Matrix Index and multiplication rules for Hermitian Pauli group products

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Given the Hermitian Pauli group products $$ \Omega_{a,b}=\{\pm 1,\pm i\}_{a,b}\cdot \{I,X,Y,Z\}_{a,b}^{\otimes n} $$ composed of $n$ 2x2 pauli matrices $(I,X,Y,Z)$ in tensor product, such that they span a $2^n$ by $2^n$ matrix space, and $a,b=0,1,\cdots , 2^n-2, 2^n-1$, does there exist a reference for their matrix index representation $$ \Omega_{a,b} = \sum_{k=0}^{2^n-1}\sum_{l=0}^{2^n-1}A_{a,b,k,l}\left|k\right>\left<l\right| $$ to find the complex coefficients $A_{a,b,k,l}$, where I write them using bras and kets, and matrix multiplication $$ \Omega_{a,b}\Omega_{g,h}=f_{a,b,g,h,u,v}\Omega_{u,v} $$ to find the complex structure constants $f_{a,b,g,h,u,v}$?

I have seen references, such as ZX, that use $\Omega_{a,b}\approx Z^aX^b$ notation, but those representations are never scaled by complex values to ensure they are Hermitian, nor have I seen an explicit matrix index representation for them.

linuxfreebird

Posted 2019-10-20T11:58:22.907

Reputation: 143

Why do you describe the reference to quantum error correction as 'ZX'? – Niel de Beaudrap – 2019-10-21T16:45:54.883

Answers

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If you write $Z^aX^b$, there's an implicit "add a phase $i$ to make it Hermitian if necessary", although I guess there are a couple of different conventions you might use the determine the sign used. So long as you're clear about the convention it doesn't really matter because you've got the extra $\pm 1$ freedom to add in to adjust for it.

As for an explicit matrix index representation, $$ A_{a,b,k,l}=i^{a\cdot b}(-1)^{a\cdot k}\delta_{b\oplus l=k} $$ where $\delta$ is the kronecker delta. Here I've chosen a slightly different convention for introducing the $i$ which makes the representation easier to write down: wherever there's a pair $ZX$, assume it's $iZX=Y$.

If you want to work out the effect of the product of two terms, you could do it long-hand, but I think it's easier to work it our with the Pauli operators first and translate it back: $$ Z^aX^bZ^gX^h=(-1)^{b\cdot g}Z^aZ^gX^bX^h=(-1)^{b\cdot g}Z^{a\oplus g}X^{b\oplus h}. $$

DaftWullie

Posted 2019-10-20T11:58:22.907

Reputation: 35 722

Thanks for the answer. Is there a reference for $A_{a,b,k,l}$ and $f_{a,b,k,l,u,v}$ or is it simply easy to derive and is common knowledge? Also, computing $f_{a,b,k,l,u,v}$ looks more complicated now, because the hamming weight has to be evaluated over modulo 4 instead of modulo 2 arithmetic, e.i. $(i)^{a\cdot b}$ versus $(-1)^{a\cdot b}$. If you could provide $f_{a,b,k,l,u,v}$ using $(i)^{a\cdot b}$, that would be welcomed. – linuxfreebird – 2019-10-21T13:30:08.557

@linuxfreebird it’s easy to derive. I just stared at it for about 30 secs and wrote down the answer. – DaftWullie – 2019-10-21T16:41:41.183