# Why does $(XZ\otimes I)|\Phi^+\rangle$ equal the Bell state $|\Psi^-\rangle$?

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}

• 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. Jun 13 at 7:54
• please use latex to format equations
– glS
Jun 13 at 12:50
• @glS shouldn't it be {\sqrt{2} in the 2nd last step? Jun 13 at 13:06
• @VanPeer yes, but it shouldn't be a screenshot in the first place. The same edit with latex would have been fine
– glS
Jun 13 at 13:12
• @KennethGoodenough I see! Could you please explain how does this phase factor concept work here? Jun 14 at 2:18

$$XZ |\Phi^+\rangle = XZ\bigg(\dfrac{|00\rangle + |11\rangle}{\sqrt{2}} \bigg) = \dfrac{|10\rangle - |01\rangle}{\sqrt{2}} = - \bigg( \dfrac{|01\rangle - |10\rangle}{\sqrt{2}} \bigg) = -|\Psi^-\rangle$$
But there is no distinction between the state $$-|\Psi^-\rangle$$ and $$|\Psi^-\rangle$$ quantum mechanically. That is, they are equivalent.