The mechanics of calculating Reduced Density Matrices

Click For Summary
SUMMARY

The discussion focuses on the calculation of reduced density matrices (RDMs) in quantum mechanics, specifically for a bipartite state involving Alice's and Bob's particles. The density matrix is defined as p(ab) = 1/3 (|bo> PREREQUISITES

  • Understanding of quantum mechanics concepts, particularly bipartite states.
  • Familiarity with density matrices and their properties.
  • Knowledge of the partial trace operation in linear algebra.
  • Basic grasp of quantum state representations using basis states.
NEXT STEPS
  • Study the mathematical definition and properties of the partial trace in quantum mechanics.
  • Learn about the construction and interpretation of density matrices in quantum systems.
  • Explore examples of reduced density matrices in various quantum states.
  • Investigate the implications of reduced density matrices in quantum entanglement and measurement.
USEFUL FOR

Quantum physicists, graduate students in quantum mechanics, and researchers working on quantum information theory will benefit from this discussion.

QITStudent
Messages
2
Reaction score
0
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,
 
Physics news on Phys.org
QITStudent said:
Does this mean that we take the trace over Bob's matrix and multiply this scalar value by all of Alice's matrix.
Your useage of the term Bob's matrix is wrong. There is no Bob's matrix before you trace, there is only the whole density matrix. In your case, this is a 9x9 matrix.

Bob's and Alice's density matrices are the reduced density matrices, which you get by taking the partial trace of the whole density matrix with respect to Alice's resp. Bob's state space. These matrices are 3x3.

As far as the result is concerned, the second one is correct. I'm not sure where your misunderstanding comes from, so let's start with the basics. What is the trace of an arbitrary operator ρ using the basis states |ai>?
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K