I'm having trouble seeing how this paper claims the following identity to be true for a three qubit system (labelled by $A$, $B$, and $C$) under a pure state $|\psi \rangle$ with real coefficients

$$\langle X_{A} X_{B}\rangle ^2 + \langle Z_{A} Z_{B}\rangle ^2 + \langle Z_{A} X_{B}\rangle ^2 + \langle X_{A} Z_{B}\rangle ^2 = 1 + \langle Y_{A} Y_{B}\rangle ^2 - \langle Y_{A} Y_{C}\rangle ^2 - \langle Y_{B} Y_{C}\rangle ^2$$

where $X$, $Y$, and $Z$ are the three Pauli matrices and the subscript is the qubit label. The expectation values have the usual meaning $\langle X_A X_B \rangle \equiv \text{tr} \rho X_A \otimes X_B \otimes 1$ where $\rho \equiv \text{tr}_C |\psi \rangle \langle \psi |$.

My Guess:

I'm thinking that this identity arises out of something like $X^2 + Y^2 + Z^2=3$ but that approach leads me nowhere since the square is outside the expectation brackets. Another problem is that while on the LHS the $C$ space is being traced over, on the RHS it's either $A$ or $B$ that's being traced over, and the square on top of everything just makes it worse. I suspect this is a well known but tediously derived identity, in which case I'll be happy to be pointed to some source material which grinds this out.

Thanks in advance. Cheers.

  • 2
    $\begingroup$ Is the last expectation value supposed to be $\left<Y_B Y_C\right>$? $\endgroup$ – Joseph Geipel Dec 20 '20 at 1:16
  • $\begingroup$ @JosephGeipel Yes, thanks. Fixed it. $\endgroup$ – Tuneer Chakraborty Dec 20 '20 at 6:00
  • $\begingroup$ I'd like to mention that I was able to "prove" this with the help of Mathematica by explicitly evaluating either side for an arbitrary state. Still it would be nice to know how to do this by hand. $\endgroup$ – Tuneer Chakraborty Dec 20 '20 at 19:37
  • $\begingroup$ @TuneerChakraborty This would constitute a proof if you were to write out the steps by hand. It's not very slick but its a proof. $\endgroup$ – Rammus Dec 21 '20 at 10:19

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