by two complex numbers $$\alpha$$ and $$\beta$$, so we say the state of a qubit is $$|\psi\rangle=\pmatrix{\alpha\\ \beta}=\alpha|0\rangle+\beta|1\rangle\in\mathbb{C}^2$$ Since a complex number $$z\in\mathbb{C}$$ may be written as $$z=a+bi$$ where $$a,b\in\mathbb{R}$$, the qubit can be represented by four real numbers, also known as a quaternion. You might expect the state of a qubit then to be represented by any point in $$\mathbb{R}^4$$, however all valid states of a qubit are constrained by the fact that every state has unit norm: $$\langle\psi|\psi\rangle=|\alpha|^2+|\beta|^2=1$$. The set of operators that maintain unit norm on quaternions is the unitary group $$U(2)$$. However, we also don't care about a global phase, which corresponds to restricting the operators to the special unitary group $$SU(2)$$. It is well known that $$SU(2)$$ is identified with the group of 3D rotations $$SO(3)$$, the symmetry group of a sphere. This is why we can represent qubit states on spheres.
• The well-known result that you linked does not say that $SU(2)$ is identified with $SO(3)$. It says that $SU(2)$ is the universal cover of $SO(3)$. Dec 27, 2023 at 10:49