# What's the outcome of measuring photons that are entangled in phi + and - states in RL, HV and +- basis? [closed]

Please, I'm trying to understand a delayed choice entanglement swapping and I'm having trouble at understanding the outcomes of measuring photons that are enangled in phi + and - states in RL, HV and +- basis?

I know that entanglement in phi + state mean that:

If we measure 0 in one qubit, we will measure 0 in the other qubit (and if we measure 1, the other will be 1).

If we measure + in one qubit, we will measure + in the other qubit (and if we measure -, the other will be -).

And if we have an entanglement in phi - state:

If we measure 0 in one qubit, we will measure 0 in the other qubit (and if we measure 1, the other will be 1).

If we measure + in one qubit, we will measure - in the other qubit (and if we measure -, the other will be +).

So, I expect that: If we have and entanglement in phi + state:

If we measure R in one qubit, we must ALWAYS measure R in the other qubit.

If we measure L in one qubit, we must ALWAYS measure L in the other qubit.

If we measure H in one qubit, we must ALWAYS measure H in the other qubit.

If we measure V in one qubit, we must ALWAYS measure V in the other qubit.

If we measure + in one qubit, we must ALWAYS measure + in the other qubit.

If we measure - in one qubit, we must ALWAYS measure - in the other qubit.

AND:

If we have and entanglement in phi - state:

If we measure R in one qubit, we must ALWAYS measure R in the other qubit.

If we measure L in one qubit, we must ALWAYS measure L in the other qubit.

If we measure H in one qubit, we must ALWAYS measure H in the other qubit.

If we measure V in one qubit, we must ALWAYS measure V in the other qubit. If we measure + in one qubit, we must ALWAYS measure - in the other qubit.

If we measure - in one qubit, we must ALWAYS measure + in the other qubit.