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Trace of a subsystem of a two qubit system

  • Thread starter Haorong Wu
  • Start date
Problem 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.
 
You might want to think again about the result you got for ##\rho^{AB}##
 
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|##
 
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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