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- A photon can have various types of polarization states (horizontal, vertical, circular, elliptical, linear at an angle ##\theta##). Any valid polarization basis is two-dimensional and can represent any state of polarization. What are the actual eigenvectors of the polarization observable? My understanding is that polarization is an observable, hence an operator.
$$|\Psi>_{12}= a |H>_1 |V>_2+b |V>_1 |V>_2+c|V>_1 |H>_2+d|H>_1 |H>_2$$

- Because the polarization bases are two dimensional, it is convenient to entangle the states of polarization of two photons. I spent some time studying the tensor product and learned that the most general two-photon polarization state is given by

where each term contemplates all possible combinations of polarization H and V for the two photons.

I know the state below is an entangle state because it cannot be expressed as a product of single photon states:

$$ (\frac {1}{\sqrt{2}}[ |H>_1 |H>_2+|V>_1 |V>_2]$$

Because of entanglement, once the polarization of the first photon is measured, the polarization of the second photon is automatically determined. For instance, if one photon turns out H the other turns out H as well and if it was measured as V the other would be V too. But wouldn't that happen also when the the system is in a state like

$$ |V>_1 |V>_2]$$

which we know is not entangled? Why not?

thanks.

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# I Entanglement and photon polarization

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