What is the actual Hilbert space of a $N$-qubit system?

Question

This question seems slightly naive. The Hilbert pace of any 2-level quantum system is given by the Bloch sphere and the algebra of observables arises from $SU(2)$, the Lie group generated by the three Pauli matrices $\sigma_x,\sigma_y,\sigma_z$.

When considering spin chains, relevant for example in the NISQ quantum devicesand for algrithms such as the VQAs, the corresponding state space as far as I understand is $\mathcal{H}_{m} = SU(2) \times \ldots \times SU(2)$, that is for a length $N$ spin chain, $N$ copies of $SU(2)$.

If my understanding is correct, I struggle to see why many people claim that a $N$ qubit system's Hilbert space is described by $SU(2^N)$. Naturally, this is quite differnt to $\mathcal{H}_N$ described before, not only as a manifold, but also in terms of its generators which are definitely not matching the dimensions of a 2-level system for $N>1$.

For example, in this talk by a very senior researcher (at 0:11:00).

Therefore my questions boils down to:

1. Is there some map between $f: \times_N SU(2) \to SU(2^N)$ for $N>1$?
2. If not, isn't the claim that the Hilbert space of $N$ qubits is $SU(2^N)$ false?

Answer 1

Just a small remark for part of the question: Letting two Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ (this can be generalized to any linear space) the tensor product $\mathcal{H}_1 \otimes \mathcal{H}_2$ is again a space. The dimension is the product $\dim(\mathcal{H}_1 \otimes \mathcal{H}_2) = \dim(\mathcal{H}_1)\dim(\mathcal{H}_2)$ and this becomes general for arbitrary tensor of spaces. So, for instance, we get that $\dim(\mathcal{H}_N) = \dim(\mathcal{H_{qubit}}^{\otimes N}) = \dim(\mathcal{H}_{qubit})^N$.

So, the claim that Hilbert space of $N$ qubits is $\mathcal{H_{qubit}}^{\otimes N}$ is true in the standard formalism. But I think that there is a distinction between Hilbert spaces and Lie Groups to be made as well, and I shall leave someone else that is an expert to do so.

Answer 2

First off, the Bloch sphere is the complex projective line $\mathbb{C}P^1$, which is homeomorphic to $S^2$, while $SU(2)$ is homeomorphic to $S^3$. $SU(2^N)$ is the group of operators on pure states, not the state space itself.

The state space of an $N$-qubit system is complex projective space, $\mathbb{C}P^{2^N - 1}$; the elements of this space are equivalence classes of vectors. To see this, let's start with a single qubit, which can be represented by a vector $(z, w) \in \mathbb{C}^2$; since the magnitude of this vector is normalized to 1, the space is actually $\mathbb{C}P^1$ rather than $\mathbb{C}^2$. Multiple qubits are represented by tensor products of single qubit spaces, which we first take to be arbitrary vectors: $$ \mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \cdots \otimes \mathbb{C}^2 (N \text{ times}) \cong \mathbb{C}^{2^N} $$ Normalizing the elements, we end up with $\mathbb{C}P^{2^N - 1}$.