5.5 Separable or entangled?

Most” vectors in \mathcal{H}_a\otimes \mathcal{H}_b are entangled: they cannot be written as product states |a\rangle\otimes|b\rangle with |a\rangle\in\mathcal{H}_a and |b\rangle\in\mathcal{H}_b.

In order to see this, let us write any joint state |\psi\rangle of \mathcal{A} and \mathcal{B} in a product basis as \begin{aligned} |\psi\rangle &= \sum_{ij} c_{ij}|a_i\rangle\otimes|b_j\rangle \\&= \sum_i|a_i\rangle\otimes\left(\sum_j c_{ij}|b_j\rangle\right) \\&= \sum_i|a_i\rangle\otimes|\phi_i\rangle \end{aligned} \tag{$\ddagger$} where the |\phi_i\rangle=\sum_j c_{ij}|b_j\rangle are vectors in \mathcal{H}_{\mathcal{B}} that need not be normalised.

Now, for any product state, these vectors have a special form. Indeed, if |\psi\rangle= |a\rangle\otimes|b\rangle then, after expanding the first state in the |a_i\rangle basis, we obtain |\psi\rangle = \sum_{i}|a_i\rangle\otimes\left(\sum_i\alpha_i|b\rangle\right). This expression has the same form as Equation (\ddagger) with |\phi_i\rangle=\alpha_i|b\rangle, i.e. each of the |\phi_i\rangle vectors in this expansion is a multiple of the same vector |b\rangle.

Conversely, if |\phi_i\rangle = \alpha_i|b\rangle for all i in Equation (\ddagger), then |\psi\rangle must be a product state.91 So if we want to identify which joint states are product states and which are not, we simply write the joint state according to Equation (\ddagger) and check if all the vectors |\phi_i\rangle are multiples of a single vector. Needless to say, if we choose the states |\phi\rangle randomly, it is very unlikely that this condition is satisfied, and we almost certainly pick an entangled state. In general, given n qubits, we need 2(2^n-1) real parameters to describe their state vector, but only 2n to describe separable states; as n grows larger, 2n becomes much much smaller than 2(2^n-1).

The problem of deciding whether or not a given state is separable is, in general, a hard problem (i.e. NP-hard). Because of this, it is interesting to try to understand the notion of separability from different points of view, and it turns out that algebraic geometry yet again has something interesting to say. The theory relies on the notion of projective space, which is a non-trivial topic to try to introduce here, so we do so only briefly, and at a very high speed.

We have repeatedly said that we only really care about state vectors up to global phase, i.e. that |\psi\rangle and |\psi'\rangle are “the same” if there exists some \theta such that |\psi\rangle=e^{i\theta}|\psi'\rangle. Combining this with our unitality requirement (that we want |\langle\psi|\psi\rangle|^2=1), we are led to studying the equivalence relation v\sim w \iff v=\lambda w \text{ for some }\lambda\in\mathbb{C}\setminus\{0\} on our Hilbert space. Geometrically, this can be understood as the space of lines through the origin, i.e. of 1-dimensional subspaces, but the geometry of projective space is a subject that really deserves many many pages to delve into, and so we won’t talk about this point of view here.

Algebraically, it turns out that we can describe the space of such equivalence classes using homogeneous coordinates. Defining projective n-space as \mathbb{P}^n \coloneqq \mathbb{C}^{n+1}/\sim (where \sim is the equivalence relation defined above), it turns out that points in \mathbb{P}^n are described by coordinates [a_0:a_1:\ldots:a_n] where a_i\in\mathbb{C} are not all simultaneously zero (i.e. there exists at least one i\in\{0,\ldots,n\} such that a_i\neq0) and where we impose that [a_0:a_1:\ldots:a_n] = [\lambda a_0:\lambda a_1:\ldots:\lambda a_n] for any \lambda\in\mathbb{C}\setminus\{0\}.

Why is this useful? Well, given any pure state \alpha_0|0\rangle+\alpha_1|1\rangle of a qubit, we obtain a unique point in \mathbb{P}^1, namely [\alpha_0:\alpha_1] (since |\langle\psi|\psi\rangle|^2=1 tells us that at least one of \alpha_0 and \alpha_1 is non-zero); conversely, given any point [a_0:a_1]\in\mathbb{P}^1, we can multiply by an appropriate \lambda\in\mathbb{C}\setminus\{0\} to assume that |a_0|^2+|a_1|^2=1, and thus obtain a unique (up to global phase) pure state a_0|0\rangle+a_1|1\rangle. That is, points in the (complex) projective line \mathbb{P}^1 correspond to pure states of a qubit.

Next, we can always express a pure state of two qubits in the form \beta_0|00\rangle + \beta_1|01\rangle + \beta_2|10\rangle + \beta_3|11\rangle and we similarly find a correspondence with points [z_0:z_1:z_2:z_3] in \mathbb{P}^3 (given, in one direction, by setting z_i\coloneqq\beta_i).

What is of interest to us here is a particular map known as the Segre embedding: \begin{aligned} \sigma\colon\mathbb{P}^1\times\mathbb{P}^1 &\longrightarrow\mathbb{P}^3 \\([a_0:a_1],[b_0:b_n]) &\longmapsto [a_0b_0:a_0b_1:a_1b_0:a_1b_1]. \end{aligned} First of all, one needs to check that this does indeed give a well defined function (i.e. that the resulting coordinate always has at least one non-zero component, and that it is invariant under multiplication by a non-zero scalar \lambda\in\mathbb{C}\setminus\{0\}). But it turns out that, not only is this a well defined function, but it is actually a “geometric” function, in that it respects the “geometric structure” of projective space. We won’t concern ourselves here with what that means, but we note that it is even more well behaved than this: as its name suggests, it actually gives an embedding (i.e. a “geometric” injection) of the 2-dimensional space \mathbb{P}^1\times\mathbb{P}^1 into the 3-dimensional space \mathbb{P}^3.

The image of the Segre embedding is called the Segre variety, and you can check that it is given by the set of points \Sigma \coloneqq \operatorname{Im}(\sigma) = \big\{ [z_0:z_1:z_2:z_3]\in\mathbb{P}^3 \mid z_0z_3-z_1z_2=0 \big\} (in algebraic-geometry language, it is the zero-locus of a single polynomial).

Now here is the punchline to all this geometric meandering: a state |\phi\rangle of two qubits is separable if and only if its corresponding point in \mathbb{P}^3 lies in the Segre variety \Sigma.

For more, see e.g. Cirici, Salvadó, and Taron’s “Characterization of quantum entanglement via a hypercube of Segre embeddings”, arXiv:2008.09583).

Quantum entanglement is one of the most fascinating aspects of quantum theory. We will now explore some of its computational implications.

  1. Even though an entangled state cannot be written as a single tensor product, it can always be written as a linear combination of tensor products, since these form a basis.↩︎