# 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

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 $\alpha^B\ne\beta^B$.