For bosons, the commutation relation between the creation operator $a_i^\dagger$ and the annihilation operator $a_j$ for modes $i$ and $j$ is given by:

$[a_i, a_j^\dagger] = a_i a_j^\dagger - a_j^\dagger a_i = \delta_{ij}$

And the anticommutation relation between two annihilation operators or two creation operators is zero:

$[a_i, a_j] = a_i a_j - a_j a_i = 0$ $[a_i^\dagger, a_j^\dagger] = a_i^\dagger a_j^\dagger - a_j^\dagger a_i^\dagger = 0$

These relations ensure that the number operator $N_i = a_i^\dagger a_i$ has integer eigenvalues, and that applying a creation operator multiple times to the vacuum state $|0\rangle$ results in a state with multiple bosons in mode $i$, without any sign changes.

For fermions, the anticommutation relation between the creation operator $a_i^\dagger$ and the annihilation operator $a_j$ for modes $i$ and $j$ is:

$\{a_i, a_j^\dagger\} = a_i a_j^\dagger + a_j^\dagger a_i = \delta_{ij}$

And the anticommutation relation between two annihilation operators or two creation operators is:

$\{a_i, a_j\} = a_i a_j + a_j a_i = 0$ $\{a_i^\dagger, a_j^\dagger\} = a_i^\dagger a_j^\dagger + a_j^\dagger a_i^\dagger = 0$

The anticommutator $\{a_i, a_j^\dagger\} = \delta_{ij}$ implies that if $i=j$, then $2 a_i^\dagger a_i = 1$, so $a_i^\dagger a_i = 1/2$. However, in standard second quantization, the number operator is defined as $N_i = a_i^\dagger a_i$, and its eigenvalues are either 0 or 1, reflecting the exclusion principle. The anticommutation relations ensure that applying a creation operator twice to any state results in zero, meaning a state can only have at most one fermion in any given mode.