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?

  • 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


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.


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