Local measurements on an entangled pair. (My understanding)

[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?

Here are my answers:
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>?

 

[azntoon]

Yes, it is possible to get only |B00> experimentally. To do this, you need to apply a phase shift of pi/2 radians to the second qubit after applying the CNOT gate. This will produce the state (cos(a)|00> + sin(a)|11>).

 

[denian]

Your conclusions are mostly correct. Just a few clarifications:

- The measurement Pm is not just measuring the particle in W along the 0-1 axis, but it is measuring the joint state of QW along the 0-1 axis. This means that both particles are being measured together, not just the one in W.
- The probability of getting a certain outcome is not solely dependent on the angle a, but it also depends on the measurement basis chosen (in this case, the 0-1 basis). So the probabilities for getting 0 or 1 may not be exactly cos(a)^2 and sin(a)^2, but they will still be related to the angle a.
- It is not accurate to say that the particles are no longer entangled after the measurement. They are still entangled, but the entanglement has been "used up" in the measurement process. This means that the particles can no longer be described by a single joint state, but rather by two separate states (one for Q and one for W). However, if you were to measure both particles again in the same basis, you would still see correlations between their outcomes.
- It is not necessary to have a classical channel to communicate information from q to w. The measurement itself can affect the state of the particle in W, so w can gain information about the measurement outcome without any classical communication.

As for your experimental question, it is indeed possible to prepare the state |B00> experimentally. One way to do this is to start with two entangled particles in the state |00> (for example, two photons in a polarization-entangled state) and then apply operations to create the desired state. For example, you could use a polarizing beam splitter and half-wave plates to create the state (|00> + |11>)/sqrt2 and then use a CNOT gate to get the state (|00> + |11>)/sqrt2. It is also possible to use other methods such as quantum teleportation to create the state |B00>.