Define $$G = I \otimes I - X \otimes X - Y \otimes Y - Z \otimes Z,$$ where $X,Y$ and $Z$ denote the Pauli matrices, and $I$ the identity. I can plug this matrix in my computer and note that $$G = \frac{1}{4} G^2 \succeq 0,$$ which proves that $G$ is PSD. But I cannot get any intuition for the above equality. I worked out $$ G^2 = 4 I \otimes I - 2 X \otimes X - 2 Y \otimes Y - 2 Z \otimes Z + 2 \left( XY \otimes XY + XZ \otimes XZ + YZ \otimes YZ \right),$$ but am unable to reduce this further to $4G$. Can anyone provide me the necessary steps to complete this computation?
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Observe that $$XY\otimes XY=-Z\otimes Z$$ and similarly for the other two terms in parenthesis. This is just using $\sigma_j\sigma_k=i\epsilon_{jk\ell}\sigma_\ell$, see https://en.wikipedia.org/wiki/Pauli_matrices#Algebraic_properties.
As for the positivity, you can also more directly observe that $G$ is a (multiple of) the projection onto the pure state $|01\rangle-|10\rangle$.
glS
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I found this out later, but one may also notice that $$ G = 2(I-S),$$ where $S$ is the swap matrix, see e.g., What is the tensorial representation of the quantum swap gate?. Then, using $S^2 = I$, we have $$ G^2 = 4 (I -S)^2 = 4 (I - 2S + S^2) = 8 (I - S) = 4G,$$ which proves that $G \succeq 0$, since $$G = \frac{1}{4} G^2 \succeq 0.$$
Matteo
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