# If the dimension of a space is prime, are quantum states in it guaranteed to be entangled?

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?

• How does the spatial dimension factor in to this at all? – J. Murray Nov 19 '18 at 13:28
• I mean the dimension of the Hilbert space. If this dimension is a prime, there are no subspaces that have the right dimensions for the tensor product – user1936752 Nov 19 '18 at 13:42
• @user1936752 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 seems to be implied by the question. As a side note, 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: physics.stackexchange.com/a/435864/206691 – Chiral Anomaly Nov 19 '18 at 23:24
• @DanYand That's an interesting point! You're right, I also assumed entangled = not a product state but I thought that for mixed states, these were equivalent definitions i.e. either you can write the state as $\rho = \sum_i p_i \rho_a(i)\otimes\rho_b(i)$ or you can't and this means you must have some nonclassical correlations? If you'd like to expand on this point with an answer (or with an addendum to your linked answer) showing states that satisfy one definition but not the other, I'd really appreciate it! – user1936752 Nov 20 '18 at 10:44

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.

• Interesting post, but the interesting thing about violating Bell's inequality is the part where things are spatially separated. This, again, requires a notion of tensor product, correct? (What I mean to say is, to spatially separate 'something' from 'something else', both of these 'somethings' need to have well-defined Hilbert spaces. Then again... I guess there are anyons. Ouch this post is making my brain hurt.) – Ruben Verresen Nov 22 '18 at 0:08
• @RubenVerresen I guess it depends on what you mean by "requires". All empirical successes of quantum theory so far are approximations to quantum field theory (QFT), so maybe we should ask how these things would be most naturally formulated in QFT. Although spatially separated regions can be expressed using a tensor product in QFT (with caveats as explained in arxiv.org/abs/1803.04993), this isn't explicitly assumed in the postulates of QFT. In this sense, the postulates of QFT seem conceptually more general. At least, that's my current view, but I still have lots to learn. – Chiral Anomaly Nov 22 '18 at 3:28