# State of the entangled particles at the event horizon of black hole

Looking through the article of Samir Mathur http://arxiv.org/abs/0909.1038 he assumes that the entangled particles created at the event horizon have all the state $$\frac{|0\rangle_{c_n}|0\rangle_{b_n}+\:|1\rangle_{c_n}|1\rangle_{b_n}}{\sqrt{2}}$$ at each step $n$ , where {$c_n$} are the particles falling into the black hole and {$b_n$} are the ones that escape (i.e Hawking radiation). More than this, on page 21 of the above article it says quote

the state produced at the interaction point is the same regardless of the state of the black hole.

I would have thought that the choice of same state is for computational ease only but this might say that it isnt. This does not seem to be an obvious thing.

• Why do the particles created at the event horizon have to be in the same state?
• Wouldnt that mean that in fact there is some correlation between them since at each stage the same state is prepared?
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