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Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and the free real scalar field Lagrangian density $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. Its conjugate canonical momentum is $\pi =\dot{\varphi }$. Spatial translation symmetry and Noether's theorem give the canonical stress-energy tensorVanishing boundary flux then makes the four-momentum of a free real scalar field constant. In particular the contravariant spatial momentum isThe minus sign is from ${\partial }^i=-{\partial }_i$; it is essential for the following mode calculation.

In the quantized theory, insert the mode expansions into this integral, using a Hermitian symmetric ordering or normal ordering. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The spatial integral is $(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}\pm \mathbf{k})$, leavingEach last term integrates to zero: replace $\mathbf{p}$ by $-\mathbf{p}$ and commute the two annihilation operators or the two creation operators. The mixed terms reorder by the canonical commutation relation. Their vacuum contribution is proportional to the integral of the odd vector $\mathbf{p}$ and vanishes with an inversion-symmetric regulator; normal ordering removes it directly. Thus the normal-ordered free scalar four-momentum has spatial partThese manipulations can be made first with box modes and a symmetric momentum cutoff, then used on finite-particle wave packets.

The commutator derivation identity givesFor $U(\mathbf{y})=e^{-i\mathbf{P}\cdot \mathbf{y}}$, the commutator expansion for exponential conjugation therefore gives the translation of a scalar creation operatorThe vacuum has zero momentum, so $U|0⟩=|0⟩$. Hence the one-particle momentum eigenstate obeys $U|\mathbf{q}⟩=e^{-i\mathbf{q}\cdot \mathbf{y}}|\mathbf{q}⟩$: translation preserves its momentum and multiplies it by the associated phase.

The annihilation operator receives the opposite phase. Substituting both phases into the field expansion givesThis is the specified conjugation convention for the spatial translation operator. On states, the active translation moves a wave packet by $\mathbf{y}$, giving a position wavefunction $\psi (\mathbf{x}-\mathbf{y})$. Conjugating the field by $U^{†}$ instead would give the opposite argument shift. Infinitesimally the displayed result is $[P_i,\varphi ]=i{\partial }_i\varphi$.

Use natural units and the Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. For the real scalar field, choose ${\mathcal{L}}_0=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. The conjugate momentum is $\pi =\dot{\varphi }$. Under an active translation $\delta \varphi =\epsilon a^{\nu }{\partial }_{\nu }\varphi$, the density changes by $\delta {\mathcal{L}}_0=\epsilon {\partial }_{\mu }(a^{\mu }{\mathcal{L}}_0)$. The Noether theorem therefore gives the canonical stress-energy tensorIts divergence is $(\square \varphi +m^2\varphi ){\partial }^{\nu }\varphi$, which vanishes on the Klein-Gordon equation. With vanishing flux at spatial infinity, the charges are conserved. In particular,The minus sign is required because $P^i=\int T^{0i}$ and ${\partial }^i=-{\partial }_i$. These are the four-momentum of a free real scalar field.

For canonical quantization, impose $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$, with both equal-time field-field commutators zero. Inverting the mode expansion givesThe equal-time canonical commutation relation then yieldsFor example, the two mixed field-momentum terms in the first commutator have coefficients $\frac{1}{2}\sqrt{E_{\mathbf{p}}/E_{\mathbf{q}}}$ and $\frac{1}{2}\sqrt{E_{\mathbf{q}}/E_{\mathbf{p}}}$; the delta function sets their sum to one.

Insert the mode expansion into the quadratic energy. Spatial integration supplies $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. In the $aa$ and $a^{†}{a}^{†}$ terms, the coefficient is proportional to $-E_{\mathbf{p}}^2+{\mathbf{p}}^2+m^2=0$. The remaining terms giveHere $E_0$ is the divergent zero-point energy, formally $\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}{\delta }^3(0)$. Thus the vacuum-free energy formula requires normal ordering, or equivalently subtraction of this constant. In a finite box with a cutoff this is the ordinary sum $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$, so the subtraction is explicit before passing to the continuum.

Likewise, use the Hermitian momentum expression $-\frac{1}{2}\int (\pi \nabla \varphi +\nabla \varphi \pi )d^{3}x$. Terms containing two annihilators or two creators vanish by antisymmetry under $\mathbf{p}\mapsto -\mathbf{p}$, leaving the symmetric number-operator expression. Its vacuum term is zero with an inversion-symmetric regulator. Therefore the normal-ordered free scalar four-momentum isThe creation operator commutators follow directly from $[a_{\mathbf{q}}^{†}{a}_{\mathbf{q}},a_{\mathbf{p}}^{†}]=a_{\mathbf{q}}^{†}(2\pi {)}^3{\delta }^3(\mathbf{q}-\mathbf{p})$:Starting from a vacuum annihilated by every $a_{\mathbf{p}}$, each creation adds a particle of mass $m$, energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$. The real scalar field has one spin-zero species: its antiparticle is the same species. Products of creation operators commute, so multiparticle states are invariant under exchange of their labels. In a normalized discrete mode, $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exists for every $r\ge 0$; there is no exclusion restriction. These are bosonic statistics from commuting creation operators and prove Bose-Einstein statistics. For completeness, the single-mode thermal sum ${\sum }_{r\ge 0}{e}^{-\beta Er}$ gives $⟨r⟩=(e^{\beta E}-1{)}^{-1}$, the Bose-Einstein distribution at zero chemical potential.