How to obtain the following equality: $EE' = (-1)^{a.b' + a'.b}E'E$

Question

I am having difficulty understanding why: $$EE' = (-1)^{a.b' + a'.b}E'E$$

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$E$ and $E'$ are error operators of the form:

$$E = i^{\lambda} X(a)Z(b)$$

$$E'=i^{\lambda '} X(a')Z(b')$$

where $\lambda \in 0,1,2,3$.

Additionally:

$$a= (a_{1}, \dots, a_{n})$$ $$b=(b_{1}, \dots, b_{n})$$ $$\lambda \in \{ 0,1,2,3 \}$$

$$X(a) |v \rangle = |a + v\rangle$$ $$Z(b) |v\rangle = (-1)^{b.v} |v\rangle$$ where $v \in \mathbb{F}_{2} ^{n}$ and $|v\rangle \in \mathbb{C}^{2^{n}}$

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I understand that $$EE' = i^{\lambda} X(a)Z(b) i^{\lambda '}X(a')Z(b')= i^{\lambda + \lambda'}X(a)Z(b)X(a')Z(b')$$ but I just can't understand why this is equal to $(-1)^{a.b' + a'.b}E'E$.

Am I missing something obvious? I think part of my problem is that I don't understand exactly what $a.b' +a'.b$ represents? Perhaps $EE'$ and $E'E$ commute iff $(a|b)$ and $(a'|b')$ are orthogonal, although I am not sure where to go from here...

Answer 1

I’m going to assume you are working from the “QECC via GF(4)” paper, link here: https://arxiv.org/abs/quant-ph/9608006, your question being on page 5. I’ll try to answer in the notation of the paper, but know that since it’s an older paper, there’s better sources to learn QECC from.

You had a good start, pulling out the constants $i^\lambda$ and $i^{\lambda’}$, because in doing so we can see the main challenge is how to deal with re-ordering the terms $X(a), X(a’), Z(b), Z(b’)$.

There are three keys to understanding how to flip-flop those terms:

• $a, b, a’, b’$ refer to vectors. So $a \cdot b’$ means the dot product of the vectors $a$ and $b’$.
• Thinking of bit-flip $\sigma_x$ and phase-flip $\sigma_z$ as matrices, recall that they anti-commute with each other (e.g. $\sigma_x \sigma_z = -\sigma_z \sigma_x$).
• Interpreting $X(a)$ physically, it means introducing a bit-flip $\sigma_x$ for the qubits in each component where $a$ has an entry of $1$. (Similarly for $Z(b)$, but with phase-flips.)

With that, we can see how to go about switching the order of $Z(b)X(a’)$. By qubit, one of these two cases occurs:

1. If the qubit in a component has either no error, or only a $\sigma_x$ from $X(a’)$, or only a $\sigma_z$ from $Z(b)$, then no negative is introduced by the reordering.
2. If the qubit has both a bit flip from $X(a’)$ and a phase flip from $Z(b)$, then anti-commutation occurs in that component, and so a negative is introduced.

So the number of negatives introduced is the places where both an $\sigma_x$ and $\sigma_z$ error are introduced; this is exactly the dot product $a’ \cdot b$. So, $$Z(b’) X(a’) = (-1)^{a’ \cdot b} X(a’) Z(b).$$

The other negatives can be picked up with similar work, and then the $i^\lambda$ and $i^{\lambda’}$ can be redistributed as needed (so they were never the important part of the problem).