State tomography on a subsystem of the GHZ state

Question

Premise: I am not sure whether I am missing something theoretically.

Given a circuit creating a GHZ state over 3 qubits, say q1, q2 and q3. If I do not consider q3 and perform a state tomography over q1 and q2, then intuitively, I should get a Bell state as output, i.e. $|\Phi^+\rangle\langle\Phi^+|$.

Nevertheless the output is the following matrix:

[[ 0.5, -0. ,  0. , -0. ],
   [-0. ,  0. ,  0. , -0. ],
   [ 0. ,  0. ,  0. ,  0. ],
   [-0. , -0. ,  0. ,  0.5]]

I am not observing q3. I also tried to reset it with the command circuit.reset(2).

This is how may look one of the 9 circuits prepared for the tomography:

EDIT: here there is a partial answer, however, I'd still like to understand what is the meaning of my output matrix.

Answer 1

Quantum state tomography returns the expected result.

State of a subsystem

In quantum mechanics the state of a subsystem is given by the partial trace of the state of the full system over the complement of the subsystem. In this case, the full system consists of qubits q1, q2 and q3 in the state $\rho_{123}=|GHZ\rangle\langle GHZ|$ and we wish to find the state $\rho_{12}$ of qubits q1 and q2. We calculate $$ \begin{align} \rho_{12}&=\mathrm{tr}_3(\rho_{123})\\ &=\mathrm{tr}_3(|GHZ\rangle\langle GHZ|)\\ &=\frac12\mathrm{tr}_3(|000\rangle\langle 000|+|000\rangle\langle 111|+|111\rangle\langle 000|+|111\rangle\langle 111|)\\ &=\frac12(|00\rangle\langle 00|\mathrm{tr}(|0\rangle\langle 0|)+|00\rangle\langle 11|\mathrm{tr}(|0\rangle\langle 1|)+\\ &\quad\,+|11\rangle\langle 00|\mathrm{tr}(|1\rangle\langle 0|)+|11\rangle\langle 11|\mathrm{tr}(|1\rangle\langle 1|)\\ &=\frac12(|00\rangle\langle 00|+|11\rangle\langle 11|)\\ &=\frac12\begin{bmatrix}1&&&\\&0&&\\&&0&\\&&&1\end{bmatrix} \end{align} $$ which matches what you have observed.

Subsystem of a pure entangled state is never in a pure state

It is the hallmark of entanglement that a state of a subsystem of a pure entangled state is not pure$^1$. Therefore, your original expectation to see a Bell pair on two of the qubits was wrong.

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^{$^1$ This can be proved for example using Schmidt decomposition.}

Answer 2

I asked a similar question several weeks ago. The answer seems to be that three entangled qubits behave differently than two entangled qubits because that's how the matrix multiplication tells them to behave. Examining the third qubit is unnecessary.

Here is a quirk implementation using both two-qubit and three-qubit entanglement.

You can see that the measurements of the two qubits gives different results even though the third qubit is never looked at or touched. The entanglement of the third qubit causes a different underlying state vector.