Trace of a subsystem of a two qubit system

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The discussion revolves around the calculation of the density operator for a two-qubit system and its reduced density operator for subsystem A. The initial density operator is presented as a mixture of three states with equal probabilities, leading to confusion about the inclusion of cross terms in the outer product. Participants emphasize the need to correctly apply the definition of the density operator, which involves considering all pure states and their corresponding outer products. The misunderstanding is clarified, prompting a reevaluation of the calculations. The conversation concludes with an acknowledgment to revisit the computations based on this new understanding.
Haorong Wu
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Homework Statement
From Nielsen's QC&QI, in page 109 (The schmidt decomposition), it reads that:
As an example, consider the state of two qubits, ##\left( \left | 00 \right> +\left | 01 \right> +\left | 11 \right> \right) / \sqrt 3##. This has no obvious symmetry property, yet if you calculate ##tr \left ( {\left( \rho ^A \right )}^2 \right )## and ##tr \left ( {\left( \rho ^B \right )}^2 \right )## you will discover that they have the same value, ##\frac 7 9## in each case.
Relevant Equations
The density operator for a system is ## \rho \equiv \sum_i p_i \left |\psi _i \right> \left < \psi _i \right |##.
Also, ## tr \left( A \left | \psi \right > \left < \psi \right | \right) =\left < \psi \left |A \right | \psi \right > ##
Consider the first qubit (subsystem A):

First, the density operator for the system AB is ## \rho ^{AB} =\frac { \left |00 \right > \left <00 \right |+ \left |01 \right > \left < 01\right |+\left | 11\right > \left < 11\right | } 3 ##.

Then, the reduced density operator of subsystem A is ## \rho ^A = \frac { \left |0 \right > \left <0 \right |+ \left |0 \right > \left < 0\right |+\left | 1\right > \left < 1\right | } 3 =\frac { 2\left |0 \right > \left <0 \right |+\left | 1\right > \left < 1\right | } 3 ##.

Thus, ## \left ( \rho ^A \right ) ^2=\frac { 4\left |0 \right > \left <0 \right |+\left | 1\right > \left < 1\right | } 9##.

So, ## tr \left ( {\left( \rho ^A \right )}^2 \right ) =\frac 5 9##.

I overchecked the procedure several times, but I can't see where am I wrong.

Thanks for reading.
 
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You might want to think again about the result you got for ##\rho^{AB}##
 
cpt_carrot said:
You might want to think again about the result you got for ##\rho^{AB}##

Hi, cpt_carrot. I still can't figure the mistake.

Here is my reasoning:

There are three possible states : ## \left | 00 \right> , \left | 01 \right> , \left | 11 \right> ## all with probabilitities of ##\frac 1 3##.

So ##\rho \equiv \sum_i p_i \left | \psi _i \right > \left < \psi _i \right |=\frac { \left |00 \right > \left <00 \right |+ \left |01 \right > \left < 01\right |+\left | 11\right > \left < 11\right | } 3 ##.

Maybe I understand the definition of density operator in a wrong way.

Could you help me point the mistake? Thanks!
 
You need to include the cross terms in the outer product. The density matrix for your pure state ##|\psi\rangle## is ##\rho= |\psi\rangle\langle\psi|## which includes terms like ##|00\rangle\langle 01|##
 
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cpt_carrot said:
You need to include the cross terms in the outer product. The density matrix for your pure state ##|\psi\rangle## is ##\rho= |\psi\rangle\langle\psi|## which includes terms like ##|00\rangle\langle 01|##
Oh, so ##
\rho \equiv \sum_i p_i \left | \psi _i \right > \left < \psi _i \right |## should be applied to a collect of pure states.
I am going to redo the calculation again.
Thanks, cpt_carrot!
 
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