Understanding the Impossibility of a Bound State of Two Identical Nucleons

[Hurkyl]

What you mean is that if you calculate the trace "for particle one", you will get a mixed state of |0> and |1> and when you calculate the trace "for particle two", you will get that SAME mixed state of |0> and |1>. Got that, but are the coefficients of |0> and |1> also the same after taking the two traces ?

The result is "50% |0> and 50% |1>": it's a statistical mixture, not a pure state. It cannot be written as "a|0> + b|1>".