Prove that qubits can be represented on a unit sphere, avoiding the density matrix formalism

Question

The Bloch Sphere is regarded as the most "intuitive" way of explaining a 2-level quantum system in computation and rotations of states described on Bloch sphere provides a really easy picture. Despite that, I've been having some problems with understanding this representation of quantum states. I read quite a few articles and answers on StackExchange too, but most of them try to explain things using Pauli Matrices and Density Matrices. But the real problem I've been struggling to solve and understand would be better stated as follows-

Prove that a 2-level Quantum State (qubits) is mathematically equivalent to a point on the Unit Sphere in $\mathbb{R}^3$.

Most of the explanations I've seen are based on hand-wavy arguments and lack proper formalism. I've been trying to prove this without involving Pauli matrices or Density matrices. By using the arguments of normalization and global-phase invariance, I've been able to deduce that we can write an arbitrary state as follows- $$|\psi \rangle=r|0\rangle + (a+ib)|1\rangle $$ where $r\in \mathbb{R}^{\geq0}$ and $ a,b\in \mathbb{R}$ with the constraint $r^2+a^2+b^2=1.$ If $a,b,r$ are the co-ordinates in the Cartesian system, then since we have $r\geq0$, the equation results in a hemisphere.

If anyone has an explanation or a proof of how qubits have an equivalent description on the Bloch sphere, or some suggestion, I would be grateful.

Answer 1

The general state is a superposition of $| 0 \rangle$ and $| 1 \rangle$ state: $$ \psi = \alpha | 0 \rangle + \beta | 1 \rangle \qquad \alpha, \beta \in \mathbb{C} $$ However, states, which are related by multiplication on arbitrary complex number $\lambda \neq 0$, are identified : $\psi \sim \lambda \psi$. So the space of quantum states is: $$ \mathbb{C}^2 / \mathbb{C}^{*} = \mathbb{CP}^{1} \simeq S^2 $$

Answer 2

The idea is to find a nice "real" representation for pure qubit states, that is, vectors $(\alpha,\beta)\in\mathbb C^2$ defined up to global phase and norm. Remember you can always write such a unit complex vector as $$\cos(\theta)\lvert0\rangle+\sin(\theta) e^{i\phi}\lvert 1\rangle$$ for some $\theta,\phi\in\mathbb R$. This comes from observing that if $x,y\in\mathbb R$ satisfy $x^2+y^2=1$, then there is some angle $\theta$ such that $x=\cos\theta$ and $y=\sin\theta$, and that $\lvert\alpha\rvert^2,\lvert\beta\rvert^2$ are an example of such real numbers. You then observe that rescaling by a global phase you can always reduce to a case in which $\alpha\in\mathbb R$.

Keeping in mind the physical interpretation of quantum states, an obvious starting point for such parametrisation it to use the probability $p_0\equiv \lvert\alpha\rvert^2=\cos(\theta)^2$. We don't need $p_1=\lvert\beta\rvert^2$ as the norm constraint tells us that $p_1=1-p_0$.

Fixed $p_0$, what freedom is left? Well there is the phase difference to take into account. This is encoded into the phase $\phi$ in the formula above. A natural way to represent geometrically such a phase is via a corresponding unit vector in $\mathbb R^2$, that is, through the mapping $\phi\to(\cos\phi,\sin\phi)$. However, this phase is attached to the chosen value of $p_0$. In particular, for $p_0=0$ and $p_0=1$ all phases $\phi$ should represent the same state, which makes our looking for a nice representation a bit trickier.

So we just found that we can represent a state using (1) the parameter $p_0$, and (2) the phase $\phi$ represented as $(\cos\phi,\sin\phi)$, with the added requirement that (3) the phase $\phi$ shouldn't matter for $p_0=0$ and $p_0=1$. An easy solution is then to use a three-dimensional representation in which $p_0\in[0,1]$ is on one axis (say the $z$-axis), and at each value of $p_0$ we have a circle around it. The radius of such circles depends on $p_0$, and is vanishing for $p_0=0,1$.

The most natural choice is then obviously to use a sphere of radius $1/2$ centred around $p_0=1/2$. By simply translating the $p_0$ axis and rescaling, you get your unit sphere centred around the origin, i.e. the Bloch sphere.

It's worth stressing that you cannot really derive the Bloch sphere representation from the sole definition of quantum state. You can prove that the Bloch sphere representation works to faithfully represent states, and show that it is convenient for all sorts of reasons, but if the only requirement is to represent states as points in $\mathbb R^3$, it is not uniquely defined: other equivalent (though probably highly inconvenient) representations are possible.

Answer 3

Parametrizing a general state as \begin{align} \cos(\vartheta/2)\vert +\rangle + e^{i\varphi}\sin(\vartheta/2)\vert -\rangle \tag{1} \end{align} (any state up to a global phase can be written that way) and computing \begin{align} \langle \sigma_x\rangle =\sin(\varphi)\sin(\vartheta)\, ,\qquad \langle \sigma_y \rangle =\cos(\varphi)\sin(\vartheta)\, ,\qquad \langle \sigma_z\rangle =\cos(\vartheta) \end{align} gives a vector \begin{align} \hat n=\left(\langle \sigma_x\rangle ,\langle \sigma_y\rangle,\langle\sigma_z\rangle \right) \end{align} which points to a point on the Bloch sphere.