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In a one-dimensional identical-particle model, a contact interaction has two-body potential $U(x,x^{\prime })=g\delta (x-x^{\prime })$, using the Dirac delta distribution. For bosons, its second quantization is the displayed normally ordered Hamiltonian operator. Projection onto orthonormal modes gives $H_{\mathrm{int}}=(g/2){\sum }_{ijkl}{I}_{ijkl}{a}_i^{†}{a}_j^{†}{a}_k{a}_l$, where $I_{ijkl}=\int {\phi }_i^{\ast }{\phi }_j^{\ast }{\phi }_k{\phi }_{l}dx$. The overlap integrals determine which mode-changing processes survive. A finite-mode model describes that projection; discarded modes can still matter to the full physical system.

For a free intrinsic-even real scalar field, momentum reflection $Rf(\mathbf{p})=f(-\mathbf{p})$ is a self-adjoint involution on the one-particle space. Its odd orthogonal projector is $(1-R)/2$. The corresponding second quantization is $N_{\mathrm{odd}}=\frac{1}{2}\int d^{3}p/(2\pi {)}^3{a}_{\mathbf{p}}^{†}(a_{\mathbf{p}}-a_{-\mathbf{p}})$, with integration over all momentum space. Conjugation by $e^{i\pi N_{\mathrm{odd}}}$ fixes even modes and reverses odd modes, hence sends $a_{\mathbf{p}}$ to $a_{-\mathbf{p}}$ and implements the parity operator. Its square is one because $N_{\mathrm{odd}}$ has integer eigenvalues. The equivalent positive expression is $N_{\mathrm{odd}}=\frac{1}{4}\int d^{3}p/(2\pi {)}^3(a_{\mathbf{p}}^{†}-a_{-\mathbf{p}}^{†})(a_{\mathbf{p}}-a_{-\mathbf{p}})$.

Take parity symmetry in quantum field theory to preserve free time evolution, so the conjugate initial data reflect along with the field. Substitute the mode expansion of a free field into the defining parity operator relation and change the integration variable $\mathbf{p}\mapsto -\mathbf{p}$. Since $E_{-\mathbf{p}}=E_{\mathbf{p}}$, coefficient comparison, or comparison of both the field and its conjugate initial data, gives the intrinsic-even conventionHere parity symmetry in quantum field theory reverses spatial momentum and preserves time. The creation operator formulas follow by taking Hermitian adjoints. The given field relation has fixed the intrinsic phase to $+1$.

For one real scalar field, let $R$ act on one-particle wave functions by $(Rf)(\mathbf{p})=f(-\mathbf{p})$. It is a Hermitian operator with $R^2=1$, so $P_{-}=(1-R)/2$ is the orthogonal projector onto odd wave functions. Its second quantization is the odd-mode number operatorThe full momentum space is integrated in both expressions; the factors prevent double counting the pairs $\{\mathbf{p},-\mathbf{p}\}$. A suitable parity operator isFor verification, $[N_{\mathrm{odd}},a_{\mathbf{p}}]=-(a_{\mathbf{p}}-a_{-\mathbf{p}})/2$. Thus the even combination $a_{\mathbf{p}}+a_{-\mathbf{p}}$ is unchanged, and the odd combination $a_{\mathbf{p}}-a_{-\mathbf{p}}$ changes sign under conjugation by $e^{i\pi N_{\mathrm{odd}}}$, exactly reproducing $a_{\mathbf{p}}\mapsto a_{-\mathbf{p}}$. The same holds for creation operators. The odd-mode number operator has integer eigenvalues, so ${\mathcal{P}}^2=1$, and $\mathcal{P}|0⟩=|0⟩$. Multiplying by an overall constant phase would not change its conjugation action. This is parity as odd-mode number parity.

Write $h_0=-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})$ for the one-particle Hamiltonian operator. The second quantization of the kinetic energy, external potential energy and repulsive contact interaction isThe factor $1/2$ counts each unordered particle pair once. Normal ordering excludes self-interaction: the interaction acts on two distinct bosons. The fixed particle number sector has $\hat{N}=\int d^{3}r{\hat{\psi }}^{†}\hat{\psi }=N$. In three dimensions, this contact interaction is understood as a regulated low-energy interaction with a matched coupling, as in renormalization of a three-dimensional dilute-gas contact coupling; the finite-mode calculation below uses its projected matrix elements.

The normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

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.

Second quantization describes identical-particle quantum states in Fock space using creation operators and annihilation operators. Relative to an orthonormal basis of one-particle states, a one-body operator becomes ${\sum }_{\alpha \beta }{h}_{\alpha \beta }{a}_{\alpha }^{†}{a}_{\beta }$. The canonical commutation relation implements bosonic statistics, while the canonical anticommutation relations implement fermionic statistics. Expanding the quantized field in a complete basis gives a delta-function equal-time bracket; restricting the expansion to finitely many modes replaces that delta function by the corresponding projection operator kernel.