# Local measurements on an entangled pair. (My understanding)

1. Mar 7, 2008

### ouacc

Say we have a BELL state |B00> = cos(a) |00>+ sin(a) |11>

then for a (just one) pair of entangled particles, we keep one on location q, the other one in location w; the one in q forms a system Q, the one in w forms W. ( 'x' in a function like |v> x |w> indicates a tensor product). Since they are entangled, they form a joint system QW, according to some tensor operations.

Define, on both systems, a projector : M0= |0><0|, M1=|1><1|. (they satisfy M0*M0+M1*M1= I)

and we use a projective measurement: Pm = Iq x (|Mm) = Iq x (|m><m|) (Iq is the identity operator on Q. m=0 or 1, is the outcome from M0 and M1, respectively), on the joint state |B00>

What is the result?

Conclusion (0)
Put it in plain English, the measurement Pm (m=0 and1) is to measure the particle in W along 0-1 axis. And this is done at location w locally.

Conclusion (1)
at location w, we know if the outcome is 0, then the particle in Q is in 0, too. with 100% certainty. (outcome 1 means the same)
After the measurement, if we measure the particle in q(the un-measured particle), say if we get Q=|0>, we know that the particle in W(the previously measured one) is 0, too, with 100% certainty.

Conclusion (2)
For the measured pair, after the measurement, they are NOT entangled any more. Q and M are decoherenced. Any later measurements on the particle in Q has no influence on the particle in W, and vise verse.

SUMMARY:
There are two possible outcomes: (ALL knowledge is at q locally, w would not know these, unless the information is sent from q to w via a classical channel)
if we find the particle at w is |0>, (that is m=0).
then we know that after the measurement, Q is in |0>, W is in |0> and the probability of this happening is cos(a)^2;
if we find the particle at w is |1>, (that is m=1)
then we know that after the measurement, Q is in |1>, W is in |1> and the probability of this happening is sin(a)^2;

Are the conclusions right or wrong?

Thanks

BTW, is it possible to get |B00> experimentally? I am thinking take |00>, use Hadamard gate to get (|0>+|1>)/sqrt2, then use CNOT to produce (|00>+|11>)/sqrt2. but I get other Bell states, ie. (|01>-|10>)/sqrt2 or (|01>+|10>)/sqrt2. Is is possible to get ONLY |B00>?