Spin operator in tight-binding model While reading Altland and Simons (Condensed Matter Field Theroy, p. 60), I came across the following problem. In tight-binding models, the exchange interaction contributes to the Hamiltonian in a form composed of “spin operators $S_i$”, and the authors give the definition of spin operators as
$$S_i = a^{\dagger}_{i\alpha} \sigma_{\alpha \beta} a_{i\beta} /2.$$
Before this I only know the definition for single particle spin operator $S = \sigma /2$ (assuming $\hbar = 1$). So I wonder what is the origin of this spin operator or the intuition behind it.
 A: The intuition is as following: when we work in second-quantization, $a^{\dagger}_{i,\beta}$ creates a particle with spin projection $\beta$ at site $i$, and $a_{i,\beta}$ annihilates such a particle. So, for example $a^{\dagger}_{i,\uparrow}a_{i,\uparrow}$ counts the number of particles with spin-up, and $a^{\dagger}_{i,\downarrow}a_{i,\downarrow}$ counts the number of down-particles. So it is quite intuitive to have the $S^z$ operator at site $i$ as
$$ S^z_i = \frac{\hbar}{2}(a^{\dagger}_{i,\uparrow}a_{i,\uparrow}-a^{\dagger}_{i,\downarrow}a_{i,\downarrow})$$
which is $\hbar/2$ if there is one spin-up particle and zero spin downs, and $(-\hbar/2)$ if the opposite is true. Another way to write it is using the Pauli matrix $\sigma^z$
$$ S^z_i = \frac{\hbar}{2}a^{\dagger}_{i,\alpha}(\sigma^z)_{\alpha,\beta}a_{i,\beta}$$
where the sum over the indexes is implied.
Now, we can generalize, as the projections to $S^x$ and $S^y$ are just a basis change. The important point is to make sure that the new definition of the spin operators maintain the algebra of spins, namely $\left[S^\alpha, S^\beta\right]=i\hbar \epsilon_{\alpha,\beta,\gamma}S^{\gamma}$. You can make sure that this indeed the case by using the commutation relations $\left[a_{i,\alpha},a^{\dagger}_{i,\beta}a_{i,\gamma}\right]=\delta_{\alpha,\beta}a_{i,\gamma}$, $\left[a^{\dagger}_{i,\alpha},a^{\dagger}_{i,\beta}a_{i,\gamma}\right]=-\delta_{\alpha,\gamma}a^{\dagger}_{i,\beta}$.
