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Let a $n-$length pauli string represent any tensor product of finitely many pauli matrices, Ex: $X\otimes Z\otimes \mathbb{I}\otimes Y\otimes \mathbb{I}\otimes X\otimes\cdots\otimes Z$ where the number of matrices (involved in the tensor product) are $n$.

We know that all $n-$length pauli strings form a group, the Pauli Group of size $4^n$.

Let the weight of an $n-$length pauli string be equal to the number of non-identity pauli matrices in it. The number of $n-$length pauli strings with weight $w$ is just: $\binom{n}{w}3^{w}$.

Can we have a bound (different from the one above) for the number of commuting pauli strings of weight $w$ ?

If someone could give me some hints for estimating this, it'll be nice.

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up vote 1 down vote accepted

I understand your question to be asking for the maximal size of a set of mutually commuting Pauli strings of weight $w$ and length $n$.

Two Pauli strings of length $n$ commute if and only if they don't have different non-identity entries in any slot. Thus, in a given slot the elements of a set of mutually commuting Pauli strings of length $n$ must have either the identity or one fixed non-identity matrix. Thus a set of mutually commuting Pauli strings of weight $w$ and length $n$ of maximal size is one that contains every possible choice of $w$ non-identity matrices, for which there is only one choice per slot, so the size of such a set is $\binom nw$.

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