Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch sphere

Question

The Bloch sphere is homeomorphic to the Riemann sphere, and there exists a stereographic projection $\Bbb S^2\to \Bbb C_\infty$. But this only holds for pure states. To quote Wikipedia:

Quantum mechanics is mathematically formulated in Hilbert space or projective Hilbert space. The space of pure states of a quantum system is given by the one-dimensional subspaces of the corresponding Hilbert space (or the "points" of the projective Hilbert space). For a two-dimensional Hilbert space, this is simply the complex projective line $\Bbb C P^1$. This is the Bloch sphere.

Essentially the set of all possible pure states of a qubit is homeomorphic to $\Bbb C P^1$.

Questions:

1. Is the set of mixed states homeomorphic to any geometrical structure (or complex projective space) in particular? Is there any similar stereographic projection corresponding to the mixed states?

2. Can we assign a manifold structure to the set of mixed states? For instance, $\Bbb C P^1$ is basically the manifold constructed by quotient as Lie group ($\Bbb C^\times$) action on $\Bbb C^2 \setminus \{0\}$. But that only corresponds to pure states.

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Edit:

Moretti says in his Physics SE answer that:

In finite dimension, barring the trivial case $\text{dim}({\cal H})=2$ where the structure of the space of the states is pictured by the Poincaré-Bloch ball as a manifold with boundary, $S(\cal H)$ has a structure which generalizes that of a manifold with boundary.

So, I suppose, in this context of qubits, the boundary is $\Bbb C P^1$, and the manifold without the boundary is the set of mixed states with some additional structure (?).

How do we describe this manifold without boundary in mathematical terms? I guess it would trivially be a 3-ball.

Answer 1

Almost. You get a manifold with boundary with the Bloch ball. The radius from the origin parameterizing how pure it is. The origin being maximally mixed. This isn't a manifold because a point on the boundary has a neighborhood that is homeomorphic to a half space but not one homeomorphic to $\Bbb R^n$.

Answer 2

I'm late to the party, but here's my take:

Pure qubit states

As you said, the space of pure states of a single qubit can be described as a complex projective line $\mathbb{C}P^1$, which is homeomorphic to a Bloch sphere.

Similarly, the space of pure states of $N$ qubits can be described as the cartesian product of $N$ complex projective lines. Infact, every pure state of $N$ qubits can be described as $N$ pure states $(\psi_1,...,\psi_N)$.

In projective geometry, we can describe this cartesian product with a Segre embedding

Mixed states of two qubits

Considering the case of just two qubits, the Segre embedding is the map:

$$\sigma: \mathbb{C}P^1\times\mathbb{C}P^1\to\mathbb{C}P^3$$ $$\sigma: ((x_0:x_1), (y_0:y_1))\mapsto(x_0y_0:x_0y_1:x_1y_0:x_1y_1)$$ $$\Sigma := Im(\sigma)$$

where the Segre variety $\Sigma$ is a quadratic surface.

Note: I'm talking about a two-dimensional surface in a three-dimensional space, but you must understand: this is a two-dimensional complex surface in a three-dimensional complex space, which means that in "real" terms it would be a four-dimensional manifold embedded into a six-dimensional space. The Bloch Sphere, instead, is a one-dimensional complex manifold, and that's why we can describe it as a two-dimensional real manifold.

Given this premise, the following statements hold:

• Every point $P\in\Sigma$ represents a pure state, by definition
• Two pure states $P_1, P_2$ generate a line $l$. Every element $Q\in l, Q\neq P_1, P_2$ is a complex linear combination of $P_1, P_2$, i.e. it is a mixed state of pure states $P_1, P_2$.
• If $Q$ is a mixed state, any line $l$ passing from $Q$ intersects $\Sigma$ in two points $P_1, P_2$, i.e. $Q$ is a mixed state of pure states $P_1, P_2$
• As a corollary, every point $Q\notin\Sigma$ represents a mixed state

In other words, the space of mixed states is exactly $\mathbb{C}P^3$, i.e. the codomain of our Segre embedding.

Mixed states of $N$ qubits

In the case of $N$ qubits we would have:

$$\sigma: (\mathbb{C}P^1)^N \to \mathbb{C}P^{2^N-1}$$

In this case, the Segre variety $\Sigma$ is an $N$-dimensional variety of degree $N$ inside a $2^N-1$-dimensional projective space. When $N$ is big, this is a very very small variety.

We can modify the list of statements above in the following way:

• Every point $P\in\Sigma$ represents a pure state, by definition
• Two pure states $P_1, P_2$ generate a line $l$. Every element $Q\in l, Q\notin \Sigma$ is a complex linear combination of $P_1, P_2$, i.e. it is a mixed state of pure states $P_1, P_2$.
• In general, $M$ pure states $P_1, ..., P_M$ generate an $M$-dimensional plane $\pi$. Every element $Q\in\pi, Q\notin\Sigma$ is a mixed state of pure states $P_1, ..., P_M$
• If $Q$ is a mixed state, any hyperplane $\pi$ passing from $Q$ intersects $\Sigma$ in at least $2^N-2$ points $P_1, ..., P_{2^N-2}$, i.e. $Q$ is a mixed state of pure states $P_1, ..., P_{2^N-2}$
• As a corollary, every point $Q\notin\Sigma$ represents a mixed state

As an addendum, I would like to note that, in this case where we have more than two qubits, not every mixed state $Q\notin\Sigma$ can be written as the mix of two pure states. In fact, with probability $1$, a mixed state $Q$ must be written as the mix of no less than $2^N-N$ pure states.

To answer your question

The set of all states (mixed and pure) of $N$ qubits can be described as the complex projective $2^N-1$ dimensional space $\mathbb{C}P^{2^N-1}$. This can be constructed, similar to what you said, by quotient as Lie Group $(\mathbb{C}^\times)$ action on $\mathbb{C}^{2^N}\backslash\{0\}$.

The set of all pure states, in this visualization, corresponds to the Segre variety $\Sigma\subset\mathbb{C}P^{2^N-1}$, which can itself be seen as the cartesian product of $N$ Bloch spheres.

If you need proofs, I can put something together.