Why are there equally many commuting and non-commuting Pauli operators with respect to a given Pauli operator?

Question

I'm working with $\mathbf{P}_n$, the set of all $n$-fold tensor products of Pauli matrices, including the identity. For a given Pauli operator $Q \in \mathbf{P}_n$ (excluding the identity), I want to understand why there are an equal number of Pauli operators $P \in \mathbf{P}_n$ that commute with $Q$ as there are that do not commute with $Q$.

Specifically, I want to show that the number of $P \in \mathbf{P}_n$ for which $PQ = QP$ is equal to the number of $P$ for which $PQ = -QP$.

I'd like a clearer and more concise explanation. Could someone provide a simple and intuitive proof or reasoning for this property?

Any help would be greatly appreciated!

Answer 1

Lets fix our initial pauli $P$. We can define the two sets of operators for $P$ as $Q_{+/-} = \{\sigma\in P_n: P\sigma = (+/-) \sigma P\}$. We can show if $Q_-$ is non empty, then there is a bijection between $Q_+$ and $Q_-$.

Let $q\in Q_-$, the bijection between the two sets is $\sigma \rightarrow q\sigma$. Clearly this is a bijection on the paulis and if $\sigma \in Q_{+/-}$ then $q\sigma P = q (+/-)P\sigma = (-/+) Pq \sigma $, so $q\sigma \in Q_{-/+}$.

The only state this doesn't work for is $I$ as $Q_-$ is empty.

Answer 2

For a single qubit, $X$ commutes with $\{I,X\}$ and anticommutes with $\{Y,Z\}$.

For two qubits, it's easy to check that $X\otimes I$ commutes with all Pauli operators of the form $X\otimes\sigma$ and $I\otimes\sigma$, where $\sigma\in\{I,X,Y,Z\}$. It is similarly easy to check that it anticommutes with Paulis of the form $Y\otimes \sigma$ and $Z\otimes \sigma$, as e.g. $$(X\otimes I)(Y\otimes \sigma)=i Z\otimes \sigma=-(Y\otimes \sigma)(X\otimes I).$$

To generalise this, observe the following:

• Any $n$-qubit Pauli operator except for $I\otimes\cdots\otimes I$ has eigenvalues $\pm1$, each one with multiplicity $2^{n/2}$. That's a consequence of each of $X,Y,Z$ having eigenvalues $\pm1$, together with the eigenvalues of a tensor product of diagonalisable matrices having eigenvalues given by the products of individual eigenvalues.

• Any two-qubit Pauli operator (except $I\otimes I$) can be written as $(U\otimes V)(X\otimes I)(U\otimes V)^\dagger$ for some unitaries $U,V$. More generally, any $n$-qubit Pauli operators $\sigma\neq I_n$ is unitarily equivalent to $X\otimes I\otimes \cdots \otimes I$ (or any other "easy" Pauli operator we want to use for reference).

• For any pair of matrices $\sigma$ and $\sigma'$, $[\sigma,\sigma']=0$ iff $[U\sigma U^\dagger,U\sigma' U^\dagger]=0$, and similarly $\{\sigma,\sigma'\}=0$ iff $\{U\sigma U^\dagger,U\sigma' U^\dagger\}=0$, for any unitary $U$.

• The above tells us that it's sufficient to prove the statement at hand for a single choice of Pauli matrix $\sigma\neq I_n$. But again, that's easy enough to do for something like $X\otimes I\otimes \cdots\otimes I$, as the set of Pauli operators commuting to it is everything of the form $I\otimes \sigma$ and $X\otimes \sigma$ for $\sigma$ any $(n-1)$-qubit Pauli operator, and the set of anticommuting Pauli operators is given by all those of the form $Y\otimes\sigma$ and $Z\otimes \sigma$.

Answer 3

Consider $P \in \mathbf{P}_n$ and the projector $\Pi = \frac{1}{|\mathbf{P}_n|} \sum_{Q \in \mathbf{P}_n} {\rm Ad}_Q$. One knows/can show that $$\Pi X = \frac{{\rm tr}(X)}{d} I$$ for any operator $X$ and the identity operator $I$ and the Hilbert space dimension $d$. Now if $P \neq I$, then $\Pi P = 0$. On the other hand $\Pi P$ is just a constant times $P$, because $QPQ = \pm P$ for every $Q \in \mathbf{P}_n$. The constant is the number of commuting elements minus the number of anticommuting elements, which has to be zero to satisfy $\Pi P = 0$.