# Entanglement of coherent states

I am new to quantum optics and I am trying to understand whether coherent states of the field could be entangled. Let's consider, for example, a two-mode state:

$|\psi^{AB} \rangle = \frac{1}{c} \left(| \alpha^A \beta^B \rangle + | \beta^A \alpha^B \rangle \right)$

where c - is a normalization constant and $\alpha$ and $\beta$ are the displacements (they can be different or the same). This state looks like a singlet state:

$|\psi^{AB}_{s} \rangle = \frac{1}{c}(| 0^A 1^B \rangle + | 1^A 0^B \rangle)$

which is entangled. However, the procedure of checking entanglement for $|\psi \rangle$, which implies calculation of covariance matrix:

$\sigma_{i,j} = \frac{1}{2} \langle R_{i} R_{j} + R_{j} R_{i} \rangle + \langle R_{i} \rangle \langle R_{j} \rangle$, $\quad (R = \{\hat{x}^A,\hat{p}^A,\hat{x}^B,\hat{p}^B\}$ in this case, $\hat{x}, \hat{p}$ - quadrature operators)

shows no entanglement in this state, moreover the covariance matrix is such, that just refers to a product vacuum two-mode state:

$\sigma = \frac{1}{2} I$, ($I$ is a 4$\times$4 identity matrix).

On the other hand if $| \psi \rangle$ is a product state, I should be able to write it as a product: and I can't imagine how to do that. Where am I wrong? Do I calculate the covariance matrix improperly and $|\psi \rangle$ is entangled, or is $|\psi \rangle$ a product state, then how to actually write it as a product? I realize that when applying squeezing to these two modes, I can get entanglement, but is it possible to make non-squeezed coherent (i. e. displaced vacuum) states entangled?

• Your state is entangled. However, it is not Gaussian, so you cannot characterize it by solely by its covariance matrix (as you have indeed observed). – Norbert Schuch Apr 22 '15 at 15:42
• Note that your state lives in a 2-dimensional space spanned by $\vert\alpha\rangle$ and $\vert\beta\rangle$. It is thus (at least regarding its entanglement properties) equivalent to a problem of a two-qubit state with the correct overlap $\langle\alpha\vert\beta\rangle$. (In particular, it converges exponentially to a maximally entangled state as $\vert\alpha-\beta\vert\rightarrow\infty$.) – Norbert Schuch Apr 22 '15 at 19:45
• Thank you! I didn't realize that it is non-Gaussian! Please correct me if I am wrong: To check the entanglement I need to take the density matrix $\rho = |\psi \rangle \langle \psi|$, calculate the partial trace $\rho^B = Tr_A(\rho) = \frac{1}{c^2}\sum_n \langle \alpha_n^A|\psi\rangle\langle \psi| \alpha_n^A\rangle$ and check if $Tr[(\rho^{B})^2] = \sum_n \langle \alpha_n^B|(\rho^B)^2|\alpha^B_n \rangle < 1$ – Ilya Apr 22 '15 at 21:11
• Also I wonder if such kind of states are ever met in experiments. Everything I saw about entanglement of continuous variable states (not too much though) is about squeezed or single photon states. – Ilya Apr 22 '15 at 21:18
• Interesting! At first glance, even I thought that this would be a Gaussian state. A useful rule of thumb would be to start off in a coherent state and imagine trying to create the state with just displacement and squeezing operators acting on it. @Ilya, it is true that they are entangled if a subsystem is mixed, but the connection between local purity and amount of entanglement between two subsystems is not so obvious (it's not, in general, a monotone), see here: maths.nottingham.ac.uk/personal/ga/papers/2251.pdf – Abhinav Apr 23 '15 at 9:51

## 2 Answers

Your state is entangled. However, it is not Gaussian, so you cannot characterize it by solely by its covariance matrix, as you have indeed observed.

On the other hand, note that for either party $X=A,B$, your state lives in a 2-dimensional space spanned by $\vert\alpha^X\rangle$ and $\vert\beta^X\rangle$. It is thus, at least regarding its entanglement properties) equivalent to a problem of a two-qubit state with the same overlap $\langle\alpha^X\vert\beta^X\rangle$, as it can be transformed to such a state bu local unitaries.

In order to determine the entanglement, we can thus consider the reduced density matrix \begin{align*} \rho_A &= \mathrm{tr}_B(\vert\psi_{AB}\rangle\langle\psi_{AB}\vert) \\ &\propto \vert\alpha^A\rangle\langle\alpha^A\vert + \vert\beta^A\rangle\langle\beta^A\vert+ \langle\alpha^B\vert\beta^B\rangle\,\vert\alpha^A\rangle\langle\beta^A\vert + \langle\beta^B\vert\alpha^B\rangle\,\vert\beta^A\rangle\langle\alpha^A\vert\ . \end{align*} (Note that by doing unitaries on $A$ and $B$, we can additionally make $a:=\langle\alpha^A\vert\beta^A\rangle$ and $b:=\langle\alpha^B\vert\beta^B\rangle$ non-negative, so that we can e.g. choose $\vert\alpha^A\rangle=\vert0\rangle$ and $\vert\beta^A\rangle = a\vert 0\rangle + \sqrt{1-a^2}\vert1\rangle$.)

Note also that the state will be entangled whenever $\alpha^A\ne\beta^A$ and $\alpha^B\ne\beta^B$.

I think you'd normally characterize an entangled (pure) bipartitioned state by looking at its reduced density matrix in diagonal form. It ought to have more than one non-zero element along the diagonal.