Why do we need non-trivial fibrations?

Question

I am currently reading this paper. I understand how the Bloch sphere $S^2$ is presented as a geometric representation of the observables of a two-state system:

$$ \alpha |0\rangle + \beta |1\rangle \quad \alpha,\beta\in\mathbb{C} \quad \longrightarrow \{(\langle\sigma_x\rangle ,\langle\sigma_y\rangle ,\langle\sigma_z\rangle )\} = S^2 \subset\mathbb{R}^3$$

I also understand how $S^3$ is presented as a geometric representation of the same two-state system:

$$ \alpha |0\rangle + \beta |1\rangle \quad \alpha,\beta\in\mathbb{C} \quad \longrightarrow {(Re(\alpha),Im(\alpha),Re(\beta),Im(\beta))}=S^3 \subset \mathbb{R}^4$$

I see that the Bloch sphere representation loses the information about a global phase, but I do not understand why we need a non-trivial (Hopf) fibration of $S^3$ to preserve this phase information. Why can't we simply assign to every point of the Bloch sphere a phase $e^{i\phi}$? Why does the two-state system demand a non-trivial fibration (and why specifically the Hopf fibration)?

PS (and a possible partial answer): Since the Bloch sphere contains all of the observable information about the system, the $S^3$ representation has to be "decomposable" into pieces such that one of the pieces is the Bloch sphere. Since $S^3\not = S^2\otimes S^1$, we need a more complicated decomposition, e.g. the Hopf fibration. But this still leaves the question: is there no other decomposition of $S^3$ that will do the trick?

Answer 1

Why can't we simply assign to every point of the Bloch sphere a phase $e^{i\phi}$?

This is the idea of a section of a fibre bundle. You are considering in this case a base space $S^{2}$ with fibre $S^{1}$. Locally the fibre bundle looks like $S^{2}×S^{1}$. However you want to consider a fibration such that the global space is not the trivial product but $S^{3}$. The Hopf fibration allows you to twist and glue the spaces to obtain the identification you want.

Is this the only fibration that gives you $S^{3}$? The answer is yes, but to see it requires some knowledge in algebraic topology and characteristic classes. Basically circle bundles are classified up to bundle isomorphism by the first Chern class. The next step is to see that the Hopf fibration belongs to the class of $1\in H^{2}(\cal{S^{2},\mathbb{Z}})$. You can do that by constructing long fibre sequences as done here. Finally you have to prove that any other circle bundle with total space $S^{3}$ belongs to this class. I asked that in mathstackexchnage and this was the answer.

Another answer can be formulated in terms of the Hopf invariant which is an homotopy invariant. A theorem by Frank Adams and subsequently by Michael Atiyah with methods of topological K-theory proved the only maps of Hopf invariant 1 are the sphere bundles in dimensions 2, 4 and 8. This settle the question if you are happy to consider the same as meaning the same up to homotopy equivalence. A proof of considering the same as the same up to homeomorphism for the base, fiber and total space can be found in this paper. (Corollary 3.9, Theorem 6.1) This also cover the other cases you are asking about.

Answer 2

The hopf-fibration is a projection from the 3-sphere to the 2-sphere. This is a dimensional reduction. The "extra" 4th-dimensional information from the 4d space of the quaternion is encoded in the lower 3d space via the global phase. The global phase is a natural hidden variable of the qubit in the "lower" 3d space - for more info see;

Unit Quaternions and the Bloch Sphere: https://arxiv.org/abs/1411.4999

The hopf-fibration $$\mathbb{S}^3\mapsto^{\mathbb{S}^1}\mathbb{S}^2$$ is defined by $$\hat{\mathcal{R}}\;=\;\hat{\Psi}\frac{\hat{\sigma}_i}{2}\hat{\Psi}^\dagger$$

$\bullet$ $\mathbb{S}^3$ The quaternion (spinor) $\hat{\Psi}(\theta,\phi,\omega)$ describes the 3-sphere embedded in $\mathbb{R}^4$.

$\bullet$ $\mathbb{S}^2$ The bloch vector $\hat{\mathcal{R}}(\theta,\phi)$ describes the 2-sphere embedded in $\mathbb{R}^3$.

$\bullet$ $\mathbb{S}^1$ The global phase $e^{i\frac{\omega}{2}}$ describes the 1-sphere embedded in $\mathbb{R}^2$.

The global phase is a natural hidden variable of the quaternion (spinor/qubit) when viewed in the Bloch sphere representation - $\mathbb{S}^1$ is the fiber bundle connecting the 3-sphere and 2-sphere. This non-trivial hopf-fibration is parameterized by the global phase, which is defined:

$$\omega\;=\;\int_0^tdt'\left[\frac{\mathcal{R}^j\mathcal{H}^j+\mathcal{R}^k\mathcal{H}^k}{(\mathcal{R}^j)^2+(\mathcal{R}^k)^2}\right]$$

where $\mathcal{R}^\bullet$ are the elements of the Bloch vector and $\mathcal{H}^\bullet$ are the elements of the Hamiltonian - which satisfy the von neumann equation $$\dot{\hat{\mathcal{R}}}\;=\;[\mathcal{\hat{H}},\hat{\mathcal{R}}]$$ expressed in the cayley basis.

The global phase tells you where you are "globally" on the 3-sphere. For example if we choose some closed path on the Bloch sphere and calculate that the global phase of one orbit is $\omega=2\pi$, then the $\mathbb{S}^1$ fibration
$$e^{i\frac{2\pi}{2}}=-1$$ The negative coefficient tells us that we have only traveled half of the total path on the 3-sphere. While it appears we have reached our starting point after one orbit, in actuality we require a second orbit of the path on the Bloch sphere to return to the initial point.

Interpreted physically this accounts for the intrinsic spin of the spin-$\frac{1}{2}$ fermions. The fermions require 2 orbits to return to their inital point - a $4\pi$ rotation. for more info see;

The Hopf-Fibration and Hidden Variables in Quantum and Classical Mechanics: http://arxiv.org/abs/1601.02569

(I authored the above article.)

Answer 3

Mantras like “the Bloch sphere contains all of the observable information about the system” are junk. This implicitly refers to the concept of a (pure) quantum state, and confuses it with an important concept of observable, but quantum states are a tricky stuff and the word “information” may become especially treacherous.

Let us start from mathematics. Relevant formalisms are: Hilbert spaces, their projectivizations, and observables. Yes, when you get the ℂ² Hilbert space (for two states) and then restrict yourself to norm-1 state vectors, then you give a 3-sphere. Why vectors different only by a U(1) multiplication represent the same state? Because values of observables (2 × 2 complex matrices) on them and their probabilities do not depend on phase. So, it is convenient to consider the projectivization; eigenstates of (non-degenerate self-conjugated) operators become uniquely defined, without annoying U(1) freedom for each of them. Not this concrete Hopf fibration does matter; the projectivization on a Hilbert space, in general, does matter. There is no choice: all Hilbert space of the same dimension are isomorphic, and projectivization is unique. It happens that all such fibrations (except the case of one-state quantum system, ℂ projectivized to singleton) are non-trivial; just a mathematical fact. I emphasize: these are slightly different mathematical formalisms for states; each has both advantages and shortcomings. The usefulness of projectivizations does not permit you to ignore U(1) in any calculation. It means that if you consider a quantum system in isolation, then the U(1) action on its quantum states may be ignored in descriptions of its behaviour.

You can “simply assign to every point of the Bloch sphere a phase”, but it just doesn’t make any sense. You will not obtain the structure of a linear space, not to say such mapping to ℂ²\{0} may not be continuous. You lose invariance and symmetry of projective formalism without gaining linearity of the Hilbert spaces’ one.

I could also explain in details the C*-algebra of observables on a qubit and its relation the sphere/ball, but this thing isn’t asked.

Now physics. If you want to understand something about quantum states, then you might forbid yourself to think that any qubit has some definite pure quantum state (hint: consider an EPR pair). If one ever said you such thing, then s/he was a liar. Our world does not contain qubits in isolation; all quantum systems are potentially entangled.