Qubits Entanglement: Calculate & Interpret

[amgo100]

Homework Statement

Determine which qubits are entangled:

$|\psi\rangle=\frac{1}{2}(|000\rangle+i|010\rangle+i|101\rangle-|111\rangle)$

Homework Equations

The Attempt at a Solution

[/B]
My idea was to first calculate the density operator
$\rho = |\psi\rangle \langle\psi|$
and then find the partial trace over the second and the third qubit. Then from Schmidt rank I would know whether the first qubit is entangled with the rest of the system. Then I could repeat the procedure for the other qubits. However the result seams to be 0 and I don't even know how to interpret this result, nor how to find which of the three qubits are entangled.

 

[mfb]

Can you split the wavefunction into a product?

 

[amgo100]

Ok, I've tried separating one of the qubits from the rest to obtain a product state and succeded for the second one (B):
$|\psi\rangle = \frac{1}{2}(|0\rangle_B + i|1\rangle_B)(|00\rangle_{AC} + i|11\rangle_{AC})$,
so it seams that qubit A is entangled with C (the first and the third).

However I'm still left with a question why the method with the partial trace gave me 0. I would expect it to give the same result.

 

[mfb]

Ok, I've tried separating one of the qubits from the rest to obtain a product state and succeded for the second one (B):
$|\psi\rangle = \frac{1}{2}(|0\rangle_B + i|1\rangle_B)(|00\rangle_{AC} + i|11\rangle_{AC})$,
so it seams that qubit A is entangled with C (the first and the third).

I agree.

Concerning the other method: Can you show your calculations?

 

[paranormal]

Your approach is correct. The partial trace over the second and third qubits will give you the reduced density operator for the first qubit. The Schmidt rank of this reduced density operator will tell you whether the first qubit is entangled with the rest of the system. If the Schmidt rank is 1, then the first qubit is not entangled. If the Schmidt rank is greater than 1, then the first qubit is entangled with the rest of the system.

In this case, the result of the partial trace over the second and third qubits is 0, which means that the first qubit is not entangled with the rest of the system. This is because the reduced density operator for the first qubit is a pure state, which means that it can be written as $|\psi'\rangle \langle\psi'|$ for some state $|\psi'\rangle$. In other words, the state of the first qubit is completely determined by the state of the other two qubits, and there is no entanglement between them.

To find which qubits are entangled, you can repeat the procedure for the other two qubits. The partial trace over the first and third qubits will give you the reduced density operator for the second qubit, and the partial trace over the first and second qubits will give you the reduced density operator for the third qubit. If the Schmidt rank of either of these reduced density operators is greater than 1, then the corresponding qubit is entangled with the rest of the system. In this case, you will find that the second and third qubits are entangled with each other.