Entangled Photons: What Happens to Photon B? Can Two Combine?

[MaxwellDemon]

If a photon A is entangled with photon B and one somehow destroys photon A, what will happen to photon B? Will it also get destroyed? And can two entangled photons combine into one?

 

[zonde]

If a photon A is entangled with photon B and one somehow destroys photon A, what will happen to photon B? Will it also get destroyed?

No. Otherwise we wouldn't be able to perform Bell experiments with photons (not even mentioning that it would allow FTL communication).

 

[vanhees71]

If the two photons are detected at well separated places and you make sure that A has absorbed only 1 of the two photons, nothing happens to the other photon. As long as A doesn't take notice of the polarization state (I guess you talk about the usual polarization-entangled two-photon states used in Bell experiments), B's photon will be completely unpolarized, i.e., its polarization state is described by the statistical operator $\hat{\rho}=1/2 \hat{1}$.

If, however A has measured the polarization state of her photon, she knows that B will find the photon in the perpendicular polarization (if B measures in the same polarization direction), i.e., then she'd associate the corresponding pure polarization state with B's photon. If, however, she hasn't B told her result, B will just not now, what he will measure and thus will simply get a random polarization result (with probability 1/2 horizontal and with probability 1/2 vertical polarization).

 

[MaxwellDemon]

If the two photons are detected at well separated places and you make sure that A has absorbed only 1 of the two photons, nothing happens to the other photon. As long as A doesn't take notice of the polarization state (I guess you talk about the usual polarization-entangled two-photon states used in Bell experiments), B's photon will be completely unpolarized, i.e., its polarization state is described by the statistical operator $\hat{\rho}=1/2 \hat{1}$.

If, however A has measured the polarization state of her photon, she knows that B will find the photon in the perpendicular polarization (if B measures in the same polarization direction), i.e., then she'd associate the corresponding pure polarization state with B's photon. If, however, she hasn't B told her result, B will just not now, what he will measure and thus will simply get a random polarization result (with probability 1/2 horizontal and with probability 1/2 vertical polarization).

That answered many questions I had. Thanks for the detailed answer!