I have been reading the article review for Optical Quantum Memory (http://arxiv.org/abs/1002.4659) when I came across this section about DLCZ protocol. I don't understand about Beam splitter. For what I can interpret, the beam splitter overlaps both photons which emitted simultaneously by two atomic ensembles. My question is how does the beam splitter mixes two photons together so they become entangled?
P.S. My English is not good. I am sorry.
You have $2$ atomic ensembles.
For each atomic ensemble, the ground state is $|\psi_0\rangle$, and the excited state is $|\psi_1\rangle$. The excited state is signaled by an idler photon.
Without beam splitter, if you detect only one photon (for the whole set), you know which atomic ensemble is excited, so you have the state $|\psi_0\rangle |\psi_1\rangle$ or the state $|\psi_1\rangle |\psi_0\rangle$
However, if you merge the optical path of the two idlers photons (with a beam splitter), and if you detect a photon, you are not able to say which atomic ensemble is in the excited state. The global state is entangled :
$$|\psi\rangle = \frac{1}{\sqrt{2}} ( |\psi_0\rangle |\psi_1\rangle + e^{i\phi} |\psi_1\rangle |\psi_0\rangle) \tag{4}$$ where $\phi$ is a phase factor due to the difference of optical path.
I feel the original question from the post is two-fold: (1) why the beam splitter changes the photon basis, and (2) why does a measurement in such a basis project and create entanglement?
The second question is answered well by others. The answer to the first question is actually a direct generalization of the classical understanding of beam splitter into quantum optics. The classical formula for electric fields passing a BS is $E_{\pm}=\frac{1}{\sqrt{2}}(E_0\pm E_1)$ for the two ports (up to a difference in the phase convention). In the quantized form, the electric field can be written as a linear combination of the field's creation and annihilation operator. Thus the BS's effects on annihilation operators equal to make $a_\pm=\frac{1}{\sqrt{2}}(a_0\pm a_1)$.
