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QITStudent
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Hi all. I'm having a little trouble in understanding precisely how to calculate reduced density matrices. No literature I've been able to get my hands on has made it clear how precisely to work out partial matrices.
For example, if we have a bi-partite state for Alice's and Bob's particles:
Phi(ab) = 3-1/2(|bo> + |a1> + |c2>)
Where |b>, |a> and |c> are superpositions of basis states |0>, |1> and |2>. Then we must have density matrix
p(ab) = 1/3 (|bo><bo| + |bo><a1| + |bo><c2| + |a1><bo| + |a1><a1| + |a1><c2| + |c2><bo| + |c2><a1| + |c2><c2|)
So then working out the reduced density matrix means that we must take the partial Trace over Bob's. However, I'm not completely clear about this part.
Does this mean that we take the trace over Bob's matrix and multiply this scalar value by all of Alice's matrix. i.e, the trace here is 3 so Alice's state will be:
p(a) = |b><b| + |b><a| + |b><c| + |a><b| + |a><a| + |a><c| + |c><b| + |c><a| + |c><c|
Or, when taking the partial trace do we eliminate any of Alice's terms that do not have a diagonal term in Bob's matrix. I.e. getting the reduced matrix as
p(a) = |b><b| + |a><a| + |c><c|
The latter seems like it should not be the case, because that would mean every reduced density matrix is diagonal. But I don't seem to be getting the required answers whenever I perform the former method.
Any help would be greatly appreciated,
For example, if we have a bi-partite state for Alice's and Bob's particles:
Phi(ab) = 3-1/2(|bo> + |a1> + |c2>)
Where |b>, |a> and |c> are superpositions of basis states |0>, |1> and |2>. Then we must have density matrix
p(ab) = 1/3 (|bo><bo| + |bo><a1| + |bo><c2| + |a1><bo| + |a1><a1| + |a1><c2| + |c2><bo| + |c2><a1| + |c2><c2|)
So then working out the reduced density matrix means that we must take the partial Trace over Bob's. However, I'm not completely clear about this part.
Does this mean that we take the trace over Bob's matrix and multiply this scalar value by all of Alice's matrix. i.e, the trace here is 3 so Alice's state will be:
p(a) = |b><b| + |b><a| + |b><c| + |a><b| + |a><a| + |a><c| + |c><b| + |c><a| + |c><c|
Or, when taking the partial trace do we eliminate any of Alice's terms that do not have a diagonal term in Bob's matrix. I.e. getting the reduced matrix as
p(a) = |b><b| + |a><a| + |c><c|
The latter seems like it should not be the case, because that would mean every reduced density matrix is diagonal. But I don't seem to be getting the required answers whenever I perform the former method.
Any help would be greatly appreciated,