Skip to main content
Physics

Return to Question

added 3 characters in body
Source Link
garyp
garyp
  • 22.5k
  • 1
  • 45
  • 87

I'm reading various introductions to SPDC, and they all seem to take the state of the entangled photons to be the Bell state that is asymmetric under exchange: $$\left(\left| HV\right>\left| VH\right>-\left| VH\right>\left| HV\right>\right)\frac{1}{\sqrt2}$$$$\Big( \left| H\right>_1\left| V\right>_2-\left| V\right>_1\left| H\right>_2 \Big)\frac{1}{\sqrt2}$$ but I haven't run across anyone explaining why. I wouldn't expect the asymmetric case for photons (bosons).

I'm reading various introductions to SPDC, and they all seem to take the state of the entangled photons to be the Bell state that is asymmetric under exchange: $$\left(\left| HV\right>\left| VH\right>-\left| VH\right>\left| HV\right>\right)\frac{1}{\sqrt2}$$ but I haven't run across anyone explaining why. I wouldn't expect the asymmetric case for photons (bosons).

I'm reading various introductions to SPDC, and they all seem to take the state of the entangled photons to be the Bell state that is asymmetric under exchange: $$\Big( \left| H\right>_1\left| V\right>_2-\left| V\right>_1\left| H\right>_2 \Big)\frac{1}{\sqrt2}$$ but I haven't run across anyone explaining why. I wouldn't expect the asymmetric case for photons (bosons).

Source Link
garyp
garyp
  • 22.5k
  • 1
  • 45
  • 87

Spontaneous Parametric Down Conversion (SPDC): why the asymmetric Bell state?

I'm reading various introductions to SPDC, and they all seem to take the state of the entangled photons to be the Bell state that is asymmetric under exchange: $$\left(\left| HV\right>\left| VH\right>-\left| VH\right>\left| HV\right>\right)\frac{1}{\sqrt2}$$ but I haven't run across anyone explaining why. I wouldn't expect the asymmetric case for photons (bosons).