# What is the control input of Electro-optic modulator (EOM)

I have tried to implement the quantum photonic gates using a quantum teleportation mechanism based in this paper. I have stuck-upon at the last stage, where the measured results from two homodyne detectors(for x and p) are given to electro-optic modulator(EOM), refer to the diagram attached. Whats is the actual value given to the EOM here. We know the external control parameters like (theta and phi) can be given to EOM. I have doubt on how these (x and p) given to EOM as input, what is the purpose here, and how it controls the EOM. Pls, clarify doubts.

Hi @Mubarak and welcome to the forum!

Based on what I read in the paper, in this circuit, the input state is mixed with two squeezed light beams that are sent to homodyne detectors for measuring \hat{x} and \hat{p}. These measurement results are then used to determine the amount of displacement in the last beam which is performed by an EOM. The circuit seems to be similar to the one in this demo. Please have a look at the tutorial and the circuit implementation explained there and let me know if you have any other questions.

Hi @sjahangiri,

How the measured values x and p are used in EOM. Any mathematical relationship between the parameters (x,p) with EOM parameters like (theta, phi), if you provide the mathematical relationship between them then it would be a great help.

From the reply by @Nicolas_Quesada in a post, it is clear that the relationship between the factors (x,p) to the displacement factor. I got the pointâ€¦