The commutation relation $[H_d, sum_{i = 1}^n sigma_i^z] = 0$ from the paper about the constrained quantum annealing (CQA)

Question

A quote from the paper "Quantum annealing for constrained optimization" by I. Hen, F. M. Spedalieri:

Let us now consider the driver Hamiltonian
$$H_d = - sum_{i=1}^n left( sigma_i^x sigma_{i+1}^x + sigma_i^y sigma_{i+1}^y right)$$
where the label $i = n + 1$ is identified with $i = 1$ ... This driver has the following atractive properties (i) as can easily be verified, it obeys $[H_d, sum_{i = 1}^n sigma_i^z] = 0$; (ii)...

I don't see why $[H_d, sum_{i = 1}^2 sigma_i^z] = 0$. Note that for $n=2$ from this commutation relation we have:
$$H_d left(sigma_1^z + sigma_2^z right) = left(sigma_1^z + sigma_2^z right) H_d$$
and
$$H_d = -2left(sigma_1^x sigma_{2}^x + sigma_1^y sigma_{2}^y right)$$
but actually:
$$H_d left(sigma_1^z + sigma_2^z right) = -left(sigma_1^z + sigma_2^z right) H_d$$
because $sigma^x sigma^z = -sigma^z sigma^x$ and $sigma^y sigma^z = -sigma^z sigma^y$, hence $sigma_1^x sigma_{2}^x sigma_1^z = -sigma_1^z sigma_1^x sigma_{2}^x$ and the similar for other terms. So, in contradiction, instead of commuting operators, we have anticommuting operators ${H_d, sum_{i = 1}^2 sigma_i^z } = 0$. Where is my mistake(s)?

Edit
According to the answers below indeed $[H_d, sum_{i = 1}^2 sigma_i^z] = 0$. Also, the operators $H_d$ and $ sum_{i = 1}^2 sigma_i^z$ (for $n=2$) commute and anticommute at the same time and there is no contradiction (as I have wrongly stated above). This is because:
$$H_d left(sigma_1^z + sigma_2^z right) = left(sigma_1^z + sigma_2^z right) H_d = 0$$
Let's prove for the first part:
$$H_d left(sigma_1^z + sigma_2^z right) = -2left(sigma_1^x sigma_{2}^x + sigma_1^y sigma_{2}^y right) left(sigma_1^z + sigma_2^z right) =

=-2left(-i sigma_1^y sigma_{2}^x + i sigma_1^x sigma_{2}^y right) -2left(-i sigma_1^x sigma_{2}^y + isigma_1^y sigma_{2}^x right) = 0
$$

DaftWullie · Accepted Answer

There's the mathematical way of doing this (which I'll do in a moment), and there's a more conceptual side to it. Note that the operator $sum Z_i$ has a bunch of eigenspaces, with eigenvalues $-N,2-N,4-N,ldots,N-2,N$. The eigenspace $-N+2k$ is spanned by all basis states comprising $N-k$ $|0rangle$s and $k$ $|1rangle$s. Now, observe that a term $X_nX_{n+1}+Y_nY_{n+1}$ is a hopping term that preserves the number of excitations (the value $k$). Specifically it changes the state of the pair of qubits based on
$$
00rightarrow00,quad 01rightarrow 10,quad 10rightarrow 01,quad 11rightarrow 11.
$$
So, you can see that the eigenspace of the $sum_iZ$ operator does not change under that action. Similarly, the eigenvectors of $H_d$ can be grouped in terms of excitation number. An eigenvector of $H_d$ within a particular excitation subspace is unchanged by the action $sum_iZ_i$ (because, within that excitation subspace, $sum_iZ_i$ is just a multiple of the identity).
That should make commutation quite clear. For any eigenvector $|lambdarangle$,
$$
[H_d,sum_iZ_i]|lambdarangle=0,
$$
and if it's true for every eigenvector, it's true for every state. If it's true for every state, the whole operator is 0.

Let's return to doing things the more obvious way. Unfortunately, just applying commutation properties is not very revealing. We want to calculate
$$
(X_1X_2+Y_1Y_2)(Z_1+Z_2)-(Z_1+Z_2)(X_1X_2+Y_1Y_2).
$$
What you want to compare are the two terms
$$
X_1X_2Z_1-Z_2Y_1Y_2.
$$
We recall that $X_1Z_1=-iY_1$ and $Z_2Y_2=-iX_2$,
so this is
$$
-iY_1X_2+iY_1X_2=0.
$$
Other pairs of terms cancel in a similar way.

Jonathan Trousdale · Answer

Since $[sigma_i^k,sigma_j^k]=0$, you can expand the first product in $[H_d, sum_{i = 1}^2 sigma_i^z]$ as
$$begin{align}
(sigma_1^x sigma_2^x + sigma_1^y sigma_2^y) (sigma_1^z + sigma_2^z)= & sigma_2^x sigma_1^x sigma_1^z +sigma_1^x sigma_2^x sigma_2^z + sigma_2^y sigma_1^y sigma_1^z +sigma_1^y sigma_2^y sigma_2^z
 = & i(-sigma_2^x sigma_1^y - sigma_1^x sigma_2^y + sigma_2^y sigma_1^x + sigma_1^y sigma_2^x)
=&i([sigma_1^y,sigma_2^x]+[sigma_2^y,sigma_1^x])=0
end{align}$$
and the second product as
$$begin{align}
(sigma_1^z + sigma_2^z)(sigma_1^x sigma_2^x + sigma_1^y sigma_2^y)= & sigma_1^z sigma_1^x sigma_2^x +sigma_2^z sigma_2^x sigma_1^x + sigma_1^z sigma_1^y sigma_2^y +sigma_2^z sigma_2^y sigma_1^y
 = &i(sigma_1^y sigma_2^x + sigma_2^y sigma_1^x -sigma_1^x sigma_2^y - sigma_2^x sigma_1^y)
=&i([sigma_1^y,sigma_2^x]+[sigma_2^y,sigma_1^x])=0
end{align}$$
showing that $[H_d, sum_{i = 1}^2 sigma_i^z]=0$ as claimed.

The commutation relation $[H_d, sum_{i = 1}^n sigma_i^z] = 0$ from the paper about the constrained quantum annealing (CQA)

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