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Given an abstract quantum system, for example, the four-state system (ququart), is it possible to calculate all different ways to split the system state space into tensor product of subsystems' state spaces? In case of infinite number of such ways, by "calculate" I mean "provide continuous parametrization and topology".

Is it possible to calculate how many (i.e. describe topology and continuous parametrization of the set of) different qubits "live" in a quqaurt and how many qubits and qutrits are in the six-state system?

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There isn't really a lot of "topology" or interesting structure here: Whenever there are non-trivial possible splits into subsystems there are infinitely many of them most of which you physically won't care about at all.

The dimension of a tensor product is the product of the dimensions of the factors, i.e. $$ \mathrm{dim}(H_1\otimes H_2) = \mathrm{dim}(H_1)\cdot \mathrm{dim}(H_2).$$ So if the Hilbert space dimension is a prime number, there are no non-trivial splits into subsystems. If it isn't, then for any $d\mathop{\vert} \mathrm{dim}(H)$ we have that $\mathbb{C}^d \otimes \mathbb{C}^{\mathrm{dim}(H) / d}\cong \mathbb{C}^{\mathrm{dim}(H)}\cong H$ simply because all Hilbert spaces with the same cardinality are isomorphic. The space of different isomorphisms between $\mathbb{C}^{\mathrm{dim}(H)}$ and $H$ is of course given by $\mathrm{GL}(\mathrm{dim}(H),\mathbb{C})$, the space of all automorphisms of a complex vector space with dimension $\dim(H)$.

So this is the split into two systems for a given divisor $d$ of the dimension of $H$, and whenever $d$ or $\mathrm{dim}(H)/d$ aren't prime you can apply this recursively to the respective factor to get all possible splits into factors.

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  • $\begingroup$ "all Hilbert spaces with the same cardinality are isomorphic" — they are all isomorphic indeed, but they still can be in different position relative to the system. For example, every system basis induce different decomposition. $\endgroup$
    – warlock
    Dec 25, 2022 at 15:32
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    $\begingroup$ @warlock Sure...that's why I enumerate the different possible isomorphisms between the "canonical" result of the tensor product $\mathbb{C}^{\mathrm{dim}(H)}$ and $H$ as $\mathrm{GL}(\mathrm{dim}(H),\mathbb{C})$. Each element of $\mathrm{GL}$ corresponds to a particular choice of basis. I'm not sure what exactly you want to tell me with your comment. $\endgroup$
    – ACuriousMind ♦
    Dec 25, 2022 at 15:38
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    $\begingroup$ An alternative perspective is that you might want to mod out the GL(n) of the individual subspaces H1 and H2 - this would parametrize the different ways to split up the space, without fixing a basis within each space. $\endgroup$
    – Norbert Schuch
    Dec 25, 2022 at 15:38
  • $\begingroup$ So, in the end, the answer to my question is $\operatorname{GL}(mn,\mathbb C)/ \operatorname{GL}(m, \mathbb C) \times \operatorname{GL}(n, \mathbb C)$, right (for prime $m$, $n$, and $\operatorname{dim}(H)=mn$)? Something like grassmannian, but different. $\endgroup$
    – warlock
    Dec 25, 2022 at 16:12
  • $\begingroup$ @warlock If you want to preserve orthogonality - which I would consider natural when splitting up Hilbert spaces - the correct group would rather be the projective unitary group. $\endgroup$
    – Norbert Schuch
    Dec 27, 2022 at 18:19

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