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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!

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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.

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  • $\begingroup$ Thanks Nick In the Qubit Hilbert space ℂ𝑑, does the 𝑑 variable represent 'dimensions'? And can you elaborate on why d = 2 in this particular instance? $\endgroup$
    – chickenj0
    Commented Dec 28, 2023 at 0:49
  • $\begingroup$ 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. $\endgroup$ Commented Dec 28, 2023 at 2:19
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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.

While it is just a convenient way to represent the qubit state graphically, there is some interesting group symmetry that allows us to do it.

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  • $\begingroup$ 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 $SO(3)$. $\endgroup$ Commented Dec 27, 2023 at 10:49
  • $\begingroup$ Forgive my loose terminology, Nick is correct. But the point remains. $\endgroup$ Commented Dec 27, 2023 at 21:14
  • $\begingroup$ Thank you Dylan! $\endgroup$
    – chickenj0
    Commented Dec 28, 2023 at 0:51

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