 Solution

In second quantization, expand the annihilating quantized field in a complete orthonormal basis of one-particle states, $\hat{\psi }(x)={\sum }_{\alpha }{\phi }_{\alpha }(x)a_{\alpha }$. At equal times, the canonical commutation relation for bosons and the canonical anticommutation relations for fermions are, respectively,andHere $\delta$ is a Dirac delta distribution. Multiplying by the basis wave functions and integrating gives $[a_{\alpha },a_{\beta }^{†}]={\delta }_{\alpha \beta }$ for bosons, or $\{a_{\alpha },a_{\beta }^{†}\}={\delta }_{\alpha \beta }$ for fermions; the two-annihilator and two-creator brackets vanish in the corresponding convention.

The mode number operator and total particle number operator areThe bosonic occupation numbers are $0,1,2,\dots$. For fermions, $a_{\alpha }^2=0$ and the canonical anticommutation relations imply $n_{\alpha }^2=n_{\alpha }$, so the fermionic occupation numbers are only $0$ and $1$, as required by the Pauli exclusion principle.

Keeping only two modes later is a projection: the projected field bracket is ${\sum }_{j=1}^2{\phi }_j(x){\phi }_j^{\ast }(x^{\prime })$, rather than the full Dirac delta distribution.