0
$\begingroup$

I struggle to see the difference between $|\Phi^+\rangle = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B)$ and the mixed state defined with probabilities {0.5, 0.5} and pure states $|00\rangle$ and $|11\rangle$.

In the case of a single qubit, it is easy to see the difference between $|+\rangle$ and the mixed state with probabilities {0.5, 0.5} and pure states $|0\rangle$ and $|1\rangle$ because when measured in the Z basis they behave the same (same probability distribution) but not in the X basis.

So what is the equivalent for Bell state?

$\endgroup$
2
  • 2
    $\begingroup$ What if you measure both systems in Z basis vs measuring both systems in the X basis? You'll again see a difference. $\endgroup$
    – Rammus
    Jul 6, 2022 at 16:07
  • $\begingroup$ possible duplicate: quantumcomputing.stackexchange.com/q/1461/55 $\endgroup$
    – glS
    Jul 6, 2022 at 18:49

1 Answer 1

0
$\begingroup$

As you properly state, a single qubit in superposition can be distinguished from a maximally mixed qubit, and from pure states in other but known bases, with the appropriate projective measurement. Applying this perspective to EPR pairs, I sometimes find it useful to think of the Bell states not as a superposition of two entangled qubits, but rather as a superposition of a single qudit (with $d=4$).

The Wikipedia article on the Bell basis measurement gives the circuit for such a Bell measurement (the adjoint of the one that creates it).

Applying this circuit to the Bell state $|\Phi^+\rangle$ gives the output $|00\rangle$ with certainty, while a maximally mixed state outputs any of $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ with uniform probability (1/4).

Furthermore it can be seen from the circuit that the Bell measurement applied to a state like $|00\rangle$ will give $\vert+0\rangle$ with certainty.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.