3

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?

glS
  • 24,708
  • 5
  • 34
  • 108
Matteo
  • 161
  • 4

2 Answers2

3

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
  • 24,708
  • 5
  • 34
  • 108
  • Thank you for the quick response! How would I see that $G$ is such a projection? – Matteo Mar 08 '24 at 09:33
  • i'm not sure there's a much smarter approach than direct observation to observe it gives the singlet. You can however more generally see it must be a pure state observing that $G$ is essentially in the form $\frac1 d(I+c_i\sigma_i)$ with $d=4$ and $\sigma_i$ operators such that $\operatorname{tr}(\sigma_i\sigma_j)=4\delta_{ij}$, and with $c_i=1$, which means $|c|^2=3=4-1$. Then use the general considerations found eg in https://quantumcomputing.stackexchange.com/a/6047/55. This approach will tell you that eg changing signs in your expr still gives projections onto (different) pure states – glS Mar 08 '24 at 09:41
1

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
  • 161
  • 4