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A rather obvious question perhaps but if I have a Hilbert space of dimension $d$ and $d$ is prime, I cannot possibly write my state as $$\rho = \sum_i p_i\rho_a\otimes\rho_b$$ simply because the dimensions don't add up correctly, yes?
In general, is the number of factors of $d$ useful if one wants to check if a state is entangled or not?
If the dimension is prime, there is no point in talking of two subsystems (which would be entangled), and thus no point in talking about entanglement (which is a property between two subsystems). Since, if subsystem 1 has dimension $d_1$ and subsystem 2 has dimension $d_2$, the total Hilbert space dimension would be $d_1d_2$, which is not prime.
If the dimension of the Hilbert space is prime then the state is guaranteed to be unentangled, simply because it can't have any tensor factor spaces to count as subsystems.
(This answer was added to address a comment from the OP.)
The accepted answer (which is correct) assumes that "entangled" is defined to mean "not a product state with respect to the given factorization of the Hilbert space". This definition is implied by the question.
However, if "entangled" is defined instead as something like "able to violate a Bell inequality", then it is no longer sensitive to the precise dimension of the Hilbert space. An example of such an alternate definition/quantification of entanglement is shown in this post: https://physics.stackexchange.com/a/435864/206691, which was inspired by section 3 in the paper http://arxiv.org/abs/1702.04924, "Entanglement measures and their properties in quantum field theory".
Here's a simple example to illustrate the idea. The CHSH inequality can be violated by a pair of qubits, which can be implemented using a 4-dimensional Hilbert space of the form $H_2\otimes H_2$, where $H_2$ is a 2-dimensional single-qubit Hilbert space. The CHSH inequality involves four observables $a_1,a_2,b_1,b_2$, two for each qubit, and an appropriate state-vector $|\psi\rangle$. Even though we may have constructed the two-qubit Hilbert space as a tensor product $H_2\otimes H_2$, it's really just a 4-dimensional Hilbert space, and the observables $a_n$ and $b_n$ can be represented by $4\times 4$ matrices. Clearly we can embed this same setup in a 5-dimensional Hilbert space, with the extra dimension "unused". The given state-vector $|\psi\rangle$ is still "entangled" in the operational sense that the CHSH inequality is still violated with respect to the given observables, even though a 5-dimensional Hilbert space cannot be factorized and therefore cannot be "entangled" according to the traditional definition.
The point of this trivial example is not to debate how the word "entanglement" should be used, but rather to illustrate the fact that when we construct a Hilbert space as a tensor product of smaller Hilbert spaces, what we're really doing is suggesting a particular set of preferred observables — namely those that affect only one (or a small number of) factors at a time. We can also select the same preferred observables (I mean, with the same algebraic relationships) without relying on a tensor product construction — and therefore without making any assumptions about the dimension of the Hilbert space, as long as it's sufficiently large.
