= Solution
In <second quantization>, expand the annihilating <quantized field> in a complete <orthonormal basis> of <one-particle states>, $\hat\psi(x)=\sum_\alpha\varphi_\alpha(x)a_\alpha$. At equal times, the <canonical commutation relation> for <bosons> and the <canonical anticommutation relations> for <fermions> are, respectively,
$$
[\hat\psi(x),\hat\psi^\dagger(x')]=\delta(x-x'),\qquad
[\hat\psi(x),\hat\psi(x')]=[\hat\psi^\dagger(x),\hat\psi^\dagger(x')]=0,
$$
and
$$
\{\hat\psi(x),\hat\psi^\dagger(x')\}=\delta(x-x'),\qquad
\{\hat\psi(x),\hat\psi(x')\}=\{\hat\psi^\dagger(x),\hat\psi^\dagger(x')\}=0.
$$
Here $\delta$ is a <Dirac delta distribution>. Multiplying by the basis <wave functions> and integrating gives $[a_\alpha,a_\beta^\dagger]=\delta_{\alpha\beta}$ for <bosons>, or $\{a_\alpha,a_\beta^\dagger\}=\delta_{\alpha\beta}$ for <fermions>; the two-annihilator and two-creator brackets vanish in the corresponding convention.
The mode <number operator> and total <particle number> operator are
$$
\boxed{n_\alpha=a_\alpha^\dagger a_\alpha,\qquad \hat N=\sum_\alpha n_\alpha.}
$$
The <bosonic occupation numbers> are $0,1,2,\ldots$. 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 $\sum_{j=1}^2\varphi_j(x)\varphi_j^*(x')$, rather than the full <Dirac delta distribution>.
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