# What is the difference between $|\Phi^+\rangle=\frac1{\sqrt2}(|0,0\rangle+|1,1\rangle)$ and a mixture of $|00\rangle$ and $|11\rangle$?

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?

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

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 $$d=4$$).
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.