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I'm slightly confused by the solution provided below by a suggested solution online to convert |$\phi^+$⟩ to |$\psi^-$⟩.

I tried doing the operation XZ but I got $\frac{1}{\sqrt2}$(|10⟩-|01⟩) instead of |$\psi^-$⟩.

However, applying ZX seems to provide me with the right answer.

Would appreciate the verification!

$$ \begin{align} (XZ \otimes I) |\Phi^+\rangle &= \frac{1}{\sqrt{2}}(XZ|0\rangle \otimes I |0\rangle + XZ |1\rangle \otimes |1\rangle) \\ &= \frac{1}{\sqrt{2}}(X|0\rangle \otimes I|0\rangle - X|1\rangle \otimes I |1\rangle) \\ &= \frac{1}{\sqrt{2}}(|1\rangle \otimes |0\rangle - |0\rangle \otimes |1\rangle) \\ &= |\Psi^-\rangle \end{align} $$

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    $\begingroup$ Note that $XZ$ is $ZX$, up to a phase factor. Since global (!) phase factors are unobservable in quantum mechanics, applying either $ZX$ or $XZ$ will lead to the same physical outcome. $\endgroup$ Jun 13 at 7:54
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    $\begingroup$ please use latex to format equations $\endgroup$
    – glS
    Jun 13 at 12:50
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    $\begingroup$ @glS shouldn't it be {\sqrt{2} in the 2nd last step? $\endgroup$
    – Van Peer
    Jun 13 at 13:06
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    $\begingroup$ @VanPeer yes, but it shouldn't be a screenshot in the first place. The same edit with latex would have been fine $\endgroup$
    – glS
    Jun 13 at 13:12
  • $\begingroup$ @KennethGoodenough I see! Could you please explain how does this phase factor concept work here? $\endgroup$
    – Colin Hong
    Jun 14 at 2:18
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@Kenneth comment is right.

Note that:

$$ XZ |\Phi^+\rangle = XZ\bigg(\dfrac{|00\rangle + |11\rangle}{\sqrt{2}} \bigg) = \dfrac{|10\rangle - |01\rangle}{\sqrt{2}} = - \bigg( \dfrac{|01\rangle - |10\rangle}{\sqrt{2}} \bigg) = -|\Psi^-\rangle $$

But there is no distinction between the state $-|\Psi^-\rangle $ and $|\Psi^-\rangle $ quantum mechanically. That is, they are equivalent.

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  • $\begingroup$ Thanks! How can we prove that these two states are not distinguishable quantum mechanically? $\endgroup$
    – Colin Hong
    Jun 14 at 2:19
  • $\begingroup$ @Colin Hong In term of why overall phase doesn't matter, check out this answer here: quantumcomputing.stackexchange.com/a/9630/9858 $\endgroup$
    – KAJ226
    Jun 14 at 4:44

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