How Does the 3-Sphere Map to the Bloch Sphere in Quantum Mechanics?

[Kreizhn]

I'm having a bit of a brain fart here, so hopefully someone can help.

Consider a closed, two-level quantum system. We know we can describe pure states as
\alpha |0\rangle + \beta |1 \rangle
for some orthonormal basis |0\rangle, |1 \rangle. The normalization conditions means we can associate this space with the unit 3-sphere
S^3 = \left\{ (z_1,z_2) \in \mathbb C^2 : |z_1|^2 + |z_2|^2 = 1 \right\}

Now on the other hand, we also use the Bloch sphere to describe such states, where geometrically the Bloch sphere is just \mathbb{PC}^1 the complex projective line.

So I'm wondering, how do we map S^3 \to \mathbb{PC}^1 quantum mechanically? What does this map look like? Is this occurring because we're throwing away the global phase in our transition to the Bloch sphere?

Something is telling me that Hopf fibrations are important here, but I just can't seem to put two and two together.

 

[espen180]

I haven't studied this in detail, but this is my understanding. Since \mathbb{R}^2 and \mathbb{C}^1 are isomorphic, so are their projective spaces \mathbb{PC}^1 and \mathbb{PR}^2.

Now, since the Hilbert space of states is a projective space, the state space of any two-state system will also be isomorphic to these. There is nothing inherently quantum mechanical about the map from one to the other. This is pure mathematics.

 

[Kreizhn]

Thanks for the reply Espen.

I agree entirely, though perhaps I was unclear in what I meant. The normalization condition alone implies that the two-level states occupy the 3-sphere. Similarly we can view two-level states as occupying the Bloch sphere. These are inherently different geometric objects. I want to know what the mapping S^3 \to \mathbb{PC}^1 looks like, and what it means physically.

Now, I think that physically, the map is removing the global phase term. That is, define the relation
|\psi_1 \rangle \sim |\psi_2 \rangle \qquad \Leftrightarrow \qquad \exists \phi \in [0,2\pi), |\psi_1 \rangle = e^{i\phi} |\psi_2 \rangle.
This is easily reflexive, transitive, and symmetric, so it's an equivalence relation. Then
\mathbb{PC}^1 \cong S^3/\sim.
To me, this looks like a Hopf fibration since we've quotiented circle fibres from each element. However, I'm not sure if this is correct.